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Authors:
Bland, Martin
Title: Introduction to Medical Statistics, An, 3rd Edition
Copyright Â©2000 Oxford University Press
> Front of Book > Authors
Author
Martin Bland
Professor of Medical Statistics
St George's Hospital Medical School, London
Authors: Bland, Martin
> Front of Book > Dedication
Dedication
To the memory of Ernest and Phyllis Bland, my parents

Copyright ©2000 Oxford University Press
> Front of Book > Preface to the Third Edition
Preface to the Third Edition
In preparing this third edition of An Introduction to Medical Statistics, I

have taken the opportunity to correct a number of mistakes and
typographical errors, and to change some of the examples and add a few
more. I have extended the treatment of several topics and introduced
some new ones, previously omitted through lack of space or energy, or
because they were then rarely seen in the medical literature. In one case,
number needed to treat, the concept had not even been invented when
the second edition was written. Other new topics include consent in
clinical trials, design and analysis of cluster-randomized trials, ecological
studies, conditional probability, repeated testing, random effects models,
intraclass correlation, and conditional odds ratios. Thanks to the wonders
of computerized typesetting, I have managed to extend the contents of
the book with a very small increase in the number of pages.
This book is for medical students, doctors, medical researchers, nurses,
members of professions allied to medicine, and all others concerned with
medical data. The range of statistical methods used in the medical and
health care literature, and hence described in this book, continues to
grow, but the time available in the undergraduate curriculum does not.
Some of the topics covered here are beyond the needs of many students,
so I have indicated by an asterisk sections which would not usually be
included in first courses. These are intended for postgraduate students
and medical researchers.
This third edition is being published with a companion volume, Statistical
Questions in Evidence-based Medicine (Bland and Peacock 2000). This
book of questions and answers includes no calculations and is
complementary to the exercises given here. In the solutions given we
make many references to An Introduction to Medical Statistics. Because
we wanted Statistical Questions in Evidence-based Medicine to be
usable with the second edition of An Introduction to Medical Statistics
(Bland 1995), I have kept the same order and numbering of the sections
in the third edition. New material has all been added at the ends of the
chapters. If the structure sometimes seems a little unwieldy, that is why.
This is a book about data, not statistical theory. The fundamental
concepts of study design, data collection and data analysis are explained
by illustration and example. Only enough mathematics and formulae are
given to make clear what is going on. For those who wish to go a little
further in their understanding, some of the more mathematical
background to the techniques described is given as appendices to the
chapters rather than in the main text.
The material covered includes all the statistical work that would be
required for a course in medicine and for the examinations of most of the
royal colleges. It includes the design of clinical trials and epidemiological
studies, data collection. summarizing and presenting data, probability, the
Binomial, Normal, Poisson. t and Chi-squared distributions, standard
errors, confidence intervals, tests of significance, large sample and small
sample comparisons of means, the use of transformations, regression
and correlation, methods based on ranks, contingency tables, odds
ratios, measurement error, reference ranges, mortality data, vital
statistics, analysis of variance, multiple and logistic regression, survival
analysis, sample size estimation, and the choice of the statistical method.
The book is firmly grounded in medical data, particularly in medical
research, and the interpretation of the results of calculations in their
medical context is emphasized. Except for a few obviously invented
numbers used to illustrate the mechanics of calculations, all the data in
the examples and exercises are real, from my own research and
statistical consultation or from the medical literature.
There are two kinds of exercise in this book. Each chapter has a set of
multiple choice questions of the ‘true or false’ type, 100 in all. Multiple
choice questions can cover a large amount of material in a short time, so
are a useful tool for revision. As MCQs are widely used in postgraduate
examinations, these exercises should also be useful to those preparing
for memberships. All the MCQs have solutions, with reference to an
appropriate part of the text or a detailed explanation for most of the
answers. Each chapter also has one long exercise. Although these
usually involve calculation, I have tried to avoid merely slotting figures
into formulae. These exercises include not only the application of
statistical techniques, but also the interpretation of the results in the light
of the source of the data.
I wish to thank many people who have contributed to the writing of this
book. First, there are the many medical students, doctors, research
workers, nurses, physiotherapists, and radiographers whom it has been
my pleasure to teach, and from whom I have learned so much. Second,
the book contains many examples drawn from research carried out with
other statisticians, epidemiologists, and social scientists, particularly
Douglas Altman, Ross Anderson, Mike Banks, Barbara Butland, Beulah
Bewley, and Walter Holland. These studies could not have been done
without the assistance of Patsy Bailey, Bob Harris. Rebecca McNair.
Janet Peacock, Swatee Patel, and Virginia Pollard. Third, the clinicians
and scientists with whom I have collaborated or who have come to me for
statistical advice not only taught me about medical data but many of them
have left me with data which are used here, including Naib Al-Saady,
Thomas Bewley, Frances Boa, Nigel Brown, Jan Davies, Peter Fish,
Caroline Flint, Nick Hall, Tessi Hanid. Michael Hutt, Riahd Jasrawi, Ian
Johnston, Moses Kipembwa, Pam Luthra, Hugh Mather, Daram Maugdal,
Douglas Maxwell, Charles Mutoka, Tim Northfield, Andreas
Papadopoulos, Mohammed Raja, Paul Richardson, and Alberto Smith. I
am particularly indebted to John Morgan, as Chapter 16 is partly based
on his work.
The original manuscript was typed by Sue Nash, Sue Fisher, Susan
Harding, Sheilah Skipp, and myself. This edition has been set by me
using LATEX, so any errors which remain are definitely my own. All the
graphs have been drawn using Stata except for the pie charts, done
using Harvard Graphics.
I thank Douglas Altman, David Jones, Robin Prescott, Klim McPherson.
Janet Peacock, and Stuart Pocock for their helpful comments on earlier
drafts. I have corrected a number of errors from the first and second
editions, and I am grateful to colleagues who have pointed them out to
me, in particular to Daniel Heitjan. I am very grateful to Janet Peacock,
who proof-read this edition. Special thanks are due to my head of
department, Ross Anderson, for all his support, and to the staff of Oxford
University Press. Most of all I thank my wife, Pauline Bland, for her
unfailing confidence and encouragement, and my children, Emily and
Nicholas Bland, for keeping my feet firmly on the ground.
M. B.
London, March 2000
> Table of Contents > Sections marked * contain material usually found only in postgraduate
courses
Sections marked * contain material

usually found only in postgraduate
courses

> Table of Contents > 1 - Introduction
1
Introduction
1.1 Statistics and medicine

Evidence-based practice is the new watchword in every profession
concerned with the treatment and prevention of disease and promotion of
health and well-being. This requires both the gathering of evidence and
its critical interpretation. The former is bringing more people into the
practice of research, and the latter is requiring of all health professionals
the ability to evaluate the research carried out. Much of this evidence is in
the form of numerical data. The essential skill required for the collection,
analysis, and evaluation of numerical data is statistics. Thus Statistics,
the science of assembling and interpreting numerical data, is the core
science of evidence-based practice.
In the past forty years medical research has become deeply involved with
the techniques of statistical inference. The work published in medical
journals is full of statistical jargon and the results of statistical
calculations. This acceptance of statistics, though gratifying to the
medical statistician, may even have gone too far. More than once I have
told a colleague that he did not need me to prove that his difference
existed, as anyone could see it, only to be told in turn that without the
magic of the P value he could not have his paper published.
Statistics has not always been so popular with the medical profession.
Statistical methods were first used in medical research in the 19th
century by workers such as Pierre-Charles-Alexandre Louis, William Farr,
Florence Nightingale and John Snow. Snow's studies of the modes of
communication of cholera, for example, made use of epidemiological
techniques upon which we have still made little improvement. Despite the
work of these pioneers, however, statistical methods did not become
widely used in clinical medicine until the middle of the twentieth century. It
was then that the methods of randomized experimentation and statistical
analysis based on sampling theory, which had been developed by Fisher
and others, were introduced into medical research, notably by Bradford
Hill. It rapidly became apparent that research in medicine raised many
new problems in both design and analysis, and much work has been
done since towards solving these by clinicians, statisticians and
epidemiologists.
Although considerable progress has been made in such fields as the
design of clinical trials, there remains much to be done in developing
research methodology in medicine. It seems likely that this will always be
so, for every research project is something new, something which has
never been done before. Under
these circumstances we make mistakes. No piece of research can be

perfect and there will always be something which, with hindsight, we
would have changed. Furthermore, it is often from the flaws in a study
that we can learn most about research methods. For this reason, the
work of several researchers is described in this book to illustrate the
problems into which their designs or analyses led them. I do not wish to
imply that these people were any more prone to error than the rest of the
human race, or that their work was not a valuable and serious
undertaking. Rather I want to learn from their experience of attempting
something extremely difficult, trying to extend our knowledge, so that
researchers and consumers of research may avoid these particular
pitfalls in the future.
1.2 Statistics and mathematics

Many people are discouraged from the study of statistics by a fear of
being overwhelmed by mathematics. It is true that many professional
statisticians are also mathematicians, but not all are, and there are many
very able appliers of statistics to their own fields. It is possible, though
perhaps not very useful, to study statistics simply as a part of
mathematics, with no concern for its application at all. Statistics may also
be discussed without appearing to use any mathematics at all (e.g. Huff
1954).
The aspects of statistics described in this book can be understood and
applied with the use of simple algebra. Only the algebra which is
essential for explaining the most important concepts is given in the main
text. This means that several of the theoretical results used are stated
without a discussion of their mathematical basis. This is done when the
derivation of the result would not aid much in understanding the
application. For many readers the reasoning behind these results is not
of great interest. For the reader who does not wish to take these results
on trust, several chapters have appendices in which simple mathematical
proofs are given. These appendices are designed to help increase the
understanding of the more mathematically inclined reader and to be
omitted by those who find that the mathematics serves only to confuse.
1.3 Statistics and computing

Practical statistics has always involved large amounts of calculation.
When the methods of statistical inference were being developed in the
first half of the twentieth century, calculations were done using pencil,
paper, tables, slide rules and, with luck, a very expensive mechanical
adding machine. Older books on statistics spend much time on the
details of carrying out calculations and any reference to a ‘computer’
means a person who computes, not an electronic device. The
development of the digital computer has brought changes to statistics as
to many other fields. Calculations can be done quickly, easily and, we
hope, accurately with a range of machines from pocket calculators with
built-in statistical functions to powerful computers analysing data on
many thousands of subjects. Many statistical methods would not be
contemplated without computers, and the development of new methods
goes hand in hand with the development of
software to carry them out. The theory of multilevel modelling (Goldstein

1995) and the programs MLn and MLWin are a good example. Most of
the calculations in this book were done using a computer and the graphs
were produced with one.
As an added bonus, my little MSDOS program Clinstat (not to be
confused with any commercial package of the same name) can be
downloaded free from my website at
http://www.sghms.ac.uk/depts/phs/staff/jmb/. It does most of the
calculations in this book, including sample size calculations and random
sampling and allocation. It does not do any multifactorial analyses, sorry.
There is also a little program to find some exact confidence intervals.
There is therefore no need to consider the problems of manual
calculation in detail. The important thing is to know why particular
calculations should be done and what the results of these calculations
actually mean. Indeed, the danger in the computer age is not so much
that people carry out complex calculations wrongly, but that they apply
very complicated statistical methods without knowing why or what the
computer output means. More than once I have been approached by a
researcher bearing a two inch thick computer printout, and asking what it
all means. Sadly, too often, it means that another tree has died in vain.
The widespread availability of computers means that more calculations
are being done, and being published, than ever before, and the chance of
inappropriate statistical methods being applied may actually have
increased. This misuse arises partly because people regard their data
analysis problems as computing problems, not statistical ones, and seek
advice from computer experts rather than statisticians. They often get
good advice on how to do it, but rather poor advice about what to do, why
to do it and how to interpret the results afterwards. It is therefore more
important than ever that the consumers of research understand
something about the uses and limitations of statistical techniques.
1.4 The scope of this book

This book is intended as an introduction to some of the statistical ideas
important to medicine. It does not tell you all you need to know to do
medical research. Once you have understood the concepts discussed
here, it is much easier to learn about the techniques of study design and
statistical analysis required to answer any particular question. There are
several excellent standard works which describe the solutions to
problems in the analysis of data (Armitage and Berry 1994, Snedecor
and Cochran 1980, Altman 1991) and also more specialized books to
which reference will be made where required.
What I hope the book will do is to give enough understanding of the
statistical ideas commonly used in medicine to enable the health
professional to read the medical literature competently and critically. It
covers enough material (and more) for an undergraduate course in
statistics for students of medicine, nursing, physiotherapy, etc. At the time
of writing, as far as can be established, it covers the material required to
answer statistical questions set in the examinations of
most of the Royal Colleges, except for the MRCPsych. I have indicated
by an asterisk in the subheading those sections which I think will be
required only by the postgraduate or the researcher.
When working through a textbook, it is useful to be able to check your
understanding of the material covered. Like most such books, this one
has exercises at the end of each chapter, but to ease the tedium most of
these are of the multiple choice type. There is also one long exercise,
usually involving calculations, for each chapter. In keeping with the
computer age, where laborious calculation would be necessary
intermediate results are given to avoid this. Thus the exercises can be
completed quite quickly and the reader is advised to try them. You can
also download some of the data sets from my website
(http://www.sghms.ac.uk/depts/phs/staff/jmb). Solutions are given at the
end of the book, in full for the long exercises and as brief notes with
references to the relevant sections in the text for MCQs. Readers who
would like more numerical exercises are recommended to Osborn
(1979). For a wealth of exercises in the understanding and interpretation
of statistics in medical research, drawn from the published literature and
popular media, you should try the companion volume to this one,
Statistical Questions in Evidence-based Medicine (Bland and Peacock
2000).
Finally, a question many students of medicine ask as they struggle with
statistics: is it worth it? As Altman (1982) has argued, bad statistics leads
to bad research and bad research is unethical. Not only may it give
misleading results, which can result in good therapies being abandoned
and bad ones adopted, but it means that patients may have been
exposed to potentially harmful new treatments for no good reason.
Medicine is a rapidly changing field. In ten years' time, many of the
therapies currently prescribed and many of our ideas about the causes
and prevention of disease will be obsolete. They will be replaced by new
therapies and new theories, supported by research studies and data of
the kind described in this book, and probably presenting many of the
same problems in interpretation. The practitioner will be expected to
decide for her- or himself what to prescribe or advise based on these
studies. So a knowledge of medical statistics is one of the most useful
things any doctor could acquire during her or his training.
> Table of Contents > 2 - The design of experiments
2
The design of experiments
2.1 Comparing treatments

There are two broad types of study in medical research: observational
and experimental. In observational studies, aspects of an existing
situation are observed, as in a survey or a clinical case report. We then
try to interpret our data to give an explanation of how the observed state
of affairs has come about. In experimental studies, we do something,
such as giving a drug, so that we can observe the result of our action.
This chapter is concerned with the way statistical thinking is involved in
the design of experiments. In particular, it deals with comparative
experiments where we wish to study the difference between the effects of
two or more treatments. These experiments may be carried out in the
laboratory in vitro or on animals or human volunteers, in the hospital or
community on human patients, or, for trials of preventive interventions, on
currently healthy people. We call trials of treatments on human subjects
clinical trials. The general principles of experimental design are the
same, although there are special precautions which must be taken when
experimenting with human subjects. The experiments whose results most
concern clinicians are clinical trials, so the discussion will deal mainly
with them.
Suppose we want to know whether a new treatment is more effective
than the present standard treatment. We could approach this in a number
of ways.
First, we could compare the results of the new treatment on new patients
with records of previous results using the old treatment. This is seldom
convincing, because there may be many differences between the
patients who received the old treatment and the patients who will receive
the new. As time passes, the general population from which patients
come may become healthier, standards of ancillary treatment and nursing
care may improve, or the social mix in the catchment area of the hospital
may change. The nature of the disease itself may change. All these
factors may produce changes in the patients' apparent response to
treatment. For example, Christie (1979) showed this by studying the
survival of stroke patients in 1978, after the introduction of a C-T head
scanner, with that of patients treated in 1974, before the introduction of
the scanner. He took the records of a group of patients treated in 1978,
who received a C-T scan, and matched each of them with a patient
treated in 1974 of the same age, diagnosis and level of consciousness on
admission. As the first column of Table 2.1 shows, patients in 1978
clearly tended to have better survival than similar patients in 1974.
The scanned 1978 patient did better than the unscanned 1974 patient in
31% of pairs. whereas the unscanned 1974 patient did better that the
scanned 1978 patient in only 7% of pairs. However, he also compared
the survival of patients in 1978 who did not receive a C-T scan with
matched patients in 1974. These patients too showed a marked
improvement in survival from 1974 to 1978 (Table 2.1). The 1978 patients
did better in 38% of pairs and the 1974 patients in only 19% of pairs.
There was a general improvement in outcome over a fairly short period of
time. If we did not have the data on the unscanned patients from 1978 we
might be tempted to interpret these data as evidence for the
effectiveness of the C-T scanner. Historical controls like this are seldom
very convincing, and usually favour the new treatment. We need to
compare the old and new treatments concurrently.
Table 2.1. Analysis of the difference in survival

for matched pairs of stroke patients (Christie
1979)
C-T scan No C-T scan

in 1978 in 1978
Pairs with 1978 9 (31%) 34 (38%)

better than 1974
Pairs with same 18 (62%) 38 (43%)

outcome
Pairs with 1978 2 (7%) 17 (19%)

worse than 1974
Second, we could obtain concurrent groups by comparing our own

patients, given the new treatment, with patients given the standard
treatment in another hospital or clinic, or by another clinician in our own
institution. Again, there may be differences between the patient groups
due to catchment, diagnostic accuracy, preference by patients for a
particular clinician, or you might just be a better therapist. We cannot
separate these differences from the treatment effect.
Third, we could ask people to volunteer for the new treatment and give
the standard treatment to those who do not volunteer. The difficulty here
is that people who volunteer and people who do not volunteer are likely
to be different in many ways apart from the treatments we give them.
They might be more likely to follow medical advice, for example. We will
consider an example of the effects of volunteer bias in §2.4.
Fourth, we can allocate patients to the new treatment or the standard
treatment and observe the outcome. The way in which patients are
allocated to treatments can influence the results enormously, as the
following example (Hill 1962) shows. Between 1927 and 1944 a series of
trials of BCG vaccine were carried out in New York (Levine and Sackett
1946). Children from families where there was a case of tuberculosis
were allocated to a vaccination group and given BCG vaccine, or to a
control group who were not vaccinated. Between 1927 and 1932
physicians vaccinated half the children, the choice of which children to
vaccinate being left to them. There was a clear advantage in survival for
the BCG group (Table 2.2). However, there was also a clear tendency for
the physician to vaccinate the children of more cooperative parents, and
to leave those of less cooperative parents as controls. From 1933
allocation to treatment or control was done centrally, alternate children
being assigned to control and vaccine.
The difference in degree of cooperation between the parents of the two

groups of children disappeared, and so did the difference in mortality.
Note that these were a special group of children, from families where
there was tuberculosis. In large trials using children drawn from the
general population, BCG was shown to be effective in greatly reducing
deaths from tuberculosis (Hart and Sutherland 1977)
Table 2.2. Results of studies of BCG vaccine in New Yo

City (Hill 1962)
Average
no. of Proportio
visits to of parent
No. of
clinic giving goo
Period No. of deaths Death
during cooperatio
of trial children from rate
1st year as judge
TB
of by visitin
follow-
up
1927–32 Selection made by physician
BCG 445 3 0.67% 3.6 43%

group
Control 545 18 3.30% 1.7 24%

group
1933–44 Alternative selection carried out centrally
BCG 566 8 1.41% 2.8 40%

group
Control 528 8 1.52% 2.4 34%

group
Different methods of allocation to treatment can produce different results.

This is because the method of allocation may not produce groups of
subjects which are comparable, similar in every respect except the
treatment. We need a method of allocation to treatments in which the
characteristics of subjects will not affect their chance of being put into any
particular group. This can be done using random allocation.
2.2 Random allocation

If we want to decide which of two people receive an advantage, in such a
way that each has an equal chance of receiving it, we can use a simple,
widely accepted method. We toss a coin. This is used to decide the way
football matches begin, for example, and all appear to agree that it is fair.
So if we want to decide which of two subjects should receive a vaccine,
we can toss a coin. Heads and the first subject receives the vaccine, tails
and the second receives it. If we do this for each pair of subjects we build
up two groups which have been assembled without any characteristics of
the subjects themselves being involved in the allocation. The only
differences between the groups will be those due to chance. As we shall
see later (Chapters 8 and 9), statistical methods enable us to measure
the likely effects of chance. Any difference between the groups which is
larger than this should be due to the treatment, since there will be no
other differences between the groups. This method of dividing subjects
into groups is called random allocation or randomization.
Several methods of randomizing have been in use for centuries, including
coins, dice, cards, lots, and spinning wheels. Some of the theory of
probability which we shall use later to compare randomized groups was
first developed as
an aid to gambling. For large randomizations we use a different, non-

physical randomizing method: random number tables. Table 2.3 provides
an example, a table of 1000 random digits. These are more properly
called pseudo-random numbers, as they are generated by a
mathematical process. They are available in tables (Kendall and
Babington Smith 1971) or can be produced by computer and some
calculators. We can use tables of random numbers in several ways to
achieve random allocation. For example, let us randomly allocate 20
subjects to two groups, which I shall label A and B. We choose a random
starting point in the table, using one of the physical methods described
above. (I used decimal dice. These are 20-sided dice, numbered 0 to 9
twice, which fit our number system more conveniently than the traditional
cube. Two such dice give a random number between 1 and 100, counting
‘0,0’ as 100.) The random starting point was row 22, column 20, and the
first 20 digits were 3, 4, 6, 2, 9, 7, 5, 3, 2, 6, 9, 7, 9, 3, 9, 2, 3, 3, 2 and 4.
We now allocate subjects corresponding to odd digits to group A and
those corresponding to even digits to B. The first digit, 3, is odd, so the
first subject goes into group A. The second digit, 4, is even, so the
second subject goes into group B, and so on. We get the allocation
shown in Table 2.4. We could allocate into three groups by assigning to A
if the digit is 1, 2, or 3, to B if 4, 5, or 6, and to C if 7, 8, or 9, ignoring 0.
There are many possibilities.
Table 2.3. The 1000 random digits
Column
Row 1– 5– 9– 13– 17– 21– 25– 29– 33–
4 8 12 16 20 24 28 32
1 36 88 28 59 46 00 67 32
45 31 73 43 32 32 15 49
2 90 40 18 95 65 16 95 15
51 66 46 54 89 80 33 88
3 98 90 48 80 91 33 40 38
41 22 37 31 39 80 82 26
4 55 71 14 64 99 82 73 92
25 27 68 04 24 30 43 68
5 02 10 77 88 79 70 59 75
99 75 21 55 97 32 87 35
6 79 55 63 08 04 18 53 58
85 66 84 63 00 34 94 01
7 33 95 06 34 13 37 95 15
53 28 81 95 93 16 06 91
8 74 13 22 37 15 42 96 90
75 13 16 76 57 38 23 24
9 06 30 00 32 36 46 17 66
66 43 66 60 60 05 31 80
10 92 31 87 76 17 31 13 17
83 60 30 83 85 48 23 32
11 61 31 98 77 72 35 69 14
21 49 29 70 11 23 47 27
12 27 01 74 38 53 53 55 35
82 01 41 77 68 26 16 66
13 61 50 94 86 10 95 88 72
05 10 85 32 72 67 21 09
14 11 85 94 49 35 39 80 54
57 67 91 48 49 41 17 45
15 15 08 92 13 26 20 72 94
16 90 86 32 01 02 45 74
16 22 29 15 76 94 48 75 81
09 66 44 74 92 13 85 28
17 69 53 35 43 83 79 92 83
13 55 87 23 32 40 20 76
18 08 79 00 35 86 10 18 43
29 37 33 34 55 91 86 50
19 37 99 55 32 71 85 31 63
29 85 63 66 98 20 93 91
20 65 14 88 28 04 42 87 20
11 04 86 92 03 99 08 55
21 66 81 30 21 15 26 33 51
22 58 80 10 53 90 77 19
22 37 77 69 20 67 46 75 69
21 13 31 22 13 29 32 79
23 51 09 68 05 14 89 37 25
43 72 38 77 62 07 89 30
24 31 37 92 15 21 03 35 84
59 83 55 31 24 93 97 61
25 79 43 52 00 44 91 11 25
05 69 93 77 82 65 71 37
Table 2.4. Allocation of 20 subjects to two

groups
Subject Digit Group
1 3 A
2 4 B
3 6 B
4 2 B
5 9 A
6 7 A
7 5 A
8 3 A
9 2 B
10 6 B
11 9 A
12 7 A
13 9 A
14 3 A
15 9 A
16 2 B
17 3 A
18 3 A
19 2 B
20 4 B
The system described above gave us unequal numbers in the two

groups, 12 in A and 8 in B. We sometimes want the groups to be of equal
size. One way to do this would be to proceed as above until either A or B
has 10 subjects in it, all the remaining subjects going into the other
groups. This is satisfactory in that each subject has an equal chance of
being allocated to A or B, but it has a disadvantage. There is a tendency
for the last few subjects all to have the same treatment. This
characteristic sometimes worries researchers, who feel that the
randomization is not quite right. In statistical terms the possible
allocations are not equally likely. If we use this method for the random
allocation described above, the 10th subject in group A would be reached
at subject 15 and the last five subjects would all be in group B. We can
ensure that all randomizations are equally likely by using the table of
random numbers in a different way. For example, we can use the table to
draw a random sample of 10 subjects from 20, as described in §3.4.
These would form group A, and the remaining 10 group B. Another way
is to put our subjects into small equal-sized groups, called blocks, and
within each block to allocate equal numbers to A and B. This gives
approximately equal numbers on the two treatments and will do so
whenever the trial stops.
The use of random numbers and the generation of the random numbers
themselves are simple mathematical operations well suited to the
computers which are now readily available to researchers. It is very easy
to program a computer to carry out random allocation, and once a
program is available it can be used over and over again for further
experiments. My program Clinstat (§1.3) does several different
randomization schemes, even offering blocks of random size.
The trial carried out by the Medical Research Council (MRC 1948) to test
the efficacy of streptomycin for the treatment of pulmonary tuberculosis is
generally considered to have been the first randomized experiment in
medicine. In this study the target population was patients with acute
progressive bilateral pulmonary tuberculosis, aged 15–30 years. All
cases were bacteriologically proved and were considered unsuitable for
other treatments then available. The trial took place in three centres and
allocation was by a series of random numbers, drawn up for each sex at
each centre. The streptomycin group contained 55
patients and the control group 52 cases. The condition of the patients on
admission is shown in Table 2.5. The frequency distributions of
temperature and sedimentation rate were similar for the two groups; if
anything the treated (S) group were slightly worse. However, this
difference is no greater than could have arisen by chance, which, of
course, is how it arose. The two groups are certain to be slightly different
in some characteristics, especially with a fairly small sample, and we can
take account of this in the analysis (Chapter 17).
Table 2.5. Condition of patients on admission

to trial of streptomycin (MRC 1948)
Group

S C
General condition Good 8 8
Fair 17 20
Poor 30 24
Max. evening temperature 98- 4 4

in first week (°F) 98.9
99- 13 12
99.9
100- 15 17
100.9
101+ 24 19
Sedimentation rate 0-10 0 0
11-20 3 2
21-50 16 20
51+ 36 29
Table 2.6. Survival at six months in the MRC

streptomycin trial, stratified by initial condition
(MRC 1948)
Maximum Group
evening
temperature
Outcome Streptomycin Control
during first
observation group group
week
98-98.9°F Alive 3 4
Dead 0 0
99-99.9°F Alive 13 11
Dead 0 1
100- Alive 15 12
100.9°F
Dead 0 5
101°F and Alive 20 11

above
Dead 4 8
After six months, 93% of the S group survived, compared to 73% of the
control group. There was a clear advantage to the streptomycin group.
The relationship of survival to initial condition is shown in Table 2.6.
Survival was more likely for patients with lower temperatures, but the
difference in survival between the S and C groups is clearly present
within each temperature category where deaths occurred.
Randomized trials are not restricted to two treatments. We can compare
several treatments. A drug trial might include the new drug, a rival drug,
and
no drug at all. We can carry out experiments to compare several factors

at once. For example, we might wish to study the effect of a drug at
different doses in the presence or absence of a second drug, with the
subject standing or supine. This is usually designed as a factorial
experiment, where every possible combination of treatments is used.
These designs are unusual in clinical research but are sometimes used in
laboratory work. They are described in more advanced texts (Armitage
and Berry 1994, Snedecor and Cochran 1980). For more on randomized
trials in general, see Pocock (1983) and Johnson and Johnson (1977).
Randomized experimentation may be criticized because we are
withholding a potentially beneficial treatment from patients. Any
biologically active treatment is potentially harmful, however, and we are
surely not justified in giving potentially harmful treatments to patients
before the benefits have been demonstrated conclusively. Without
properly conducted controlled clinical trials to support it, each
administration of a treatment to a patient becomes an uncontrolled
experiment, whose outcome, good or bad, cannot be predicted.
2.3 * Methods of allocation without random

numbers
In the second stage of the New York studies of BCG vaccine, the children
were allocated to treatment or control alternately. Researchers often ask
why this method cannot be used instead of randomization, arguing that
the order in which patients arrive is random, so the groups thus formed
will be comparable. First, although the patients may appear to be in a
random order, there is no guarantee that this is the case. We could never
be sure that the groups are comparable. Second, this method is very
susceptible to mistakes, or even to cheating in the patients' perceived
interest. The experimenter knows what treatment the subject will receive
before the subject is admitted to the trial. This knowledge may influence
the decision to admit the subject, and so lead to biased groups. For
example, an experimenter might be more prepared to admit a frail patient
if the patient will be on the control treatment than if the patient would be
exposed to the risk of the new treatment. This objection applies to using
the last digit of the hospital number for allocation.
Knowledge of what treatment the next patient will receive can certainly
lead to bias. For example, Schulz et al. (1995) looked at 250 controlled
trials. They compared trials where treatment allocation was not
adequately concealed from researchers with trials where there was
adequately concealment. They found an average treatment effect 41%
larger in the trials with inadequate concealment.
There are several examples reported in the literature of alterations to
treatment allocations. Holten (1951) reported a trial of anticoagulant
therapy for patients with coronary thrombosis. Patients who presented on
even dates were to be treated and patients arriving on odd dates were to
form the control group. The author reports that some of the clinicians
involved found it ‘difficult to remember’ the criterion for allocation. Overall
the treated patients did better than the controls (Table 2.7). Curiously, the
controls on the even dates (wrongly allocated) did considerably better
than control patients on the odd dates (correctly
allocated) and even managed to do marginally better than those who

received the treatment. The best outcome, treated or not, was for those
who were incorrectly allocated. Allocation in this trial appears to have
been rather selective.
Table 2.7. Outcome of a clinical trial using

systematic allocation, with errors in allocation
(Holten 1951)
Even dates Odd dates

Outcome
Treated Control Treated Control
Survived 125 39 10 125
Died 39 11 0 (0%) 81
(25%) (22%) (36%)
Total 164 50 10 206
Other methods of allocation set out to be random but can fall into this sort
of difficulty. For example, we could use physical mixing to achieve
randomization. This is quite difficult to do. As an experiment, take a deck
of cards and order them in suits from ace of clubs to king of spades. Now
shuffle them in the usual way and examine them. You will probably see
many runs of several cards which remain together in order. Cards must
be shuffled very thoroughly indeed before the ordering ceases to be
apparent. The physical randomization method can be applied to an
experiment by marking equal numbers on slips of paper with the names
of the treatments, sealing them into envelopes and shuffling them. The
treatment for a subject is decided by withdrawing an envelope. This
method was used in another study of anticoagulant therapy by Carleton
et al. (1960). These authors reported that in the latter stages of the trial
some of the clinicians involved had attempted to read the contents of the
envelopes by holding them up to the light, in order to allocate patients to
their own preferred treatment.
Interfering with the randomization can actually be built into the allocation
procedure, with equally disastrous results. In the Lanarkshire Milk
Experiment, discussed by Student (1931), 10000 school children
received three quarters of a pint of milk per day and 10000 children acted
as controls. The children were weighed and measured at the beginning
and end of the six-month experiment. The object was to see whether the
milk improved the growth of children. The allocation to the ‘milk’ or control
group was done as follows:
The teachers selected the two classes of pupils, those getting milk and
those acting as controls, in two different ways. In certain cases they
selected them by ballot and in others on an alphabetical system. In any
particular school where there was any group to which these methods had
given an undue proportion of well-fed or ill-nourished children, others
were substituted to obtain a more level selection.
The result of this was that the control group had a markedly greater
average height and weight at the start of the experiment than did the milk
group. Student interpreted this as follows:
Presumably this discrimination in height and weight was not made
deliberately, but it would seem probable that the teachers, swayed by the
very human feeling that the poorer children needed the milk more than
the comparatively well to do, must have unconsciously made too large a
substitution for the ill-nourished among the (milk group) and too few
among the controls and that this unconscious selection affected
secondarily, both measurements.

Whether the bias was conscious or not, it spoiled the experiment, despite
being from the best possible motives.
There is one non-random method which can be used successfully in
clinical trials: minimization. In this method, new subjects are allocated to
treatments so as to make the treatment groups as similar as possible in
terms of the important prognostic factors. It is beyond the scope of this
book, but see Pocock (1983) for a description.
2.4 Volunteer bias

People who volunteer for new treatments and those who refuse them
may be very different. An illustration is provided by the field trial of Salk
poliomyelitis vaccine carried out in 1954 in the USA (Meier 1977). This
was carried out using two different designs simultaneously, due to a
dispute about the correct method. In some districts, second grade
schoolchildren were invited to participate in the trial, and randomly
allocated to receive vaccine or an inert saline injection. In other districts,
all second grade children were offered vaccination and the first and third
grade left unvaccinated as controls. The argument against this ‘observed
control’ approach was that the groups may not be comparable, whereas
the argument against the randomized control method was that the saline
injection could provoke paralysis in infected children. The results are
shown in Table 2.8. In the randomized control areas the vaccinated group
clearly experienced far less polio than the control group. Since these
were randomly allocated, the only difference between them should be the
treatment, which is clearly preferable to saline. However, the control
group also had more polio than those who had refused to participate in
the trial. The difference between the control and not inoculated group is
in both treatment (saline injection) and selection; they are self-selected
as volunteers and refusers. The observed control areas enable us to
distinguish between these two factors. The polio rates in the vaccinated
children are very similar in both parts of the study, as are the rates in the
not inoculated second grade children. It is the two control groups which
differ. These were selected in different ways: in the randomized control
areas they were volunteers, whereas in the observed control areas they
were everybody eligible, both potential volunteers and potential refusers.
Now suppose that the vaccine were saline instead, and that the
randomized vaccinated children had the same polio experience as those
receiving saline. We would expect 200745 × 57/100000 = 114 cases,
instead of the 33 observed. The total number of cases in the randomized
areas would be 114 + 115 + 121 = 350 and the rate per 100000 would be
47. This compares very closely with the rate of 46 in the observed control
first and third grade group. Thus it seems that the principal difference
between the saline control group of volunteers and the not inoculated
group of refusers is selection, not treatment.
There is a simple explanation of this. Polio is a viral disease transmitted
by the faecal—oral route. Before the development of vaccine almost
everyone in the
population was exposed to it at some time, usually in childhood. In the

majority of cases, paralysis does not result and immunity is conferred
without the child being aware of having been exposed to polio. In a small
minority of cases, about 1 in 200, paralysis or death occurs and a
diagnosis of polio is made. The older the exposed individual is, the
greater the chance of paralysis developing. Hence, children who are
protected from infection by high standards of hygiene are likely to be
older when they are first exposed to polio than those children from homes
with low standards of hygiene, and thus more likely to develop the clinical
disease. There are many factors which may influence parents in their
decision as to whether to volunteer or refuse their child for a vaccine trial.
These may include education, personal experience, current illness, and
others, but certainly include interest in health and hygiene. Thus in this
trial the high risk children tended to be volunteered and the low risk
children tended to be refused. The higher risk volunteer control children
experienced 57 cases of polio per 100000, compared to 36 per 100000
among the lower risk refusers.
Table 2.8. Result of the field trial of Salk

poliomyelitis vaccine (Meier 1977)
Paralytic polio
Number
Study group Number Rate per
in group
of cases 100000
Randomized control
Vaccinated 200745 33 16
Control 201229 115 57
Not 338778 121 36

inoculated
Observed control
Vaccinated 221998 38 17
2nd grade
Control 1st 725173 330 46

and 3rd
grade
Unvaccinated 123605 43 35
2nd grade
In most diseases, the effect of volunteer bias is opposite to this. Poor
conditions are related both to refusal to participate and to high risk,
whereas volunteers tend to be low risk. The effect of volunteer bias is
then to produce an apparent difference in favour of the treatment. We
can see that comparisons between volunteers and other groups can
never be reliable indicators of treatment effects.
2.5 Intention to treat

In the observed control areas of the Salk trial (Table 2.8), quite apart from
the non-random age difference, the vaccinated and control groups are
not comparable. However, it is possible to make a reasonable
comparison in this study by comparing all second grade children, both
vaccinated and refused, to the control group. The rate in the second
grade children is 23 per 100000, which is less than the rate of 46 in the
control group, demonstrating the effectiveness of the vaccine. The
‘treatment’ which we are evaluating is not vaccination itself, but a policy
of offering vaccination and treating those who accept. A similar problem
can arise in a randomized trial, for example in evaluating the
effectiveness
of health checkups (South-east London Screening Study Group 1977).

Subjects were randomized to a screening group or to a control group.
The screening group were invited to attend for an examination, some
accepted and were screened and some refused. When comparing the
results in terms of subsequent mortality, it was essential to compare the
controls to the screening groups containing both screened and refusers.
For example, the refusers may have included people who were already
too ill to come for screening. The important point is that the random
allocation procedure produces comparable groups and it is these we
must compare, whatever selection may be made within them. We
therefore analyse the data according to the way we intended to treat
subjects, not the way in which they were actually treated. This is analysis
by intention to treat. The alternative, analysing by treatment actually
received, is called on treatment analysis.
Analysis by intention to treat is not free of bias. As some patients may
receive the other group's treatment, the difference may be smaller than it
should be. We know that there is a bias and we know that it will make the
treatment difference smaller, by an unknown amount. On treatment
analyses, on the other hand, are biased in favour of showing a difference,
whether there is one or not. Statisticians call methods which are biased
against finding any effect conservative. If we must err, we like to do so in
the conservative direction.
2.6 Cross-over designs

Sometimes it is possible to use a subject as her or his own control. For
example, when comparing analgesics in the treatment of arthritis,
patients may receive in succession a new drug and a control treatment.
The response to the two treatments can then be compared for each
patient. These designs have the advantage of removing variability
between subjects. We can carry out a trial with fewer subjects than would
be needed for a two group trial.
Although all subjects receive all treatments, these trials must still be
randomized. In the simplest case of treatment and control, patients may
be given two different regimes: control followed by treatment or treatment
followed by control. These may not give the same results, e.g. there may
be a long-term carry-over effect or time trend which makes treatment
followed by control show less of a difference than control followed by
treatment. Subjects are, therefore, assigned to a given order at random.
It is possible in the analysis of cross-over studies to estimate the size of
any carry-over effects which may be present.
As an example of the advantages of a cross-over trial, consider a trial of
pronethalol in the treatment of angina pectoris (Pritchard et al. 1963).
Angina pectoris is a chronic disease characterized by attacks of acute
pain. Patients in this trial received either pronethalol or an inert control
treatment (or placebo, see §2.8) in four periods of two weeks, two
periods on the drug and two on the control treatment. These periods
were in random order. The outcome measure was the number of attacks
of angina experienced. These were recorded by the patient in a diary.
Twelve patients took part in the trial. The results are shown
in Table 2.9. The advantage in favour of pronethalol is shown by 11 of the

12 patients reporting fewer attacks of pain while on pronethalol than while
on the control treatment. If we had obtained the same data from two
separate groups of patients instead of the same group under two
conditions, it would be far from clear that pronethalol is superior because
of the huge variation between subjects. Using a two group design, we
would need a much larger sample of patients to demonstrate the efficacy
of the treatment.
Table 2.9. Results of a trial of pronethalol for

the treatment of angina pectoris (Pritchard et
al. 1963)
Number of attacks Difference

Patient while on placebo –
number
Placebo Pronethalol pronethalol
1 71 29 42
2 323 348 –25
3 8 1 7
4 14 7 7
5 23 16 7
6 34 25 9
7 79 65 14
8 60 41 19
9 2 0 2
10 3 0 3
11 17 15 2
12 7 2 5
Cross-over designs can be useful for laboratory experiments on animals

or human volunteers. They should only be used in clinical trials where the
treatment will not affect the course of the disease and where the patient's
condition would not change appreciably over the course of the trial. A
cross-over trial could be used to compare different treatments for the
control of arthritis or asthma, for example, but not to compare different
regimes for the management of myocardial infarction. However, a cross-
over trial cannot be used to demonstrate the long-term action of a
treatment, as the nature of the design means that the treatment period
must be limited. As most treatments of chronic disease must be used by
the patient for a long time, a two sample trial of long duration is usually
required to investigate fully the effectiveness of the treatment.
Pronethalol, for example, was later found to have quite unacceptable side
effects in long term use.
For more on cross-over trials, see Senn (1993) and Jones and Kenward
(1989).
2.7 Selection of subjects for clinical trials

I have discussed the allocation of subjects to treatments at some length,
but we have not considered where they come from. The way in which
subjects are selected for an experiment may have an effect on its
outcome. In practice, we are usually limited to subjects which are easily
available to us. For example, in an animal experiment we must take the
latest batch from the animal house. In a clinical trial of the treatment of
myocardial infarction, we must be content with patients who are brought
into the hospital. In experiments on human volunteers
we sometimes have to use the researchers themselves.

As we shall see more fully in Chapter 3, this has important consequences
for the interpretation of results. In trials of myocardial infarction, for
example, we would not wish to conclude that, say, the survival rate with a
new treatment in a trial in London would be the same as in a trial in
Edinburgh. The patients may have a different history of diet, for example,
and this may have a considerable effect on the state of their arteries and
hence on their prognosis. Indeed, it would be very rash to suppose that
we would get the same survival rate in a hospital a mile down the road.
What we rely on is the comparison between randomized groups from the
same population of subjects, and hope that if a treatment reduces
mortality in London it will also do so in Edinburgh. This may be a
reasonable supposition, and effects which appear in one population are
likely to appear in another, but it cannot be proved on statistical grounds
alone. Sometimes in extreme cases it turns out not to be true. BCG
vaccine has been shown, by large, well conducted randomized trials, to
be effective in reducing the incidence of tuberculosis in children in the
UK. However, in India it appears to be far less effective (Lancet 1980).
This may be because the amount of exposure to tuberculosis is so
different in the two populations.
Given that we can use only the experimental subjects available to us,
there are some principles which we use to guide our selection from them.
As we shall see later, the lower the variability between the subjects in an
experiment is. the better chance we have of detecting a treatment
difference if it exists. This means that uniformity is desirable in our
subjects. In an animal experiment this can be achieved by using animals
of the same strain raised under controlled conditions. In a clinical trial we
usually restrict our attention to patients of a defined age group and
severity of disease. The Salk vaccine trial (§2.4) only used children in
one school year. In the streptomycin trial (§2.2) the subjects were
restricted to patients with acute bilateral pulmonary tuberculosis,
bacteriologically proved, aged between 15 and 30 years, and unsuitable
for other current therapy. Even with this narrow definition there was
considerable variation among the patients, as Tables 2.5 and 2.6 show.
Tuberculosis had to be bacteriologically proved because it is important to
make sure that everyone has the disease we wish to treat. Patients with
a different disease are not only potentially being wrongly treated
themselves, but may make the results difficult to interpret. Restricting
attention to a particular subset of patients, though useful, can lead to
difficulties. For example, a treatment shown to be effective and safe in
young people may not necessarily be so in the elderly. Trials have to be
carried out on the sort of patients it is proposed to treat.
2.8 Response bias and placebos

The knowledge that she or he is being treated may alter a patient's
response to treatment. This is called the placebo effect. A placebo is a
pharmacologically inactive treatment given as if it were an active
treatment. This effect may take many forms, from a desire to please the
doctor to measurable biochemical
changes in the brain. Mind and body are intimately connected, and
unless the psychological effect is actually part of the treatment we usually
try to eliminate such factors from treatment comparisons. This is
particularly important when we are dealing with subjective assessments,
such as of pain or well-being.
Fig. 2.1. Pain relief in relation to drug and to colour of placebo (after
Huskisson 1974)
A fascinating example of the power of the placebo effect is given by

Huskisson (1974). Three active analgesics, aspirin, Codis and
Distalgesic, were compared with an inert placebo. Twenty two patients
each received the four treatments in a cross-over design. The patients
reported pain relief on a four point scale, from 0 = no relief to 3 =
complete relief. All the treatments produced some pain relief, maximum
relief being experienced after about two hours (Figure 2.1). The three
active treatments were all superior to placebo, but not by very much. The
four drug treatments were given in the form of tablets identical in shape
and size, but each drug was given in four different colours. This was
done so that patients could distinguish the drugs received, to say which
they preferred. Each patient received four different colours, one for each
drug, and the colour combinations were allocated randomly. Thus some
patients received red placebos, some blue, and so on. As Figure 2.1
shows, red placebos were markedly more effective than other colours,
and were just as effective as the active drugs! In this study not only is the
effect of a pharmacologically inert placebo in producing reported pain
relief demonstrated, but also the wide variability and unpredictability of
this response. We must clearly take account of this in trial design.
Incidentally, we should not conclude that red placebos always work best.
There is, for example, some evidence that patients being treated for
anxiety prefer tablets to be in a soothing green, and depressive
symptoms respond best to a lively yellow (Schapira et al. 1970).
In any trial involving human subjects it is desirable that the subjects
should not be able to tell which treatment is which. In a study to compare
two or more treatments this should be done by making the treatments as
similar as possible. Where subjects are to receive no treatment an
inactive placebo should be used if possible. Sometimes when two very
different active treatments are compared a double placebo or double
dummy can be used. For example, when comparing a drug given a
single dose with a drug taken daily for seven days, subjects on
the single dose drug may receive a daily placebo and those on the daily
dose a single placebo at the start.
Placebos are not always possible or ethical. In the MRC trial of
streptomycin. where the treatment involved several injections a day for
several months, it was not regarded as ethical to do the same with an
inert saline solution and no placebo was given. In the Salk vaccine trial,
the inert saline injections were placebos. It could be argued that paralytic
polio is not likely to respond to psychological influences, but how could
we be really sure of this? The certain knowledge that a child had been
vaccinated may have altered the risk of exposure to infection as parents
allowed the child to go swimming, for example. Finally, the use of a
placebo may also reduce the risk of assessment bias as we shall see in
§2.9.
2.9 Assessment bias and double blind studies

The response of subjects is not the only thing affected by knowledge of
the treatment. The assessment by the researcher of the response to
treatment may also be influenced by the knowledge of the treatment.
Some outcome measures do not allow for much bias on the part of the
assessor. For example, if the outcome is survival or death, there is little
possibility that unconscious bias may affect the observation. However, if
we are interested in an overall clinical impression of the patient's
progress, or in changes in an X-ray picture, the measurement may be
influenced by our desire (or otherwise) that the treatment should
succeed. It is not enough to be aware of this danger and allow for it, as
we may have the similar problem of ‘bending over backwards to be fair’.
Even such an apparently objective measure as blood pressure can be
influenced by the expectations of the experimenter, and special
measuring equipment has been devised to avoid this (Rose et al. 1964).
We can avoid the possibility of such bias by using blind assessment,
that is, the assessor does not know which treatment the subject is
receiving. If a clinical trial cannot be conducted in such a way that the
clinician in charge does not know the treatment, blind assessment can
still be carried out by an external assessor. When the subject does not
know the treatment and blind assessment is used, the trial is said to be
double blind. (Researchers on eye disease hate the terms ‘blind’ and
‘double blind’, prefering ‘masked’ and ‘double masked’ instead.)
Placebos may be just as useful in avoiding assessment bias as for
response bias. The subject is unable to tip the assessor off as to
treatment, and there is likely to be less material evidence to indicate to an
assessor what it is. In the anticoagulant study by Carleton et al. (1960)
described above, the treatment was supplied through an intravenous
drip. Control patients had a dummy drip set up, with a tube taped to the
arm but no needle inserted, primarily to avoid assessment bias. In the
Salk trial, the injections were coded and the code for a case was only
broken after the decision had been made as to whether the child had
polio and if so of what severity.
In the streptomycin trial, one of the outcome measures was radiological
change. X-ray plates were numbered and then assessed by two

radiologists and a clinician, none of whom knew to which patient and
treatment the plate belonged. The assessment was done independently,
and they only discussed a plate if they had not all come to the same
conclusion. Only when a final decision had been arrived at was the link
between plate and patient made. The results are shown in Table 2.10.
The clear advantage of streptomycin is shown in the considerable
improvement of over half the S group, compared to only 8% of the
controls.
Table 2.10. Assessment of radiological

appearance at six months as compared with
appearance on admission (MRC 1948)
Radiological
S Group C Group
assessment
Considerable 28 51% 4 8%
improvement
Moderate or slight 10 18% 13 25%

improvement
No material change 2 4% 3 6%
Moderate or slight 5 9% 12 23%

deterioration
Considerable 6 11% 6 11%

deterioration
Deaths 4 7% 14 27%
Total 55 100% 52 100%
2.10 * Laboratory experiments

So far we have looked at clinical trials, but exactly the same principles
apply to laboratory research on animals. It may well be that in this area
the principles of randomization are not so well understood and even more
critical attention is needed from the reader of research reports. One
reason for this may be that great effort has been put into producing
genetically similar animals, raised in conditions as close to uniform as is
practicable. The researcher using such animals as subjects may feel that
the resulting animals show so little biological variability that any natural
differences between them will be dwarfed by the treatment effects. This is
not necessarily so, as the following examples illustrate.
A colleague was looking at the effect of tumour growth on macrophage
counts in rats. The only significant difference was between the initial
values in tumour induced and non-induced rats, that is, before the
tumour-inducing treatment was given. There was a simple explanation for
this surprising result. The original design had been to give the tumour-
inducing treatment to each of a group of rats. Some would develop
tumours and others would not, and then the macrophage counts would
be compared between the two groups thus defined. In the event, all the
rats developed tumours. In an attempt to salvage the experiment my
colleague obtained a second batch of animals, which he did not treat, to
act as controls. The difference between the treated and untreated
animals was thus due to differences in parentage or environment, not to
treatment.
That problem arose by changing the design during the course of the
experiment. Problems can arise from ignoring randomization in the
design of a comparative experiment. Another colleague wanted to know
whether a treatment would affect weight gain in mice. Mice were taken
from a cage one by one
and the treatment given, until half the animals had been treated. The
treated animals were put into smaller cages, five to a cage, which were
placed together in a constant environment chamber. The control mice
were in cages also placed together in the constant environment chamber.
When the data were analysed, it was discovered that the mean initial
weights was greater in the treated animals than in the control group. In a
weight gain experiment this could be quite important! Perhaps larger
animals were easier to pick up, and so were selected first. What that
experimenter should have done was to place the mice in the boxes, give
each box a place in the constant environment chamber, then allocate the
boxes to treatment or control at random. We would then have two groups
which were comparable, both in initial values and in any environmental
differences which may exist in the constant environment chamber.
2.11 * Experimental units

In the weight gain experiment described above, each box of mice
contained five animals. These animals were not independent of one
another, but interacted. In a box the other four animals formed part of the
environment of the fifth, and so might influence its growth. The box of five
mice is called an experimental unit. An experimental unit is the smallest
group of subjects in an experiment whose response cannot be affected
by other subjects. We need to know the amount of natural variation which
exists between experimental units before we can decide whether the
treatment effect is distinguishable from this natural variation. In the
weight gain experiment, the mean weight gain in each box should be
calculated, and the mean difference estimated using the two-sample t
method (§10.3). In human studies, the same thing happens when groups
of patients, such as all those in a hospital ward or a general practice are
randomized as a group. This might happen in a trial of health promotion,
for example, where special clinics are advertised and set up in GP
surgeries. It would be impractical to exclude some patients from the clinic
and impossible to prevent patients from the practice interacting with and
influencing one another. All the practice patients must be treated as a
single unit. Trials where experimental units contain more than one
subject are called cluster randomized.
The question of the experimental unit arises when the treatment is
applied to the provider of care rather than to the patient directly. For
example, White et al. (1989) compared three randomly allocated groups
of GPs, the first given an intensive programme of small group education
to improve their treatment of asthma, the second a lesser intervention,
and the third no intervention at all. For each GP, a sample of her or his
asthmatic patients was selected. These patients received questionnaires
about their symptoms, the research hypothesis being that the intensive
programme would result in fewer symptoms among their patients. The
experimental unit was the GP, not the patient. The asthma patients
treated by an individual GP were used to monitor the effect of the
intervention on that GP. The proportion of patients who reported
symptoms was used as a measure of the GP's effectiveness, and the
mean of these proportions was compared between
the groups using one-way analysis of variance (§10.9). Another example

would be a trial of population screening for a disease (§15.3), where
screening centres were set up in some health districts and not in others.
We should find the mortality rate for each district separately and then
compare the mean rate in the group of screening districts with that in the
group of control districts.
The most extreme case arises when there is only one experimental unit
per treatment. For example, consider a health education experiment
involving two schools. In one school a special health education
programme was mounted, aimed to discourage children from smoking.
Both before and afterwards, the children in each school completed
questionnaires about cigarette smoking. In this example the school is the
experimental unit. There is no reason to suppose that two schools should
have the same proportion of smokers among their pupils, or that two
schools which do have equal proportions of smokers will remain so. The
experiment would be much more convincing if we had several schools
and randomly allocated them to receive the health education programme
or to be controls. We would then look for a consistent difference between
the treated and control schools, using the proportion of smokers in the
school as the variable.
2.12 * Consent in clinical trials

I started my research career in agriculture. Our experimental subjects,
being barley plants, had no rights. We sprayed them with whatever
chemicals we chose and burnt them after harvest and weighing. We
cannot treat human subjects in the same way. We must respect the rights
of our research subjects and their welfare must be our primary concern.
This has not always been the case, most notoriously in the Nazi death
camps (Leaning 1996). The Declaration of Helsinki (BMJ 1996a), which
lays down the principles which govern research on human subjects, grew
out of the trials in Nuremburg of the perpetrators of these atrocities (BMJ
1996b).
If there is a treatment, we should not leave patients untreated if this in
any way affects their well-being. The world was rightly outraged by the
Tuskegee Study, where men with syphilis were left untreated to see what
the long-term effects of the disease might be (Brawley 1998, Ramsay
1998). This is an extreme example but it is not the only one. Women with
dysplasia found at cervical cytology have been left untreated to see
whether cancer developed (Mudur 1997). Patients are still being asked to
enter pharmaceutical trials where they may get a placebo, even though
an effective treatment is available, allegedly because regulators insist on
it.
People should not be treated without their consent. This general principle
is not confined to research. Patients should also be asked whether they
wish to take part in a research project and whether they agree to be
randomized. They should know to what they are consenting, and usually
recruits to clinical trials are given information sheets which explain to
them randomization, the alternative treatments, and the possible risks
and benefits. Only then can they give informed or valid consent. For
children who are old enough to understand, both child and
parent should be informed and give their consent, otherwise parents

must consent (Doyal 1997). People get very upset and angry if they think
that they have been experimented on without their knowledge and
consent, or if they feel that they have been tricked into it without being
fully informed. A group of women with cervical cancer were given an
experimental radiation treatment, which resulted in severe damage,
without proper information (Anon 1997). They formed a group which they
called RAGE, which speaks for itself.
Patients are sometimes recruited into trials when they are very distressed
and very vulnerable. If possible they should have time to think about the
trial and discuss it with their family. Patients in trials are often not at all
clear about what is going on and have wrong ideas about what is
happening (Snowdon et al. 1997). They may be unable to recall giving
their consent, and deny having given it. They should always be asked to
sign consent forms and should be given a separate patient information
sheet and a copy of the form to keep.
A difficulty arises with the randomized consent design (Zelen 1979,
1992). In this, we have a new, active treatment and either no control
treatment or usual care. We randomize subjects to active or control. We
then offer the new treatment to the active group, who may refuse, and the
control group gets usual care. The active group is asked to consent to the
new treatment and all subjects are asked to consent to any measurement
required. They might be told that they are in a research study, but not that
they have been randomized. Thus only patients in the active group can
refuse the trial, though all can refuse measurement. Analysis is then by
intention to treat (§2.5). For example, Dennis et al. (1997) wanted to
evaluate a stroke family care worker. They randomized patients without
their knowledge, then asked them to consent to follow-up consisting of
interviews by a researcher. The care worker visited those patients and
their families who had been randomized to her. McLean (1997) argued
that if patients could not be informed about the randomization without
jeopardizing the trial, the research should not be done. Dennis (1997)
argued that to ask for consent to randomization might bias the results,
because patients who did not receive the care worker might be resentful
and be harmed by this. My own view is that we should not allow one
ethical consideration, informed consent, to outweigh all others and this
design can be acceptable (Bland 1997).
There is a special problem in cluster randomized trials. Patients cannot
consent to randomization, but only to treatment. In a trial where general
practices are allocated to offer health checks, for example, patients can
consent to the health checks only if they are in a health check practice,
though all would have to consent to an end of trial assessment.
Research on human subjects should always be approved by an
independent ethics committee, whose role is to represent the interests of
the research subject. Where such a system is not in place, terrible things
can happen. In the USA, research can be carried out without ethical
approval if the subjects are private patients in a private hospital without
any public funding, and no new drug or device is used. Under these
circumstances, plastic surgeons carried out a trial comparing two
methods performing face-lifts, one on each side of the face,
without patients' consent (Bulletin of Medical Ethics 1998).
2M Multiple choice questions 1 to 6

(Each branch is either true or false)
1. When testing a new medical treatment, suitable control groups
include patients who:
(a) are treated by a different doctor at the same time;
(b) are treated in a different hospital;
(c) are not willing to receive the new treatment;
(d) were treated by the same doctor in the past;
(e) are not suitable for the new treatment.
View Answer
2. In an experiment to compare two treatments, subjects are

allocated using random numbers so that:
(a) the sample may be referred to a known population;
(b) when deciding to admit a subject to the trial, we do not know
which treatment that subject would receive;
(c) the subjects will get the treatment best suited to them;
(d) the two groups will be similar, apart from treatment;
(e) treatments may be assigned according to the characteristics of
the subject.
View Answer
3. In a double blind clinical trial:

(a) the patients do not know which treatment they receive;
(b) each patient receives a placebo;
(c) the patients do not know that they are in a trial;
(d) each patient receives both treatments;
(e) the clinician making assessment does not know which
treatment the patient receives.
View Answer
4. In a trial of a new vaccine, children were assigned at random to a

‘vaccine’ and a ‘control’ group. The ‘vaccine’ group were offered
vaccination, which two-thirds accepted. The control group were
offered nothing:
(a) the group which should be compared to the controls is all
children who accepted vaccination;
(b) those refusing vaccination should be included in the control
group;
(c) the trial is double blind;
(d) those refusing vaccination should be excluded;
(e) the trial is useless because not all the treated group were
vaccinated.
View Answer
Table 2.11. Method of delivery in the KYM study
Accepted Refused Control

Method of KYM KYM women
delivery
% n % n % n
Normal 80.7 352 69.8 30 74.8 354
Instrumental 12.4 54 14.0 6 17.8 84

Caesarian 6.9 30 16.3 7 7.4 35
5. Cross-over designs for clinical trials:

(a) may be used to compare several treatments;
(b) involve no randomization;
(c) require fewer patients than do designs comparing independent
groups;
(d) are useful for comparing treatments intended to alleviate
chronic symptoms;
(e) use the patient as his own control.
View Answer
6. Placebos are useful in clinical trials:

(a) when two apparently similar active treatments are to be
compared;
(b) to guarantee comparability in non-randomized trials;
(c) because the fact of being treated may itself produce a
response;
(d) because they may help to conceal the subject's treatment from
assessors;
(e) when an active treatment is to be compared to no treatment.
View Answer
2E Exercise: The ‘Know Your Midwife’ trial

The Know Your Midwife (KYM) scheme was a method of delivering
maternity care for low-risk women. A team of midwives ran a clinic, and
the same midwife would give all antenatal care for a mother, deliver the
baby, and give postnatal care. The KYM scheme was compared to
standard antenatal care in a randomized trial (Flint and Poulengeris
1986). It was thought that the scheme would be very attractive to women
and that if they knew it was available they might be reluctant to be
randomized to standard care. Eligible women were randomized without
their knowledge to KYM or to the control group, who received the
standard antenatal care provided by St. George's Hospital. Women
randomized to KYM were sent a letter explaining the KYM scheme and
inviting them to attend. Some women declined and attended the standard
clinic instead. The mode of delivery for the women is shown in Table
2.11. Normal obstetric data were recorded on all women, and the women
were asked to complete questionnaires (which they could refuse) as part
of a study of antenatal care, though they were not told about the trial.
1. The women knew what type of care they were receiving. What
effect might this have on the outcome?
View Answer
2. What comparison should be made to test whether KYM has any

effect on method of delivery?
View Answer
3. Do you think it was ethical to randomize women without their

knowledge?
View Answer
> Table of Contents > 3 - Sampling and observational studies
3
Sampling and observational studies
3.1 Observational studies

In this chapter we shall be concerned with observational studies. Instead
of changing something and observing the result, as in an experiment or
clinical trial, we observe the existing situation and try to understand what
is happening. Most medical studies are observational, including research
into human biology in healthy people, the natural history of disease, the
causes and distribution of disease, the quality of measurement, and the
process of medical care.
One of the most important and difficult tasks in medicine is to determine
the causes of disease, so that we may devise methods of prevention. We
are working in an area where experiments are often neither possible nor
ethical. For example. to determine that cigarette smoking caused cancer,
we could imagine a study in which children were randomly allocated to a
‘twenty cigarettes a day for fifty years’ group and a ‘never smoke in your
life’ group. All we would have to do then would be to wait for the death
certificates. However, we could not persuade our subjects to stick to the
treatment and deliberately setting out to cause cancer is hardly ethical.
We must therefore observe the disease process as best we can. by
watching people in the wild rather than under laboratory conditions.
We can never come to an unequivocal conclusion about causation in
observational studies. The disease effect and possible cause do not exist
in isolation but in a complex interplay of many intervening factors. We
must do our best to assure ourselves that the relationship we observe is
not the result of some other factor acting on both ‘cause’ and ‘effect’. For
example, it was once thought that the African fever tree, the yellow-
barked acacia, caused malaria, because those unwise enough to camp
under them were likely to develop the disease. This tree grows by water
where mosquitos breed, and provides an ideal day-time resting place for
these insects, whose bite transmits the plasmodium parasite which
produces the disease. It was the water and the mosquitos which were the
important factors, not the tree. Indeed, the name ‘malaria’ comes from a
similar incomplete observation. It means ‘bad air’ and comes from the
belief that the disease was caused by the air in low-lying, marshy places,
where the mosquitos bred. Epidemiological study designs must try to
deal with the complex interrelationships between different factors in order
to deduce the true mechanism of disease causation. We also use a
number of different approaches to the study of these problems, to see
whether all produce the same answer.
There are many problems in interpreting observational studies, and the
medical consumer of such research must be aware of them. We have no
better way to tackle many questions and so we must make the best of
them and look for consistent relationships which stand up to the most
severe examination. We can also look for confirmation of our findings
indirectly, from animal models and from dose-response relationships in
the human population. However, we must accept that perfect proof is
impossible and it is unreasonable to demand it. Sometimes, as with
smoking and health, we must act on the balance of the evidence.
We shall start by considering how to get descriptive information about
populations in which we are interested. We shall go on to the problem of
using such information to study disease processes and the possible
causes of disease.
3.2 Censuses
One simple question we can ask about any group of interest is how many
members it has. For example, we need to know how many people live in
a country and how many of them are in various age and sex categories,
in order to monitor the changing pattern of disease and to plan medical
services. We can obtain it by a census. In a census, the whole of a
defined population is counted. In the United Kingdom, as in many
developed countries, a population census is held every ten years. This is
done by dividing the entire country into small areas called enumeration
districts, usually containing between 100 and 200 households. It is the
responsibility of an enumerator to identify every household in the district
and ensure that a census form is completed, listing all members of the
household and providing a few simple pieces of information. Even though
completion of the census form is compelled by law, and enormous effort
goes into ensuring that every household is included, there are
undoubtedly some who are missed. The final data, though extremely
useful, are not totally reliable.
The medical profession takes part in a massive, continuing census of
deaths, by providing death certificates for each death which occurs,
including not only the name of the deceased and cause of death, but also
details of age, sex, place of residence and occupation. Census methods
are not restricted to national populations. They can be used for more
specific administrative purposes too. For example, we might want to
know how many patients are in a particular hospital at a particular time,
how many of them are in different diagnostic groups, in different age/sex
groups, and so on. We can then use this information together with
estimates of the death and discharge rates to estimate how many beds
these patients will occupy at various times in the future (Bewley et al.
1975, 1981).
3.3 Sampling
A census of a single hospital can only give us reliable information about
that hospital. We cannot easily generalize our results to hospitals in
general. If we want to obtain information about the hospitals of the United
Kingdom, two courses are open to us: we can study every hospital, or we
can take a representative sample of hospitals and use that to draw
conclusions about hospitals as a whole.
Most statistical work is concerned with using samples to draw

conclusions about some larger population. In the clinical trials described
in Chapter 2, the patients act as a sample from a larger population
consisting of all similar patients and we do the trial to find out what would
happen to this larger group were we to give them a new treatment.
The word ‘population’ is used in common speech to mean ‘all the people
living in an area’, frequently of a country. In statistics, we define the term
more widely. A population is any collection of individuals in which we
may be interested, where these individuals may be anything, and the
number of individuals may be finite or infinite. Thus, if we are interested
in some characteristics of the British people, the population is ‘all people
in Britain’. If we are interested in the treatment of diabetes the population
is ‘all diabetics’. If we are interested in the blood pressure of a particular
patient, the population is ‘all possible measurements of blood pressure in
that patient’. If we are interested in the toss of two coins, the population is
‘all possible tosses of two coins’. The first two examples are finite
populations and could in theory if not practice be completely examined;
the second two are infinite populations and could not. We could only ever
look at a sample, which we will define as being a group of individuals
taken from a larger population and used to find out something about that
population.
How should we choose a sample from a population? The problem of
getting a representative sample is similar to that of getting comparable
groups of patients discussed in §2.1,2,3. We want our sample to be
representative, in some sense, of the population. We want it to have all
the characteristics in terms of the proportions of individuals with particular
qualities as has the whole population. In a sample from a human
population, for example, we want the sample to have about the same
proportion of men and women as in the population, the same proportions
in different age groups, in occupational groups, with different diseases,
and so on. In addition, if we use a sample to estimate the proportion of
people with a disease, we want to know how reliable this estimate is, how
far from the proportion in the whole population the estimate is likely to be.
It is not sufficient to choose the most convenient group. For example, if
we wished to predict the results of an election, we would not take as our
sample people waiting in bus queues. These may be easy to interview, at
least until the bus comes, but the sample would be heavily biased
towards those who cannot afford cars and thus towards lower income
groups. In the same way, if we wanted a sample of medical students we
would not take the front two rows of the lecture theatre. They may be
unrepresentative in having an unusually high thirst for knowledge, or poor
eyesight.
How can we choose a sample which does not have a built-in bias? We
might divide our population into groups, depending on how we think
various characteristics will affect the result. To ask about an election, for
example, we might group the population according to age, sex and social
class. We then choose a number of people in each group by knocking on
doors until the quota is made up, and interview them. Then, knowing the
distributions of these categories in the population (from census data, etc.)
we can get a far better picture of the
views of the population. This is called quota sampling. In the same way
we could try to choose a sample of rats by choosing given numbers of
each weight, age, sex, etc. There are difficulties with this approach. First,
it is rarely possible to think of all the relevant classifications. Second, it is
still difficult to avoid bias within the classifications, by picking interviewees
who look friendly, or rats which are easy to catch. Third, we can only get
an idea of the reliability of findings by repeatedly doing the same type of
survey, and of the representativeness of the sample by knowing the true
population values (which we can actually do in the case of elections), or
by comparing the results with a sample which does not have these
drawbacks. Quota sampling can be quite effective when similar surveys
are made repeatedly as in opinion polls or market research. It is less
useful for medical problems, where we are continually asking new
questions. We need a method where bias is avoided and where we can
estimate the reliability of the sample from the sample itself. As in §2.2, we
use a random method: random sampling.
3.4 Random sampling

The problem of obtaining a sample which is representative of a larger
population is very similar to that of allocating patients into two
comparable groups. We want a way of choosing members of the sample
which does not depend on their own characteristics. The only way to be
sure of this is to select them at random, so that whether or not each
member of the population is chosen for the sample is purely a matter of
chance.
For example, to take a random sample of 5 students from a class of 80,
we could write all the names on pieces of paper, mix them thoroughly in a
hat or other suitable container, and draw out five. All students have the
same chance of being chosen, and so we have a random sample. All
samples of 5 students are equally likely, too, because each student is
chosen quite independently of the others. This method is called simple
random sampling.
As we have seen in §2.2, physical methods of randomizing are often not
very suitable for statistical work. We usually use tables of random digits,
such as Table 2.3. or random numbers generated by a computer
program. We could use Table 2.3 to draw our sample of 5 from 80
students in several ways. For example, we could list the students,
numbered from 1 to 80. This list from which the sample is to be drawn is
called the sampling frame. We choose a starting point in the random
number table (Table 2.3), say row 20, column 5. This gives us the
following pairs of digits:
14 04 88 86 28 92 04 03 42 99 87 08
We could use these pairs of digits directly as subject numbers. We
choose subjects numbered 14 and 4. There is no subject 88 or 86, so the
next chosen is number 28. There is no 92, so the next is 4. We already
have this subject in the sample, so we carry on to the next pair of digits,
03. The final member of the sample is number 42. Our sample of 5
students is thus numbers 3, 4, 14, 28 and 42.
There appears to be some pattern in this sample. Two numbers are

adjacent (3 and 4) and 3 are divisible by 14 (14, 28 and 42). Random
numbers often appear to us to have pattern, perhaps because the human
mind is always looking for it. On the other hand, if we try to make the
sample ‘more random’ by replacing either 3 or 4 by a subject near the
end of the list, we are imposing a pattern of uniformity on the sample and
destroying its randomness. All groups of five are equally likely and may
happen, even 1, 2, 3, 4, 5.
This method of using the table is fine for drawing a small sample, but it
can be tedious for drawing large samples, because of the need to check
for duplicates. There are many other ways of doing it. For example, we
can drop the requirement for a sample of fixed size, and only require that
each member of the population will have a fixed probability of being in the
sample. We could draw a 5/80 = 1/16 sample of our class by using the
digits in groups to give a decimal number. say,
0.1404 0.8886 0.2892 0.0403 0.4299 0.8708
We then choose the first member of the population if 0.1404 is less than
1/16. It is not, so we do not include this member, nor the second,
corresponding to 0.8886, nor the third, corresponding to 0.2892. The
fourth corresponds to 0.0403. which is less than 1/16 (0.0625) and so the
fourth member is chosen as a member of the sample, and so on. This
method is only suitable for fairly large samples, as the size of the sample
obtained can be very variable in small sampling problems. In the example
there is a higher than 1 in 10 chance of finishing with a sample of 2 or
fewer.
As with random allocation (§2.2), random sampling is an operation ideally
suited to computers. My free program Clinstat (§1.3) provides two
random sampling schemes. Some computer programs for managing
primary care practices actually have the capacity to take a random
sample for any defined group of patients built in.
Random sampling ensures that the only ways in which the sample differs
from the population will be those due to chance. It has a further
advantage. because the sample is random, we can apply the methods of
probability theory to the data obtained. As we shall see in Chapter 8, this
enables us to estimate how far from the population value the sample
value is likely to be.
The problem with random sampling is that we must have a list of the
population from which the sample is to be drawn. Lists of populations
may be hard to find, or they may be very cumbersome. For example, to
sample the adult population in the UK, we could use the electoral roll. But
a list of some 40000000 names would be difficult to handle, and in
practice we would first take a random sample of electoral wards, and
then a random sample of electors within these wards. This is, for obvious
reasons, a multi-stage random sample. This approach contains the
element of randomness, and so samples will be representative of the
populations from which they are drawn. However, not all samples have
an equal chance of being chosen, so it is not the same as simple random
sampling.
We can also carry out sampling without a list of the population itself,
provided we have a list of some larger units which contain all the
members of the population. For example, we can obtain a random
sample of school children in an area by starting with a list of schools,
which is quite easy to come by. We then draw a simple random sample of
schools and all the children within our chosen schools form the sample of
children. This is called a cluster sample, because we take a sample of
clusters of individuals. Another example would be sampling from any
age/sex group in the general population by taking a sample of addresses
and then taking everyone at the chosen addresses who matched our
criteria.
Sometimes it is desirable to divide the population into different strata, for
example into age and sex groups, and take random samples within
these. This is rather like quota sampling, except that within the strata we
choose at random. If the different strata have different values of the
quantity we are measuring, this stratified random sampling can
increase our precision considerably. There are many complicated
sampling schemes for use in different situations. For example, in a study
of cigarette smoking and respiratory disease in Derbyshire
schoolchildren, we drew a random sample of schools, stratified by school
type (single-sex/mixed, selective/non-selective, etc.). Some schools
which took children to age 13 then fed into the same 14+ school were
combined into one sampling unit. Our sample of children was all children
in the chosen schools who were in their first secondary school year
(Banks et al. 1978). We thus had a stratified random cluster sample.
These sampling methods affect the estimate obtained. Stratification
improves the precision, cluster sampling worsens it. The sampling
scheme should be taken into account in the analysis (Cochran 1977, Kish
1994). Often it is ignored, as was done by Banks et al. (1978) (that is, by
me), but it should not be and results may be reported as being more
precise than they really are.
In §2.3 I looked at the difficulties which can arise using methods of
allocation which appear random but do not use random numbers. In
sampling, two such methods are often suggested by researchers. One is
to take every tenth subject from the list, or whatever fraction is required.
The other is to use the last digit of some reference number, such as the
hospital number, and take as the sample subjects where this is, say, 3 or
4. These sampling methods are systematic or quasi-random. It is not
usually obvious why they should not give ‘random’ samples, and it may
be that in many cases they would be just as good as random sampling.
They are certainly easier. To use them, we must be very sure that there is
no pattern to the list which could produce an unrepresentative group. If it
is possible, random sampling seems safer.
Volunteer bias can be as serious a problem in sampling studies as it is in
trials (§2.4). If we can only obtain data from a subset of our random
sample, then this subset will not be a random sample of the population.
Its members will be self selected. It is often very difficult to get data from
every member of a sample. The proportion for whom data is obtained is
called the response rate and in a sample survey of the general
population is likely to be between
70% and 80%. The possibility that those lost from the sample are
different in some way must be considered. For example, they may tend to
be ill, which can be a serious problem in disease prevalence studies. In
the school study of Banks et al. (1978), the response rate was 80%, most
of those lost being absent from school on the day. Now, some of these
absentees were ill and some were truants. Our sample may thus lead us
to underestimate the prevalence of respiratory symptoms, by omitting
sufferers with current acute disease, and the prevalence of cigarette
smoking by omitting those who have gone for a quick smoke behind the
bike sheds.
One of the most famous sampling disasters, the Literary Digest poll of
1936, illustrates these dangers (Bryson 1976). This was a poll of voting
intentions in the 1936 US presidential election, fought by Roosevelt and
Landon. The sample was a complex one. In some cities every registered
voter was included, in others one in two, and for the whole of Chicago
one in three. Ten million sample ballots were mailed to prospective
voters, but only 2.3 million, less than a quarter, were returned. Still, two
million is a lot of Americans, and these predicted a 60% vote to Landon.
In fact, Roosevelt won with 62% of the vote. The response was so poor
that the sample was most unlikely to be representative of the population,
no matter how carefully the original sample was drawn. Two million
Americans can be wrong! It is not the mere size of the sample, but its
representativeness which is important. Provided the sample is truly
representative, 2000 voters is all you need to estimate voting intentions
to within 2%, which is enough for election prediction if they tell the truth
and do not change their minds (see §18E).
3.5 Sampling in clinical and epidemiological

studies
Having extolled the virtues of random sampling and cast doubt on all
other sampling methods, I must admit that most medical data are not
obtained in this way. This is partly because the practical difficulties are
immense. To obtain a reasonable sample of the population of the UK,
anyone can get a list of electoral wards, take a random sample of them,
buy copies of the electoral rolls for the chosen wards and then take a
random sample of names from it. But suppose you want to obtain a
sample of patients with carcinoma of the bronchus. You could get a list of
hospitals easily enough and get a random sample of them, but then
things would become difficult. The names of patients will only be released
by the consultant in charge should he so wish, and you will need his
permission before approaching them. Any study of human patients
requires ethical approval, and you will need this from the ethics
committee of each of your chosen hospitals. Getting the cooperation of
so many people is a task to daunt the hardiest, and obtaining ethical
approval alone can take more than a year. In the UK, we now have a
system of multi-centre research ethics committees, but as local approval
must also be obtained the delays may still be immense.
The result of this is that clinical studies are done on the patients to hand.
I have touched on this problem in the context of clinical trials (§2.7) and
the
same applies to other types of clinical study. In a clinical trial we are

concerned with the comparison of two treatments and we hope that the
superior treatment in Stockport will also be the superior treatment in
Southampton. If we are studying clinical measurement, we can hope that
a measurement method which is repeatable in Middlesbrough will be
repeatable in Maidenhead, and that two different methods giving similar
results in one place will give similar results in another. Studies which are
not comparative give more cause for concern. The natural history of a
disease described in one place may differ in unpredictable ways from that
in another, due to differences in the environment and the genetic make
up of the local population. Reference ranges for quantities of clinical
interest, the limits within which values from most health people will lie,
may well differ from place to place.
Studies based on local groups of patients are not without value. This is
particularly so when we are concerned with comparisons between
groups, as in a clinical trial, or relationships between different variables.
However, we must always bear the limitations of the sampling method in
mind when interpreting the results of such studies.
In general, most medical research has to be carried out using samples
drawn from populations which are much more restricted than those about
which we wish to draw conclusions. We may have to use patients in one
hospital instead of all patients, or the population of a small area rather
than that of the whole country or planet. We may have to rely on
volunteers for studies of normal subjects, given most people's dislike of
having needles pushed into them and disinclination to spend hours
hooked up to batteries of instruments. Groups of normal subjects contain
medical students, nurses and laboratory staff far more often than would
be expected by chance. In animal research the problem is even worse,
for not only does one batch of one strain of mice have to represent the
whole species, it often has to represent members of a different order,
namely humans.
Findings from such studies can only apply to the population from which
the sample was drawn. Any conclusion which we come to about wider
populations, such as all patients with the disease in question, depends on
evidence which is not statistical and often unspecified, namely our
general experience of natural variability and experience of similar studies.
This may let us down, and results established in one population may not
apply to another. We have seen this in the use of BCG vaccine in India
(§2.7). It is very important wherever possible that studies should be
repeated by other workers on other populations, so that we can sample
the larger population at least to some extent.
In two types of study, case reports and case series, the subjects come
before the research, as it is suggested by their existence. There is no
sampling. They are used to raise questions rather than to answer them.
A case report is a description of a single patient whose case displays
interesting features. This is used to generate ideas and raise questions,
rather than to answer them. It clearly cannot be planned in advance; it
arises from the case. For example, Velzeboer et al. (1997) reported the
case of an 11-month-old Pakistani
girl was admitted to hospital because of drowsiness, malaise and

anorexia. She had stopped crawling or standing up and scratched her
skin continuously. All investigations were negative. Her 6-year-old sister
was then brought in with similar symptoms. (Note that there are two
patients here, but they are part of the same case.) The doctors guessed
that exposure to mercury might be to blame. When asked, the mother
reported that 2 weeks before the younger child's symptoms started,
mercury from a broken thermometer had been dropped on the carpet in
the children's room. Mercury concentration in a urine sample taken on
admission was 12.6 µg/1 (slightly above the accepted normal value of 10
µg/1). Exposure was confirmed by a high mercury concentration in her
hair. After 3 months treatment the symptoms had disappeared totally and
urinary mercury had fallen below the detection limit of 1 µg/1. This case
called into question the normal values for mercury in children.
A case series is similar to a case report, except that a number of similar
cases have been observed. For example, Shaker et al. (1997) described
15 patients examined for hypocalcaemia or skeletal disease, in whom the
diagnosis of coeliac disease was subsequently made. In 11 of them
gastrointestinal symptoms were absent or mild. They concluded that
bone loss may be a sign of coeliac disease and this diagnosis should be
considered. The design does not allow them to draw any conclusions
about how often this might happen. To do that they would have to collect
data systematically, using a cohort design (§3.7) for example.
3.6 Cross-sectional studies

One possible approach to the sampling problem is the cross-sectional
study. We take some sample or whole narrowly defined population and
observe them at one point in time. We get poor estimates of means and
proportions in any more general population, but we can look at
relationships within the sample. For example, in an epidemiological study,
Banks et al. (1978) gave question-naires to all first year secondary school
boys in a random sample of schools in Derbyshire (§3.4). Among boys
who had never smoked, 3% reported a cough first thing in the morning,
compared to 19% of boys who said that they smoked one or more
cigarettes per week. The sample was representative of boys of this age
in Derbyshire who answer questionnaires, but we want our conclusions to
apply at least to the United Kingdom, if not the developed world or the
whole planet. We argue that although the prevalence of symptoms and
the strength of the relationship may vary between populations, the
existence of the relationship is unlikely only to occur in the population
studied. We cannot conclude that smoking causes respiratory symptoms.
Smoking and respiratory symptoms may not be directly related, but may
both be related to some other factor. A factor related to both possible
cause and possible effect is called confounding. For example, children
whose parents smoke may be more likely to develop respiratory
symptoms, because of passive inhalation of their parent's smoke, and
also be more influenced to try smoking themselves. We can test this by
looking separately at the relationship between the child's smoking and
symptoms for those
whose parents are not smokers, and for those whose parents are
smokers. As Figure 3.1 shows, this relationship in fact persisted and
there was no reason to suppose that a third causal factor was at work.
Fig. 3.1. Prevalence of self-reported morning cough in Derbyshire
schoolboys, by their own and their parents' cigarette smoking (Bland
et al. 1978)
Most diseases are not suited to this simple cross-sectional approach,

because they are rare events. For example, lung cancer accounts for 9%
of male deaths in the UK (OPCS, DH2 No.7), and so is a very important
disease. However the proportion of people who are known to have the
disease at any given time, the prevalence, is quite low. Most deaths from
lung cancer take place after the age of 45, so we will consider a sample
of men aged 45 and over. The average remaining life span of these men,
in which they could contract lung cancer, will be about 30 years. The
average time from diagnosis to death is about a year, so of those who will
contract lung cancer only 1/30 will have been diagnosed when the
sample is drawn. Only 9% of the sample will develop lung cancer
anyway, so the proportion with the disease at any time is 1/30 × 9% =
0.3% or 3 per thousand. We would need a very large sample indeed to
get a worthwhile number of lung cancer cases.
Cross-sectional designs are used in clinical studies also. For example,
Rodin et al. (1998) studied polycystic ovary disease (PCO) in a random
sample of Asian women from the lists of local general practices and from
a local translating service. We found that 52% of the sample had PCO,
very high compared to that found in other UK samples. However, this
would not provide a good estimate for Asian women in general, because
there may be many differences between this sample, such as their
regions of origin, and Asian women living elsewhere. We also found that
PCO women had higher fasting glucose levels than non-PCO women. As
this is a comparison within the sample, it seems plausible to conclude
that among Asian women PCO tends to be associated with raised
glucose. We cannot say whether PCO raises glucose or whether raised
glucose increases the risk of PCO, because they are measured at the
same time.
Table 3.1. Standardized death rates per year per 1000

men aged 35 or more in relation to most recent amount
smoked, 53 months follow-up (Doll and Hill 1956)
Death rate among

Men smoking a daily
Cause of average weight of
death Non- tobacco of
Smokers
smokers
15–
1–14g
24g
Lung 0.07 0.90 0.47 0.86

cancer
Other 2.04 2.02 2.01 1.56

cancer
Other 0.81 1.13 1.00 1.11

respiratory
Coronary 4.22 4.87 4.64 4.60

thrombosis
Other 6.11 6.89 6.82 6.38

causes
All causes 13.25 15.78 14.92 14.49
3.7 Cohort studies

One way of getting round the problem of the small proportion of people
with the disease of interest is the cohort study. We take a group of
people, the cohort, and observe whether they have the suspected causal
factor. We then follow them over time and observe whether they develop
the disease. This is a prospective design, as we start with the possible
cause and see whether this leads to the disease in the future. It is also
longitudinal, meaning that subjects are studied at more than one time. A
cohort study usually takes a long time, as we must wait for the future
event to occur. It involves keeping track of large numbers of people,
sometimes for many years, and often very large numbers must be
included in the sample to ensure sufficient numbers will develop the
disease to enable comparisons to be made between those with and
without the factor.
A noted cohort study of mortality in relation to cigarette smoking was
carried out by Doll and Hill (1956). They sent a questionnaire to all
members of the medical profession in the UK, who were asked to give
their name, address, age and details of current and past smoking habits.
The deaths among this group were recorded. Only 60% of doctors
cooperated, so in fact the cohort does not represent all doctors. The
results for the first 53 months are shown in Table 3.1.
The cohort represents doctors willing to return questionnaires, not people
as a whole. We cannot use the death rates as estimates for the whole
population, or even for all doctors. What we can say is that, in this group,
smokers were far more likely than non-smokers to die from lung cancer. It
would be surprising if this relationship were only true for doctors, but we
cannot definitely say that this would be the case for the whole population,
because of the way the sample has been chosen.
We also have the problem of other intervening variables. Doctors were
not allocated to be smokers or non-smokers as in a clinical trial; they
chose for themselves. The decision to begin smoking may be related to
many things (social factors, personality factors, genetic factors) which
may also be related to lung cancer. We must consider these alternative
explanations very carefully before coming to any conclusion about the
causes of cancer. In this study there were no data to test such
hypotheses.
The same technique is used, usually on a smaller scale, in clinical

studies. For example, Casey et al. (1996) studied 55 patients with very
severe rheumatoid arthritis affecting the spine and the use of all four
limbs. These patients were operated on in an attempt to improve their
condition and their subsequent progress was monitored. We found that
only 25% had a favourable outcome. We could not conclude from this
that surgery would be worthwhile in 25% of such patients generally. Our
patients might have been particularly ill or unusually fit, our surgeons
might be the best or they might be (relatively speaking) ham-fisted
butchers. However, we compared these results with other studies
published in the medical literature, which were similar. These studies
together gave a much better sample of such patients than any study
alone could do (see §17.11, meta-analysis). We looked at which
characteristics of the patients predicted a good or bad outcome and
found that the area of cross-section of the spinal cord was the important
predictor. We were much more confident of this finding, because it arose
from studying relationships between variables within the sample. It
seems quite plausible from this study alone that patients whose spinal
cords have already atrophied are unlikely to benefit from surgery.
3.8 Case-control studies

Another solution to the problem of the small number of people with the
disease of interest is the case-control study. In this we start with a
group of people with the disease, the cases. We compare them to a
second group without the disease, the controls. In an epidemiological
study, we then find the exposure of each subject to the possible
causative factor and see whether this differs between the two groups.
Before their cohort study, Doll and Hill (1950) carried out a case-control
study into the aetiology of lung cancer. Twenty London hospitals notified
all patients admitted with carcinoma of the lung, the cases. An interviewer
visited the hospital to interview the case, and, at the same time, selected
a patient with diagnosis other than cancer, of the same sex and within the
same 5 year age group as the case, in the same hospital at the same
time, as a control. When more than one suitable patient was available,
the patient chosen was the first in the ward list considered by the ward
sister to be fit for interview. Table 3.2 shows the relationship between
smoking and lung cancer for these patients. A smoker was anyone who
had smoked as much as one cigarette a day for as much as one year. It
appears that cases were more likely than controls to smoke cigarettes.
Doll and Hill concluded that smoking is an important factor in the
production of carcinoma of the lung.
The case-control study is an attractive method of investigation, because
of its relative speed and cheapness compared to other approaches.
However, there are difficulties in the selection of cases, the selection of
controls, and obtaining the data. Because of these, case-control studies
sometimes produce contradictory and conflicting results.
The first problem is the selection of cases. This usually receives little
consideration beyond a definition of the type of disease and a statement
about the
confirmation of the diagnosis. This is understandable, as there is usually

little else that the investigators can do about it. They start with the
available set of patients. However, these patients do not exist in isolation.
They are the result of some process which has led to them being
diagnosed as having the disease and thus being available for study. For
example, suppose we suspect that oral contraceptives might cause
cancer of the breast. We have a group of patients diagnosed as having
cancer of the breast. We must ask ourselves whether any of these were
detected at a medical examination which took place because the woman
was seeing a doctor to receive a prescription. If this were so, the risk
factor (pill) would be associated with the detection of the disease rather
than its cause. This is called ascertainment bias.
Table 3.2. Numbers of smokers and non-

smokers among lung cancer patients and age
and sex matched controls with diseases other
than cancer (Doll and Hill 1950)
Non-
Smokers Total
smokers
Males
Lung cancer 2 (0.3%) 647 649

patients (99.7%)
Control 27 (4.2%) 622 649

patients (95.8%)
Females
Lung cancer 19 41 60
patients (31.7%) (68.3%)
Control 32 28 60
patients (53.3%) (46.7%)
Far more difficulty is caused by the selection of controls. We want a

group of people who do not have the disease in question, but who are
otherwise comparable to our cases. We must first decide the population
from which they are to be drawn. There are two main sources of controls:
the general population and patients with other diseases. The latter may
be preferred because of its accessibility. Now these two populations are
clearly not the same. For example, Doll and Hill (1950) gave the current
smoking habits of 1014 men and women with diseases other than cancer,
14% of whom were currently non-smokers. They commented that there
was no difference between smoking in the disease groups respiratory
disease, cardiovascular disease, gastro-intestinal disease and others.
However, in the general population the percentage of current non-
smokers was 18% for men and 59% for women (Todd 1972). The
smoking rate in the patient group as a whole was high. Since their report,
of course, smoking has been associated with diseases in each group.
Smokers get more disease and are more likely to be in hospital than non-
smokers.
Intuitively, the comparison we want to make is between people with the
disease and healthy people, not people with a lot of other diseases. We
want to find out how to prevent disease, not how to choose one disease
or another! However, it is much easier to use hospital patients as
controls. There may then be a bias because the factor of interest may be
associated with other diseases. Suppose we want to investigate the
relationship between a disease and cigarette smoking using hospital
controls. Should we exclude patients with lung cancer from the control
group? If we include them, our controls may have more smokers
than the general population, but if we exclude them we may have fewer.
This problem is usually resolved by choosing specific patient groups,
such as fracture cases, whose illness is thought to be unrelated to the
factor being investigated. In case-control studies using cancer registries,
controls are sometimes people with other forms of cancer. Sometimes
more than one control group is used.
Having defined the population we must choose the sample. There are
many factors which affect exposure to risk factors, such as age and sex.
The most straightforward way is to take a large random sample of the
control population, ascertain all the relevant characteristics, and then
adjust for differences during the analysis, using methods described in
Chapter 17. The alternative is to try to match a control to each case, so
that for each case there is a control of the same age, sex, etc. Having
done this, then we can compare our cases and controls knowing that the
effects of these intervening variables are automatically adjusted for. If we
wish to exclude a case we must exclude its control, too, or the groups will
no longer be comparable. We can have more than one control per case,
but the analysis becomes complicated.
Matching on some variables does not ensure comparability on all.
Indeed, if it did there would be no study. Doll and Hill matched on age,
sex and hospital. They recorded area of residence and found that 25% of
their cases were from outside London, compared to 14% of controls. If
we want to see whether this influences the smoking and lung cancer
relationship we must make a statistical adjustment anyway. What should
we match for? The more we match for, the fewer intervening variables
there are to worry about. On the other hand, it becomes more and more
difficult to find matches. Even matching on age and sex, Doll and Hill
could not always find a control in the same hospital, and had to look
elsewhere. Matching for more than age and sex can be very difficult.
Having decided on the matching variables we then find in the control
population all the possible matches. If there are more matches than we
need, we should choose the number required at random. Other methods,
such as that used by Doll and Hill who allowed the ward sister to choose,
have obvious problems of potential bias. If no suitable control can be
found, we can do two things. We can widen the matching criteria, say
age to within ten years rather than five, or we can exclude the case.
There are difficulties in interpreting the results of case-control studies.
One is that the case-control design is often retrospective, that is, we are
starting with the present disease state, e.g. lung cancer, and relating it to
the past, e.g. history of smoking. We may have to rely on the unreliable
memories of our subjects. This may lead both to random errors among
cases and controls and systematic recall bias, where one group, usually
the cases, recalls events better than the other. For example, the mother
of a handicapped child may be more likely than the mother of a normal
child to remember events in pregnancy which may have caused damage.
There is a problem of assessment bias in such studies, just as in clinical
trials (§2.9). Interviewers will very often know whether the interviewee is
a case or control and this may well affect the way questions are asked.
These and other considerations make case-control studies extremely
difficult to interpret. The evidence from such studies can be useful, but
data from other types of investigation must be considered, too, before
any firm conclusions are drawn.
The case-control design is used clinically to investigate the natural history
of disease by comparing patients with healthy subjects or patients with
another disease. For example, Kiely et al. (1995) were interested in
lymphatic function in inflammatory arthritis. We compared arthritis
patients (the cases) with healthy volunteers (the controls). Lymphatic flow
was measured in the arms of these subjects and the groups compared.
We found that lymphatic drainage was less in the cases than in the
control group, but this was only so for arms which were swollen
(oedematous).
3.9 * Questionnaire bias in observational studies

In observational studies, much data may have to be supplied by the
subjects themselves. The way in which a question is asked may influence
the reply. Sometimes the bias in a question is obvious. Compare these:
(a) Do you think people should be free to provide the best medical care
possible for themselves and their families, free of interference from a
State bureaucracy?
(b) Should the wealthy be able to buy a place at the head of the queue
for medical care, pushing aside those with greater need, or should
medical care be shared solely on the basis of need for it?
Version (a) expects the answer yes, version (b) expects the answer no.
We would hope not to be misled by such blatant manipulation, but the
effects of question wording can be much more subtle than this. Hedges
(1978) reports several examples of the effects of varying the wording of
questions. He asked two groups of about 800 subjects one of the
following:
(a) Do you feel you take enough care of your health, or not?
(b) Do you feel you take enough care of your health, or do you think you
could take more care of your health?
In reply to question (a), 82% said that they took enough care, whereas
only 68% said this in reply to question (b). Even more dramatic was the
difference between this pair:
(a) Do you think a person of your age can do anything to prevent ill-
health in the future or not?
(b) Do you think a person of your age can do anything to prevent ill-
health in the future, or is it largely a matter of chance?
Not only was there a difference in the percentage who replied that they
could do something, but as Table 3.3 shows this answer was related to
age for version (a) but not for version (b). Here version (b) is ambiguous,
as it is quite possible to think that health is largely a matter of chance but
that there is still something one can do about it. Only if it is totally a
matter of chance is there nothing one can do.
Table 3.3. Replies to two similar questions

about ill health, by age (Hedges 1978)
Age (years)
16– 35– Total
34 54 55+
Can do 75% 64% 56% 65%

something (a)
Can do 45% 49% 50% 49%

something (b)
Sometimes the respondents may interpret the question in a different way

from the questioner. For example, when asked whether they usually
coughed first thing in the morning, 3.7% of the Derbyshire schoolchildren
replied that they did. When their parents were asked about the child's
symptoms 2.4% replied positively, not a dramatic difference. Yet when
asked about cough at other times in the day or at night 24.8% of children
said yes, compared to only 4.5% of their parents (Bland et al. 1979).
These symptoms all showed relationships to the child's smoking and
other potentially causal variables, and also to one another. We are forced
to admit that we are measuring something, but that we are not sure what!
Another possibility is that respondents may not understand the question
at all, especially when it includes medical terms. In an earlier study of
cigarette smoking by children, we found that 85% of a sample agreed
that smoking caused cancer, but that 41% agreed that smoking was not
harmful (Bewley et al. 1974). There are at least two possible explanations
for this: being asked to agree with the negative statement ‘smoking is not
harmful’ may have confused the children, or they may not see cancer as
harmful. We have evidence for both of these possibilities. In a repeat
study in Kent we asked a further sample of children whether they agreed
that smoking caused cancer and that ‘smoking is bad for your health’
(Bewley and Bland 1976). In this study 90% agreed that smoking causes
cancer and 91% agreed that smoking is bad for your health. In another
study (Bland et al. 1975), we asked children what was meant by the term
‘lung cancer’. Only 13% seemed to us to understand and 32% clearly did
not, often saying ‘I don't know’. They nearly all knew that lung cancer was
caused by smoking, however.
The setting in which a question is asked may also influence replies.
Opinion pollsters International Communications and Market Research
conducted a poll in which half the subjects were questioned by
interviewers about their voting preference and half were given a secret
ballot (McKie 1992). By each method 33% chose ‘Labour’, but 28%
chose ‘Conservative’ at interview and 7% would not say, whereas 35%
chose ‘Conservative’ by secret ballot and only 1% would not say. Hence
the secret method produced a Conservative majority, as at the then
recent general election, and the open interview a Labour majority. For
another example, Sibbald et al. (1994) compared two random samples of
GPs. One sample were approached by post and then by telephone if they
did not reply after two reminders, and the other were contacted directly
by telephone. Of the predominantly postal sample, 19% reported that
they provided counselling themselves, compared to 36% of the telephone
sample, and 14% reported that
their health visitor provided counselling compared to 30% of the

telephone group. Thus the method of asking the question influenced the
answer. One must be very cautious when interpreting questionnaire
replies.
Fig. 3.2. Volatile substance abuse mortality and unemployment in
the counties of Great Britain (The area of the circle is proportional to
the population of the county, so reflects the importance of the
observation)
Often the easiest and best method, if not the only method, of obtaining
data about people is to ask them. When we do it, we must be very careful
to ensure that questions are straightforward, unambiguous and in
language the respondents will understand. If we do not do this then
disaster is likely to follow.
3.10 * Ecological studies

Ecology is the study of living things in relation to their environment. In
epidemiology, an ecological study is one where the disease is studied in
relation to characteristics of the communities in which people live. For
example, we might take the death rates from heart disease in several
countries and see whether this is related to the national annual
consumption of animal fat per head.
Esmail et al. (1977) carried out an ecological study of factors related to
deaths from volatile substance abuse (VSA, also called solvent abuse,
inhalant abuse or glue sniffing). The observational units were the
administrative counties of Great Britain. The deaths were obtained from a
national register of deaths held at St. George's and the age and sex
distribution in each county from national census data. These were used
to calculate an index of mortality adjusted for age, the standardized
mortality ratio (§16.3). Indicators of social deprivation were also obtained
from census data. Figure 3.2 shows the relationship between VSA
mortality and unemployment in the counties. Clearly, there is a
relationship. The mortality is higher in counties where unemployment is
high.
Relationships found in ecological studies are indirect. We must not
conclude that there is a relationship at the level of the person. This is the
ecological
fallacy. For example, we cannot conclude from Figure 3.2 that
unemployed people are at a greater risk of dying from VSA than the
employed. The peak age for VSA death is among schoolchildren, who
are not included in the unemployment figures. It is not the unemployed
people who are dying. Unemployment is just one indicator of social
deprivation, and VSA deaths are associated with many of them.
Ecological studies can be useful to generate hypotheses. For example,
the observation that hypertension is common in countries where there is
a high intake of dietary salt might lead us to investigate the salt
consumption and blood pressure of individual people, and a relationship
there might in turn lead to dietary interventions. These leads often turn
out to be false, however, and the ecological study alone is never enough.

7. In statistical terms, a population:
(a) consists only of people;
(b) may be finite;
(c) may be infinite;
(d) can be any set of things in which we are interested;
(e) may consist of things which do not actually exist.
View Answer
8. A one day census of in-patients in a psychiatric hospital could:

(a) give good information about the patients in that hospital at that
time;
(b) give reliable estimates of seasonal factors in admissions;
(c) enable us to draw conclusions about the psychiatric hospitals
of Britain;
(d) enable us to estimate the distribution of different diagnoses in
mental illness in the local area;
(e) tell us how many patients there were in the hospital.
View Answer
9. In simple random sampling:

(a) each member of the population has an equal chance of being
chosen;
(b) adjacent members of the population must not be chosen;
(c) likely errors cannot be estimated;
(d) each possible sample of the given size has an equal chance of
being chosen;
(e) the decision to include a subject in the sample depends only on
the subject's own characteristics.
View Answer
10. Advantages of random sampling include:

(a) it can be applied to any population;
(b) likely errors can be estimated;
(c) it is not biassed;
(d) it is easy to do;
(e) the sample can be referred to a known population.
View Answer
11. In a case-control study to investigate whether eczema in

children is related to cigarette smoking by their parents:
(a) parents would be asked about their smoking habits at the
child's birth and the child observed for subsequent development of
eczema;
(b) children of a group of parents who smoke would be compared
to children of a group of parents who are non-smokers;
(c) parents would be asked stop to smoking to see whether their
children's eczema was reduced;
(d) the smoking habits of the parents of a group of children with
eczema would be compared to the smoking habits of the parents
of a group of children without eczema;
(e) parents would be randomly allocated to smoking or non-
smoking groups.
View Answer
12. To examine the relationship between alcohol consumption and

cancer of the oesophagus, feasible studies include:
(a) questionnaire survey of a random sample from the electoral
role;
(b) comparison of history of alcohol consumption between a group
of oesophageal cancer patients and a group of healthy controls
matched for age and sex;
(c) comparison of current oesophageal cancer rates in a group of
alcoholics and a group of teetotallers;
(d) comparison by questionnaire of history of alcohol consumption
between a group of oesophageal cancer patients and a random
sample from the electoral role in the surrounding district;
(e) comparison of death rates due to cancer of the oesophagus in
a large sample of subjects whose alcohol consumption has been
determined in the past.
View Answer
13. * In a study of hospital patients, 20 hospitals were chosen at

random from a list of all hospitals. Within each hospital, 10% of
patients were chosen at random:
(a) the sample of patients is a random sample;
(b) all hospitals had an equal chance of being chosen;
(c) all hospital patients had an equal chance of being chosen at
the outset;
(d) the sample could be used to make inferences about all hospital
patients at that time;
(e) all possible samples of patients had an equal chance of being
chosen.
View Answer
Table 3.4. Doorstep delivery of milk bottles

and exposure to bird attack
No. (%) exposed

Cases Controls
Doorstep milk delivery 29 47

(91%) (73%)
Previous milk bottle attack 26 25

by birds (81%) (39%)
Milk bottle attack in week 26 5 (8%)

before illness (81%)
Protective measures taken 6 14

(19%) (22%)
Handling attacked milk 17 5 (8%)

bottle in week before illness (53%)
Drinking milk from attacked 25 5 (8%)

bottle in week before illness (80%)
Table 3.5. Frequency of bird attacks on milk

bottles
Number of days of week

Cases Controls
when attacks took place
0 3 42
1–3 11 3
4–5 5 1
6–7 10 1
3E Exercise: Campylobacter jejuni infection

Campylobacter jejuni is a bacterium causing gastro-intestinal illness,
spread by the faecal-oral route. It infects many species, and human
infection has been recorded from handling pet dogs and cats, handling
and eating chicken and other meats, and via milk and water supplies.
Treatment is by antibiotics.
In May, 1990, there was a fourfold rise in the isolation rate of C. jejuni in
the Ogwr District, Mid-Glamorgan. The mother of a young boy admitted
to hospital with febrile convulsions resulting from C. jejuni infection
reported that her milk bottles had been attacked by birds during the week
before her son's illness, a phenomenon which had been associated with
campylobacter infection in another area. This observation, with the rise in
C. jejuni, prompted a case-control study (Southern et al. 1990).
A ‘case’ was defined as a person with laboratory confirmed C. jejuni
infection with onset between May 1 and June 1 1990, resident in an area
with Bridgend at its centre. Cases were excluded if they had spent one or
more nights away from this area in the week before onset, if they could
have acquired the infection elsewhere, or were members of a household
in which there had been a case of diarrhea in the preceding four weeks.
The controls were selected from the register of the general practice of the
case, or in a few instances from practices serving the same area. Two
controls were selected for each case, matched for sex, age (within 5
years), and area of residence.
Cases and controls were interviewed by means of a standard
questionnaire at home or by telephone. Cases were asked about their
exposure to various
factors in the week before the onset of illness. Controls were asked the
same questions about the corresponding week for their matched cases. If
a control or member of his or her family had had diarrhea lasting more
than 3 days in the week before or during the illness of the respective
case, or had spent any nights during that week away from home, another
control was found. Evidence of bird attack included the pecking or tearing
off of milk bottle tops. A history of bird attack was defined as a previous
attack at that house.
Fifty-five people with Campylobacter infection resident in the area were
reported during the study period. Of these, 19 were excluded and 4 could
not be interviewed, leaving 32 cases and 64 matched controls. There
was no difference in milk consumption between cases and controls, but
more cases than controls reported doorstep delivery of bottled milk,
previous milk bottle attack by birds, milk bottle attack by birds in the index
week, and handling or drinking milk from an attacked bottle (Table 3.4).
Cases reported bird attacks more frequently than controls (Table 3.5).
Controls were more likely to have protected their milk bottles from attack
or to have discarded milk from attacked bottles. Almost all subjects
whose milk bottles had been attacked mentioned that magpies and
jackdaws were common in their area, though only 3 had actually
witnessed attacks and none reported bird droppings near bottles.
None of the other factors investigated (handling raw chicken; eating
chicken bought raw; eating chicken, beef or ham bought cooked; eating
out; attending barbecue; cat or dog in the house; contact with other cats
or dogs; and contact with farm animals) were significantly more common
in controls than cases. Bottle attacks seemed to have ceased when the
study was carried out, and no milk could be obtained for analysis.
1. What problems were there in selecting cases?
View Answer
2. What problems were there in the selection of controls?

View Answer
3. Are there any problems about data collection?

View Answer
4. From the above, do you think there is convincing evidence that

bird attacks on milk bottles cause campylobacter infection?
View Answer
5. What further studies might be carried out?

View Answer
> Table of Contents > 4 - Summarizing data
4
Summarizing data
4.1 Types of data

In Chapters 2 and 3 we looked at ways in which data are collected. In this
chapter we shall see how data can be summarized to help to reveal
information they contain. We do this by calculating numbers from the
data which extract the important material. These numbers are called
statistics. A statistic is anything calculated from the data alone.
It is often useful to distinguish between three types of data: qualitative,
discrete quantitative and continuous quantitative. Qualitative data arise
when individuals may fall into separate classes. These classes may have
no numerical relationship with one another at all, e.g. sex: male, female;
types of dwelling: house, maisonette, flat, lodgings; eye colour: brown,
grey, blue, green, etc. Quantitative data are numerical, arising from
counts or measurements. If the values of the measurements are integers
(whole numbers), like the number of people in a household, or number of
teeth which have been filled, those data are said to be discrete. If the
values of the measurements can take any number in a range, such as
height or weight, the data are said to be continuous. In practice there is
overlap between these categories. Most continuous data are limited by
the accuracy with which measurements can be made. Human height, for
example, is difficult to measure more accurately than to the nearest
millimetre and is more usually measured to the nearest centimetre. So
only a finite set of possible measurements is actually available, although
the quantity ‘height’ can take an infinite number of possible values, and
the measured height is really discrete. However, the methods described
below for continuous data will be seen to be those appropriate for its
analysis.
We shall refer to qualities or quantities such as sex, height, age, etc. as
variables, because they vary from one member of a sample to another. A
qualitative variable is also termed a categorical variable or an attribute.
We shall use these terms interchangeably.
4.2 Frequency distributions

When data are purely qualitative, the simplest way to deal with them is to
count the number of cases in each category. For example, in the analysis
of the census of a psychiatric hospital population (§3.2), one of the
variables of interest was the patient's principal diagnosis (Bewley et al.
1975). To summarize these data,
we count the number of patients having each diagnosis. The results are
shown in Table 4.1. The count of individuals having a particular quality is
called the frequency of that quality. For example, the frequency of
schizophrenia is 474. The proportion of individuals having the quality is
called the relative frequency or proportional frequency. The relative
frequency of schizophrenia is 474/1467 = 0.32 or 32%. The set of
frequencies of all the possible categories is called the frequency
distribution of the variable.
Table 4.1. Principal diagnosis of patients in

Tooting Bec Hospital
Diagnosis Number of patients
Schizophrenia 474
Affective disorders 277

Organic brain syndrome 405
Subnormality 58
Alcoholism 57
Other and not known 196
Total 1467
Table 4.2. Likelihood of discharge of patients in Tooting

Hospital
Relative Cumulative
Discharge Frequency
frequency frequency
Unlikely 871 0.59 871
Possible 339 0.23 1210
Likely 257 0.18 1467
Total 1467 1.00 1467

In this census we assessed whether patients were ‘likely to be
discharged’, ‘possibly to be discharged’ or ‘unlikely to be discharged’.
The frequencies of these categories are shown in Table 4.2. Likelihood of
discharge is a qualitative variable, like diagnosis, but the categories are
ordered. This enables us to use another set of summary statistics, the
cumulative frequencies. The cumulative frequency for a value of a
variable is the number of individuals with values less than or equal to that
value. Thus, if we order likelihood of discharge from ‘unlikely’, through
‘possibly’ to ‘likely’ the cumulative frequencies are 871, 1210 (= 871 +
339) and 1467. The relative cumulative frequency for a value is the
proportion of individuals in the sample with values less than or equal to
that value. For the example they are 0.59 (= 871/1467), 0.82 and 1.00.
Thus we can see that the proportion of patients for whom discharge was
not thought likely was 0.82 or 82%.
As we have noted, likelihood of discharge is a qualitative variable, with
ordered categories. Sometimes this ordering is taken into account in
analysis, sometimes not. Although the categories are ordered these are
not quantitative data. There is no sense in which the difference between
‘likely’ and ‘possibly’ is the same as the difference between ‘possibly’ and
‘unlikely’.
Table 4.3. Parity of 125 women attending antenatal

clinics at St. George's Hospital
Relative
Relative
Cumulative cumulative
Parity Frequency frequency
frequency frequency
(per cent)
(per cent)
0 59 47.2 59 47.2
1 44 35.2 103 82.4
2 14 11.2 117 93.6
3 3 2.4 120 96.0
4 4 3.2 124 99.2
5 1 0.8 125 100.0
Total 125 100.0 125 100.0
Table 4.4. FEV1 (litres) of 57 male medical stude
2.85 3.19 3.50 3.69 3.90 4.14 4.32 4.50
2.85 3.20 3.54 3.70 3.96 4.16 4.44 4.56
2.98 3.30 3.54 3.70 4.05 4.20 4.47 4.68
3.04 3.39 3.57 3.75 4.08 4.20 4.47 4.70
3.10 3.42 6.60 3.78 4.10 4.30 4.47 4.71

3.10 3.48 3.60 3.83 4.14 4.30 4.50 4.78
Table 4.3 shows the frequency distribution of a quantitative variable,

parity. This shows the number of previous pregnancies for a sample of
women booking for delivery at St. George's Hospital. Only certain values
are possible, as the number of pregnancies must be an integer, so this
variable is discrete. The frequency of each separate value is given.
Table 4.4 shows a continuous variable, forced expiratory volume in one
second (FEV1) in a sample of male medical students. As most of the
values occur only once, to get a useful frequency distribution we need to
divide the FEV1 scale into class intervals, e.g. from 3.0 to 3.5, from 3.5 to
4.0, and so on, and count the number of individuals with FEV1s in each
class interval. The class intervals should not overlap, so we must decide
which interval contains the boundary point to avoid it being counted
twice. It is usual to put the lower boundary of an interval into that interval
and the higher boundary into the next interval. Thus the interval starting
at 3.0 and ending at 3.5 contains 3.0 but not 3.5. We can write this as
‘3.0 - ’ or ‘3.0 - 3.5-’ or ‘3.0 - 3.499’. Including the lower boundary in the
class interval has this advantage. Most distributions of measurements
have a zero point below which we cannot go, whereas few have an exact
upper limit. If we were to include the upper boundary in the interval
instead of the lower, we would have two possible ways of dealing with
zero. It could be left as an isolated point, not in an interval. Alternatively, it
could be included in the lowest interval, which would then not be exactly
comparable to the others as it would include both boundaries while all the
other intervals only included the upper.
If we take a starting point of 2.5 and an interval of 0.5 we get the
frequency distribution shown in Table 4.5. Note that this is not unique. If
we take a
starting point of 2.4 and an interval of 0.2 we get a different set of

frequencies.
Table 4.5. Frequency distribution of FEV1 in 57
male medical students
Relative frequency (per

FEV1 Frequency
cent)
2.0 0 0.0
2.5 3 5.3
3.0 9 15.8
3.5 14 24.6
4.0 15 26.3
4.5 10 17.5
5.0 6 10.5
5.5 0 0.0
Total 57 100.0
Table 4.6. Tally system for finding the
frequency distribution of FEV1
FEV1 Frequency
2.0 0
2.5 /// 3
3.0 ///// //// 9
3.5 ///// ///// //// 14
4.0 ///// ///// ///// 15
4.5 ///// ///// 10
5.0 ///// / 6
5.5 0
Total 57
The frequency distribution can be calculated easily and accurately using

a computer. Manual calculation is not so easy and must be done carefully
and systematically. One way recommended by many texts (e.g. Hill 1977)
is to set up a tally system, as in Table 4.6. We go through the data and
for each individual make a tally mark by the appropriate interval. We then
count up the number in each interval. In practice this is very difficult to do
accurately, and it needs to be checked and double-checked. Hill (1977)
recommends writing each number on a card and dealing the cards into
piles corresponding to the intervals. It is then easy to check that each pile
contains only those cases in that interval and count them. This is
undoubtedly superior to the tally system. Another method is to order the
observations from lowest to highest before marking the interval
boundaries and counting, or to use the stem and leaf plot described
below. Personally, I always use a computer.
4.3 Histograms and other frequency graphs

Graphical methods are very useful for examining frequency distributions.
Figure 4.1 shows a graph of the cumulative frequency distribution for the
FEV1
data. This is called a step function. We can smooth this by joining

successive points where the cumulative frequency changes by straight
lines, to give a cumulative frequency polygon. Figure 4.2 shows this
for the cumulative relative frequency distribution of FEV1. This plot is
very useful for calculating some of the summary statistics referred to in
§4.5.
Fig. 4.1. Cumulative frequency distribution of FEV1 in a sample of
male medical students
Fig. 4.2. Cumulative frequency polygon of FEV1
The most common way of depicting a frequency distribution is by a

histogram. This is a diagram where the class intervals are on an axis
and rectangles with heights or areas proportional to the frequencies
erected on them. Figure 4.3 shows the histogram for the FEV1
distribution in Table 4.5. The vertical scale shows frequency, the number
of observations in each interval.
Sometimes we want to show the distribution of a discrete variable (e.g.
Table 4.3) as a histogram. If our intervals are 0–1-, 1–2-, etc., the actual
observations will all be at one end of the interval. Making the starting
point of the interval as a fraction rather than an integer gives a slightly
better picture (Figure 4.5). This can also be helpful for continuous data
when there is a lot of digit preference (§15.2). For example, where most
observations are recorded as integers or as something point five, starting
the interval at something point seven five can give a more accurate
picture.
Fig. 4.3. Histogram of FEV1: frequency scale

Fig. 4.4. Histogram of FEV1: frequency per unit FEV1 or frequency
density scale
Fig. 4.5. Histograms of parity (Table 4.3) using integer and fractional
cut-off points for the intervals
Table 4.7. Distribution of age in people

suffering accidents in the home (Whittington
1977)
Relative
Age Relative frequency
frequency (per
group per year (per cent)
cent)
0–4 25.3 5.06
5–14 18.9 1.89

15– 30.3 1.01
44
45– 13.6 0.68

64
65+ 11.7 0.33
Fig. 4.6. Histograms of age distribution of home accident victims,

using the relative frequency scale and the relative frequency density
scale
Figure 4.4 shows a histogram for the same distribution as Figure 4.3, with
frequency per unit FEV1 (or frequency density) shown on the vertical
axis. The distributions appear identical and we may well wonder whether
it matters which method we choose. We see that it does matter when we
consider a frequency distribution with unequal intervals, as in Table 4.7. If
we plot the histogram using the heights of the rectangles to represent
relative frequency in the interval we get the left-hand histogram in Figure
4.6, whereas if we use the relative frequency per year we get the right-
hand histogram. These histograms tell different stories. The left-hand
histogram in Figure 4.6 suggests that the most common age for accident
victims is between 15 and 44 years, whereas the right-hand histogram
suggests it is between 0 and 4. The right-hand histogram is correct, the
left-hand histogram being distorted by the unequal class intervals. It is
therefore preferable in general to use the frequency per unit (frequency
density) rather than per class interval when plotting a histogram. The
frequency for a particular interval is then represented by the area of the
rectangle on that interval. Only when the class intervals are all equal can
the frequency for the class interval be represented
by the height of the rectangle. The computer programmer finds equal

intervals much easier, however, and histograms with unequal intervals
are now uncommon.
Fig. 4.7. Frequency polygons of FEV1 and PEF in medical students

Fig. 4.8. Stem and leaf plot for the FEV1 data, rounded down to one
decimal place
Rather than a histogram consisting of vertical rectangles, we can plot a

frequency polygon instead. To do this we join the centre points of the
tops of the rectangles, then omit the rectangles (Figure 4.7(a)). Where a
cell of the histogram is empty we join the line to the centre of the cell at
the horizontal axis (Figure 4.7(b), males). This can be useful if we want to
show two or more frequency distributions on the same graph, as in
(Figure 4.7(b)). When we do this, the comparison is easier if we use
relative frequency or relative frequency density rather than frequency.
This makes it easier to compare distributions with different numbers of
subjects.
A different version of the histogram has been developed by Tukey (1977),
the stem and leaf plot (Figure 4.8). The rectangles are replaced by the
numbers themselves. The ‘stem’ is the first digit or digits of the number
and the ‘leaf the trailing digit. The first row of Figure 4.8 represents the
numbers 2.8, 2.8, and 2.9, which in the data are 2.85, 2.85, and 2.98.
The plot provides a good summary of data structure while at the same
time we can see other characteristics such as a tendency to prefer some
trailing digits to others, called digit preference (§15.1). It is also easy to
construct and much less prone to error than the tally method of finding a
frequency distribution.
4.4 Shapes of frequency distribution

Figure 4.3 shows a frequency distribution of a shape often seen in
medical data. The distribution is roughly symmetrical about its central
value and has frequency concentrated about one central point. The most
common value is called the
mode of the distribution and Figure 4.3 has one such point. It is
unimodal. Figure 4.9 shows a very different shape. Here there are two
distinct modes, one near 5 and the other near 8.5. This distribution is
bimodal. We must be careful to distinguish between the unevenness in
the histogram which results from using a small sample to represent a
large population and those which result from genuine bimodality in the
data. The trough between 6 and 7 in Figure 4.9 is very marked and might
represent a genuine bimodality. In this case we have children, some of
whom have a condition which raises the cholesterol level and some of
whom do not. We actually have two separate populations represented
with some overlap between them. However, almost all distributions
encountered m medical statistics are unimodal.
Fig. 4.9. Serum cholesterol in children from kinships with familial

hypercholesterolaemia (Leonard et al 1977)
Fig. 4.10. Serum triglyceride in cord blood from 282 babies (Table
4.8)
Figure 4.10 differs from Figure 4.3 in a different way. The distribution of
serum triglyceride is skew, that is, the distance from the central value to
the extreme is much greater on one side than it is on the other. The parts
of the histogram near the extremes are called the tails of the distribution.
If the tails are equal the distribution is symmetrical, as in Figure 4.3. If
the tail on the right is longer than the tail on the left as in Figure 4.10, the
distribution is skew to the right or positively skew. If the tail on the left
is longer, the distribution is skew to the left or negatively skew. This is
unusual, but Figure 4.11 shows an example. The negative skewness
comes about because babies can be born alive at any gestational age
from about 20 weeks, but soon after 40 weeks the baby will have to be
born. Pregnancies will not be allowed to go on for more than 44 weeks;
the birth would be induced artificially. Most distributions encountered in
medical work are symmetrical or skew to the right, for reasons we shall
discuss later (§7.4).
Table 4.8. Serum triglyceride measurements in cord b
0.15 0.29 0.32 0.36 0.40 0.42 0.46 0.50
0.16 0.29 0.33 0.36 0.40 0.42 0.46 0.50
0.20 0.29 0.33 0.36 0.40 0.42 0.47 0.52
0.20 0.29 0.33 0.36 0.40 0.44 0.47 0.52
0.20 0.29 0.33 0.36 0.40 0.44 0.47 0.52
0.20 0.29 0.33 0.36 0.40 0.44 0.47 0.52
0.21 0.30 0.33 0.36 0.40 0.44 0.47 0.52
0.22 0.30 0.33 0.36 0.40 0.44 0.48 0.52
0.24 0.30 0.33 0.37 0.40 0.44 0.48 0.52
0.25 0.30 0.34 0.37 0.40 0.44 0.48 0.53
0.26 0.30 0.34 0.37 0.40 0.44 0.48 0.54
0.26 0.30 0.34 0.37 0.40 0.44 0.48 0.54

0.26 0.30 0.34 0.38 0.40 0.45 0.48 0.54
0.27 0.30 0.34 0.38 0.40 0.45 0.48 0.54
0.27 0.30 0.34 0.38 0.41 0.45 0.48 0.54
0.27 0.31 0.34 0.38 0.41 0.45 0.48 0.54
0.28 0.31 0.34 0.38 0.41 0.45 0.48 0.55
0.28 0.32 0.35 0.39 0.41 0.45 0.48 0.55
0.28 0.32 0.35 0.39 0.41 0.46 0.48 0.55
0.28 0.32 0.35 0.39 0.41 0.46 0.49 0.55
0.28 0.32 0.35 0.39 0.41 0.46 0.49 0.55
0.28 0.32 0.35 0.39 0.42 0.46 0.49 0.55
0.28 0.32 0.35 0.40 0.42 0.46 0.50 0.55
0.28 0.32 0.36 0.40 0.42 0.46 0.50 0.55
4.5 Medians and quantiles

We often want to summarize a frequency distribution in a few numbers,
for ease of reporting or comparison. The most direct method is to use
quantiles. The quantiles are values which divide the distribution such
that there is a given proportion of observations below the quantile. For
example, the median is a quantile. The median is the central value of the
distribution, such that half the points are less than or equal to it and half
are greater than or equal to it. We can estimate any quantiles easily from
the cumulative frequency distribution
or a stem and leaf plot. For the FEV1 data the median is 4.1, the 29th
value in Table 4.4. If we have an even number of points, we choose a
value midway between the two central values.
Fig. 4.11. Gestational age at birth for 1749 deliveries at St. George's
Hospital
In general, we estimate the q quantile, the value such that a proportion q

will be below it, as follows. We have n ordered observations which divide
the scale into n + 1 parts: below the lowest observation, above the
highest and between each adjacent pair. The proportion of the
distribution which lies below the ith observation is estimated by i/(n+1).
We set this equal to q and get i = q(n + 1). If i is an integer, the ith
observation is the required quantile estimate. If not, let j be the integer
part of i, the part before the decimal point. The quantile will lie between
the jth and j + 1th observations. We estimate it by
For the median, for example, the 0.5 quantile, i = q(n+1) = 0.5 × (57+1) =
29, the 29th observation as before.
Other quantiles which are particularly useful are the quartiles of the
distribution. The quartiles divide the distribution into four equal parts,
called fourths. The second quartile is the median. For the FEV1 data the
first and third quartiles are 3.54 and 4.53. For the first quartile, i = 0.25 ×
58 = 14.5. The quartile is between the 14th and 15th observations, which
are both 3.54. For the third quartile, i = 0.75 × 58 = 43.5, so the quartile
lies between the 42nd and 43rd observations, which are 4.50 and 4.56.
The quantile is given by 4.50 + (4.56 - 4.50) × (43.5 - 43) = 4.53. We
often divide the distribution at 99 centiles or percentiles. The median is
thus the 50th centile. For the 20th centile of FEV1, i = 0.2 × 58 = 11.6, so
the quantile is between the 11th and 12th observation, 3.42 and 3.48,
and can be estimated by 3.42 + (3.48 - 3.42) × (11.6 - 11) = 3.46. We can
estimate these easily from Figure 4.2 by finding the position of the
quantile on the vertical axis, e.g. 0.2 for
the 20th centile or 0.5 for the median, drawing a horizontal line to
intersect the cumulative frequency polygon, and reading the quantile off
the horizontal axis.
Fig. 4.12. Box and whisker plots for FEV1 and for serum triglyceride
Fig. 4.13. Box plots showing a roughly symmetrical variable in four

groups, with an outlying point (data in Table 10.8)
Tukey (1977) used the median, quartiles, maximum and minimum as a

convenient five figure summary of a distribution. He also suggested a
neat graph, the box and whisker plot, which represents this (Figure
4.12). The box shows the distance between the quartiles, with the
median marked as a line, and the ‘whiskers’ show the extremes. The
different shapes of the FEV1 and serum triglyceride distributions is clear
from the graph. For display purposes, an observation whose distance
from the edge of the box (i.e. the quartile) is more than 1.5 times the
length of the box (i.e. the interquartile range, §4.7) may be called an
outlier. Outliers may be shown as separate points (Figure 4.13). The plot
can be useful for showing the comparison of several groups (Figure
4.13).
4.6 The mean
The median is not the only measure of central value for a distribution.
Another is the arithmetic mean or average, usually referred to simply as
the mean. This is found by taking the sum of the observations and
dividing by their number. For example, consider the following hypothetical
data:
2 3 9 5 4 0 6 3 4
The sum is 36 and there are 9 observations, so the mean is 36/9 = 4.0.
At this point we will need to introduce some algebraic notation, widely
used in statistics. We denote the observations by
x1, x2, …, xi, …, xn
There are n observations and the ith of these is xi = For the example, x4
= 5 and n = 9. The sum of all the xi is
The summation sign is an upper case Greek letter, sigma, the Greek S.
When it is obvious that we are adding the values of x1, for all values of i,
which runs from 1 to n, we abbreviate this to ∑xi or simply to ∑x. The
mean of the xi is denoted by [x with bar above] (‘x bar’), and
The sum of the 57 FEV1s is 231.51 and hence the mean is 231.51/57 =
4.06. This is very close to the median, 4.1, so the median is within 1% of
the mean. This is not so for the triglyceride data. The median triglyceride
(Table 4.8) is 0.46 but the mean is 0.51, which is higher. The median is
10% away from the mean. If the distribution is symmetrical the sample
mean and median will be about the same, but in a skew distribution they
will not. If the distribution is skew to the right, as for serum triglyceride,
the mean will be greater, if it is skew to the left the median will be greater.
This is because the values in the tails affect the mean but not the
median.
The sample mean has much nicer mathematical properties than the
median and is thus more useful for the comparison methods described
later. The median is a very useful descriptive statistic, but not much used
for other purposes.
4.7 Variance, range and interquartile range

The mean and median are measures of the position of the middle of the
distribution, which we call the central tendency. We shall also need a
measure of the spread or variability of the distribution, called the
dispersion.
One obvious measure is the range, the difference between the highest
and lowest values. For the data of Table 4.4, the range is 5.43–2.85 =
2.58 litres. The
range is often presented as the two extremes. 2.85–5.43 litres, rather

than their difference. The range is a useful descriptive measure, but has
two disadvantages. Firstly, it depends only on the extreme values and so
can vary a lot from sample to sample. Secondly, it depends on the
sample size. The larger the sample is, the further apart the extremes are
likely to be. We can see this if we consider a sample of size 2. If we add
a third member to the sample the range will only remain the same if the
new observation falls between the other two, otherwise the range will
increase. We can get round the second of these problems by using the
interquartile range, the difference between the first and third quartiles.
For the data of Table 4.4, the interquartile range is 4.53 -- 3.54 = 0.99
litres. The interquartile range, too, is often presented as the two
extremes, 3.54–4.53 litres. However, the interquartile range is quite
variable from sample to sample and is also mathematically intractable.
Although a useful descriptive measure, it is not the one preferred for
purposes of comparison.
Table 4.9. Deviations from the mean of 9

observations
Deviations Squared
Observations from the mean deviations (xi
xi xi - [x with bar - [x with bar
above] above])2
2 -2 4
3 -1 1
9 5 25
5 1 1
4 0 0
0 -4 16
6 2 4
3 -1 1
4 0 0
36 0 52
The most commonly used measures of dispersion are the variance and
standard deviation. We start by calculating the difference between each
observation and the sample mean, called the deviations from the
mean, Table 4.9. If the data are widely scattered, many of the
observations xi will be far from the mean [x with bar above] and so many
deviations xi - [x with bar above] will be large. If the data are narrowly
scattered, very few observations will be far from the mean and so few
deviations xi - [x with bar above] will be large. We need some kind of
average deviation to measure the scatter. If we add all the deviations
together, we get zero, because ∑(xi - [x with bar above]) = ∑xi - ∑[x with
bar above] = ∑xi - n[x with bar above] and n[x with bar above] = ∑xi.
Instead we square the deviations and then add them, as shown in Table
4.9. This removes the effect of sign; we are only measuring the size of
the deviation not the direction. This gives us ∑(xi - [x with bar above])2, in
the example equal to 52, called the sum of squares about the mean,
usually abbreviated to sum of squares.
Clearly, the sum of squares will depend on the number of observations as
well as the scatter. We want to find some kind of average squared
deviation. This leads to a difficulty. Although we want an average squared
deviation, we divide the sum of squares by n - 1, not n. This is not the
obvious thing to do and puzzles
many students of statistical methods. The reason is that we are

interested in estimating the scatter of the population, rather than the
sample, and the sum of squares about the sample mean is proportional
to n - 1 (§4A, §6B), Dividing by n would lead to small samples producing
lower estimates of variability than large samples. The minimum number
of observations from which the variability can be estimated is 2, a single
observation cannot tell us how variable the data are. If we used n as our
divisor, for n - I the sum of squares would be zero, giving a variance of
zero. With the correct divisor of n - 1, n = 1 gives the meaningless ratio
0/0, reflecting the impossibility of estimating variability from a single
observation. The estimate of variability is called the variance, defined by
We have already said that ∑(xi - [x with bar above])2 is called the sum of
squares. The quantity n - 1 is called the degrees of freedom of the
variance estimate (§7A). We have:
We shall usually denote the variance by s2. In the example, the sum of
squares is 52 and there are 9 observations, giving 8 degrees of freedom.
Hence s2 = 52/8 = 6.5.
The formula ∑(xi - [x with bar above])2 gives us a rather tedious
calculation. There is another formula for the sum of squares, which
makes the calculation easier to carry out. This is simply an algebraic
manipulation of the first form and give exactly the same answers. We
thus have two formulae for variance:
The algebra is quite simple and is given in §4B. For example, using the
second formula for the nine observations, we have:
as before. On a calculator this is a much easier formula than the first, as

the numbers need only be put in once. It can be inaccurate, because we
subtract one large number from another to get a small one. For this
reason the first formula would be used in a computer program.
4.8 Standard deviation

The variance is calculated from the squares of the observations. This
means that it is not in the same units as the observations, which limits its
use as a descriptive statistic. The obvious answer to this is to take the
square root, which will then have the same units as the observations and
the mean. The square root of the variance is called the standard
deviation, usually denoted by s. Thus,
Returning to the FEV data, we calculate the variance and standard
deviation as follows. We have n = 57, ∑xi 231.51, = ∑xi2 = 965.45:
Figure 4.14 shows the relationship between mean, standard deviation

and frequency distribution. For FEV1, we see that the majority of
observations are within one standard deviation of the mean, and nearly
all within two standard deviations of the mean. There is a small part of
the histogram outside the [x with bar above] - 2s to [x with bar above] +
2s interval, on either side of this symmetrical histogram. As
Figure 4.14 also shows, this is true for the highly skew triglyceride data,
too. In this case, however, the outlying observations are all in one tail of
the distribution. In general, we expect roughly 2/3 of observations to lie
within one standard deviation of the mean and 95% to lie within two
standard deviations of the mean.
Fig. 4.14. Histograms of FEV1 and triglyceride with mean and
standard deviation
Table 4.10. Population of 100 random digits for a sam
9 1 0 7 5 6 9 5 8 8 1 0 5
1 8 8 8 5 2 4 8 3 1 6 5 5
2 8 1 8 5 8 4 0 1 9 2 1 6
1 9 7 9 7 2 7 7 0 8 1 6 3
7 0 2 8 8 7 2 5 4 1 8 6 8
Appendices
4A Appendix: The divisor for the
variance
The variance is found by dividing the sum of squares about the sample
mean by n - 1, not by n. This is because we want the scatter about the
population mean, and the scatter about the sample mean is always less.
The sample mean is ‘closer’ to the data points than is the population
mean. We shall try a little sampling experiment to show this. Table 4.10
shows a set of 100 random digits which we shall take as the population to
be sampled. They have mean 4.74 and the sum of squares about the
mean is 811.24. Hence the average squared difference from the mean is
8.1124. We can take samples of size two at random from this population
using a pair of decimal dice, which will enable us to choose any digit
numbered from 00 to 99. The first pair chosen was 5 and 6 which has
mean 5.5. The sum of squares about the population mean 4.74 is (5 -
4.74)2 + (6 - 4.74)2 = 1.655. The sum of squares about the sample mean
is (5 - 5.5)2 + (6 - 5.5)2 = 0.5.
The sum of squares about the population mean is greater than the sum
of squares about the sample mean, and this will always be so. Table 4.11
shows this for 20 such samples of size two. The average sum of squares
about the population mean is 13.6, and about the sample mean it is 5.7.
Hence dividing by the sample size (n = 2) we have mean square
differences of 6.8 about the population mean and 2.9 about the sample
mean. Compare this to 8.1 for the population as a whole. We see that the
sum of squares about the population
mean is quite close to 8.1, while the sum of squares about the sample
mean is much less. However, if we divide the sum of squares about the
sample mean by n - 1, i.e. 1, instead of n we have 5.7, which is not much
different to the 6.8 from the sum of squares about the population mean.
Table 4.11. Sampling pairs from Table 4.10

Sample ∑(xi - µ)2 ∑(xi - [x with bar above])2
5 6 1.655 0.5
8 8 21.255 0.0
6 1 15.575 12.5
9 3 21.175 18.0
5 5 0.135 0.0
7 7 10.215 0.0
1 7 19.095 18.0
9 8 28.775 0.5
3 3 6.055 0.0
5 1 14.055 8.0
8 3 13.655 12.5
5 7 5.175 2.0
5 2 5.575 4.5
5 7 5.175 2.0
8 8 21.255 0.0
3 2 10.535 0.5
0 4 23.015 8.0
9 3 21.175 18.0
5 2 7.575 4.5
6 9 19.735 4.5
Mean 13.6432 5.7
Table 4.12. Mean sums of squares aobut the

sample mean for sets of 100 random samples
from Table 4.11
Mean variance
Number in sample, estimates
n
2 4.5 9.1
3 5.4 8.1
4 5.9 7.9
5 6.2 7.7
10 7.2 8.0
Table 4.12 shows the results of a similar experiment with more samples
being taken. The table shows the two average variance estimates using n
and n - 1 as the divisor of the sum of squares, for sample sizes 2, 3, 4, 5
and 10. We see that the sum of squares about the sample mean divided
by n increases steadily with sample size, but if we divide it by n - 1
instead of n the estimate does not change as the sample size increases.
The sum of squares about the sample mean is proportional to n - 1.
4B Appendix: Formulae for the sum of

squares
The different formulae for sums of squares are derived as follows:

because [x with bar above] has the same value for each of the n
observations. Now, so
We thus have three formulae for variance:

14. Which of the following are qualitative variables:
(a) sex;
(b) parity;
(c) diastolic blood pressure;
(d) diagnosis;
(e) height.
View Answer
15. Which of the following are continuous variables:

(a) blood glucose;
(b) peak expiratory flow rate;
(c) age last birthday;
(d) exact age;
(e) family size.
View Answer
16. When a distribution is skew to the right:

(a) the median is greater than the mean;
(b) the distribution is unimodal;
(c) the tail on the left is shorter than the tail on the right;
(d) the standard deviation is less than the variance;
(e) the majority of observations are less than the mean.
View Answer
17. The shape of a frequency distribution can be described using:

(a) a box and whisker plot;
(b) a histogram:
(c) a stem and leaf plot;
(d) mean and variance;
(e) a table of frequencies.
View Answer
18. For the sample 3, 1, 7, 2, 2:

(a) the mean is 3:
(b) the median is 7:
(c) the mode is 2:
(d) the range is 1:
(e) the variance is 5.5.
View Answer
19. Diastolic blood pressure has a distribution which is slightly

skew to the right. If the mean and standard deviation were
calculated for the diastolic pressures of a random sample of men:
(a) there would be fewer observations below the mean than above
it;
(b) the standard deviation would be approximately equal to the
mean;
(c) the majority of observations would be more than one standard
deviation from the mean:
(d) the standard deviation would estimate the accuracy of blood
pressure measure ment:
(e) about 95% of observations would be expected to be within two
standard deviations of the mean.
View Answer
4E Exercise: Mean and standard deviation

This exercise gives some practice in one of the most fundamental
calculations in statistics, that of the sum of squares and standard
deviation. It also shows the relationship of the standard deviation to the
frequency distribution. Table 4.13 shows blood glucose levels obtained
from a group of medical students.
1. Make a stem and leaf plot for these data.
View Answer
2. Find the minimum, maximum and quartiles and sketch a box and
whisker plot.
View Answer
3. Find the frequency distribution, using a class interval of 0.5.

View Answer
Table 4.13. Random blood glucose levels from a group

of first year medical students (mmol/litre)
4.7 3.6 3.8 2.2 4.7 4.1 3.6 4.0 4.4
4.2 4.1 4.4 5.0 3.7 3.6 2.9 3.7 4.7
3.9 4.8 3.3 3.3 3.6 4.6 3.4 4.5 3.3
3.4 4.0 3.8 4.1 3.8 4.4 4.9 4.9 4.3
4. Sketch the histogram of this frequency distribution. What term

best describes the shape: symmetrical, skew to the right or skew to
the left?
View Answer
5. For the first column only, i.e. for 4,7, 4.2, 3.9, and 3.4, calculate the
standard deviation using the deviations from the mean formula
First calculate the sum of the observations and the sum of the
observations squared. Hence calculate the sum of squares about
the mean. Is this the same as that found in 4 above? Hence
calculate the variance and the standard deviation.
View Answer
6. For the same four numbers, calculate the standard deviation

using the formula
First calculate the sum of the observations and the sum of the
observations squared. Hence calculate the sum of squares about
the mean. Is this the same as that found in 4 above ? Hence
calculate the variance and the standard deviation.
View Answer
7. Use the following summations for the whole sample: ∑xi = 162.2,
∑xi2 = 676.74. Calculate the mean of the sample, the sum of squares
about the mean, the degrees of freedom for this sum of squares,
and hence estimate the variance and standard deviation.
View Answer
8. Calculate the mean ± one standard deviation and mean ± two

standard deviations. Indicate these points and the mean on the
histogram. What do you observe about their relationship to the
frequency distribution?
View Answer
> Table of Contents > 5 - Presenting data
5
Presenting data
5.1 Rates and proportions

Having collected our data as described in Chapters 2 and 3 and
extracted information from it using the methods of Chapter 4, we must
find a way to convey this information to others. In this chapter we shall
look at some of the methods of doing that. We begin with rates and
proportions.
When we have data in the form of frequencies, we often need to compare
the frequency with certain conditions in groups containing different totals.
In Table 2.1, for example, two groups of patient pairs were compared, 29
where the later patient had a C-T scan and 89 where neither had a C-T
scan. The later patient did better in 9 of the first group and 34 of the
second group. To compare these frequencies we compare the
proportions 9/29 and 34/89. These are 0.31 and 0.38, and we can
conclude that there is little difference. In Table 2.1, these were given as
percentages, that is, the proportion out of 100 rather than out of 1, to
avoid the decimal point. In Table 2.8, the Salk vaccine trial, the
proportions contracting polio were presented as the number per 100000
for the same reason.
A rate expresses the frequency of the characteristic of interest per 1000
(or per 100000, etc.) of the population, per unit of time. For example, in
Table 3.1, the results of the study of smoking by doctors, the data were
presented as the number of deaths per 1 000 doctors per year. This is
not a proportion, as a further adjustment has been made to allow for the
time period observed. Furthermore, the rate has been adjusted to take
account of any differences in the age distributions of smokers and non-
smokers (§16.2). Sometimes the actual denominator for a rate may be
continually changing. The number of deaths from lung cancer among
men in England and Wales for 1983 was 26 502. The denominator for the
death rate, the number of males in England and Wales, changed
throughout 1983, as some died, some were born, some left the country
and some entered it. The death rate is calculated by using a
representative number, the estimated population at the end of June 1983,
the middle of the year. This was 24 175 900, giving a death rate of 26
502/24 175 900, which equals 0.001 096, or 109.6 deaths per 100 000 at
risk per year. A number of the rates used in medical statistics are
described in §16.5.
The use of rates and proportions enables us to compare frequencies
obtained from unequal sized groups, base populations or time periods,
but we must beware of their use when their bases or denominators are
not given. Victora (1982)
reported a drug advertisement sent to doctors which described the

antibiotic phosphomycin as being ‘100% effective in chronic urinary
infections’. This is very impressive. How could we fail to prescribe a drug
which is 100% effective? The study on which this was based used 8
patients, after excluding ‘those whose urine contained phosphomycin-
resistant bacteria’. If the advertisement has said the drug was effective in
100% of 8 cases, we would have been less impressed. Had we known
that it worked in 100% of 8 cases selected because it might work in them,
we would have been still less impressed. The same paper quotes an
advertisement for a cold remedy, where 100% of patients showed
improvement. This was out of 5 patients! As Victora remarked, such
small samples are understandable in the study of very rare diseases, but
not for the common cold.
Sometimes we can fool ourselves as well as others by omitting
denominators. I once carried out a study of the distribution of the soft
tissue tumour Kaposi's sarcoma in Tanzania (Bland et al. 1977), and
while writing it up I came across a paper setting out to do the same thing
(Schmid 1973). One of the factors studied was tribal group, of which
there are over 100 in Tanzania. This paper reported ‘the tribal incidence
in the Wabende, Wambwe and Washirazi is remarkable… These small
tribes, each with fewer than 90 000 people, constitute the group in which
a tribal factor can be suspected’. This is based on the following rates of
tumours per 10 000 population: national, 0.1; Wabende, 1.3; Wambwe.
0.7; Washirazi, 1.3. These are very big rates compared to the national,
but the populations on which they are based are small, 8 000, 14 000 and
15 000 respectively (Egero and Henin 1973). To get a rate of 1.3/10 000
out of 8 000 Wabende people we must have 8 000 × 1.3/10 000 = 1
case! Similarly we have 1 case among the 14 000 Wambwe and 2
among the 15 000 Washirazi. We can see that there are not enough data
to draw the conclusions which the author has done. Rates and
proportions are powerful tools and we must beware of them becoming
detached from the original data.
5.2 Significant figures

When we calculated the death rate due to lung cancer among men in
1983 we quoted the answer as 0.001 096 or 109.6 per 100 000 per year.
This is an approximation. The rate to the greatest number of figures my
calculator will give is 0.001 096 215 653 and this number would probably
go on indefinitely, turning into a recurring series of digits. The decimal
system of representing numbers cannot in general represent fractions
exactly. We know that 1/2 = 0.5, but 1/3 = 0.333 333 33…, recurring
infinitely. This does not usually worry us, because for most applications
the difference between 0.333 and 1/3 is too small to matter. Only the first
few non-zero digits of the number are important and we call these the
significant digits or significant figures. There is usually little point in
quoting statistical data to more than three significant figures. After all, it
hardly matters whether the lung cancer mortality rate is 0.001 096 or
0.001 097. The value 0.001 096 is given to 4 significant figures. The
leading zeros are not significant, the first significant digit in this number
being ‘1’. To three significant
figures we get 0.001 10, because the last digit is 6 and so the 9 which
precedes it is rounded up to 10. Note that significant figures are not the
same as decimal places. The number 0.001 10 is given to 5 decimal
places, the number of digits after the decimal point. When rounding to the
nearest digit, we leave the last significant digit, 9 in this case, if what
follows it is less than 5, and increase by one if what follows is greater
than 5. When we have exactly 5, I would always round up, i.e. 1.5 goes
to 2. This means that 0, 1, 2, 3, 4 go down and 5, 6, 7, 8, 9 go up, which
seems unbiased. Some writers take the view that 5 should go up half the
time and down half the time, since it is exactly midway between the
preceding digit and that digit plus one. Various methods are suggested
for doing this but I do not recommend them myself. In any case, it is
usually a mistake to round to so few significant figures that this matters.
How many significant figures we need depends on the use to which the
number is to be put and on how accurate it is anyway. For example, if we
have a sample of 10 sublingual temperatures measured to the nearest
half degree, there is little point in quoting the mean to more than 3
significant figures. What we should not do is to round numbers to a few
significant figures before we have completed our calculations. In the lung
cancer mortality rate example, suppose we round the numerator and
denominator to two significant figures. We have 27 000/24 000 000 =
0.001 125 and the answer is only correct to two figures. This can spread
through calculations causing errors to build up. We always try to retain
several more significant figures than we required for the final answer.
Consider Table 5.1. This shows mortality data in terms of the exact
numbers of deaths in one year. The table is taken from a much larger
table (OPCS 1991) which shows the numbers dying from every cause of
death in the International Classification of Diseases (ICD), which gives
numerical codes to many hundreds of causes of death. The full table,
which also gives deaths by age group, covers 70 A4 pages. Table 5.1
shows deaths for broad groups of diseases called ICD chapters. This
table is not a good way to present these data if we want to get an
understanding of the frequency distribution of cause of death, and the
differences between causes in men and women. This is even more true
of the 70 page original. This is not the purpose of the table, of course. It
is a source of data, a reference document from which users extract
information for their own purposes. Let us see how Table 5.1 can be
simplified. First, we can reduce the number of significant figures. Let us
be extreme and reduce the data to one significant figure (Table 5.2). This
makes comparisons rather easier, but it is still not obvious which are the
most important causes of death. We can improve this by re-ordering the
table to put the most frequent cause, diseases of the circulatory system,
first (Table 5.3). We can also combine a lot of the smaller categories into
an ‘others’ group. I did this arbitrarily, by combining all those accounting
for less than 2% of the total. Now it is clear at a glance that the most
important causes of death in England and Wales are diseases of the
circulatory system, neoplasms and diseases of the respiratory system,
and that these dwarf all the others. Of course, mortality is not the only
indicator of the importance of a disease. ICD chapter XIII, diseases of the
musculo-skeletal
system and connective tissues, are easily seen from Table 5.2 to be only
minor causes of death, but this group includes arthritis and rheumatism,
the most important illness in its effects on daily activity.
Table 5.1. Deaths by sex and cause, England

and Wales, 1989 (OPCS 1991, DH2 No. 10)
Number of
Chapter and type of deaths
I.C.D.
disease
Males Females
I Infectious and 1 1 297

parasitic 246
II Neoplasms 75 69 948
(cancers) 172
III Endocrine, nutritional 4 5 758

and metabolic 395
diseases and
immunity disorders
IV Blood and blood 1 1 422

forming organs 002
V Mental disorders 4 9 225

493
VI Nervous system and 5 5 990

sense organs 466
VII Circulatory system 127 137

435 165
VIII Respiratory system 33 33 223

489
IX Digestive system 7 10 779

900
X Genitourinary 3 4 156
system 616
XI Complications of 0 56
pregnancy, childbirth
and the puerperium
XII Skin and 250 573

subcutaneous
tissues
XIII Musculo-skeletal 1 4 139

system and 235
connective tissues
XIV Congenital 897 869

anomalies
XV Certain conditions 122 118

originating in the
perinatal period
XVI Signs, symptoms 1 3 082

and ill-defined 582
conditions
XVII Injury and poisoning 11 6 427

073
Total 279 294

373 227
5.3 Presenting tables
Tables 5.1, 5.2 and 5.3 illustrate a number of useful points about the
presentation of tables. Like all the tables in this book, they are designed
to stand alone from the text. There is no need to refer to material buried
in some paragraph to interpret the table. A table is intended to
communicate information, so it should be easy to read and understand. A
table should have a clear title, stating clearly and unambiguously what
the table represents. The rows and columns must also be labelled clearly.
When proportions, rates or percentages are used in a table together with
frequencies, they must be easy to distinguish from one another. This can
be done, as in Table 2.10, by adding a ‘%’ symbol, or by including a place
of decimals. The addition in Table 2.10 of the ‘total’ row and the ‘100%’
makes it clear that the percentages are calculated from the number in the
treatment group, rather than the number with that particular outcome or
the total number of patients.
Table 5.2. Deaths by sex and cause, England

and Wales, 1989, rounded to one significant
figure
Number of
Chapter and type of deaths
I.C.D.
disease
Males Females
I Infectious and 1 1 000

parasitic 000
II Neoplasms 80 70 000
(cancers) 000
III Endocrine, nutritional 4 6 000

and metabolic 000
diseases and
immunity disorders
IV Blood and blood 1 1 000

forming organs 000
V Mental disorders 4 9 000

000
VI Nervous system and 5 6 000

sense organs 000
VII Circulatory system 100 100

000 000
VIII Respiratory system 30 30 000

000
IX Digestive system 8 10 000

000
X Genitourinary 4 4 000
system 000
XI Complications of 0 60
pregnancy, childbirth
and the puerperium
XII Skin and 300 600

subcutaneous
tissues
XIII Musculo-skeletal 1000 4000

system and
connective tissues
XIV Congenital 900 900

anomalies
XV Certain conditions 100 100

originating in the
perinatal period
XVI Signs, symptoms 2 3 000

and ill-defined 000
conditions
XVII Injury and poisoning 10 6 000

000
Total 300 300

000 000
Table 5.3. Deaths by sex, England and Wales,

1989, for major causes
Number of
I.C.D. Chapter and type of deaths
disease
Males Females
Circulatory system (VII) 100 100

000 000
Neoplasms (cancers) (II) 80 70 000

000
Respiratory system (VIII) 30 30 000

000
Injury and poisoning (XVII) 10 6 000

000
Digestive system (IX) 8 000 10 000
Others 20 20 000
000
Total 300 300

000 000
5.4 Pie charts

It is often convenient to present data pictorially. Information can be
conveyed much more quickly by a diagram than by a table of numbers.
This is particularly useful when data are being presented to an audience,
as here the information has to be got across in a limited time. It can also
help a reader get the salient points of a table of numbers. Unfortunately,
unless great care is taken, diagrams can also be very misleading and
should be treated only as an addition to numbers, not a replacement.
We have already discussed methods of illustrating the frequency
distribution of a qualitative variable. We will now look at an equivalent of
the histogram for
qualitative data, the pie chart or pie diagram. This shows the relative
frequency for each category by dividing a circle into sectors, the angles of
which are proportional to the relative frequency. We thus multiply each
relative frequency by 360, to give the corresponding angle in degrees.
Table 5.4. Calculations for a pie chart of the

distribution of cause of death
Cause of Relative Angle

Frequency
death frequency (degrees)
Circulatory 137 165 0.466 19 168
system
Neoplasms 69 948 0.237 73 86

(cancers)
Respiratory 33 223 0.112 92 41

system
Injury and 6 427 0.021 84 8

poisoning
Digestive 10 779 0.036 63 13

system
Nervous 5 990 0.020 36 7

system
Others 30 695 0.104 32 38
Total 294 227 1.000 00 361

Fig. 5.1. Pie chart showing the distribution of cause of death among
females, England and Wales, 1983
Table 5.4 shows the calculation for drawing a pie chart to represent the
distribution of cause of death for females, using the data of Tables 5.1
and 5.3. (The total degrees are 361 rather than 360 because of rounding
errors in the calculations.) The resulting pie chart is shown in Figure 5.1.
This diagram is said to resemble a pie cut into pieces for serving, hence
the name.
5.5 Bar charts

A bar chart or bar diagram shows data in the form of horizontal or
vertical bars. For example, Table 5.5 shows the mortality due to cancer of
the oesophagus in England and Wales over a 10 year period. Figure 5.2
shows these data in the form of a bar chart, the heights of the bars being
proportional to the mortality.
Table 5.5. Cancer of the oesophagus:

standardized mortality rate per 100 000 per
year, England and Wales, 1960--1969
Year Mortality rate Year Mortality rate
60 5.1 65 5.4
61 5.0 66 5.4
62 5.2 67 5.6
63 5.2 68 5.8
64 5.2 69 6.0
Fig. 5.2. Bar chart showing the relationship between mortality due to
cancer of the oesophagus and year, England and Wales, 1960–1969
There are many uses for bar charts. As in Figure 5.2, they can be used to
show the relationship between two variables, one being quantitative and
the other either qualitative or a quantitative variable which is grouped, as
is time in years. The values of the first variable are shown by the heights
of bars, one bar for each category of the second variable.
Bar charts can be used to represent relationships between more than two
variables. Figure 5.3 shows the relationship between children's reports of
breathlessness and cigarette smoking by themselves and their parents.
We can see quickly that the prevalence of the symptom increases both
with the child's smoking and with that of their parents. In the published
paper reporting these respiratory symptom data (Bland et al. 1978) the
bar chart was not used; the data were given in the form of tables. It was
thus available for other researchers to compare to their own or to carry
out calculations upon. The bar chart was used to present the results
during a conference, where the most important thing was to convey an
outline of the analysis quickly.
Bar charts can also be used to show frequencies. For example, Figure
5.4(a) shows the relative frequency distributions of causes of death
among men and women, Figure 5.4(b) shows the frequency distribution
of cause of death among
men. Figure 5.4(b) looks very much like a histogram. The distinction
between these two terms is not clear. Most statisticians would describe
Figures 4.3, 4.4, and 4.6 as histograms, and Figures 5.2 and 5.3 as bar
charts, but I have seen books which actually reverse this terminology and
others which reserve the term ‘histogram’ for a frequency density graph,
like Figures 4.4 and 4.6.
Fig. 5.3. Bar chart showing the relationship between the prevalence
of self-reported breathlessness among schoolchildren and two
possible causative factors
Fig. 5.4. Bar charts showing data from Table 5.1
5.6 Scatter diagrams

The bar chart would be a rather clumsy method for showing the
relationship between two continuous variables, such as vital capacity and
height (Table 5.6). For this we use a scatter diagram or scattergram
(Figure 5.5). This is made by marking the scales of the two variables
along horizontal and vertical axes. Each pair of measurements is plotted
with a cross, circle, or some other suitable symbol at the point indicated
by using the measurements as coordinates.
Table 5.7 shows serum albumin measured from a group of alcoholic
patients and a group of controls (Hickish et al. 1989). We can use a
scatter diagram to
present these data also. The vertical axis represents albumin and we
choose two arbitrary points on the horizontal axis to represent the
groups.
Table 5.6. Vital capacity (VC) and height for 44 female m

students
Height VC Height VC Height VC

(cm) (litres) (cm) (litres) (cm) (litres)
155.0 2.20 161.2 3.39 166.0 3.66
155.0 2.65 162.0 2.88 166.0 3.69
155.4 3.06 162.0 2.96 166.6 3.06
158.0 2.40 162.0 3.12 167.0 3.48

160.0 2.30 163.0 2.72 167.0 3.72
160.2 2.63 163.0 2.82 167.0 3.80
161.0 2.56 163.0 3.40 167.6 3.06
161.0 2.60 164.0 2.90 167.8 3.70
161.0 2.80 165.0 3.07 168.0 2.78
161.0 2.90 166.0 3.03 168.0 3.63
161.0 3.40 166.0 3.50 169.4 2.80
Fig. 5.5. Scatter diagram showing the relationship between vital

capacity and height for a group of female medical students
Table 5.7. Albumin measured in alcoholics and con
Alcoholics Controls
15 28 39 41 44 48 34 41 43 45
16 29 39 43 45 48 39 42 43 45
17 32 39 43 45 49 39 42 43 45
18 37 40 44 46 51 40 42 43 45
20 38 40 44 46 51 41 42 44 45
21 38 40 44 46 52 41 42 44 45
28 38 41 44 47 41 42 44 45
Fig. 5.6. Scatter diagrams showing the data of Table 5.7
In Table 5.7 there are many identical observations in each group, so we

need to allow for this in the scatter diagram. If there is more than one
observation at the same coordinate we can indicate this in several ways.
We can use the number of observations in place of the chosen symbol,
but this method is becoming obsolete. As in Figure 5.6(a), we can
displace the points slightly in a random direction (called jittering). This is
what Stata does and so what I have done in most of this book.
Alternatively, we can use a systematic sideways shift, to form a more
orderly picture as in Figure 5.6(b). The latter is often used when the
variable on the horizontal axis is categorical rather than continuous. Such
scatter diagrams are very useful for checking the assumptions of some of
the analytical methods which we shall use later. A scatter diagram where
one variable is a group is also called a dot plot. As a presentational
device, they enable us to show far more information than a bar chart of
the group means can do. For this reason, statisticians usually prefer them
to other types of graphical display.
5.7 Line graphs and time series

The data of Table 5.5 are ordered in a way that those of Table 5.6 are
not, in that they are recorded at intervals in time. Such data are called a
time series. If we plot a scatter diagram of such data, as in Figure 5.7, it
is natural to join successive points by lines to form a line graph. We do
not even need to mark the points at all; all we need is the line. This would
not be sensible in Figure 5.5, as the observations are independent of one
another and quite unrelated, whereas in Figure 5.7 there is likely to be a
relationship between adjacent points. Here the mortality rate recorded for
cancer of the oesophagus will depend on a number of things which vary
over time including possibly causal factors, such as tobacco and alcohol
consumption, and clinical factors, such as better diagnostic techniques
and methods of treatment.
Line graphs are particularly useful when we want to show the change of
more than one quantity over time. Figure 5.8 shows levels of zidovudine
(AZT) in
the blood of AIDS patients at several times after administration of the

drug, for patients with normal fat absorption and with fat malabsorption
(§10.7). The difference in response to the two treatments is very clear.
Fig. 5.7. Line graph showing changes in cancer of the oesophagus

mortality over time
Fig. 5.8. Line graph to show the response to administration of
zidovudine in two groups of AIDS patients
5.8 Misleading graphs

Figure 5.2 is clearly titled and labelled and can be read independently of
the surrounding text. The principles of clarity outlined for tables apply
equally here. After all, a diagram is a method of conveying information
quickly and this object is defeated if the reader or audience has to spend
time trying to sort out exactly what a diagram really means. Because the
visual impact of diagrams can be so great, further problems arise in their
use.
The first of these is the missing zero. Figure 5.9 shows a second bar
chart
representing the data of Table 5.5. This chart appears to show a very
rapid increase in mortality, compared to the gradual increase shown in
Figure 5.2. Yet both show the same data. Figure 5.9 omits most of the
vertical scale, and instead stretches that small part of the scale where the
change takes place. Even when we are aware of this, it is difficult to look
at this graph and not think that it shows a large increase in mortality. It
helps if we visualize the baseline as being somewhere near the bottom of
the page.
Fig. 5.9. Bar chart with zero omitted on the vertical scale
There is no zero on the horizontal axis in Figures 5.2 and 5.9, either.
There are two reasons for this. There is no practical ‘zero time’ on the
calendar; we use an arbitrary zero. Also, there is an unstated assumption
that mortality rates vary with time and not the other way round.
The zero is omitted in Figure 5.5. This is almost always done in scatter
diagrams, yet if we are to gauge the importance of the relationship
between vital capacity and height by the relative change in vital capacity
over the height range we need the zero on the vital capacity scale. The
origin is often omitted on scatter diagrams because we are usually
concerned with the existence of a relationship and the distributions
followed by the observations, rather than its magnitude. We estimate the
latter in a different way, described in Chapter 11.
Line graphs are particularly at risk of undergoing the sort of distortion of
missing zero described in §5.8. Many computer programs resist drawing
bar charts like Figure 5.9, but will produce a line graph with a truncated
scale as the default. Figure 5.10 shows a line graph with a truncated
scale, corresponding to Figure 5.9. Just as there, the message of the
graph is a dramatic increase in mortality, which the data themselves do
not really support. We can make this even more dramatic by stretching
the vertical scale and compressing the horizontal scale. The effect is now
really impressive and looks much more likely than Figure 5.7 to attract
research funds, Nobel prizes and interviews on television. Huff (1954)
aptly names such horrors ‘gee whiz’ graphs. They are even more
dramatic if we omit the scales altogether and show only the soaring line.
Fig. 5.10. Line graphs with a missing zero and with a stretched
vertical and compressed horizontal scale, a ‘gee whiz’ graph
Fig. 5.11. Figure 5.1 with three-dimensional effects
This is not to say that authors who show only part of the scale are
deliberately trying to mislead. There are often good arguments against
graphs with vast areas of boring blank paper. In Figure 5.5, we are not
interested in vital capacities near zero and can feel quite justified in
excluding them. In Figure 5.10 we certainly are interested in zero
mortality; it is surely what we are aiming for. The point is that graphs can
so easily mislead the unwary reader, so let the reader beware.
The advent of powerful personal computers led to an increase in the
ability to produce complicated graphics. Simple charts, such as Figure
5.1, are informative but not visually exciting. One way of decorating such
graphs is make them appear three-dimensional. Figure 5.11 shows the
effect. The angles are no longer proportional to the numbers which they
represent. The areas are, but because they are different shapes it is
difficult to compare them. This defeats the primary object of conveying
information quickly and accurately. Another approach to decorating
diagrams is to turn them into pictures. In a pictogram the bars of
the bar chart are replaced by pictures. Pictograms can be highly

misleading, as the height of a picture, drawn with three-dimensional
effect, is proportional to the number represented, but what we see is the
volume. Such decorated graphs are like the illuminated capitals of
medieval manuscripts: nice to look at but hard to read. I think they should
be avoided.
Fig. 5.12. Tuberculosis mortality in England and Wales, 1871–1971
(DHSS 1976)
Huff (1954) recounts that the president of a chapter of the American

Statistical Association criticized him for accusing presenters of data of
trying to deceive. The statistician argued that incompetence was the
problem. Huff's reply was that diagrams frequently sensationalize by
exaggeration and rarely minimize anything, that presenters of data rarely
distort those data to make their case appear weaker than it is. The errors
are too one-sided for us to ignore the possibility that we are being fooled.
When presenting data, especially graphically, be very careful that the
data are shown fairly. When on the receiving end, beware!
5.9 Logarithmic scales

Figure 5.12 shows a line graph representing the fall in tuberculosis
mortality in England and Wales over 100 years (DHSS 1976). We can
see a rather unsteady curve, showing the continuing decline in the
disease. Figure 5.12 also shows the mortality plotted on a logarithmic (or
log) scale. A logarithmic scale is one where two pairs of points will be
the same distance apart if their ratios are equal, rather than their
differences. Thus the distance between 1 and 10 is equal to that between
10 and 100, not to that between 10 and 19. (See §5A if you do not
understand this.) The logarithmic line shows a clear kink in the curve
about 1950, the time when a number of effective anti-TB measures,
chemotherapy with streptomycin, BCG vaccine and mass screening with
X-rays, were introduced. If we consider the properties of logarithms
(§5A), we can see how the log scale for the tuberculosis mortality data
produced such sharp changes in the curve. If the relationship is such that
the mortality is falling with a constant proportion, such as 10% per year,
the absolute fall each year depends on the absolute level in the
preceding year:
mortality in 1960 = constant × mortality in 1959
So if we plot mortality on a log scale we get:

log(mortality in 1960) = log(constant) + log(mortality in 1959)
For mortality in 1961, we have
Hence we get a straight line relationship between log mortality and time t:
log(mortality after t years) = t × log(constant) + log(mortality as start)
When the constant proportion changes, the slope of the straight line
formed by plotting log(mortality) against time changes and there is a very
obvious kink in the line.
Log scales are very useful analytic tools. However, a graph on a log scale
can be very misleading if the reader does not allow for the nature of the
scale. The log scale in Figure 5.12 shows the increased rate of reduction
in mortality associated with the anti-TB measures quite plainly, but it
gives the impression that these measures were important in the decline
of TB. This is not so. If we look at the corresponding point on the natural
scale, we can see that all these measures did was to accelerate a decline
which had been going on for a long time (see Radical Statistics Health
Group 1976)
Appendices
5A Appendix: Logarithms
Logarithms are not simply a method of calculation dating from before the
computer age, but a set of fundamental mathematical functions. Because
of their special properties they are much used in statistics. We shall start
with logarithms (or logs for short) to base 10, the common logarithms
used in calculations. The log to base 10 of a number x is y where
x = 10y
We write y = log10(x). Thus for example log10(10) = 1, log10(100) = 2,
log10(1 000) = 3, log10(10 000) = 4, and so on. If we multiply two
numbers, the log of the product is the sum of their logs:
log(xy) = log(x) + log(y)
For example.
100 × 1 000 = 102 × 103 = 102+3 = 105 = 100 000
Or in log terms:
log10(100 × 1 000) = log10(100) + log10(1 000) = 2 + 3 = 5
Hence, 100 × 1 000 = 105 = 100 000. This means that any multiplicative
relationship of the form
y = a × b × c × d
can be made additive by a log transformation:
log(y) = log(a) + log(b) + log(c) + log(d)
This is the process underlying the fit to the Lognormal Distribution
described in §7.4.
There is no need to use 10 as the base for logarithms. We can use any
number. The log of a number x to base b can be found from the log to
base a by a simple calculation:
Ten is convenient for arithmetic using log tables, but for other purposes it
is less so. For example, the gradient, slope or differential of the curve y =
log10(x) is log10(e)/x, where e = 2.718 281… is a constant which does not
depend on the base of the logarithm. This leads to awkward constants
spreading through formulae. To keep this to a minimum we use logs to
the base e, called natural or Napierian logarithms after the mathematician
John Napier. This is the logarithm usually produced by LOG(X) functions
in computer languages.
Figure 5.13 shows the log curve for three different bases, 2, e and 10.
The curves all go through the point (1,0), i.e. log(1) = 0. As x approaches
0, log(x) becomes a larger and larger negative number, tending towards
minus infinity as x tends to zero. There are no logs of negative numbers.
As x increases from 1, the curve becomes flatter and flatter. Though
log(x) continues to increase, it does so more and more slowly. The
curves all go through (base, 1) i.e. log(base) = 1. The curve for log to the
base 2 goes through (2,1), (4,2), (8,3) because 21 = 2. 22 = 4, 23 = 8. We
can see that the effect of replacing data by their logs will be to stretch out
the scale at the lower end and contract it at the upper.
We often work with logarithms of data rather than the data themselves.
This may have several advantages. Multiplicative relationships may
become additive, curves may become straight lines and skew
distributions may become symmetrical.
We transform back to the natural scale using the antilogarithm or
antilog. If y = log10(x), x = 10y is the antilog of y. If Z = loge(x), x = ez or x
= exp(z) is the antilog of z. If your computer program does not transform
back, most calculators have ex and 10x functions for this purpose.
Fig. 5.13. Logarithmic curves to three different bases

20. ‘After treatment with Wondermycin, 66.67% of patients made a
complete recovery’
(a) Wondermycin is wonderful;
(b) this statement may be misleading because the denominator is
not given;
(c) the number of significant figures used suggest a degree of
precision which may not be present;
(d) some control information is required before we can draw any
conclusions about Wondermycin;
(e) there might be only a very small number of patients.
View Answer
21. The number 1 729.543 71:
(a) to two significant figures is 1 700;
(b) to three significant figures is 1 720;
(c) to six decimal places is 1 729.54;
(d) to three decimal places is 1 729.544;
(e) to five significant figures is 1 729.5.
View Answer
Fig. 5.14. A dubious graph
22. Figure 5.14:

(a) shows a histogram;
(b) should have the vertical axis labelled;
(c) should show the zero on the vertical axis;
(d) should show the zero on the horizontal axis;
(e) should show the units for the vertical axis.
View Answer
23. Logarithmic scales used in graphs showing time trends:

(a) show changes in the trend clearly;
(b) often produce straight lines;
(c) give a clear idea of the magnitude of changes;
(d) should show the zero point from the original scale;
(e) compress intervals between large numbers compared to those
between small numbers.
View Answer
24. The following methods can be used to show the relationship

between two variables:
(a) histogram;
(b) pie chart;
(c) scatter diagram;
(d) bar chart;
(e) line graph.
View Answer
Table 5.8. Weekly geriatric admissions in

Wandsworth Health District from May to
September, 1982 and 1983 (Fish et al. 1985)
Week 1982 1983 Week 1982 1983

1 24 20 12 11 25
2 22 17 13 6 22
3 21 21 14 10 26
4 22 17 15 13 12
5 24 22 16 19 33
6 15 23 17 13 19
7 23 20 18 17 21
8 21 16 19 10 28
9 18 24 20 16 19
10 21 21 21 24 13
11 17 20 22 15 29
5E Exercise: Creating graphs

In this exercise we shall display graphically some of the data we have
studied so far.
1. Table 4.1 shows diagnoses of patients in a hospital census.
Display these data as a graph.
View Answer
2. Table 2.8 shows the paralytic polio rates for several groups of
children. Construct a bar chart for the results from the randomized
control areas.
View Answer
3. Table 3.1 shows some results from the study of mortality in

British doctors. Show these graphically.
View Answer
4. Table 5.8 shows the numbers of geriatric admissions in

Wandsworth Health District for each week from May to September in
1982 and 1983. Show these data graphically. Why do you think the
two years were different?
View Answer
> Table of Contents > 6 - Probability
6
Probability
6.1 Probability
We use data from a sample to draw conclusions about the population
from which it is drawn. For example, in a clinical trial we might observe
that a sample of patients given a new treatment respond better than
patients given an old treatment. We want to know whether an
improvement would be seen in the whole population of patients, and if so
how big it might be. The theory of probability enables us to link samples
and populations, and to draw conclusions about populations from
samples. We shall start the discussion of probability with some simple
randomizing devices, such as coins and dice, but the relevance to
medical problems should soon become apparent.
We first ask what exactly is meant by ‘probability’. In this book I shall take
the frequency definition: the probability that an event will happen under
given circumstances may be defined as the proportion of repetitions of
those circumstances in which the event would occur in the long run. For
example, if we toss a coin it comes down either heads or tails. Before we
toss it, we have no way of knowing which will happen, but we do know
that it will either be heads or tails. After we have tossed it, of course, we
know exactly what the outcome is. If we carry on tossing our coin, we
should get several heads and several tails. If we go on doing this for long
enough, then we would expect to get as many heads as we do tails. So
the probability of a head being thrown is half, because in the long run a
head should occur on half of the throws. The number of heads which
might arise in several tosses of the coin is called a random variable, that
is, a variable which can take more than one value with given probabilities.
In the same way, a thrown die can show six faces, numbered one to six,
with equal probability. We can investigate random variables such as the
number of sixes in a given number of throws, the number of throws
before the first six, and so on. There is another, broader definition of
probability which leads to a different approach to statistics, the Bayesian
school (Bland and Altman 1998), but it is beyond the scope of this book.
The frequency definition of probability also applies to continuous
measurement, such as human height. For example, suppose the median
height in a population of women is 168 cm. Then half the women are
above 168 cm in height. If we choose women at random (i.e. without the
characteristics of the woman influencing the choice) then in the long run
half the women chosen will have
heights above 168 cm. The probability of a woman having height above
168 cm is one half. Similarly, if 1/10 of the women have height greater
than 180 cm. a woman chosen at random will have height greater than
180 cm with probability 1/10. In the same way we can find the probability
of height being between any given values. When we measure a
continuous quantity we are always limited by the method of
measurement, and so when we say a woman's height is 170 cm we
mean that it is between, say, 169.5 and 170.5 cm, depending on the
accuracy with which we measure. So what we are interested in is the
probability of the random variable taking values between certain limits
rather than particular values.
6.2 Properties of probability

The following simple properties follow from the definition of probability.
1. A probability lies between 0.0 and 1.0. When the event never happens
the probability is 0.0, when it always happens the probability is 1.0.
2. Addition rule. Suppose two events are mutually exclusive, i.e. when
one happens the other cannot happen. Then the probability that one or
the other happens is the sum of their probabilities. For example, a
thrown die may show a one or a two, but not both. The probability that
it shows a one or a two = 1/6 + 1/6 = 2/6.
3. Multiplication rule. Suppose two events are independent, i.e.
knowing one has happened tells us nothing about whether the other
happens. Then the probability that both happen is the product of their
probabilities. For example, suppose we toss two coins. One coin does
not influence the other, so the results of the two tosses are
independent, and the probability of two heads occurring is 1/2 × 1/2 =
1/4. Consider two independent events A and B. The proportion of times
A happens in the long run is the probability of A. Since A and B are
independent, of those times when A happens, a proportion, equal to
probability of B, will have B happen also. Hence the proportion of times
that A and B happen together is the probability of A multiplied by the
probability of B.
6.3 Probability distributions and random

variables
Suppose we have a set of events which are mutually exclusive and which
includes all the events which can possibly happen. The sum of their
probabilities is 1.0. The set of these probabilities make up a probability
distribution. For example, if we toss a coin the two possibilities, head or
tail, are mutually exclusive and these are the only events which can
happen. The probability distribution is:
PROB(head) = 1/2
PROB(tail) = 1/2
Now, let us define a variable, which we will denote by the symbol X, such
that X = 0 if the coin shows a tail and X = 1 if the coin shows a head. X is
the number of heads shown on a single toss, which must be 0 or 1. We

do not know before the toss what X will be, but do know the probability of
it having any possible value. X is a random variable (§6.1) and the
probability distribution is also the distribution of X. We can represent this
with a diagram, as in Figure 6.1(a).
Fig. 6.1. Probability distributions for the number of heads shown in
the toss of one coin and in tosses of two coins
What happens if we toss two coins at once? We now have four possible
events: a head and a head, a head and a tail, a tail and a head, a tail and
a tail. Clearly, these are equally likely and each has probability 1/4. Let Y
be the number of heads. Y has three possible values: 0, 1, and 2. Y = 0
only when we get a tail and a tail and has probability 1/4. Similarly, Y = 2
only when we get a head and a head, so has probability 1/4. However, Y
= 1 either when we get a head and tail, or when we have a tail and a
head, and so has probability 1/4 + 1/4 = 1/2. We can write this probability
distribution as:
PROB(Y = 0) = 1/4
PROB(Y = 1) = 1/2
PROB(Y = 2) = 1/4
The probability distribution of Y is shown in Figure 6.1(b).
6.4 The Binomial distribution

We have considered the probability distributions of two random variables:
X, the number of heads in one toss of a coin, taking values 0 and 1, and
Y, the number of heads when we toss two coins, taking values 0, 1 or 2.
We can increase the number of coins; Figure 6.2 shows the distribution
of the number of heads obtained when 15 coins are tossed. We do not
need the probability of a ‘head’ to be 0.5: we can count the number of
sixes when dice are thrown. Figure 6.2 also shows the distribution of the
number of sixes obtained from 10 dice. In general, we can think of the
coin or the die as trials, which can have outcomes success (head or six)
or failure (tail or one to five). The distributions in Figures 6.1 and 6.2 are
all examples of the Binomial distribution, which arises frequently in
medical applications. The Binomial distribution is the distribution
followed
by the number of successes in n independent trials when the probability

of any single trial being a success is p. The Binomial distribution is in fact
a familiy of distributions, the members of which are defined by the values
of n and p. The values which define which member of the distribution
family we have are called the parameters of the distribution.
Fig. 6.2. Distribution of the number of heads shown when 15 coins

are tossed and of the number of sixes shown when 10 dice are
thrown, examples of the Binomial distribution
Simple randomizing devices like coins and dice are of interest in

themselves, but not of obvious relevance to medicine. However, suppose
we are carrying out a random sample survey to estimate the unknown
prevalence, p, of a disease. Since members of the sample are chosen at
random and independently from the population, the probability of any
chosen subject having the disease is p. We thus have a series of
independent trials, each with probability of success p, and the number of
successes, i.e. members of the sample with the disease, will follow a
Binomial distribution. As we shall see later, the properties of the Binomial
distribution enable us to say how accurate is the estimate of prevalence
obtained (§8.4).
We could calculate the probabilities for a Binomial distribution by listing
all the ways in which, say, 15 coins can fall. However, there are 215 = 32
768 combinations of 15 coins, so this is not very practical. Instead, there
is a formula for the probability in terms of the number of throws and the
probability of a head. This enables us to work these probabilities out for
any probability of success and any number of trials. In general, we have
n independent trials with the probability that a trial is a success being p.
The probability of r successes is
where n!. called n factorial, is n × (n - 1) × (n - 2) ×… × 2 × 1. This rather

forbidding formula arises like this. For any particular series of r
successes, each with probability p, and n - r failures, each with probability
1 - p, the probability of the series happening is pr(1 - p)(n-r), since the
trials are independent and the multiplicative rule applies. The number of
ways in which r things may be chosen from n things is n!/r!(n-r)! (§6A).
Only one combination can happen at
one time, so we have n!/r!(n-r)! mutually exclusive ways of having r

successes, each with probability pr(1 - p)(n-r). The probability of having r
successes is the sum of these n!/r!(n - r)! probabilities, giving the formula
above. Those who remember the binomial expansion in mathematics will
see that this is one term of it, hence the name Binomial distribution.
Fig. 6.3. Binomial distributions with different n, p = 0.3
We can apply this to the number of heads in tosses of two coins. The
number of heads will be from a Binomial distribution with p = 0.5 and n =
2. Hence the probability of two heads (r = 2) is:
Note that 0! = 1 (§6A), and anything to the power 0 is 1. Similarly for r = 1

and r = 0:
This is what was found for two coins in §6.3. We can use this distribution
whenever we have a series of trials with two possible outcomes. If we
treat a group of patients, the number who recover is from a Binomial
distribution. If we measure the blood pressure of a group of people, the
number classified as hypertensive is from a Binomial distribution.
Figure 6.3 shows the Binomial distribution for p = 0.3 and increasing
values of n. The distribution becomes more symmetrical as n increases.
It is converging to the Normal distribution, described in the next chapter.
6.5 Mean and variance
The number of different probabilities in a Binomial distribution can be
very large and unwieldy. When n is large, we usually need to summarize
these probabilities in some way. Just as a frequency distribution can be
described by its mean and variance, so can a probability distribution and
its associated random variable.
The mean is the average value of the random variable in the long run. It
is also called the expected value or expectation and the expectation of
a random variable X is usually denoted by E(X). For example, consider
the number of heads in tosses of two coins. We get 0 heads in 1/4 of
pairs of coins, i.e. with probability 1/4. We get 1 head in 1/2 of pairs of
coins, and 2 heads in 1/4 of pairs. The average value we should get in
the long run is found by multiplying each value by the proportion of pairs
in which it occurs and adding:
If we kept on tossing pairs of coins, the average number of heads per

pair would be 1. Thus for any random variable which takes discrete
values the mean, expectation or expected value is found by summing
each possible value multiplied by its probability.
Note that the expected value of a random variable does not have to be a
value that the random variable can actually take. For example, for the
mean number of heads in throws of one coin we have either no heads or
1 head, each with probability half, and the expected value is 0 × ½ + 1 ×
½ = ½. The number of heads must be 0 or 1, but the expected value is
half, the average which we would get in the long run.
The variance of a random variable is the average squared difference
from the mean. For the number of heads in tosses of two coins, 0 is 1
unit from the mean and occurs for 1/4 of pairs of coins, 1 is 0 units from
the mean and occurs for half of the pairs and 2 is 1 unit from the mean
and occurs for 1/4 of pairs, i.e. with probability 1/4. The variance is then
found by squaring these differences, multiplying by the proportion of
times the difference will occur (the probability) and adding:
We denote the variance of a random variable X by VAR(X). In
mathematical terms,
VAR(X) = E (X2 - E(X)2)
The square root of the variance is the standard deviation of the random
variable or distribution. We often use the Greek letter µ, pronounced ‘mu’,
and σ, ‘sigma’,
for the mean and standard deviation of a probability distribution. The

variance is then σ2.
The mean and variance of the distribution of a continuous variable, of
which more in Chapter 7, are defined in a similar way. Calculus is used to
define them as integrals, but this need not concern us here. Essentially
what happens is that the continuous scale is broken up into many very
small intervals and the value of the variable in that very small interval is
multiplied by the probability of being in it, then these are added.
6.6 Properties of means and variances

When we use the mean and variance of probability distributions in
statistical calculations, it is not the details of their formulae which we
need to know, but some of their simple properties. Most of the formulae
used in statistical calculations are derived from these. The reasons for
these properties are quite easy to see in a non-mathematical way.
If we add a constant to a random variable, the new variable so created
has a mean equal to that of the original variable plus the constant. The
variance and standard deviation will be unchanged. Suppose our random
variable is human height. We can add a constant to the height by
measuring the heights of people standing on a box. The mean height of
people plus box will now be the mean height of the people plus the
constant height of the box. The box will not alter the variability of the
heights, however. The difference between the tallest and smallest, for
example, will be unchanged. We can subtract a constant by asking the
people to stand in a constant hole to be measured. This reduces the
mean but leaves the variance unchanged as before. (My free program
Clinstat (§1.3) has a simple graphics program which illustrates this.)
If we multiply a random variable by a positive constant, the mean and
standard deviation are multiplied by the constant, the variance is
multiplied by the square of the constant. For example, if we change our
units of measurements, say from inches to centimetres, we multiply each
measurement by 2.54. This has the effect of multiplying the mean by the
constant, 2.54, and multiplying the standard deviation by the constant
since it is in the same units as the observations. However, the variance is
measured in squared units, and so is multiplied by the square of the
constant. Division by a constant works in the same way. If the constant is
negative, the mean is multiplied by the constant and so changes sign.
The variance is multiplied by the square of the constant, which is positive,
so the variance remains positive. The standard deviation, which is the
square root of the variance, is always positive. It is multiplied by the
absolute value of the constant, i.e. the constant without the negative sign.
If we add two random variables the mean of the sum is the sum of the
means, and, if the two variables are independent, the variance of the sum
is the sum of their variances. We can do this by measuring the height of
people standing on boxes of random height. The mean height of people
on boxes is the mean height of people + the mean height of the boxes.
The variability of the heights is also
increased. This is because some short people will find themselves on

small boxes, and some tall people will find themselves on large boxes. If
the two variables are not independent, something different happens. The
mean of the sum remains the sum of the means, but the variance of the
sum is not the sum of the variances. Suppose our people have decided
to stand on the boxes, not just at a statistician's whim, but for a purpose.
They wish to change a light bulb, and so must reach a required height.
Now the short people must pick large boxes, whereas tall people can
make do with small ones. The result is a reduction in variability to almost
nothing. On the other hand, if we told the tallest people to find the largest
boxes and the shortest to find the smallest boxes, the variablity would be
increased. Independence is an important condition.
If we subtract one random variable from another, the mean of the
difference is the difference between the means, and, if the two variables
are independent, the variance of the difference is the sum of their
variances. Suppose we measure the heights above ground level of our
people standing in holes of random depth. The mean height above
ground is the mean height of the people minus the mean depth of the
hole. The variability is increased, because some short people stand in
deep holes and some tall people stand in shallow holes. If the variables
are not independent, the additivity of the variances breaks down, as it did
for the sum of two variables. When the people try to hide in the holes,
and so must find a hole deep enough to hold them, the variability is again
reduced.
The effects of multiplying two random variables and of dividing one by
another are much more complicated. Fortunately we rarely need to do
this.
We can now find the mean and variance of the Binomial distribution with
parameters n and p. First consider n = 1. Then the probability distribution
is:
The mean is therefore 0 × (1 - p) + 1 × p = p. The variance is
Now, a variable from the Binomial distribution with parameters n and p is

the sum of n independent variables from the Binomial distribution with
parameters 1 and p. So its mean is the sum of n means all equal to p,
and its variance is the sum of n variances all equal to p(1 - p). Hence the
Binomial distribution has mean = np and variance = np(1 - p). For large
sample problems, these are more useful than the Binomial probability
formula.
The properties of means and variances of random variables enable us to
find a formal solution to the problem of degrees of freedom for the
sample variance discussed in Chapter 4. We want an estimate of
variance whose expected value is the population variance. The expected
value of Σ(xi - [x with bar above])2 can be shown to
be (n - 1)VAR(x) (§6B) and hence we divide by n - 1, not n, to get our

estimate of variance.
Fig. 6.4. Poisson distributions with four different means
6.7 * The Poisson distribution

The Binomial distribution is one of many probability distributions which
are used in statistics. It is a discrete distribution, that is it can take only a
finite set of possible values, and is the discrete distribution most
commonly encountered in medical applications. One other discrete
distribution is worth discussing at this point, the Poisson distribution.
Although, like the Binomial, the Poisson distribution arises from a simple
probability model, the mathematics involved is more complicated and will
be omitted.
Suppose events happen randomly and independently in time at a
constant rate. The Poisson distribution is the distribution followed by
the number of events which happen in a fixed time interval. If events
happen with rate µ events per unit time, the probability of r events
happening in unit time is
where e = 2.718…, the mathematical constant. If events happen

randomly and independently in space, the Poisson distribution gives the
probabilities for the number of events in unit volume or area.
There is seldom any need to use individual probabilities of this
distribution,
as its mean and variance suffice. The mean of the Poisson distribution for
the number of events per unit time is simply the rate, µ. The variance of
the Poisson distribution is also equal to µ. Thus the Poisson is a family of
distributions, like the Binomial, but with only one parameter, µ. This
distribution is important, because deaths from many diseases can be
treated as occuring randomly and independently in the population. Thus,
for example, the number of deaths from lung cancer in one year among
people in an occupational group, such as coal miners, will be an
observation from a Poisson distribution, and we can use this to make
comparisons between mortality rates (§16.3).
Figure 6.4 shows the Poisson distribution for four different means. You
will see that as the mean increases the Poisson distribution looks rather
like the Binomial distribution in Figure 6.3. We shall discuss this similarity
further in the next chapter.
6.8 * Conditional probability

Sometimes we need to think about the probability of an event if another
event has happened. For example, we might ask what is the probablity
that a patient has coronary artery disease if he or she has tingling pain in
the left arm. This is called a conditional probability, the probability of
the event (coronary artery disease) given a condition (tingling pain). We
write this probability thus, separating the event and the condition by a
vertical bar:
PROB(coronary artery disease | tingling pain)
Conditional probablities are useful in statistical aids to diagnosis (§15.7).
For a simpler example, we can go back to tosses of two coins. If we toss
one coin then the other, the first toss alters the probabilities for the
possible outcomes for the two coins:
PROB(both coins heads | first coin head) = 0.5
PROB(head and tail | first coin head) = 0.5
PROB(both coins tails | first coin head) = 0.0
and
PROB(both coins heads | first coin tail) = 0.0
PROB(head and tail | first coin tail) = 0.5
PROB(both coins tails | first coin tail) = 0.5
The multiplicative rule (§6.2) can be extended to deal with events which
are not independent. For two events A and B:
PROB(A and B) = PROB(A | B)PROB(B) = PROB(B | A)PROB(A).
It is important to understand that PROB(A | B) and PROB(B | A) are not
the same. For example, Table 6.1 shows the relationship between two
diseases, hay fever and eczema in a large group of children. The
probability that in this group a child with hay fever will have eczema also
is
PROB(eczema | hay fever) = 141/1 069 = 0.13
the proportion of children with hayfever who have eczema also. This is
clearly much less than the probablity that a child with eczema will have
hay fever,
PROB(hay fever | eczema) = 141/561 = 0.25
the proportion of children with eczema who have hayfever also.
Table 6.1. Relationship between hay fever and

eczema at age 11 in the National Child
Development Study
Hay fever
Eczema Total
Yes No
Yes 141 420 561
No 928 13 525 14 453
Total 1 069 13 945 15 522
This may look obvious, but confusion between conditional probabilities is

common and can cause serious problems, for example in the
consideration of forensic evidence. Typically, this will produce the
probability that a material found a crime scene (DNA, fibres, etc.) will
match the suspect as closely as it does given that the material did not
come from the subject. This is
PROB(evidence | suspect not at crime scene).
It is not the same as
PROB(suspect not at crime scene | evidence),
but this is often how it is interpreted, an inversion known as the
prosecutor's fallacy.
Appendices
6A Appendix: Permutations and
combinations
For those who never knew, or have forgotten, the theory of combinations,
it goes like this. First, we look at the number of permutations, i.e. ways of
arranging a set of objects. Suppose we have n objects. How many ways
can we order them? The first object can be chosen n ways, i.e. any
object. For each first object there are n - 1 possible second objects, so
there are n × (n - 1) possible first and second permutations. There are
now only n - 2 choices for the third object, n - 3 choices for the fourth,
and so on, until there is only one choice for the last. Hence, there are n ×
(n - 1) × (n - 2) × … × 2 × 1 permutations of n objects. We call this
number the factorial of n and write it ‘n!’.
Now we want to know how many ways there are of choosing r objects
from n objects. Having made a choice of r objects, we can order those in
r! ways. We can also order the n - r not chosen in (n - r)! ways. So the
objects can be ordered in r!(n - r)! ways without altering the objects
chosen. For example, say we choose the first two from three objects, A,
B and C. Then if these are A and B, two permutations give this choice,
ABC and BAC. This is, of course, 2! × 1! = 2 permutations. Each
combination of r things accounts for r!(n - r)! of the n! permutations
possible, so there are
possible combinations. For example, consider the number of

combinations of two objects out of three, say A, B and C. The possible
choices are AB, AC and BC. There is no other possibility. Applying the
formula, we have n = 3 and r = 2 so
Sometimes in using this formula we come across r = 0 or r = n leading to
0!. This cannot be defined in the way we have chosen, but we can
calculate its only possible value, 0! = 1. Because there is only one way of
choosing n objects from n, we have
so 0! = 1.
6B Appendix: Expected value of a sum of

squares
The properties of means and variances described in §6.6 can be used to

answer the question raised in §4.7 and §4A about the divisor in the
sample variance. We ask why the variance from a sample is
and not
We shall be concerned with the general properties of samples of size n,

so we shall treat n as a constant and xi and [x with bar above] as random
variables. We shall suppose xi has mean µ and variance σ2.
The expected value of the sum of square is
because the expected value of the difference is the difference between

the expected values and n is a constant. Now, the population variance σ2
is the average squared distance from the population mean µ, so
because µ is a constant. Because E(xi) = µ, we have
and so we find E(x2i) = σ2 + µ2 and so E(Σx2i) = n(σ2 + µ2), being the sum
of n numbers all of which are σ2 + µ2. We now find the value of E((Σxi)2).
We need
Just as E(x2i) = σ2 + µ2 = VAR(xi) + (E(xi))2 so
So
So the expected value of the sum of squares is (n - 1)σ2 and we must

divide the sum of squares by n - 1, not n, to obtain the estimate of the
variance, σ2.
We shall find the variance of the sample mean, [x with bar above], useful
later (§8.2):

(Each branch is either true or false.)
25. The events A and B are mutually exclusive, so:
(a) PROB(A or B) = PROB(A) + PROB(B);
(b) PROB(A and B) = 0;
(c) PROB(A and B) = PROB(A) PROB(B);
(d) PROB(A) = PROB(B);
(e) PROB(A) + PROB(B) = 1.
View Answer
26. The probability of a woman aged 50 having condition X is 0.20

and the probability of her having condition Y is 0.05. These
probabilities are independent:
Fig. 19.5. Pie chart showing the distribution of patients in

Tooting Bec Hospital by diagnostic group
Fig. 19.6. Bar chart showing the results of the Salk vaccine trial
(a) the probability of her having both conditions is 0.01;

(b) the probability of her having both conditions is 0.25;
(c) the probability of her having either X, or Y, or both is 0.24;
(d) if she has condition X, the probability of her having Y also is
0.01;
(e) if she has condition Y, the probability of her having X also is
0.20.
View Answer
27. The following variables follow a Binomial distribution:

(a) number of sixes in 20 throws of a die;
(b) human weight;
(c) number of a random sample of patients who respond to a
treatment;
(d) number of red cells in 1 ml of blood;
(e) proportion of hypertensives in a random sample of adult men.
View Answer
28. Two parents each carry the same recessive gene which each
transmits to their child with probability 0.5. If their child will develop
clinical disease if it inherits the gene from both parents and will be a
carrier if it inherits the gene from one parent only then:
(a) the probability that their next child will have clinical disease is
0.25;
(b) the probability that two successive children will both develop
clinical disease is 0.25 × 0.25;
(c) the probability their next child will be a carrier without clinical
disease is 0.50:
(d) the probability of a child being a carrier or having clinical
disease is 0.75;
(e) if the first child does not have clinical disease, the probability
that the second child will not have clinical disease is 0.752.
View Answer
Table 6.2. Number of men remaining alive at

ten year intervals (from English Life Table No.
11, Males)
Number Number
Age in Age in
surviving, surviving,
years, x years, x
lx lx
0 1 000 60 758
10 959 70 524
20 952 80 211
30 938 90 22
40 920 100 0
50 876
29. If a coin is spun twice in succession:

(a) the expected number of tails is 1.5;
(b) the probability of two tails is 0.25;
(c) the number of tails follows a Binomial distribution;
(d) the probability of at least one tail is 0.5;
(e) the distribution of the number of tails is symmetrical.
View Answer
30. If X is a random variable, mean µ and variance σ2:

(a) E(X + 2) = µ;
(b) VAR(X + 2) = σ2;
(c) E(2X) = 2µ;
(d) VAR(2X) = 2σ2;
(e) VAR(X/2) = σ2/4.
View Answer
31. If X and Y are independent random variables:

(a) VAR(X + Y) = VAR(X) + VAR(Y);
(b) E(X + Y) = E(X) + E(Y);
(c) E(X - Y) = E(X) - E(Y);
(d) VAR(X - Y) = VAR(X) - VAR(Y);
(e) VAR(-X) = - VAR(X).
View Answer
6E Exercise: Probability and the life table

In this exercise we shall apply some of the basic laws of probability to a
practical exercise. The data are based on a life table. (I shall say more
about these in §16.4.) Table 6.2 shows the number of men, from a group
numbering 1000 at birth, who we would expect to be alive at different
ages. Thus, for example, after 10 years, we see that 959 survive and so
41 have died, at 20 years 952 survive and so 48 have died, 41 between
ages 0 and 9 and 7 between ages 10 and 19.
1. What is the probability that an individual chosen at random will
survive to age 10?
View Answer
2. What is the probability that this individual will die before age 10?
Which property of probability does this depend on?
View Answer
3. What are the probabilities that the individual will survive to ages
10, 20. 30, 40, 50, 60, 70. 80, 90, 100? Is this set of probabilities a
probability distribution?
View Answer
4. What is the probability that an individual aged 60 years survives

to age 70?
View Answer
5. What is the probability that two men aged 60 will both survive to
age 70? Which property of probability is used here?
View Answer
6. If we had 100 individuals aged 60, how many would we expect to

attain age 70?
View Answer
7. What is the probability that a man dies in his second decade? You
can use the fact that PROB(death in 2nd) + PROB(survives to 3rd) =
PROB(survives to 2nd).
View Answer
8. For each decade, what is the probability that a given man will die
in that decade? This is a probability distribution—why? Sketch the
distribution.
View Answer
9. As an approximation, we can assume that the average number of

years lived in the decade of death is 5. Thus, those who die in the
2nd decade will have an average life span of 15 years. The
probability of dying in the 2nd decade is 0.007, i.e. a proportion
0.007 of men have a mean lifetime of 15 years. What is the mean
lifetime of all men? This is the expectation of life at birth.
View Answer
> Table of Contents > 7 - The Normal distribution
7
The Normal distribution
7.1 Probability for continuous variables

When we derived the theory of probability in the discrete case, we were
able to say what the probability was of a random variable taking a
particular value. As the number of possible values increases, the
probability of a particular value decreases. For example, in the Binomial
distribution with p = 0.5 and n = 2, the most likely value, 1, has probability
0.5. In the Binomial distribution with p = 0.5 and n = 100 the most likely
value, 50, has probability 0.08. In such cases we are usually more
interested in the probability of a range of values than one particular value.
For a continuous variable, such as height, the set of possible values is
infinite and the probability of any particular value is zero (§6.1). We are
interested in the probability of the random variable taking values between
certain limits rather than taking particular values. If the proportion of
individuals in the population whose values are between given limits is p,
and we choose an individual at random, the probability of choosing an
individual who lies between these limits is equal to p. This comes from
our definition of probability, the choice of each individual being equally
likely. The problem is finding and giving a value to this probability.
When we find the frequency distribution for a sample of observations, we
count the number of values in which fall within certain limits (§4.2). We
can represent this as a histogram such as Figure 7.1 (§4.3). One way of
presenting the histogram is as relative frequency density, the proportion
of observations in the interval per unit of X (§4.3), Thus, when the interval
size is 5, the relative frequency density is the relative frequency divided
by 5 (Figure 7.1). The relative frequency in an interval is now represented
by the width of the interval multiplied by the density, which gives the area
of the rectangle. Thus, the relative frequency between any two points can
be found from the area under the histogram between the points. For
example, to estimate the relative frequency between 10 and 20 in Figure
7.1 we have the density from 10 to 15 as 0.05 and between 15 and 20 as
0.03. Hence the relative frequency is
0.05 × (15 - 10) + 0.03 × (20 - 15) = 0.25 + 0.15 = 0.40
If we take a larger sample we can use smaller intervals. We get a
smoother looking histogram, as in Figure 7.2, and as we take larger and
larger samples, and so smaller and smaller intervals, we get a shape
very close to a smooth curve (Figure 7.3). As the sample size
approaches that of the population, which we can assume to be very
large, this curve becomes the relative frequency density of the whole
population. Thus we can find the proportion of observations between any
two limits by finding the area under the curve, as indicated in Figure 7.3.
Fig. 7.1. Histogram showing relative frequency density

Fig. 7.2. The effect on a frequency distribution of increasing sample
size
If we know the equation of this curve, we can find the area under it.
(Mathematically we do this by integration, but we do not need to know
how to integrate to use or to understand practical statistics—all the
integrals we need have been done and tabulated.) Now, if we choose an
individual at random, the probability that X lies between any given limits
is equal to the proportion of individuals who fall between these limits.
Hence, the relative frequency distribution for the whole population gives
us the probability distribution of the variable. We call this curve the
probability density function.
Fig. 7.3. Relative frequency density or probability density function,
showing the probability of an observation between 10 and 20
Fig. 7.4. Mean, µ, standard deviation, σ, and a probability density

function
Probability density functions have a number of general properties. For
example, the total area under the curve must be one, since this is the
total probability of all possible events. Continuous random variables have
means, variances and standard deviations defined in a similar way to
those for discrete random variables and possessing the same properties
(§6.5). The mean will be somewhere near the middle of the curve and
most of the area under the curve will be between the mean minus two
standard deviations and the mean plus two standard deviations (Figure
7.4).
The precise shape of the curve is more difficult to ascertain. There are
many possible probability density functions and some of these can be
shown to arise from simple probability situations, as were the Binomial
and Poisson distributions. However, most continuous variables with
which we have to deal, such as
height, blood pressure, serum cholesterol, etc., do not arise from simple
probability situations. As a result, we do not know the probability
distribution for these measurements on theoretical grounds. As we shall
see, we can often find a standard distribution whose mathematical
properties are known, which fits observed data well and which enables us
to draw conclusions about them. Further, as sample size increases the
distribution of certain statistics calculated from the data, such as the
mean, become independent of the distribution of the observations
themselves and follow one particular distribution form, the Normal
distribution. We shall devote the remainder of this chapter to a study of
this distribution.
Fig. 7.5. Binomial distributions for p = 0.3 and six different values of
n, with corresponding Normal distribution curves
7.2 The Normal distribution

The Normal distribution, also known as the Gaussian distribution, may be
regarded as the fundamental probability distribution of statistics. The
word ‘normal’ is not used here in its common meaning of ‘ordinary or
common’, or its medical meaning of ‘not diseased’. The usage relates to
its older meaning of ‘conforming to a rule or pattern’, and as we shall see,
the Normal distribution is the form to which the Binomial distribution tends
as its parameter n increases. There is no implication that most variables
follow a Normal distribution.
We shall start by considering the Binomial distribution as n increases. We
saw in §6.4 that, as n increases, the shape of the distribution changes.
The most extreme possible values become less likely and the distribution
becomes more symmetrical. This happens whatever the value of p. The
position of the distribution along the horizontal axis, and its spread, are
still determined by p, but the shape is not. A smooth curve can be drawn
which goes very close to these points. This is the Normal distribution
curve, the curve of the continuous distribution which the Binomial
distribution approaches as n increases. Any Binomial distribution may be
approximated by the Normal distribution of the same mean
and variance provided n is large enough. Figure 7.5 shows the Binomial
distributions of Figure 6.3 with the corresponding Normal distribution
curves. From n = 10 onwards the two distributions are very close.
Generally, if both np and n(1 - p) exceed 5 the approximation of the
Binomial to the Normal distribution is quite good enough for most
practical purposes. See §8.4 for an application. The Poisson distribution
has the same property, as Figure 6.4 suggests.
Fig. 7.6. Sums of observations from a Uniform distribution
The Binomial variable may be regarded as the sum of n independent

identically distributed random variables, each being the outcome of one
trial taking value 1 with probability p. In general, if we have any series of
independent, identically distributed random variables, then their sum
tends to a Normal distribution as the number of variables increases. This
is known as the central limit theorem. As most sets of measurements
are observations of such a series of random variables, this is a very
important property. From it, we can deduce that the sum or mean of any
large series of independent observations follows a Normal distribution.
For example, consider the Uniform or Rectangular distribution. This is
the distribution where all values between two limits, say 0 and 1, are
equally likely and no other values are possible. Observations from this
arise if we take random digits from a table of random numbers such as
Table 2.3. Each observation of the Uniform variable is formed by a series
of such digits placed after a decimal point. On a microcomputer, this is
usually the distribution produced by the RND(X) function in the BASIC
language. Figure 7.6 shows the histogram for the frequency distribution
of 500 observations from the Uniform distribution
between 0 and 1. It is quite different from the Normal distribution. Now

suppose we create a new variable by taking two Uniform variables and
adding them (Figure 7.6), The shape of the distribution of the sum of two
Uniform variables is quite different from the shape of the Uniform
distribution. The sum is unlikely to be close to either extreme, here 0 or 2,
and observations are concentrated in the middle near the expected
value. The reason for this is that to obtain a low sum, both the Uniform
variables forming it must be low; to make a high sum both must be high.
But we get a sum near the middle if the first is high and the second low,
or the first is low and second high, or both first and second are moderate.
The distribution of the sum of two is much closer to the Normal than is
the Uniform distribution itself. However, the abrupt cut-off at 0 and at 2 is
unlike the corresponding Normal distribution. Figure 7.6 also shows the
result of adding four Uniform variables and six Uniform variables. The
similarity to the Normal distribution increases as the number added
increases and for the sum of six the correspondence is so close that the
distributions could not easily be told apart.
The approximation of the Binomial to the Normal distribution is a special
case of the central limit theorem. The Poisson distribution is another. If
we take a set of Poisson variables with the same rate and add them, we
will get a variable which is the number of random events in a longer time
interval (the sum of the intervals for the individual variables) and which is
therefore a Poisson distribution with increased mean. As it is the sum of a
set of independent, identically distributed random variables it will tend
towards the Normal as the mean increases. Hence as the mean
increases the Poisson distribution becomes approximately Normal. For
most practical purposes this is when the mean exceeds 10. The similarity
between the Poisson and the Binomial noted in §6.7 is a part of a more
general convergence shown by many other distributions.
7.3 Properties of the Normal distribution

In its simplest form the equation of the Normal distribution curve, called
the Standard Normal distribution, is usually denoted by φ(z), where φ
is the Greek letter ‘phi’:
where π is the usual mathematical constant. The medical reader can be

reassured that we do not need to use this forbidding formula in practice.
The Standard Normal distribution has a mean of 0, a standard deviation
of 1 and a shape as shown in Figure 7.7. The curve is symmetrical about
the mean and often described as ‘bell-shaped’ (though I have never seen
a bell like it). We can note that most of the area, i.e. the probability, is
between -1 and +1, the large majority between -2 and +2, and almost all
between -3 and +3.
Although the Normal distribution curve has many remarkable properties,
it has one rather awkward one: it cannot be integrated. In other words,
there is no simple formula for the probability of a random variable from a
Normal
distribution lying between given limits. The areas under the curve can be
found numerically, however, and these have been calculated and
tabulated. Table 7.1 shows the area under the probability density curve
for different values of the Normal distribution. To be more precise, for a
value z the table shows the area under the curve to the left of z, i.e. from
minus infinity to z (Figure 7.8). Thus Φ(z) is the probability that a value
chosen at random from the Standard Normal distribution will be less than
z. Φ is the Greek capital ‘phi’. Note that half this table is not strictly
necessary. We need only the half for positive z as Φ(-z) + Φ(z) = 1. This
arises from the symmetry of the distribution. To find the probability of z
lying between two values a and b, where b > a, we find Φ(b) - Φ(a). To
find the probability of z being greater than a we find 1 - Φ(a). These
formulae are all examples of the additive law of probability. Table 7.1
gives only a few values of z, and much more extensive ones are
available (Lindley and Miller 1955, Pearson and Hartley 1970). Good
statistical computer programs will calculate these values when they are
needed.
Fig. 7.7. The Standard Normal distribution
Table 7.1. The Normal distributio
z Φ(z) z Φ(z) z Φ(z) z
-3.0 0.001 -2.0 0.023 -1.0 0.159 0.0

-2.9 0.002 -1.9 0.029 -0.9 0.184 0.1
-2.8 0.003 -1.8 0.036 -0.8 0.212 0.2
-2.7 0.003 -1.7 0.045 -0.7 0.242 0.3
-2.6 0.005 -1.6 0.055 -0.6 0.274 0.4
-2.5 0.006 -1.5 0.067 -0.5 0.309 0.5
-2.4 0.008 -1.4 0.081 -0.4 0.345 0.6
-2.3 0.011 -1.3 0.097 -0.3 0.382 0.7
-2.2 0.014 -1.2 0.115 -0.2 0.421 0.8
-2.1 0.018 -1.1 0.136 -0.1 0.460 0.9
-2.0 0.023 -1.0 0.159 0.0 0.500 1.0
There is another way of tabulating a distribution, using what are called
percentage points. The one-sided P percentage point of a distribution

is the value z such that there is a probability P% of an observation from
that distribution being greater than or equal to z (Figure 7.8). The two-
sided P percentage point is the value z such that there is a probability
P% of an observation being greater than or equal to z or less than or
equal to -z (Figure 7.8). Table 7.2 shows both one sided and two sided
percentage points for the Normal distribution. The probability is quoted as
a percentage because when we use percentage points we are usually
concerned with rather small probabilities, such as 0.05 or 0.01, and use
of the percentage form, making them 5% and 1%, cuts out the leading
zero.
Table 7.2. Percentage points of the Normal

distribution
One-sided Two-sided
P1 (z) P2 (z)
50 0.00
25 0.67 50 0.67
10 1.28 20 1.28
5 1.64 10 1.64
2.5 1.96 5 1.96
1 2.33 2 2.33
0.5 2.58 1 2.58
0.1 3.09 0.2 3.09

0.05 3.29 0.1 3.29
The table shows the probability P1 (z) of a

Normal variable with mean 0 and variance 1
being greater than z, and the probability P2 (z)
of a Normal variable with mean 0 and variance
1 being less than -z or greater than z.
Fig. 7.8. One- and two-sided percentage points (5%) of the Standard
Normal distribution
So far we have examined the Normal distribution with mean 0 and

standard deviation 1. If we add a constant µ to a Standard Normal
variable, we get a new variable which has mean µ (see §6.6). Figure 7.9
shows the Normal distribution with mean 0 and the distribution obtained
by adding 1 to it together with their two-sided 5% points. The curves are
identical apart from a shift along the axis.
On the curve with mean 0 nearly all the probability is between -3 and +3.
For the curve with mean 1 it is between -2 and +4, i.e. between the mean
-3 and the mean +3. The probability of being a given number of units
from the mean is the same for both distributions, as is also shown by the
5% points.
Fig. 7.9. Normal distributions with different means and with different
variances, showing two-sided 5% points
If we take a Standard Normal variable, with standard deviation 1, and

multiply by a constant σ we get a new variable which has standard
deviation σ. Figure 7.9 shows the Normal distribution with mean 0 and
standard deviation 1 and the distribution obtained by multiplying by 2.
The curves do not appear identical. For the distribution with standard
deviation 2, nearly all the probability is between -6 and +6, a much wider
interval than the -3 and +3 for the standard distribution. The values -6
and +6 are -3 and +3 standard deviations. We can see that the
probability of being a given number of standard deviations from the mean
is the same for both distributions. This is also seen from the 5% points,
which represent the mean plus or minus 1.96 standard deviations in each
case.
In fact if we add µ to a Standard Normal variable and multiply by σ, we
get a Normal distribution of mean µ, and standard deviation σ. Tables 7.1
and 7.2 apply to it directly, if we denote by z the number of standard
deviations above the mean, rather than the numerical value of the
variable. Thus, for example, the two sided 5% points of a Normal
distribution with mean 10 and standard deviation 5 are found by 10 - 1.96
× 5 = 0.2 and 10 + 1.96 × 5 = 19.8, the value 1.96 being found from Table
7.2.
This property of the Normal distribution, that multiplying or adding
constants still gives a Normal distribution, is not as obvious as it might
seem. The Binomial does not have it, for example. Take a Binomial
variable with n = 3, possible values 0, 1, 2, and 3, and multiply by 2, The
possible values are now 0, 2, 4, and 6. The Binomial distribution with n =
6 has also possible values 1, 3, and 5, so the distributions are different
and the one which we have derived is not a member of the Binomial
family.
We have seen that adding a constant to a variable from a Normal
distribution gives another variable which follows a Normal distribution. If
we add two variables from Normal distributions together, even with
different means and
variances, the sum follows a Normal distribution. The difference between

two variables from Normal distributions also follows a Normal distribution.
Fig. 7.10. Distribution of height in a sample of 1794 pregnant women

(data of Brooke et al. 1989)
Fig. 7.11. Distribution of serum triglyceride (Table 4.8) and log10
triglyceride in cord blood for 282 babies, with corresponding Normal
distribution curves
7.4 Variables which follow a Normal distribution

So far we have discussed the Normal distribution as it arises from
sampling as the sum or limit of other distributions. However, many
naturally occurring variables, such as human height, appear to follow a
Normal distribution very closely. We might expect this to happen if the
variable were the result of adding variation from a number of different
sources. The process shown by the central limit theorem may well
produce a result close to Normal. Figure 7.10 shows the distribution of
height in a sample of pregnant women, and the corresponding Normal
distribution curve. The fit to the Normal distribution is very good.
If the variable we measure is the result of multiplying several different
sources of variation we would not expect the result to be Normal from the
properties
discussed in §7.2, which were all based on addition of variables.

However, if we take the log transformation of such a variable (§5A) we
would then get a new variable which is the sum of several different
sources of variation and which may well have a Normal distribution. This
process often happens with quantities which are part of metabolic
pathways, the rate at which reaction can take place depending on the
concentrations of other compounds. Many measurements of blood
constituents exhibit this, for example. Figure 7.11 shows the distribution
of serum triglyceride measured in cord blood for 282 babies (Table 4.8).
The distribution is highly skewed and quite unlike the Normal distribution
curve. However, when we take the logarithm of the triglyceride
concentration, we have a remarkably good fit to the Normal distribution
(Fig. 7.11). If the logarithm of a random variable follows a Normal
distribution, the random variable itself follows a Lognormal distribution.
We often want to change the scale on which we analyse our data so as
to get a Normal distribution. We call this process of analysing a
mathematical function of the data rather than the data themselves
transformation. The logarithm is the transformation most often used, the
square root and reciprocal are others (see also §10.4). For a single
sample, transformation enables us to use the Normal distribution to
estimate centiles (§4.5). For example, we often want to estimate the 2.5th
and 97.5th centiles. which together enclose 95% of the observations. For
a Normal distribution, these can be estimated by [x with bar above]
±1.96s. We can transform the data so that the distribution is Normal,
calculate the centile, and then transform back to the original scale.
Consider the triglyceride data of Figure 7.11 and Table 4.8. The mean is
0.51 and the standard deviation 0.22. The mean for the log10 transformed
data is -0.33 and the standard deviation is 0.17. What happens if we
transform back by the antilog? For the mean, we get 10-0.33 = 0.47. This
is less than the mean for the raw data. The antilog of the mean log is not
the same as the untransformed arithmetic mean. In fact, this the
geometric mean, which is the nth root of the product of the observations.
If we add the logs of the observations we get the log of their product
(§5A). If we multiply the log of a number by a second number, we get the
log of the first raised to the power of the second. So if we divide the log
by n, we get the log of the nth root. Thus the mean of the logs is the log
of the geometric mean. On back transformation, the reciprocal
transformation also yields a mean with a special name, the harmonic
mean, the reciprocal of the mean of the reciprocals.
The geometric mean is in the original units. If triglyceride is measured in
mmol/litre, the log of a single observation is the log of a measurement in
mmol/litre. The sum of n logs is the log of the product of n measurements
in mmol/litre and is the log of a measurement in mmol/litre to the nth. The
nth root is thus again the log of a number in mmol/litre and the antilog is
back in the original units, mmol/litre (see §5A).
The antilog of the standard deviation, however, is not measured in the
original units. To calculate the standard deviation we take the difference
between each log observation and subtract the log geometric mean,
using the usual formula
Σ(xi - [x with bar above])2/(n-1) (§4.8). Thus we have the difference

between the log of two numbers each measured in mmol/litre, giving the
log of their ratio (§5A) which is the log of a dimensionless pure number. It
would be the same if the triglycerides were measured in mmol/litre or
mg/100ml. We cannot transform the standard deviation back to the
original scale.
If we want to use the standard deviation, it is easiest to do all calculations
on the transformed scale and transform back, if necessary, at the end.
For example, the 2.5th centile on the log scale is -0.33 - 1.96 × 0.17 =
-0.66 and the 97.5th centile is -0.33 + 1.96 × 0.17 = 0.00. To get these we
took the log of something in mmol/litre and added or subtracted the log of
a pure number (i.e. multiplied on the natural scale), so we still have the
log of something in mmol/litre. To get back to the original scale we antilog
to get 2.5th centile = 0.22 and 97.5th centile = 1.00 mmol/litre.
Transforming the data to a Normal distribution and then analysing on the
transformed scale may look like cheating. I do not think it is. The scale on
which we choose to measure things need not be linear, though this is
often convenient. Other scales can be much more useful. We measure
pH on a logarithmic scale, for example. Should the magnitude of an
earthquake be measured in mm of amplitude (linear) or on the Richter
scale (logarithmic)? Should spectacle lenses be measured in terms of
focal length in cm (linear) or dioptres (reciprocal)? We often choose non-
linear scales because they suit our purpose and for statistical analysis it
often suits us to make the distribution Normal, by finding a scale of
measurement where this is the case.
7.5 The Normal plot

Many statistical methods can only be used if the observations follow a
Normal distribution (see Chapters 10 and 11). There are several ways of
investigating whether observations follow a Normal distribution. With a
large sample we can inspect a histogram to see whether it looks like a
Normal distribution curve. This does not work well with a small sample,
and a more reliable method is the Normal plot. This is a graphical
method, which can be done using ordinary graph paper and a table of the
Normal distribution, with specially printed Normal probability paper, or,
much more easily, using a computer. Any good general statistical
package will give Normal plots; if it does not then it is not a good
package. The Normal plot method can be used to investigate the Normal
assumption in samples of any size, and is a very useful check when
using methods such as the t distribution methods described in Chapter
10.
The Normal plot is a plot of the cumulative frequency distribution for the
data against the cumulative frequency distribution for the Normal
distribution. First, we order the data from lowest to highest. For each
ordered observation we find the expected value of the observation if the
data followed a Standard Normal distribution. There are several
approximate formulae for this. I shall follow Armitage and Berry (1994)
and use for the ith observation z where Φ(z) = (i - 0.5)/n. Some books
and programs use Φ(z) = i/(n + 1) and there are other
more complex formulae. It does not make much difference which is used.
We find from a table of the Normal distribution the values of z which
correspond to Φ(z) = 0.5/n, 1.5/n, etc. (Table 7.1 lacks detail for practical
work, but will do for illustration.) For 5 points, for example, we have Φ(z)
= 0.1, 0.3, 0.5, 0.7, and 0.9. and z = -1.3, -0.5, 0, 0.5, and 1.3. These are
the points of the Standard Normal distribution which correspond to the
observed data. Now, if the observed data come from a Normal
distribution of mean µ and variance σ2, the observed point should equal
σz + µ, where z is the corresponding point of the Standard Normal
distribution. If we plot the Standard Normal points against the observed
values we should get something close to a straight line. We can write the
equation of this line as σz + µ = x, where x is the observed variable and z
the corresponding quantile of the Standard Normal distribution. We can
rewrite this as
which goes through the point defined by (µ, 0) and has slope 1/σ (see
§11.1). If the data are not from a Normal distribution we will not get a
straight line, but a curve of some sort. Because we plot the quantiles of
the observed frequency
distribution against the corresponding quantiles of the theoretical (here

Normal) distribution, this is also referred to as a quantile–quantile plot
or q–q plot.
Table 7.3. Vitamin D levels measured in the

blood of 26 healthy men, data of Hickish et al.
(1989)
14 25 30 42 54
17 26 31 43 54
20 26 31 46 63
21 26 32 48 67
22 27 35 52 83
24
Table 7.4. Calculation of the Normal plot for the

vitamin D data
Vit Vit
i Φ(z) z i Φ(z) z
D D
1 14 0.019 -2.07 14 31 0.519 0.05
2 17 0.058 -1.57 15 32 0.558 0.15
3 20 0.096 -1.30 16 35 0.596 0.24
4 21 0.135 -1.10 17 42 0.635 0.34
5 22 0.173 -0.94 18 43 0.673 0.45
6 24 0.212 -0.80 19 46 0.712 0.56
7 25 0.250 -0.67 20 48 0.750 0.67
8 26 0.288 -0.56 21 52 0.788 0.80

9 26 0.327 -0.45 22 54 0.827 0.94
10 26 0.365 -0.34 23 54 0.865 1.10
11 27 0.404 -0.24 24 63 0.904 1.30
12 30 0.442 -0.15 25 67 0.942 1.57
13 31 0.481 -0.05 26 83 0.981 2.07
Φ(z) = (i - 0.5)/26
Fig. 7.12. Blood vitamin D levels and log10 vitamin D for 26 normal
men, with Normal plots
Table 7.3 shows vitamin levels measured in the blood of 26 healthy men.
The calculation of the Normal plot is shown in Table 7.4. Note that the
Φ(z) = (i - 0.5)/26 and z are symmetrical, the second half being the first
half with opposite sign. The value of the Standard Normal deviate, z, can
be found by interpolation in Table 7.1, by using a fuller table, or by
computer. Figure 7.12 shows the histogram and the Normal plot for these
data. The distribution is skew and the Normal plot shows a pronounced
curve. Figure 7.12 also shows the vitamin D data after log transformation.
It is quite easy to produce the Normal plot, as the corresponding
Standard Normal deviate, z, is unchanged. We only need to log the
observations and plot again. The Normal plot for the transformed data
conforms very well to the theoretical line, suggesting that the distribution
of log vitamin D level is close to the Normal.
A single bend in the Normal plot indicates skewness. A double curve
indicates that both tails of the distribution are different from the Normal,
usually being too long, and many curves may indicate that the distribution
is bimodal (Figure 7.13). When the sample is small, of course, there will
be some random fluctuations.
There are several different ways to display the Normal plot. Some
programs plot the data distribution on the vertical axis and the theoretical
Normal distribution on the horizontal axis, which reverses the direction of
the curve. Some
plot the theoretical Normal distribution with mean [x with bar above], the
sample mean, and standard deviation s, the sample standard deviation.
This is done by calculating [x with bar above] + sz. Figure 7.14(a) shows
both these features, the Normal plot drawn by the program Stata's
‘qnorm’ command. The straight line is the line of equality. This plot is
identical to the second plot in Figure 7.12, except for the change of scale
and switching of the axes. A slight variation is the standardized Normal
probability plot or p-p plot, where we standardize the observations to
zero mean and standard deviation one, y = (x-[x with bar above])/s, and
plot the cumulative Normal
probabilities, Φ(y), against (i - 0.5)/n or ?/(n + 1) (Figure 7.14(b),

produced by the Stata command ‘pnorm’)- There is very little difference
between Figure 7.14(a) and (b) and the quantile and probability versions
of the Normal plot should be interpreted in the same way.
Fig. 7.13. Blood sodium and systolic blood pressure measured in

250 patients in the Intensive Therapy Unit at St. George's Hospital,
with Normal plots (data of Freidland et al. 1996)
Fig. 7.14. Variations on the Normal plot for the vitamin D data
Appendices
7A Appendix: Chi-squared, t, and F
Less mathematically inclined readers can skip this section, but those who
persevere should find that applications like chi-squared tests (Chapter
13) appear much more logical.
Many probability distributions can be derived for functions of Normal
variables which arise in statistical analysis. Three of these are particularly
important: the Chi-squared, t and F distributions. These have many
applications, some of which we shall discuss in later chapters.
The Chi-squared distribution is denned as follows. Suppose Z is a
Standard Normal variable, so having mean 0 and variance 1. Then the
variable formed by Z2 follows the Chi-squared distribution with 1 degree
of freedom. If we have n such independent Standard Normal variables,
Z1, Z2,…, Zn then the variable defined by
χ2 = Z21 + Z22 + … + Z2n
is defined to be the Chi-squared distribution with n degrees of
freedom. χ is the Greek letter ‘chi’, pronounced ‘ki’ as in ‘kite’. The
distribution curves for several different numbers of degrees of freedom
are shown in Figure 7.15. The mathematical description of this curve is
rather complicated, but we do not need to go into this.
Some properties of the Chi-squared distribution are easy to deduce. As
the distribution is the sum of n independent identically distributed random
variables it tends to the Normal as n increases, from the central limit
theorem (§7.2). The convergence is slow, however, (Figure 7.15) and the
square root of chi-squared converges much more quickly. The expected
value of Z2 is the variance of Z, the expected value of Z being 0, and so
E(Z2) = 1. The expected value of chi-squared with n degrees of freedom
is thus n:
The Chi-squared distribution has a very important property. Suppose we

restrict our attention to a subset of possible outcomes for the n random
variables Z1, Z2,…, Zn. The subset will be defined by those values of Z1,
Z2,…, Zn which satisfy the equation a1Z1 + a2Z2 + … + anZn = k, where
a1, a2 …, an, and k are constants. (This is called a linear constraint).
Then under this restriction, χ2 = Σ Z2i follows a Chi-squared distribution
with n - 1 degrees of freedom. If there are m such constraints such that
none of the equations can be calculated
from the others, then we have a Chi-squared distribution with n - m

degrees of freedom. This is the source of the name ‘degrees of freedom’.
Fig. 7.15. Some Chi-squared distributions
The proof of this is too complicated to give here, involving such

mathematical abstractions as n dimensional spheres, but its implications
are very important. First, consider the sum of squares about the
population mean µ of a sample of size n from a Normal distribution,
divided by σ2 · σ (xi - µ)2/σ2 will follow a Chi-squared distribution with n
degrees of freedom, as the (xi - µ)/σ have mean 0 and variance 1 and
they are independent. Now suppose we replace µ by an estimate
calculated from the data, [x with bar above]. The variables are no longer
independent, they must satisfy the relationship Σ(xi - [x with bar above]) =
0 and we now have n - 1 degrees of freedom. Hence Σ(xi - [x with bar
above])2/σ2 follows a Chi-squared distribution with n - 1 degrees of
freedom. The sum of squares about the mean of any Normal sample with
variance σ2 follows the distribution of a Chi-squared variable multiplied by
σ2. It therefore has expected value (n - 1)σ2 and we divide by n - 1 to give
the estimate of σ2.
Thus, provided the data are from a Normal distribution, not only does the
sample mean follow a Normal distribution, but the sample variance is
from a Chi-squared distribution times σ2/(n - 1). Because the square root
of the Chi-squared distribution converges quite rapidly to the Normal, the
distribution of the sample standard deviation is approximately Normal for
n > 20, provided the data themselves are from a Normal distribution.
Another important property of the variances of Normal samples is that, if
we take many random samples from the same population, the sample
variance and sample mean are independent if,
and only if, the data are from a Normal distribution.
The F distribution with m and n degrees of freedom is the distribution of

(χ2m)/(χ2n/n), the two ratio of two independent X2 variables each divided
by its degrees of freedom. This distribution is used for comparing
variances. If we have two independent estimates of the same variance
calculated from Normal data, the variance ratio will follow the F
distribution. We can use this for comparing two estimates of variance
(§10.8), but it main uses are in comparing groups of means (§10.9) and
in examining the effects of several factors together (§17.2).

32. The Normal distribution:
(a) is also called the Gaussian distribution;
(b) is followed by many variables;
(c) is a family of distributions with two parameters;
(d) is followed by all measurements made in healthy people;
(e) is the distribution towards which the Poisson distribution tends
as its mean in creases.
View Answer
33. The Standard Normal distribution:

(a) is skew to the left;
(b) has mean = 1.0;
(c) has standard deviation = 0.0;
(d) has variance = 1.0;
(e) has the median equal to the mean.
View Answer
34. The PEFRs of a group of 11-year-old girls follow a Normal

distribution with mean 300 litre/min and a standard deviation 20
litre/min:
(a) about 95% of the girls have PEFR between 260 and 340
litre/min;
(b) 50% of the girls have PEFR above 300 litre/min;
(c) the girls have healthy lungs;
(d) about 5% of girls have PEFR below 260 litre/min;
(e) all the PEFRs must be less than 340 litre/min.
View Answer
35. The mean of a large sample:

(a) is always greater than the median;
(b) is calculated from the formula Σxn/n
(c) is from an approximately Normal distribution;
(d) increases as the sample size increases;
(e) is always greater than the standard deviation.
View Answer
36. If X and Y are independent variables which follow Standard

Normal distributions, a Normal distribution is also followed by:
(a) 5X;
(b) X2;
(c) X + 5;
(d) X - Y;
(e) X/Y.
View Answer
37. When a Normal plot is drawn with the Standard Normal deviate
on the y axis:
(a) a straight line indicates that observations are from a Normal
Distribution;
(b) a curve with decreasing slope indicates positive skewness;
(c) an ‘S’ shaped curve (or ogive) indicates long tails;
(d) a vertical line will occur if all observations are equal;
(e) if there is a straight line its slope depends on the standard
deviation.
View Answer
7E Exercise: A Normal plot

In this exercise we shall return to the blood glucose data of §4E and try to
decide how well they conform to a Normal distribution.
1. From the box and whisker plot and the histogram found in
exercise §4E (if you have not tried exercise §4E see the solution in
Chapter 19), do the blood glucose levels look like a Normal
distribution?
View Answer
2. Construct a Normal plot for the data. This is quite easy as they
are ordered already. Find (i - 0.5)/n for i = 1 to 40 and obtain the
corresponding cumulative Normal probabilities from Table 7.1. Now
plot these probabilities against the corresponding blood glucose.
View Answer
3. Does the plot appear to give a straight line? Do the data follow a
Normal distribution?
View Answer
> Table of Contents > 8 - Estimation
8
Estimation
8.1 Sampling distributions

We have seen in Chapter 3 how samples are drawn from much larger
populations. Data are collected about the sample so that we can find out
something about the population. We use samples to estimate quantities
such as disease prevalence, mean blood pressure, mean exposure to a
carcinogen, etc. We also want to know by how much these estimates
might vary from sample to sample.
In Chapters 6 and 7 we saw how the theory of probability enables us to
link random samples with the populations from which they are drawn. In
this chapter we shall see how probability theory enables us to use
samples to estimate quantities in populations, and to determine the
precision of these estimates. First we shall consider what happens when
we draw repeat samples from the same population. Table 8.1 shows a
set of 100 random digits which we can use as the population for a
sampling experiment. The distribution of the numbers in this population is
shown in Figure 8.1. The population mean is 4.7 and the standard
deviation is 2.9.
The sampling experiment is done by using a suitable random sampling
method to draw repeated samples from the population. In this case
decimal dice were a convenient method. A sample of size four was
chosen: 6, 4, 6 and 1. The mean was calculated: 17/4 = 4.25. This was
repeated to draw a second sample of 4 numbers: 7, 8, 1, 8. Their mean is
6,00. This sampling procedure was done 20 times altogether, to give the
samples and their means shown in Table 8.2.
These sample means are not all the same. They show random variation.
If we were able to draw all of the 3 921 225 possible samples of size 4
and calculate their means, these means themselves would form a
distribution. Our 20 sample means are themselves a sample from this
distribution. The distribution of all possible sample means is called the
sampling distribution of the mean. In general, the sampling distribution
of any statistic is the distribution of the values of the
statistic which would arise from all possible samples.
Table 8.1. Population of 100 random digits for a sam
9 1 0 7 5 6 9 5 8 8 1 0 5
1 8 8 8 5 2 4 8 3 1 6 5 5
2 8 1 8 5 8 4 0 1 9 2 1 6
1 9 7 9 7 2 7 7 0 8 1 6 3
7 0 2 8 8 7 2 5 4 1 8 6 8
Fig. 8.1. Distribution of the population of Table 8.1
8.2 Standard error of a sample mean

For the moment we shall consider the sampling distribution of the mean
only. As our sample of 20 means is a random sample from it, we can use
this to estimate some of the parameters of the distribution. The twenty
means have their own mean and standard deviation. The mean is 5.1
and the standard deviation is 1.1. Now the mean of the whole population
is 4.7, which is close to the mean of the samples. But the standard
deviation of the population is 2.9, which is considerably greater than that
of the sample means. If we plot a histogram for the sample of means
(Figure 8.2) we see that the centre of the sampling distribution and the
parent population distribution are the same, but the scatter of the
sampling distribution is much less.
Table 8.2. Random samples drawn in a sampling
Sample 6 7 7 1 5 5
4 8 9 8 2 5
6 1 2 8 9 7
1 8 7 4 5 8
Mean 4.25 6.00 6.25 5.25 5.25 6.25
Sample 7 7 2 8 3 4
8 3 5 0 7 8
7 8 0 7 4 7
2 7 8 7 8 7
Mean 6.00 6.25 3.75 5.50 5.50 6.50
Fig. 8.2. Distribution of the population of Table 8.1 and of the sample
of the means of Table 8.2
The sample mean is an estimate of the population mean. The standard

deviation of its sampling distribution is called the standard error of the
estimate. It provides a measure of how far from the true value the
estimate is likely to be. In most estimation, the estimate is likely to be
within one standard error of the true mean and unlikely to be more than
two standard errors from it. We shall look at this more precisely in §8.3.
In almost all practical situations we do not know the true value of the
population variance σ2 but only its estimate s2 (§4.7). We can use this to
estimate the standard error by s/√n. This estimate is also referred to as
the standard error of the mean. It is usually clear from the context
whether the standard error is the true value or that estimated from the
data.
When the sample size n is large, the sampling distribution of [x with bar
above] tends to a Normal distribution. Also, we can assume that s2 is a
good estimate of σ2. So for large n [x with bar above], is, in effect, an
observation from a Normal distribution with mean µ and standard
deviation estimated by s/√n. So with probability 0.95, x is within two, or
more precisely is within 1.96 standard errors of µ. With small samples we
cannot assume either a Normal distribution or, more importantly,
that s2 is a good estimate of σ2. We shall discuss this in Chapter 10.
Fig. 8.3. Samples of means from a Standard Normal variable

The mean and standard error are often written as 4.062 ± 0.089. This is
rather misleading, as the true value may be up to two standard errors
from the mean with a reasonable probability. This practice is not
recommended.
There is often confusion between the terms ‘standard error’ and ‘standard
deviation’. This is understandable, as the standard error is a standard
deviation (of the sampling distribution) and the terms are often
interchanged in this context. The convention is this: we use the term
‘standard error’ when we measure the precision of estimates, and the
term ‘standard deviation’ when we are concerned with the variability of
samples, populations or distributions. If we want to say how good our
estimate of the mean FEV1 measurement is, we quote the standard error
of the mean. If we want to say how widely scattered the FEV1
measurements are, we quote the standard deviation, s.
8.3 Confidence intervals

The estimate of mean FEV1 is a single value and so is called a point
estimate. There is no reason to suppose that the population mean will be
exactly equal to the point estimate, the sample mean. It is likely to be
close to it, however, and the amount by which it is likely to differ from the
estimate can be found
from the standard error. What we do is find limits which are likely to
include the population mean, and say that we estimate the population
mean to lie somewhere in the interval (the set of all possible values)
between these limits. This is called an interval estimate.
Fig. 8.4. Sampling distribution of the mean of 4 observations from a
Standard Normal distribution
For instance, if we regard the 57 FEV measurements as being a large

sample we can assume that the sampling distribution of the mean is
Normal, and that the standard error is a good estimate of its standard
deviation (see §10.6 for a discussion of how large is large). We therefore
expect about 95% of such means to be within 1.96 standard errors of the
population mean, µ. Hence, for about 95% of all possible samples, the
population mean must be greater than the sample mean minus 1.96
standard errors and less than the sample mean plus 1.96 standard
errors. If we calculated x - 1.96se and x + 1.96se for all possible
samples, 95% of such intervals would contain the population mean. In
this case these limits are 4.062 - 1.96 × 0.089 to 4.062 + 1.96 × 0.089
which gives 3.89 to 4.24, or 3.9 to 4.2 litres, rounding to two significant
figures; 3.9 and 4.2 are called the 95% confidence limits for the
estimate, and the set of values between 3.9 and 4.2 is called the 95%
confidence interval. The confidence limits are the values at the ends of
the confidence interval.
Strictly speaking, it is incorrect to say that there is a probability of 0.95
that the population mean lies between 3.9 and 4.2, though it is often put
that way (even by me). The population mean is a number, not a random
variable, and has no probability. It is the probability that limits calculated
from a random sample will include the population value which is 95%.
Figure 8.5 shows confidence intervals for the mean for 20 random
samples of 100 observations from the Standard Normal distribution. The
population mean is, of course, 0.0, shown by the horizontal line. Some
sample means are close to 0.0, some further away, some above and
some below. The population mean is contained by 19 of the 20
confidence intervals. In general, for 95% of confidence intervals it will be
true to
say that the population value lies within the interval. We just don't know
which 95%. We express this by saying that we are 95% confident that the
mean lies between these limits.
Fig. 8.5. Mean and 95% confidence interval for 20 random samples
of 100 observations from the Standard Normal distribution
In the FEV1 example, the sampling distribution of the mean is Normal
and its standard deviation is well estimated because the sample is large.
This is not always true and although it is usually possible to calculate
confidence intervals for an estimate they are not all quite as simple as
that for the mean estimated from a large sample. We shall look at the
mean estimated from a small sample in §10.2.
There is no necessity for the confidence interval to have a probability of
95%. For example, we can also calculate 99% confidence limits. The
upper 0.5% point of the Standard Normal distribution is 2.58 (Table 7.2),
so the probability of a Standard Normal deviate being above 2.58 or
below -2.58 is 1% and the probability of being within these limits is 99%.
The 99% confidence limits for the mean FEV1 are therefore, 4.062 - 2.58
× 0.089 and 4.062 + 2.58 × 0.089, i.e. 3.8 and 4.3 litres. These give a
wider interval than the 95% limits, as we would expect since we are more
confident that the mean will be included. The probability we choose for a
confidence interval is thus a compromise between the desire to include
the estimated population value and the desire to avoid parts of scale
where there is a low probability that the mean will be found. For most
purposes, 95% confidence intervals have been found to be satisfactory.
Standard error is not the only way in which we can calculate confidence
intervals, although at present it is the one used for most problems. In
§8.8 I describe a different approach based on the exact probabilities of a
distribution, which requires no large sample assumption. In §8.9 I
describe a large sample method which uses the Binomial distribution
directly. There are others, which I shall omit because they are rarely
used.
8.4 Standard error and confidence interval for a

proportion
The standard error of a proportion estimate can be calculated in the
same way. Suppose the proportion of individuals who have a particular
condition in a given population is p, and we take a random sample of size
n, the number observed with the condition being r. Then the estimated
proportion is r/n. We have seen (§6.4) that r comes from a Binomial
distribution with mean np and variance np(1-p). Provided n is large, this
distribution is approximately Normal. So r/n, the estimated proportion, is
Normally distributed with mean given by np/n = p, and variance given by
since n is constant, and the standard error is
We can estimate this by replacing p by r/n.
The standard error of the proportion is only of use if the sample is large
enough for the Normal approximation to apply. A rough guide to this is
that np and n(1 - p) should both exceed 5. This is usually the case when
we are concerned with straightforward estimation. If we try to use the
method for smaller samples, we may get absurd results. For example, in
a study of the prevalence of HIV in ex-prisoners (Turnbull et al. 1992), of
29 women who did not inject drugs one was HIV positive. The authors
reported this to be 3.4%, with a 95% confidence interval -3.1% to 9.9%.
The lower limit of -3.1%, obtained from the observed proportion minus
1.96 standard errors, is impossible. As Newcombe (1992) pointed out,
the correct 95% confidence interval can be obtained from the exact
probabilities of the Binomial distribution and is 0.1% to 17.8% (§8.8).
8.5 The difference between two means

In many studies we are more interested in the difference between two
parameters than in their absolute value. These could be means,
proportions, the slopes of lines, and many other statistics. When samples
are large we can assume that sample means and proportions are
observations from a Normal distribution, and that the calculated standard
errors are good estimates of the standard deviations
of these Normal distributions. We can use this to find confidence

intervals.
For example, suppose we wish to compare the means, [x with bar
above]1 and [x with bar above]2, of two large samples, sizes n1 and n2.
The expected difference between the sample means is equal to the
difference between the population means, i.e. E([x with bar above]1 - [x
with bar above]2) = µ1 - µ2. What is the standard error of the difference?
The variance of the difference between two independent random
variables is the sum of their variances (§6.6). Hence, the standard error
of the difference between two independent estimates is the square root of
the sum of the squares of their standard errors. The standard error of a
mean is √s2/n, so the standard error of the difference between two
independent means is
For an example, in a study of respiratory symptoms in schoolchildren

(Bland et al. 1974), we wanted to know whether children reported by their
parents to have respiratory symptoms had worse lung function than
children who were not reported to have symptoms. Ninety-two children
were reported to have cough during the day or at night, and their mean
PEFR was 294.8 litre/min with standard deviation 57.1 litre/min, and 1643
children were not reported to have this symptom, their mean PEFR being
313.6 litre/min with standard deviation 55.2 litre/min. We thus have two
large samples, and can apply the Normal distribution. We have
n1 = 92, [x with bar above]1 = 294.8, s1 = 57.1, n2 = 1643, [x with bar
above]2 = 313.6, s2 = 55.2
The difference between the two groups is [x with bar above]1 - [x with bar
above]2 = 294.8 - 313.6 = -18.8. The standard error of the difference is
We shall treat the sample as being large, so the difference between the
means can be assumed to come from a Normal distribution and the
estimated standard error to be a good estimate of the standard deviation
of this distribution. (For small samples see §10.3 and §10.6.) The 95%
confidence limits for the difference are thus -18.8-1.96 × 6.11 and
-18.8+1.96 × 6.11, i.e. -6.8 and -30.8 litre/min. The confidence interval
does not include zero, so we have good evidence that, in this population,
children reported to have day or night cough have lower mean PEFR
than others. The difference is estimated to be between 7 and 31 litre/min
lower in children with the symptom, so it may be quite small.
When we have paired data, such as a cross-over trial (§2.6) or a
matched case-control study (§3.8), the two-sample method does not
work. Instead, we calculate the differences between the paired
observations for each subject, then find the mean difference, its standard
error and confidence interval as in §8.3.
Table 8.3. Cough during the day or at night at

age 14 and bronchitis before age 5 (Holland et
al. 1978)
Bronchitis at 5
Cough at 14 Total
Yes No
Yes 26 44 70
No 247 1002 1249

Total 273 1046 1319
8.6 Comparison of two proportions
Provided the conditions of Normal approximation are met (see §8.4) we

can find a confidence interval for the difference in the usual way.
For example, consider Table 8.3. The researchers wanted to know to
what extent children with bronchitis in infancy get more respiratory
symptoms in later life than others. We can estimate the difference
between the proportions reported to cough during the day or at night
among children with and children without a history of bronchitis before
age 5 years. We have estimates of two proportions, p1 = 26/273 =
0.09524 and p2 = 44/1046 = 0.04207. The difference between them is p1
- p2 = 0.09524 - 0.04207 = 0.05317. The standard error of the difference
is
The 95% confidence interval for the difference is 0.05317 - 1.96 × 0.0188
to 0.05317 + 1.96 × 0.0188 = 0.016 to 0.090. Although the difference is
not very precisely estimated, the confidence interval does not include
zero and gives us clear evidence that children with bronchitis reported in
infancy are more likely than others to be reported to have respiratory
symptoms in later life. The data on lung function in §8.5 gives us some
reason to suppose that this is not entirely due to response bias (§3.9). As
in §8.4, the confidence interval must be estimated
differently for small samples.

This difference in proportions may not be very easy to interpret. The ratio
of two proportions is often more useful. Another method, the odds ratio, is
described in §13.7. The ratio of the proportion with cough at age 14 for
bronchitis before 5 to the proportion with cough at age 14 for those
without bronchitis before 5 is p1/p2 = 0.09524/0.04207 = 2.26. Children
with bronchitis before 5 are more than twice as likely to cough during the
day or at night at age 14 than children with no such history.
The standard error for this ratio is complex, and as it is a ratio rather than
a difference it does not approximate well to a Normal distribution. If we
take the logarithm of the ratio, however, we get the difference between
two logarithms, because log(p1/p2) = log(p1) - log(p2) (§5A). We can find
the standard error for the log ratio quite easily. We use the result that, for
any random variable X with mean µ and variance σ2, the approximate
variance of log(X) is given by VAR(loge(X)) = σ2/µ2 (see Kendall and
Stuart 1969). Hence, the variance of log(p) is
For the difference between the two logarithms we get
The standard error is the square root of this. (This formula is often written
in terms of frequencies, but I think this version is clearer.) For the
example the log ratio is loge(2.26385) = 0.81707 and the standard error is
The 95% confidence interval for the log ratio is therefore 0.81707-1.96 ×
0.23784 to 0.81707+1.96 × 0.23784 = 0.35089 to 1.28324. The 95%
confidence interval for the ratio of proportions itself is the antilog of this:
e0.35089 to e1.28324 = 1.42 to 3.61. Thus we estimate that the proportion of
children reported to cough during the day or at night among those with a
history of bronchitis is between 1.4 to 3.6 times the proportion among
those without a history of bronchitis.
The proportion of individuals in a population who develop a disease or
symptom is equal to the probability that any given individual will develop
the disease, called the risk of an individual developing a disease. Thus in
Table 8.3 the risk
that a child with bronchitis before age 5 will cough at age 14 is 26/273 =
0.095 24, and the risk for a child without bronchitis before age 5 is
44/1046 = 0.04207. To compare risks for people with and without a
particular risk factor, we look at the ratio of the risk with the factor to the
risk without the factor, the relative risk. The relative risk of cough at age
14 for bronchitis before 5 is thus 2.26. To estimate the relative risk
directly, we need a cohort study (§3.7) as in Table 8.3. We estimate
relative risk for a case-control study in a different way (§13.7).
In the unusual situtation when the samples are paired, either matched or
two observations on the same subject, we use a different method (§13.9).
8.7 * Standard error of a sample standard

deviation
8.8 * Confidence interval for a proportion when
numbers are small
In §8.4 I mentioned that the standard error method for a proportion does
not work when the sample is small. Instead, the confidence interval can
be found using the exact probabilities of the Binomial distribution. The
method works like this. Given n, we find the value PL for the parameter p
of the Binomial distribution which gives a probability 0.025 of getting an
observed number of successes, r, as big as or bigger than the value
observed. We do this by calculating the probabilities from the formula in
§6.4, iterating round different possible values of p until we get the right
one. We also find the value pU for the parameter p of the Binomial
distribution which gives a probability 0.025 of getting an observed
number of successes as small as or smaller than the value observed.
The exact 95% confidence interval is PL to pU. For example, suppose we
observe 3 successes out of 10 trials. The Binomial distribution with n =
10 which has the total probability for 3 or more successes equal to 0.025
has parameter p = 0.067. The distribution which has the total probability
for 3 or fewer successes equal to 0.025 has p = 0.652. Hence the 95%
confidence interval for the proportion in the population is 0.067 to 0.652.
Figure 8.6 shows the two distributions. No large sample approximation is
required and we can use this for any size of sample. Pearson and Hartley
(1970) give a table for calculating exact Binomial confidence intervals.
Even better, you can download a free program from my website (§1.3).
Fig. 8.6. Distributions showing the calculation of the exact
confidence interval for three successes out of ten trials.
Unless the observed proportion is zero or one, these values are never
included in the exact confidence interval. The population proportion of
successes cannot be zero if we have observed a success in the sample.
It cannot be one if we have observed a failure.
8.9 * Confidence interval for a median and other

quantiles
We round j and k up to the next integer. Then the 95% confidence

interval is between the jth and the kth observations in the ordered data.
For the 57 FEV measurements of Table 4.4, the median was 4.1 litres
(§4.5). For the 95% confidence interval for the median, n = 57 and q =
0.5, and
The 95% confidence interval is thus from the 22nd to the 36th
observation, 3.75 to 4.30 litres from Table 4.4. Compare this to the 95%
confidence interval for the mean, 3.9 to 4.2 litres, which is completely
included in the interval for the median. This method of estimating
percentiles is relatively imprecise. Another example is given §15.5.
8.10 What is the correct confidence interval?
A confidence interval only estimates errors due to sampling. They do not
allow for any bias in the sample and give us an estimate for the
population of which our data can be considered a random sample. As
discussed in §3.5, it is often not clear what this population is, and we rely
far more on the estimation of differences than absolute values. This is
particularly true in clinical trials. We start with patients in one locality,
exclude some, allow refusals, and the patients cannot be regarded as a
random sample of patients in general. However, we then randomize into
two groups which are then two samples from the same population, and
only the treatment differs between them. Thus the difference is the thing
we want the confidence interval for, not for either group separately. Yet
researchers often ignore the direct comparison in favour of estimation
using each group separately.
For example, Salvesen et al. (1992) reported follow-up of two
randomized controlled trials of routine ultrasonography screening during
pregnancy. At ages 8 to 9 years, children of women who had taken part
in these trials were followed up. A subgroup of children underwent
specific tests for dyslexia. The test results classified 21 of the 309
screened children (7%, 95% confidence interval 3-10%) and 26 of the
294 controls (9%, 95% confidence interval 4–12%) as dyslexic. Much
more useful would be a confidence interval for the difference between
prevalences (-6.3 to 2.2 percentage points) or their ratio (0.44 to 1.34),
because we could then compare the groups directly.

38. The standard error of the mean of a sample:
(a) measures the variability of the observations;
(b) is the accuracy with which each observation is measured;
(c) is a measure of how far the sample mean is likely to be from
the population mean;
(d) is proportional to the number of observations;
(e) is greater than the estimated standard deviation of the
population.
View Answer
39. The 95% confidence limits for the mean estimated from a set of
observations
(a) are limits between which, in the long run, 95% of observations
fall;
(b) are a way of measuring the precision of the estimate of the
mean;
(c) are limits within which the sample mean falls with probability
0.95;
(d) are limits which would include the population mean for 95% of
possible samples;
(e) are a way of measuring the variability of a set of observations.
View Answer
40. If the size of a random sample were increased, we would expect:

(a) the mean to decrease;
(b) the standard error of the mean to decrease;
(c) the standard deviation to decrease;
(d) the sample variance to increase;
(e) the degrees of freedom for the estimated variance to increase.
View Answer
41. The prevalence of a condition in a population is 0.1. If the

prevalence is estimated repeatedly from samples of size 100, these
estimates will form a distribution which:
(a) is a sampling distribution;
(b) is approximately Normal;
(c) has mean = 0.1;
(d) have variance = 9;
(e) is Binomial.
View Answer
42. It is necessary to estimate the mean FEV1 by drawing a sample

from a large population. The accuracy of the estimate will depend
on:
(a) the mean FEV1 in the population;
(b) the number in the population;
(c) the number in the sample;
(d) the way the sample is selected;
(e) the variance of FEV1 in the population.
View Answer
43. In a study of 88 births to women with a history of

thrombocytopenia (Samuels et al. 1990), the same condition was
recorded in 20% of babies (95% confidence interval 13% to 30%,
exact method):
(a) Another sample of the same size will show a rate of
thrombocytopenia between 13% and 30%;
(b) 95% of such women have a probability of between 13% and
30% of having a baby with thrombocytopenia;
(c) It is likely that between 13% and 30% of births to such women
in the area would show thrombocytopenia;
(d) If the sample were increased to 880 births, the 95% confidence
interval would be narrower;
(e) It would be impossible to get these data if the rate for all
women was 10%.
View Answer
8E Exercise: Means of large samples

Table 8.4 summarizes data collected in a study of plasma magnesium in
diabetics. The diabetic subjects were all insulin-dependent subjects
attending a diabetic clinic over a 5 month period. The non-diabetic
controls were a mixture of blood donors and people attending day
centres for the elderly, to give a wide age
distribution. Plasma magnesium follows a Normal distribution very

closely.
Table 8.4. Plasma magnesium in insulin-

dependent diabetics and healthy controls
Standard
Number Mean
deviation
Insulin- 227 0.719 0.068

dependent
diabetics
Non-diabetic 140 0.810 0.057

controls
Fig. 8.7. Distribution of magnesium in diabetics and controls,
showing the proportion of diabetics above the lower limit of reference
interval
1. Calculate an interval which would include 95% of plasma

magnesium measurements from the control population. This is what
we call the 95% reference interval, described in detail in §15.5. It
tells us something about the distribution of plasma magnesium in
the population.
View Answer
2. What proportion of insulin-dependent diabetics would lie within

this 95% reference interval? (Hint: find how many standard
deviations from the diabetic mean the lower limit is, then use the
table of the Normal distribution, Table 7.1, to find the probability of
exceeding this. See Figure 8.7.)
View Answer
3. Find the standard error of the mean plasma magnesium for each
group.
View Answer
4. Find a 95% confidence interval for the mean plasma magnesium

in the healthy population. How does the confidence interval differ
from the 95% reference interval? Why are they different?
View Answer
5. Find the standard error of the difference in mean plasma

magnesium between insulin-dependent diabetics and healthy
people.
View Answer
6. Find a 95% confidence interval for the difference in mean plasma

magnesium between insulin-dependent diabetics and healthy
people. Is there any evidence that diabetics have lower plasma
magnesium than non-diabetics in the population from which these
data come?
View Answer
7. Would plasma magnesium be a good diagnostic test for

diabetes?
View Answer
> Table of Contents > 9 - Significance tests
9
Significance tests
9.1 Testing a hypothesis

In Chapter 8 I dealt with estimation and the precision of estimates. This is
one form of statistical inference, the process by which we use samples to
draw conclusions about the populations from which they are taken. In this
chapter I shall introduce a different form of inference, the significance test
or hypothesis test.
A significance test enables us to measure the strength of the evidence
which the data supply concerning some proposition of interest. For
example, consider the cross-over trial of pronethalol for the treatment of
angina (§2.6). Table 9.1 shows the number of attacks over four weeks on
each treatment. These 12 patients are a sample from the population of all
patients. Would the other members of this population experience fewer
attacks while using pronethalol? We can see that the number of attacks
is highly variable from one patient to another, and it is quite possible that
this is true from one period of time to another as well. So it could be that
some patients would have fewer attacks while on pronethalol than while
on placebo quite by chance. In a significance test, we ask whether the
difference observed was small enough to have occurred by chance if
there were really no difference in the population. If it were so, then the
evidence in favour of there being a difference between the treatment
periods would be weak or absent. On the other hand, if the difference
were much larger than we would expect due to chance if there were no
real population difference, then the evidence in favour of a real difference
would be strong.
To carry out the test of significance we suppose that, in the population,
there is no difference between the two treatments. The hypothesis of ‘no
difference’ or ‘no effect’ in the population is called the null hypothesis. If
this is not true, then the alternative hypothesis must be true, that there is
a difference between the treatments in one direction or the other. We
then find the probability of getting data as different from what would be
expected, if the null hypothesis were true, as are those data actually
observed. If this probability is large the data are consistent with the null
hypothesis; if it is small the data are unlikely to have arisen if the null
hypothesis were true and the evidence is in favour of the alternative
hypothesis.
Table 9.1. Trial of pronethalol for the prevention

of angina pectoris
Number of attacksDifference
while on Sign of
placebo —
difference
Placebo Pronethalol pronethalol
71 29 42 +
323 348 -25 -
8 1 7 +
14 7 7 +
23 16 7 +
34 25 9 +
79 65 14 +
60 41 19 +
2 0 2 +
3 0 3 +
17 15 2 +
7 2 5 +
9.2 An example: The sign test

I shall now describe a particular test of significance, the sign test, to test
the null hypothesis that placebo and pronethalol have the same effect on
angina. Consider the differences between the number of attacks on the
two treatments for each patient, as in Table 9.1. If the null hypothesis
were true, then differences in number of attacks would be just as likely to
be positive as negative, they would be random. The probability of a
change being negative would be equal to the probability of it being
positive, so both probabilities would be 0.5. Then the number of
negatives would be an observation from a Binomial distribution (§6.4)
with n = 12 and p = 0.5. (If there were any subjects who had the same
number of attacks on both regimes we would omit them, as they provide
no information about the direction of any difference between the
treatments. In this test, n is the number of subjects for whom there is a
difference, one way or the other.)
If the null hypothesis were true, what would be the probability of getting
an observation from this distribution as extreme as the value we have
actually observed? The expected number of negatives would be np = 6.
What is the probability of getting a value as far from expectation as is that
observed? The number of negative differences is 1. The probability of
getting one negative change is
This is not a likely event in itself. However, we are interested in the

probability of getting a value as far or further from the expected value, 6,
as is 1, and clearly 0 is further and must be included. The probability of
no negative changes is
So the probability of one or fewer negative changes is 0.00293 + 0.00024

= 0.00317. The null hypothesis is that there is no difference, so the
alternative hypothesis is that there is a difference in one direction or the
other. We must, therefore, consider the probability of getting a value as
extreme on the other side of the mean, that is 11 or 12 negatives (Figure
9.1). The probability of 11 or 12 negatives is also 0.00317, because the
distribution is symmetrical. Hence, the probability of getting as extreme a
value as that observed, in either direction, is 0.00317 + 0.00317 =
0.00634. This means that if the null hypothesis were true we would have
a sample which is so extreme that the probability of it arising by chance is
0.006, less than one in a hundred.
Fig. 9.1. Extremes of the Binomial distribution for the sign test
Thus, we would have observed a very unlikely event if the null hypothesis
were true. This means that the data are not consistent with null
hypothesis, and we can conclude that there is strong evidence in favour
of a difference between the treatments. (Since this was a double blind
randomized trial, it is reasonable to suppose that this was caused by the
activity of the drug.)
9.3 Principles of significance tests

The sign test is an example of a test of significance. The number of
negative changes is called the test statistic, something calculated from
the data which can be used to test the null hypothesis. The general
procedure for a significance test is as follows.
1. Set up the null hypothesis and its alternative.
2. Find the value of the test statistic.
3. Refer the test statistic to a known distribution which it would follow if
the null hypothesis were true.
4. Find the probability of a value of the test statistic arising which is as or
more extreme than that observed, if the null hypothesis were true.
5. Conclude that the data are consistent or inconsistent with the null
hypothesis.
We shall deal with several different significance tests in this and
subsequent chapters. We shall see that they all follow this pattern.
If the data are not consistent with the null hypothesis, the difference is
said to be statistically significant. If the data do not support the null
hypothesis, it is sometimes said that we reject the null hypothesis, and if
the data are consistent with the null hypothesis it is said that we accept it.
Such an ‘all or nothing’ decision making approach is seldom appropriate
in medical research. It is preferable to think of the significance test
probability as an index of the strength of evidence against the null
hypothesis. The term ‘accept the null hypothesis’ is also misleading
because it implies that we have concluded that the null hypothesis is true,
which we should not do. We cannot prove statistically that something,
such as a treatment effect, does not exist. It is better to say that we have
not rejected or have failed to reject the null hypothesis.
The probability of such an extreme value of the test statistic occurring if
the null hypothesis were true is often called the P value. It is not the
probability that the null hypothesis is true. This is a common
misconception. The null hypothesis is either true or it is not; it is not
random and has no probability. I suspect that many researchers have
managed to use significance tests quite effectively despite holding this
incorrect view.
9.4 Significance levels and types of error

We must still consider the question of how small is small. A probability of
0.006, as in the example above, is clearly small and we have a quite
unlikely event. But what about 0.06, or 0.1? Suppose we take a
probability of 0.01 or less as constituting reasonable evidence against the
null hypothesis. If the null hypothesis is true, we shall make a wrong
decision one in a hundred times. Deciding against a true null hypothesis
is called an error of the first kind, type I error, or α error. We get an
error of the second kind, type II error, or β error if we do not reject a
null hypothesis which is in fact false. (α and β are the Greek letters
‘alpha’ and ‘beta’.) Now the smaller we demand the probability be before
we decide against the null hypothesis, the larger the observed difference
must be, and so the more likely we are to miss real differences. By
reducing the risk of an error of the first kind we increase the risk of an
error of the second kind.
The conventional compromise is to say that differences are significant if
the probability is less than 0.05. This is a reasonable guide-line, but
should not be taken as some kind of absolute demarcation. There is not a
great difference between probabilities of 0.06 and 0.04, and they surely
indicate similar strength of evidence. It is better to regard probabilities
around 0.05 as providing some evidence against the null hypothesis,
which increases in strength as the probability falls. If we decide that the
difference is significant, the probability is sometimes referred to as the
significance level. We say that the significance level is high
if the P value is low.
Fig. 9.2. One- and two-sided tests
As a rough and ready guide, we can think of P values as indicating the

strength of evidence like this:
greater than 0.1: little or no evidence of a difference or relationship
between 0.05 and 0.1: weak evidence of a difference or relationship
between 0.01 and 0.05: evidence of a difference or relationship
less than 0.01: strong evidence of a difference or relationship
less than 0.001: very strong evidence of a difference or relationship
9.5 One- and two-sided tests of significance

In the above example, the alternative hypothesis was that there was a
difference in one direction or the other. This is called a two-sided or two-
tailed test, because we used the probabilities of extreme values in both
directions. It would have been possible to have the alternative hypothesis
that there was a decrease in the pronethalol direction, in which case the
null hypothesis would be that the number of attacks on the placebo was
less than or equal to the number on pronethalol. This would give P =
0.00317, and of course, a higher significance level than the two sided
test. This would be a one-sided or one-tailed test (Figure 9.2). The logic
of this is that we should ignore any signs that the active drug is harmful to
the patients. If what we were saying was ‘if this trial does not give a
significant reduction in angina using pronethalol we will not use it again’,
this might be reasonable, but the medical research process does not
work like that. This is one of several pieces of evidence and so we should
certainly use a method of inference which would enable us to detect
effects in either direction.
The question of whether one- or two-sided tests should be the norm has
been the subject of considerable debate among practitioners of statistical
methods. Perhaps the position taken depends on the field in which the
testing is usually done. In biological science, treatments seldom have
only one effect and relationships between variables are usually complex.
Two-sided tests are almost always preferable.
There are circumstances in which a one-sided test is appropriate. In a

study of the effects of an investigative procedure, laparoscopy and
hydrotubation, on the fertility of sub-fertile women (Luthra et al. 1982), we
studied women presenting at an infertility clinic. These women were
observed for several months, during which some conceived, before
laparoscopy was carried out on those still infertile. These were then
observed for several months afterwards and some of these women also
conceived. We compared the conception rate in the period before
laparoscopy with that afterwards. Of course, women who conceived
during the first period did not have a laparoscopy. We argued that the
less fertile a woman was the longer it was likely to take her to conceive.
Hence, the women who had the laparoscopy should have a lower
conception rate (by an unknown amount) than the larger group who
entered the study, because the more fertile women had conceived before
their turn for laparoscopy came. To see whether laparoscopy increased
fertility, we could test the null hypothesis that the conception rate after
laparoscopy was less than or equal to that before. The alternative
hypothesis was that the conception rate after laparoscopy was higher
than that before. A two-sided test was inappropriate because if the
laparoscopy had no effect on fertility the post laparoscopy rate was
expected to be lower; chance did not come into it. In fact the post
laparoscopy conception rate was very high and the difference clearly
significant.
9.6 Significant, real and important

If a difference is statistically significant, then it may well be real, but not
necessarily important. For example, we may look at the effect of a drug,
given for some other purpose, on blood pressure. Suppose we find that
the drug raises blood pressure by an average of 1 mmHg, and that this is
significant. A rise in blood pressure of 1 mmHg is not clinically important,
so, although it may be there, it does not matter. It is (statistically)
significant, and real, but not important.
On the other hand, if a difference is not statistically significant, it could
still be real. We may simply have too small a sample to show that a
difference exists. Furthermore, the difference may still be important. The
difference in mortality in the anticoagulant trial of Carleton et al. (1960),
described in Chapter 2, was not significant, the difference in percentage
survival being 5.5 in favour of the active treatment. However, the authors
also quote a confidence interval for the difference in percentage survival
of 24.2 percentage points in favour of heparin to 13.3 percentage points
in favour of the control treatment. A difference in survival of 24
percentage points in favour of the treatment would certainly be important
if it turned out to be the case. ‘Not significant’ does not imply that there is
no effect. It means that we have failed to demonstrate the existence of
one. Later studies showed that anticoagulation is indeed effective.
A particular case of misinterpretation of non-significant results occurs in
the interpretation of randomized clinical trials where there is a
measurement before treatment and another afterwards. Rather than
compare the after treatment
measure between the two groups, researchers can be tempted to test

separately the null hypotheses that the measure in the treatment group
has not changed from baseline and that the measure in the control group
has not changed from baseline. If one group shows a significant
difference and the other does not, the researchers then conclude that the
treatments are different.
For example, Kerrigan et al. (1993) assessed the effects of different
levels of information on anxiety in patients due to undergo surgery. They
randomized patients to receive either simple or detailed information about
the procedure and its risks. Anxiety was again measured after patients
had been given the information. Kerrigan et al. (1993) calculated
significance tests for the mean change in anxiety score for each group
separately. In the group given detailed information the mean change in
anxiety was not significant (P = 0.2), interpreted incorrectly as ‘no
change’. In the other group the reduction in anxiety was significant (P =
0.01). They concluded that there was a difference between the two
groups because the change was significant in one group but not in the
other. This is incorrect. There may, for example, be a difference in one
group which just fails to reach the (arbitrary) significance level and a
difference in the other which just exceeds it, the differences in the two
groups being similar. We should compare the two groups directly. It is
these which are comparable apart from the effects of treatment, being
randomized, not the before and after treatment means which could be
influenced by many other factors. An alternative analysis tested the null
hypothesis that after adjustment for initial anxiety score the mean anxiety
scores are the same in patients given simple and detailed information.
This showed a significantly higher mean score in the detailed information
group (Bland and Altman 1993). Testing within each group separately is
essentially the same error as calculating a confidence interval for each
group separately (§8.9).
9.7 Comparing the means of large samples
We can use this confidence interval to carry out a significance test of the
null hypothesis that the difference between the means is zero, i.e. the
alternative hypothesis is that µ1 and µ2 are not equal. If the confidence
interval includes zero, then the probability of getting such extreme data if
the null hypothesis were true is greater than 0.05 (i.e. 1 - 0.95). If the
confidence interval excludes zero, then the probability of such extreme
data under the null hypothesis is less
than 0.05 and the difference is significant. Another way of doing the same
thing is to note that
is from a Standard Normal distribution, i.e. mean 0 and variance 1. Under

the null hypothesis that µ1 - µ2 or µ1 = µ2 - 0, this is
This is the test statistic, and if it lies between -1.96 and +1.96 then the
probability of such an extreme value is greater than 0.05 and the
difference is not significant. If the test statistic is greater than 1.96 or less
than -1.96, there is a less than 0.05 probability of such data arising if the
null hypothesis were true, and the data are not consistent with null
hypothesis; the difference is significant at the 0.05 or 5% level. This is the
large sample Normal test or z test for two means.
For an example, in a study of respiratory symptoms in schoolchildren
(§8.5), we wanted to know whether children reported by their parents to
have respiratory symptoms had worse lung function than children who
were not reported to have symptoms. Ninety-two children were reported
to have cough during the day or at night, and their mean PEFR was
294.8 litre/min with standard deviation 57.1 litre/min; 1643 children were
reported not to have the symptom, and their mean PEFR was 313.6
litre/min with standard deviation 55.2 litre/min. We thus have two large
samples, and can apply the Normal test. We have
The difference between the two groups is [x with bar above]1 - [x with bar
above]2 = 294.8 - 313.6 = -18.8. The standard error of the difference is
The test statistic is
Under the null hypothesis this is an observation from a Standard Normal

distribution, and so P < 0.01 (Table 7.2). If the null hypothesis were true,
the data which we have observed would be unlikely. We can conclude
that there is good evidence that children reported to have cough during
the day or at night have lower PEFR than other children.
This has a probability of about 0.16, and so the data are consistent with
the null hypothesis. However, the 95% confidence interval for the
difference is -14.6-1.96 × 10.5 to -14.6+1.96 × 10.5 giving -35 to 6
litre/min. We see that the difference could be just as great as for cough.
Because the size of the smaller sample is not so great, the test is less
likely to detect a difference for the phlegm comparison than for the cough
comparison. The advantages of confidence intervals over tests of
significance are discussed by Gardner and Altman (1986). Confidence
intervals are usually more informative than P values, particularly non-
significant ones.
9.8 Comparison of two proportions

Suppose we wish to compare two proportions p1 and p2, estimated from
large independent samples size n1 and n2. The null hypothesis is that the
proportion in the populations from which the samples are drawn are the
same, p say. Since under the null hypothesis the proportions for the two
groups are the same, we can get one common estimate of the proportion
and use it to estimate the standard errors. We estimate the common
proportion from the data by
where p1 = r1/n2 - p2 = r2/n2. We want to make inferences from the
difference between sample proportions, p1 - p2, so we require the
standard error of this difference.
since the samples are independent. Hence
As p is based on more subjects than either p1 or p2, if the null hypothesis

were true then standard errors would be more reliable than those
estimated in §8.6 using p1 and p2 separately. We then find the test
statistic
In §8.6, we looked at the proportions of children with bronchitis in infancy

and with no such history who were reported to have respiratory
symptoms in later life. We had 273 children with a history of bronchitis
before age 5 years, 26 of whom were reported to have day or night
cough at age 14. We had 1046 children with no bronchitis before age 5
years, 44 of whom were reported to have day or night cough at age 14.
We shall test the null hypothesis that the prevalence of the symptom is
the same in both populations, against the alternative that it is not:
Referring this to Table 7.2 of the Normal distribution, we find the
probability of such an extreme value is less than 0.01, so we conclude
that the data are not consistent with the null hypothesis. There is good
evidence that children with a history of bronchitis are more likely to be
reported to have day or night cough at age 14.
Note that the standard error used here is not the same as that found in
§8.6. It is only correct if the null hypothesis is true. The formula of §8.6
should be used for finding the confidence interval. Thus the standard
error used for testing is not identical to that used for estimation, as was
the case for the comparison of two means. It is possible for the test to be
significant and the confidence interval include zero. This property is
possessed by several related tests and confidence intervals.
This is a large sample method, and is equivalent to the chi-squared test
for a 2 by 2 table (§13.1,2). How small the sample can be and methods
for small samples are discussed in §13.3-6.
Note that we do not need a different test for the ratio of two proportions,
as the null hypothesis that the ratio in the population is one is the same
as the null hypothesis that the difference in the population is zero.
9.9 * The power of a test

The test for comparing means in §9.7 is more likely to detect a large
difference between two populations than a small one. The probability that
a test will produce a significant difference at a given significance level is
called the power of the test. For a given test, this will depend on the true
difference between the populations compared, the sample size and the
significance level chosen. We have already noted in §9.4 that we are
more likely to obtain a significant difference with a significance level of
0.05 than with one of 0.01. We have greater power if the P value chosen
to be considered as significant is larger.
For the comparison of PEFR in children with and without phlegm (§9.7),
for
example, suppose that the population means were in fact µ1 = 310 and
µ2 = 295 litre/min, and each population had standard deviation 55
litre/min. The sample sizes were n1 = 1708 and n2 = 27, so the standard
error of the difference would be
The population difference we want to be able to detect is µ1 - µ2 = 310-

295 = 15, and so
From Table 7.1, Φ(0.55) is between 0.691 and 0.726, about 0.71. The
power of the test would be 1 - 0.71 = 0.29. If these were the population
means and standard deviation, our test would have had a poor chance of
detecting the difference in means, even though it existed. The test would
have low power. Figure 9.3 shows how the power of this test changes
with the difference between population means. As the difference gets
larger, the power increases, getting closer and closer to 1. The power is
not zero even when the population difference is zero, because there is
always the possibility of a significant difference, even when the null
hypothesis is true. 1- power = β, the probability of a Type II or beta error
(§9.4) if the population difference = 15 litres/min.
Fig. 9.3. Power curve for a comparison of two means from samples
of size 1708 and 27
9.10 * Multiple significance tests

If we test a null hypothesis which is in fact true, using 0.05 as the critical
significance level, we have a probability of 0.95 of coming to a ‘not
significant’ (i.e. correct) conclusion. If we test two independent true null
hypotheses, the probability that neither test will be significant is 0.95 ×
0.95 = 0.90 (§6.2). If we test twenty such hypotheses the probability that
none will be significant is
0.9520 = 0.36.0. This gives a probability of 1 - 0.36 = 0.64 of getting at

least one significant result; we are more likely to get one than not. The
expected number of spurious significant results is 20 × 0.05 = 1.
Many medical research studies are published with large numbers of
significance tests. These are not usually independent, being carried out
on the same set of subjects, so the above calculations do not apply
exactly. However, it is clear that if we go on testing long enough we will
find something which is ‘significant’. We must beware of attaching too
much importance to a lone significant result among a mass of non-
significant ones. It may be the one in twenty which we should get by
chance alone.
This is particularly important when we find that a clinical trial or
epidemiological study gives no significant difference overall, but does so
in a particular subset of subjects, such as women aged over 60. For
example, Lee et al. (1980) simulated a clinical trial of the treatment of
coronary artery disease by allocating 1 073 patient records from past
cases into two ‘treatment’ groups at random. They then analysed the
outcome as if it were a genuine trial of two treatments. The analysis was
quite detailed and thorough. As we would expect, it failed to show any
significant difference in survival between those patients allocated to the
two ‘treatments’. Patients were then subdivided by two variables which
affect prognosis, the number of diseased coronary vessels and whether
the left ventricular contraction pattern was normal or abnormal. A
significant difference in survival between the two ‘treatment’ groups was
found in those patients with three diseased vessels (the maximum) and
abnormal ventricular contraction. As this would be the subset of patients
with the worst prognosis, the finding would be easy to account for by
saying that the superior ‘treatment’ had its greatest advantage in the
most severely ill patients! The moral of this story is that if there is no
difference between the treatments overall, significant differences in
subsets are to be treated with the utmost suspicion. This method of
looking for a difference in treatment effect between subgroups of subjects
is incorrect. A correct approach would be to use a multifactorial analysis,
as described in Chapter 17, with treatment and group as two factors, and
test for an interaction between groups and treatments. The power for
detecting such interactions is quite low, and we need a larger sample
than would be needed simply to show a difference overall (Altman and
Matthews 1996, Matthews and Altman 1996a, b).
This spurious significant difference comes about because, when there is
no real difference, the probability of getting no significant differences in
six subgroups is 0.956 = 0.74, not 0.95. We can allow for this effect by
the Bonferroni method. In general, if we have k independent significant
tests, at the α level, of null hypotheses which are all true, the probability
that we will get no significant differences is (1 - α)k. If we make α small
enough, we can make the probability that none of the separate tests is
significant equal to 0.95. Then if any of the k tests has a P value less
than α, we will have a significant difference between the treatments at the
0.05 level. Since α will be very small, it can be shown that (1- α)k ≈ 1 - kα.
If we put kα = 0.05, so α = 0.05/k we will have probability
0.05 that one of the k tests will have a P value less than α if the null
hypotheses are true. Thus, if in a clinical trial we compare two treatments
within 5 subsets of patients, the treatments will be significantly different at
the 0.05 level if there is a P value less than 0.01 within any of the
subsets. This is the Bonferroni method. Note that they are not significant
at the 0.01 level, but at only the 0.05 level. The k tests together test the
composite null hypothesis that there is no treatment effect on any
variable.
We can do the same thing by multiplying the observed P value from the
significance tests by the number of tests, k, any kP which exceeds one
being ignored. Then if any kP is less than 0.05, the two treatments are
significant at the 0.05 level.
For example, Williams et al. (1992) randomly allocated elderly patients
discharged from hospital to two groups. The intervention group received
timetabled visits by health visitor assistants, the control patients group
were not visited unless there was perceived need. Soon after discharge
and after one year, patients were assessed for physical health, disability,
and mental state using questionnaire scales. There were no significant
differences overall between the intervention and control groups, but
among women aged 75–79 living alone the control group showed
significantly greater deterioration in physical score than did the
intervention group (P = 0.04), and among men over 80 years the control
group showed significantly greater deterioration in disability score than
did the intervention group (P = 0.03). The authors stated that ‘Two small
sub-groups of patients were possibly shown to have benefited from the
intervention…. These benefits, however, have to be treated with caution,
and may be due to chance factors.’ Subjects were cross-classified by age
groups, whether living alone, and sex, so there were at least eight
subgroups, if not more. Thus even if we consider the three scales
separately, only a P value less than 0.05/8 = 0.006 would provide
evidence of a treatment effect. Alternatively, the true P values are 8 ×
0.04 = 0.32 and 8 × 0.03 = 0.24.
A similar problem arises if we have multiple outcome measurements. For
example, Newnham et al. (1993) randomized pregnant women to receive
a series of Doppler ultrasound blood flow measurements or to control.
They found a significantly higher proportion of birthweights below the
10th and 3rd centiles (P = 0.006 and P = 0.02). These were only two of
many comparisons, however, and one would suspect that there may be
some spurious significant differences among so many. At least 35 were
reported in the paper, though only these two were reported in the
abstract. (Birthweight was not the intended outcome variable for the trial.)
These tests are not independent, because they are all on the same
subjects, using variables which may not be independent. The proportions
of birthweights below the 10th and 3rd centiles are clearly not
independent, for example. The probability that two correlated variables
both give non-significant differences when the null hypothesis is true is
greater than (1- α)2 because if the first test is not significant, the second
now has a probability greater than 1 - α of being not significant also.
(Similarly, the probability that both are significant exceeds α2, and the
probability that only one is significant is reduced.)
For k tests the probability of no significant differences is greater than (1 -

α)k and so greater than 1 - kα. Thus if we carry out each test at the α =
0.05/k level, we will have a probability of no significant differences which
is greater than 0.95. A P value less than α for any variable, or kP < 0.05,
would mean that the treatments were significantly different. For the
example, the P values could be adjusted by 35 × 0.006 = 0.21 and 35 ×
0.02 = 0.70.
Because the probability of obtaining no significant differences if the null
hypotheses are all true is greater than the 0.95 which we want it to be,
the overall P value is actually smaller than the nominal 0.05, by an
unknown amount which depends on the lack of independence between
the tests. The power of the test, its ability to detect true differences in the
population, is correspondingly diminished. In statistical terms, the test is
conservative.
Other multiple testing problems arise when we have more than two
groups of subjects and wish to compare each pair of groups (§10.9),
when we have a series of observations over time, such as blood pressure
every 15min after administration of a drug, where there may be a
temptation to test each time point separately (§10.7), and when we have
relationships between many variables to examine, as in a survey. For all
these problems, the multiple tests are highly correlated and the
Bonferroni method is inappropriate, as it will be highly conservative and
may miss real differences.
9.11 * Repeated significance tests and

sequential analysis
A special case of multiple testing arises in clinical trials, where patients
are admitted at different times. There can be a temptation to keep looking
at the data and carrying out significant tests. As described above (§9.10),
this is liable to produce spurious significant differences, detecting
treatment effects where none exist. I have heard of researchers testing
the difference each time a patient is added and stopping the trial as soon
as the difference is significant, then submitting the paper for publication
as if only one test had been carried out. I will be charitable and put this
down to ignorance.
It is quite legitimate to set up a trial where the treatment difference is
tested every time a patient is added, provided this repeated testing is
designed into the trial and the overall chance of a significant difference
when the null hypothesis is true remains 0.05. Such designs are called
sequential clinical trials. A comprehensive account is given by
Whitehead (1997).
An alternative approach which is quite often used is to take a small
number of looks at the data as the trial progresses, testing at a
predetermined P value. For example, we could test three times, rejecting
the null hypothesis of no treatment effect the first time only if P < 0.001,
the second time if P < 0.01, and the third time if P < 0.04. Then if the null
hypothesis is true, the probability that there will not be a significant
difference is approximately 0.999 × 0.99 × 0.96 = 0.949, so the overall
alpha level will be 1 - 0.949 = 0.051, i.e. approximately 0.05. (The
calculation is approximate because the tests are not independent.) If the
null hypothesis is rejected at any of these tests, the overall P value is
0.05, not the
nominal one. This approach can be used by data monitoring committees,

where if the trial shows a large difference early on the trial can be
stopped yet still allow a statistical conclusion to be drawn. This is called
the alpha spending or P-value spending approach.
Two particular methods which you might come across are the grouped
sequential design of Pocock (1977, 1982), where each test is done at the
same nominal alpha value, and the method of O'Brien and Fleming
(1979), widely used by the pharmaceutical industry, where the nominal
alpha values decrease sharply as the trial progresses.

44. In a case–control study, patients with a given disease drank
coffee more frequently than did controls, and the difference was
highly significant. We can conclude that:
(a) drinking coffee causes the disease;
(b) there is evidence of a real relationship between the disease
and coffee drinking in the sampled population;
(c) the disease is not related to coffee drinking;
(d) eliminating coffee would prevent the disease;
(e) coffee and the disease always go together.
View Answer
45. When comparing the means of two large samples using the
Normal test:
(a) the null hypothesis is that the sample means are equal;
(b) the null hypothesis is that the means are not significantly
different;
(c) standard error of the difference is the sum of the standard
errors of the means;
(d) the standard errors of the means must be equal;
(e) the test statistic is the ratio of the difference to its standard
error.
View Answer
46. In a comparison of two methods of measuring PEFR, 6 of 17

subjects had higher readings on the Wright peak flow meter, 10 had
higher readings on the mini peak flow meter and one had the same
on both. If the difference between the instruments is tested using a
sign test:
(a) the test statistic may be the number with the higher reading on
the Wright meter;
(b) the null hypothesis is that there is no tendency for one
instrument to read higher than the other;
(c) a one-tailed test of significance should be used;
(d) the test statistic should follow the Binomial distribution (n = 16
and p = 0.5) if the null hypothesis were true;
(e) the instruments should have been presented in random order.
View Answer
47. In a small randomized double blind trial of a new treatment in
acute myocardial infarction, the mortality in the treated group was
half that in the control group, but the difference was not significant.
We can conclude that:
(a) the treatment is useless;
(b) there is no point in continuing to develop the treatment;
(c) the reduction in mortality is so great that we should introduce
the treatment immediately;
(d) we should keep adding cases to the trial until the Normal test
for comparison of two proportions is significant;
(e) we should carry out a new trial of much greater size.
View Answer
48. In a large sample comparison between two groups, increasing

the sample size will:
(a) improve the approximation of the test statistic to the Normal
distribution;
(b) decrease the chance of an error of the first kind;
(c) decrease the chance of an error of the second kind;
(d) increase the power against a given alternative;
(e) make the null hypothesis less likely to be true.
View Answer
49. In a study of breast feeding and intelligence (Lucas et al. 1992),

300 children who were very small at birth were given their mother's
breast milk or infant formula, at the choice of the mother. At the age
of 8 years the IQ of these children was measured. The mean IQ in
the formula group was 92.8, compared to a mean of 103.0 in the
breast milk group. The difference was significant, P < 0.001:
(a) there is good evidence that formula feeding of very small
babies reduces IQ at age eight;
(b) there is good evidence that choosing to express breast milk is
related to higher IQ in the child at age eight;
(c) type of milk has no effect on subsequent IQ;
(d) the probability that type of milk affects subsequent IQ is less
than 0.1%;
(e) if type of milk were unrelated to subsequent IQ, the probability
of getting a difference in mean IQ as big as that observed is less
than 0.001.
View Answer
9E Exercise: Crohn's disease and cornflakes

The suggestion that cornflakes may cause Crohn's disease arose in the
study of James (1977). Crohn's disease is an inflammatory disease,
usually of the last part of the small intestine. It can cause a variety of
symptoms, including vague pain, diarrhoea, acute pain and obstruction.
Treatment may be by drugs or surgery, but many patients have had the
disease for many years. James' initial hypothesis was that foods taken at
breakfast may be associated with Crohn's disease. James studied 16
men and 18 women with Crohn's disease, aged 19–64 years, mean time
since diagnosis 4.2 years. These were compared to controls, drawn from
hospital
patients without major gastro-intestinal symptoms. Two controls were

chosen per patient, matched for age and sex. James interviewed all
cases and controls himself. Cases were asked whether they ate various
foods for breakfast before the onset of symptoms, and controls were
asked whether they ate various foods before a corresponding time (Table
9.2). There was a significant excess of eating of cornflakes, wheat and
bran among the Crohn's patients. The consumption of different cereals
was interrelated, people reporting one cereal being likely to report others.
In James' opinion the principal association of Crohn's disease was with
cornflakes, based on the apparent strength of the association. Only one
case had never eaten cornflakes.
Table 9.2. Numbers of Crohn's disease patients and
controls who ate various cereals regularly (at least onc
per week) (James 1977)
Significance
Patients Controls
Cornflakes Regularly 23 17 P < 0.000 1
Rarely or 11 51
never
Wheat Regularly 16 12 P < 0.01
Rarely or 18 56
never
Porridge Regularly 11 15 0.5 > P >

0.1
Rarely or 23 53
never
Rice Regularly 8 10 0.5 > P >

0.1
Rarely or 26 56
never
Bran Regularly 6 2 P = 0.02
Rarely or 28 66
never
Muesli Regularly 4 3 P = 0.17
Rarely or 30 65
never
Several papers soon appeared in which this study was repeated, with
variations. None was identical in design to James' study and none
appeared to support his findings. Mayberry et al. (1978) interviewed 100
patients with Crohn's disease, mean duration nine years. They obtained
100 controls, matched for age and sex, from patients and their relatives
attending a fracture clinic. Cases and controls were interviewed about
their current breakfast habits (Table 9.3). The only significant difference
was an excess of fruit juice drinking in controls. Cornflakes were eaten by
29 cases compared to 22 controls, which was not significant. In this study
there was no particular tendency for cases to report more foods than
controls. The authors also asked cases whether they knew of an
association between food (unspecified) and Crohn's disease. The
association with cornflakes was reported by 29, and 12 of these had
stopped eating them, having previously eaten them regularly. In their 29
matched controls, 3 were past cornflakes eaters. Of the 71 Crohn's
patients who were unaware of the association, 21 had discontinued
eating cornflakes compared to 10 of their 71 controls. The authors
remarked ‘seemingly patients with Crohn's disease had significantly
reduced their consumption of cornflakes compared with controls,
irrespective of whether they were aware of the possible association’.
1. Are the cases and controls comparable in either of these studies?
View Answer
2. What other sources of bias could there be in these designs?

View Answer
Table 9.3. Number of patients and controls

regularly consuming certain foods at least
twice weekly (Mayberry et al. 1978)
Crohn's
Foods at patients Controls Significance
breakfast (n = (n = 100) test
100)
Bread 91 86
Toast 59 64
Egg 31 37
Fruit or 14 30 P < 0.02

fruit juice
Porridge 20 18
Weetabix, 21 19
shreddies
or
shredded
wheat
Cornflakes 29 22
Special K 4 7
Rice 6 6
krispies
Sugar 3 1
puffs
Bran or all 13 12
bran
Muesli 3 10
Any 55 55
Cereal
3. What is the main point of difference in design between the study

of James and that of Mayberry et al.?
View Answer
4. In the study of Mayberry et al. how many Crohn's cases and how
many controls had ever been regular eaters of cornflakes? How
does this compare with James' findings?
View Answer
5. Why did James think that eating cornflakes was particularly

important?
View Answer
6. For the data of Table 9.2, calculate the percentage of cases and
controls who said that they ate the various cereals. Now divide the
proportion of cases who said that they had eaten the cereal by the
proportion of controls who reported eating it. This tells us, roughly,
how much more likely cases were to report the cereal than were
controls. Do you think eating cornflakes is particularly important?
View Answer
7. If we have an excess of all cereals when we ask what was ever

eaten, and none when we ask what is eaten now, what possible
factors could account for this?
View Answer
> Table of Contents > 10 - Comparing the means of small samples
10
Comparing the means of small samples
10.1 The t distribution

We have seen in Chapters 8 and 9 how the Normal distribution can be
used to calculate confidence intervals and to carry out tests of
significance for the means of large samples. In this chapter we shall see
how similar methods may be used when we have small samples, using
the t distribution, and go on to compare several means.
So far, the probability distributions we have used have arisen because of
the way data were collected, either from the way samples were drawn
(Binomial distribution), or from the mathematical properties of large
samples (Normal distribution). The distribution did not depend on any
property of the data themselves. To use the t distribution we must make
an assumption about the distribution from which the observations
themselves are taken, the distribution of the variable in the population.
We must assume this to be a Normal distribution. As we saw in Chapter
7, many naturally occurring variables have been found to follow a Normal
distribution closely. I shall discuss the effects of any deviations from the
Normal later.
Fig. 10.1. Student's t distribution with 1, 4 and 20 degrees of
freedom, showing convergence to the Standard Normal distribution
Table 10.1. Two-tailed probability points of the t dis
D.f. Probability D.f.

0.10 0.05 0.01 0.001 0.10
10% 5% 1% 0.1% 10%
1 6.31 12.70 63.66 636.62 16 1.75
2 2.92 4.30 9.93 31.60 17 1.74
3 2.35 3.18 5.84 12.92 18 1.73
4 2.13 2.78 4.60 8.61 19 1.73
5 2.02 2.57 4.03 6.87 20 1.72
6 1.94 2.45 3.71 5.96 21 1.72
7 1.89 2.36 3.50 5.41 22 1.72
8 1.86 2.31 3.36 5.04 23 1.71
9 1.83 2.26 3.25 4.78 24 1.71
10 1.81 2.23 3.17 4.59 25 1.71

11 1.80 2.20 3.11 4.44 30 1.70
12 1.78 2.18 3.05 4.32 40 1.68
13 1.77 2.16 3.01 4.22 60 1.67
14 1.76 2.14 2.98 4.14 120 1.66
15 1.75 2.13 2.95 4.07 ∞ 1.64
D.f. = Degrees of freedom.
∞ = infinity, same as the Standard Normal distribution.
Like the Normal distribution, the t distribution function cannot be

integrated algebraically and its numerical values have been tabulated.
Because the t distribution depends on the degrees of freedom, it is not
usually tabulated in full like the Normal distribution in Table 7.1. Instead,
probability points are given for different degrees of freedom. Table 10.1
shows two sided probability points for selected degrees of freedom.
Thus, with 4 degrees of freedom, we can see that, with probability 0.05, t
will be 2.78 or more from its mean, zero.
Because only certain probabilities are quoted, we cannot usually find the
exact probability associated with a particular value of t. For example,
suppose we want to know the probability of t on 9 degrees of freedom
being further from zero than 3.7. From Table 10.1 we see that the 0.01
point is 3.25 and the 0.001 point is 4.78. We therefore know that the
required probability lies between 0.01 and 0.001. We could write this as
0.001 < P < 0.01. Often the lower bound, 0.001, is omitted and we write P
< 0.01. With a computer it is possible to calculate the exact probability
every time, so this common practice is due to disappear.
Fig. 10.2. Sample t ratios derived from 750 samples of 4 human

heights and the t distribution, after Student (1908)
10.2 The one-sample t method

We can use the t distribution to find confidence intervals for means
estimated from a small sample from a Normal distribution. We do not
usually have small samples in sample surveys, but we often find them in
clinical studies. For example, we can use the t distribution to find
confidence intervals for the size of difference between two treatment
groups, or between measurements obtained from subjects under two
conditions. I shall deal with the latter, single sample problem first.
The population mean, µ, is unknown and we wish to estimate it using a
95% confidence interval. We can see that, for 95% of samples, the
difference between [x with bar above] and µ is at most t standard errors,
where t is the value of the t distribution such that 95% of observations will
be closer to zero than t. For a large sample this will be 1.96 as for the
Normal distribution. For small samples we must use Table 10.1. In this
table, the probability that the t distribution is further from zero than t is
given, so we must first find one minus our desired probability, 0.95. We
have 1 - 0.95 = 0.05, so we use the 0.05 column of the table to get the
value of t. We then have the 95% confidence interval: [x with bar above] -
t standard errors to [x with bar above] - t standard errors. The usual
application of this is to differences between measurements made on the
same or on matched pairs of subjects. In this application the one sample
t test is also known as the paired t test.
Consider the data of Table 10.2. (I asked the researcher why there were
so many missing data. He told me that some of the biopsies were not
usable to count the capillaries, and that some of these patients were
amputees and the foot itself was missing.) We shall estimate the
difference in capillary density
between the worse foot (in terms of ulceration, not capillaries) and the
better foot for the ulcerated patients. The first step is to find the
differences (worse – better). We then find the mean difference and its
standard error, as described in §8.2. These are in the last column of
Table 10.2.
Table 10.2. Capillary density (per mm2) in the feet of ulc

patients and a healthy control group (data supplied by
Lamah)
Controls Ulcerated patients

Average
Average
of
Right Left of right Worse Better
worse
foot foot and foot foot
and
left†
better
19 16 17.5 9 ? 9.0
25 30 27.5 11 ? 11.0
25 29 27.0 15 10 12.5
26 33 29.5 16 21 18.5
26 28 27.0 18 18 18.0
30 28 29.0 18 18 18.0
33 36 34.5 19 26 22.5
33 29 31.0 20 ? 20.0
34 37 35.5 20 20 20.0
34 33 33.5 20 33 26.5
34 37 35.5 20 26 23.0
34 ? 34.0 21 15 18.0
35 38 36.5 22 23 22.5
36 40 38.0 22 ? 22.0
39 41 40.0 23 23 23.0
40 39 39.5 25 30 27.5
41 39 40.0 26 31 28.5
41 39 40.0 27 26 26.5
56 48 52.0 27 ? 27.0
35 23 29.0
47 42 44.5
? 24 24.0
? 28 28.0
Number 19 23
Mean 34.08 22.59
Sum of 956.13 1176.32

squares
Variance 53.12 53.47
Standard 7.29 7.31

deviation
Standard 0.38 0.32

error
†When one observation is missing the average = the other
observation.
? = Missing data.
To find the 95% confidence interval for the mean difference we must
suppose that the differences follow a Normal distribution. To calculate the
interval, we first require the relevant point of the t distribution from Table
10.1. There are 16 non-missing differences and hence n - 1 = 15 degrees
of freedom associated with s2. We want a probability of 0.95 of being
closer to zero than t, so we go to Table 10.1 with probability = 1 - 0.95 =
0.05. Using the 15 d.f. row, we get t = 2.13. Hence the difference
between a sample mean and the population mean is less than 2.13
standard errors for 95% of samples, and the 95% confidence interval is
-0.81 - 2.13 × 1.51 to -0.81 + 2.13 × 1.51 = -4.03 to +2.41
capillaries/mm2.
On the basis of these data, the capillary density could be less in the
worse affected foot by as much as 4.03 capillaries/mm2, or greater by as
much as 2.41 capillaries/mm2. In the large sample case, we would use
the Normal distribution instead of the t distribution, putting 1.96 instead of
2.13. We would not then need the differences themselves to follow a
Normal distribution.
Fig. 10.3. Normal plot for differences and plot of difference against
average for the data of Table 10.2, ulcerated patients
We can also use the t distribution to test the null hypothesis that in the
population the mean difference is zero. If the null hypothesis were true,
and the differences follow a Normal distribution, the test statistic
mean/standard error would be from a t distribution with n - 1 degrees of
freedom. This is because the null hypothesis is that the mean difference
µ = 0, hence the numerator [x with bar above] - µ = [x with bar above].
We have the usual ‘estimate over standard error’ formula. For the
example, we have
If we go to the 15 degrees of freedom row of Table 10.1, we find that the
probability of such an extreme value arising is greater than 0.10, the 0.10
point of the distribution being 1.75. Using a computer we would find P =
0.6. The data are consistent with the null hypothesis and we have failed
to demonstrate the existence of a difference. Note that the confidence
interval is more informative than the significance test.
We could also use the sign test to test the null hypothesis of no
difference. This gives us 5 positives out of 12 differences (4 differences,
being zero, give no useful information) which gives a two sided
probability of 0.8, a little larger than that given by the t test. Provided the
assumption of a Normal distribution is true, the t test is preferred because
it is the most powerful test, and so most likely to detect differences
should they exist.
The validity of the paired t method described above depends on the
assumption that the differences are from a Normal distribution. We can
check the assumption of a Normal distribution by a Normal plot (§7.5).
Figure 10.3 shows a Normal plot for the differences. The points lie close
to the expected line, suggesting that there is little deviation from the
Normal.
Another plot which is a useful check here is the difference against the
subject mean (Figure 10.3). If the difference depends on magnitude, then
we should be careful of drawing any conclusion about the mean
difference. We may want to investigate this further, perhaps by
transforming the data (§10.4). In this case the difference between the two
feet does not appear to be related to the level of capillary density and we
need not be concerned about this.
The differences may look like a fairly good fit to the Normal even when
the measurements themselves do not. There are two reasons for this: the
subtraction removes variability between subjects, leaving the
measurement error which is more likely to be Normal, and the two
measurement errors are then added by the differencing, producing the
tendency of sums to the Normal seen in the Central Limit theorem (§7.3).
The assumption of a Normal distribution for the one sample case is quite
likely to be met. I discuss this further in §10.5.
10.3 The means of two independent samples

Suppose we have two samples from populations which have a Normal
distribution, with which we want to estimate the difference between the
population means. If the samples were large, the 95% confidence interval
for the difference would be the observed difference - 1.96 standard errors
to observed difference + 1.96 standard errors. Unfortunately, we cannot
simply replace 1.96 by a number from Table 10.1. This is because the
standard error does not have the simple form described in §10.1. It is not
based on a single sum of squares, but rather is the square root of the
sum of two constants multiplied by two sums of squares. Hence, it does
not follow the square root of the Chi-squared distribution as required for
the denominator of a t distributed random variable (§7A). In order to use
the t distribution we must make a further assumption about the data. Not
only must the samples be from Normal distributions, they must be from
Normal distributions with the same variance. This is not as unreasonable
an assumption as it may sound. A difference in mean but not in variability
is a common phenomenon. The PEFR data for children with and without
symptoms analysed in §8.5 and §9.6 show the characteristic very clearly,
as do the average capillary densities in Table 10.2.
We now estimate the common variance, s2. First we find the sum of
squares about the sample mean for each sample, which we can label
SS1 and SS2. We form a combined sum of squares by SS1 + SS2. The
sum of squares for the first group, SS1, has n1 - 1 degrees of freedom
and the second, SS2, has n2 - 1 degrees of freedom. The total degrees of
freedom is therefore n1 - 1 + n2 - 1 = n1 + n2 - 2. We have lost 2 degrees
of freedom because we have a sum of squares about two means, each
estimated from the data. The combined estimate of variance is
The standard error of [x with bar above]1 - [x with bar above]2 is

Now we have a standard error related to the square root of the Chi-
squared distribution and we can get a t distributed variable by
having n1 + n2 - 2 degrees of freedom. The 95% confidence interval for

the difference between population means, µ1 - µ2, is
where t is the 0.05 point with n1 + n2 - 2 degrees of freedom from Table

10.1. Alternatively, we can test the null hypothesis that in the population
the difference is zero, i.e. that µ1 = µ2, using the test statistic
which would follow the t distribution with n1 + n2 - 2 d.f. if the null

hypothesis were true.
Fig. 10.4. Scatter plot against group and Normal plot for the patient
averages of Table 10.2
For a practical example, Table 10.2 shows the average capillary density
over both feet (if present) for normal control subjects as well as ulcer
patients. We shall estimate the difference between the ulcerated patients
and controls. We can check the assumptions of Normal distribution and
uniform variance. From Table 10.2 the variances appear remarkably
similar, 53.12 and 53.47. Figure 10.4 shows that there appears to be a
shift of mean only. The Normal plot combines by groups by taking the
differences between each observation and its group mean, called the
residuals. This has a slight kink at the end but no pronounced curve,
suggesting that there is little deviation from the Normal. I therefore feel
quite happy that the assumptions of the two-sample t method are met.
First we find the common variance estimate, s2. The sums of squares
about the two sample means are 956.13 and 1176.32. This gives the
combined sum of squares about the sample means to be 956.13 +
1176.32 = 2132.45. The combined degrees of freedom are n1 + n2 - 2 =
19 + 23 - 2 = 40. Hence s2 = 2132.45/40 = 53.31. The standard error of
the difference between means is
The value of the t distribution for the 95% confidence interval is found
from the 0.05 column and 40 degrees of freedom row of Table 10.1,
giving t0.05 = 2.02. The difference between means (control – ulcerated) is
34.08 - 22.59 = 11.49. Hence the 95% confidence interval is 11.49 - 2.02
× 2.26 to 11.49 + 2.02 × 2.26, giving 6.92 to 16.06 capillaries/mm2.
Hence there is clearly a difference in capillary density between normal
controls and ulcerated patients.
To test the null hypothesis that in the population the control - ulcerated
difference is zero, the test statistic is difference over standard error,
11.49/2.26 = 5.08. If the null hypothesis were true, this would be an
observation from the t distribution with 40 degrees of freedom. From
Table 10.1, the probability of such an extreme value is less than 0.001.
Hence the data are not consistent with the null hypothesis and we can
conclude that there is strong evidence of a difference in the populations
which these patients represent.
10.4 The use of transformations

We have already seen (§7.4) that some variables which do not follow a
Normal distribution can be made so by a suitable transformation. The
same transformation can be used to make the variance similar in different
groups, called variance stabilizing transformations. Because mean and
variance in samples from the same population are independent if and
only if the distribution is Normal (§7A), stable variances and Normal
distributions tend to go together.
Often standard deviation and mean are connected by a simple
relationship of the form s = a[x with bar above]b, where a and b are
constants. If this is so, it can be shown that the variance will be stabilized
by raising the observations to the power 1 - b,
unless b = 1, when we use the log. (I shall resist the temptation to prove
this, though I can. Any book on mathematical statistics will do it.) Thus, if
the standard deviation is proportional to the square root of the mean (i.e.
variance proportional to mean), e.g. Poisson variance (§6.7), b = 0.5, 1 -
b = 0.5, and we use a square root transformation. If the standard
deviation is proportional to the mean we log. If the standard deviation is
proportional to the square of the mean we have b = 2, 1 - b = -1, and we
use the reciprocal. Another, rarely seen transformation is used when
observations are Binomial proportions. Here the standard deviation
increases as the proportion goes from 0.0 to 0.5, then decreases as the
proportion goes from 0.5 to 1.0. This is the arcsine square root
transformation. Whether it works depends on how much other variation
there is. It has now been largely superseded by logistic regression
(§17.8).
Table 10.3. Biceps skinfold thickness (mm) in

two groups of patients
Crohn's disease Coeliac disease
1.8 2.8 4.2 6.2 1.8 3.8
2.2 3.2 4.4 6.6 2.0 4.2
2.4 3.6 4.8 7.0 2.0 5.4
2.5 3.8 5.6 10.0 2.0 7.6
2.8 4.0 6.0 10.4 3.0

Fig. 10.5. Scatter plot, histogram, and Normal plot for the biceps
skinfold data
When we have several groups we can plot log(s) against log([x with bar
above]) then draw a line through the points. The slope of the line is b
(see Healy 1968). Trial and error, however, combined with scatter plots,
histograms, and Normal plots, usually suffice.
Table 10.3 shows some data from a study of anthropometry and
diagnosis in patients with intestinal disease (Maugdal et al. 1985). We
were interested in differences in anthropometrical measurements
between patients with different diagnoses, and here we have the biceps
skinfold measurements for 20 patients with Crohn's disease and 9
patients with coeliac disease. The data have been put into order of
magnitude and it is fairly obvious that the distribution is skewed to the
right. Figure 10.5 shows this clearly. I have subtracted the group mean
from each observation, giving what is called the within-group residuals,
and then found both the frequency distribution and Normal plot. The
distribution is clearly skew, and this is reflected in the Normal plot, which
shows a pronounced curvature.
Fig. 10.6. Scatter plot, histogram, and Normal plot for the biceps
skinfold data, after square root, log, and reciprocal transformations
We need a Normalizing transformation, if one can be found. The usual

best guesses are square root, log, and reciprocal, with the log being the
most likely to succeed. Figure 10.6 shows the scatter plot, histogram, and
Normal plot for the residuals after transformation. (These logarithms are
natural, to base e, rather than to base 10. It makes no difference to the
final result and the calculations are the same to the computer.) The fit to
the Normal distribution is not perfect, but for each transformation is much
better than in Figure 10.5. The log looks the best for the equality of
variance and the Normal distribution. We could use the two-sample t
method on these data quite happily.
Table 10.4 shows the results of the two sample t method used with the
raw, untransformed data and with each transformation. The t test statistic
increases and its associated probability decreases as we move closer to
a Normal distribution, reflecting the increasing power of the t test as its
assumptions are more closely met. Table 10.4 also shows the ratio of the
variances in the two samples. We can see that, as the transformed data
gets closer to a Normal distribution, the variances tend to become more
equal also.
The transformed data clearly gives a better test of significance than the
raw data. The confidence intervals for the transformed data are more
difficult to interpret, however, so the gain here is not so apparent. The
confidence limits for the difference cannot be transformed back to the
original scale. If we try it, the square root and reciprocal limits give
ludicrous results. The log gives interpretable results (0.89 to 2.03) but
these are not limits for the difference in
millimetres. How could they be, for they do not contain zero yet the
difference is not significant? They are in fact the 95% confidence limits
for the ratio of the Crohn's disease geometric mean to the coeliac
disease geometric mean (§7.4). If there were no difference, of course,
the expected value of this ratio would be one, not zero, and so lies within
the limits. The reason is that when we take the difference between the
logarithms of two numbers, we get the logarithm of their ratio, not of their
difference (§5A).
Table 10.4. Biceps skinfold thickness compared for tw

groups of patients, using different transformations
Two-sample 95%
t test, 27 d.f. Confidence
interval for Variance
Transformation difference
t P on larger/sma
transformed
scale
None, raw data 1.28 0.21 -0.71 to
3.07 mm
Square root 1.38 0.18 -0.140 to

0.714
Logarithm 1.48 0.15 -0.114 to

0.706
Reciprocal -1.65 0.11 -0.203 to

0.022
Because the log transformation is the only one which gives useful
confidence intervals, I would use it unless it were clearly inadequate for
the data, and another transformation clearly superior. When this happens
we are reduced to a significance test only, with no meaningful estimate.
10.5 Deviations from the assumptions of t

methods
The methods described in this chapter depend on some strong
assumptions about the distributions from which the data come. This often
worries users of statistical methods, who feel that these assumptions
must limit greatly the use of t distribution methods and find the attitude of
many statisticians, who often use methods based on Normal
assumptions almost as a matter of course, rather sanguine. We shall look
at some consequences of deviations from the assumptions.
First we shall consider a non-Normal distribution. As we have seen, some
variables conform very closely to the Normal distribution, others do not.
Deviations occur in two main ways: grouping and skewness. Grouping
occurs when a continuous variable, such as human height, is measured
in units which are fairly large relative to the range. This happens, for
example, if we measure human height to the nearest inch. The heights in
Figure 10.2 were to the nearest inch, and the fit to the t distribution is
very good. This was a very coarse grouping, as the standard deviation of
heights was 2.5 inches and so 95% of the 3000 observations had values
over a range of 10 inches, only 10 or 11 possible values in all. We can
see from this that if the underlying distribution is Normal, rounding the
measurement is not going to affect the application of the t distribution by
much.
The other assumption of the two-sample t method is that the variances in

the two populations are the same. If this is not correct, the t distribution
will not necessarily apply. The effect is usually small if the two
populations are from a Normal distribution. This situation is unusual
because, for samples from the same population, mean and variance are
independent if the distribution is Normal (§7A). There is an approximate t
method, as we noted in §10.3. However, unequal variance is more often
associated with skewness in the data, in which case a transformation
designed to correct one fault often tends to correct the other as well.
Both the paired and two-sample t methods are robust to most deviations
from the assumptions. Only large deviations are going to have much
effect on these methods. The main problem is with skewed data in the
one-sample method, but for reasons given in §10.2, the paired test will
usually provide differences with a reasonable distribution. If the data do
appear to be non-Normal, then a Normalizing transformation will improve
matters. If this does not work, then we must turn to methods which do not
require these assumptions (§9.2, §12.2, §12.3).
10.6 What is a large sample?

In this chapter we have looked at small sample versions of the large
sample methods of §8.5 and §9.7. There we ignored both the distribution
of the variable and the variability of s2, on the grounds that they did not
matter provided the samples were large. How small can a large sample
be? This question is critical to the validity of these methods, but seldom
seems to be discussed in text books.
Provided the assumptions of the t test apply, the question is easy enough
to answer. Inspection of Table 10.1 will show that for 30 degrees of
freedom the 5% point is 2.04, which is so close to the Normal value of
1.96 that it makes little difference which is used. So for Normal data with
uniform variance we can forget the t distribution when we have more than
30 observations.
When the data are not in this happy state, things are not so simple. If the
t method is not valid, we cannot assume that a large sample method
which approximates to it will be valid. I recommend the following rough
guide. First, if in doubt, treat the sample as small. Second, transform to a
Normal distribution if possible. In the paired case you should transform
before subtraction. Third, the more non-Normal the data, the larger the
sample needs to be before we can
ignore errors in the Normal approximation.
Table 10.5. Blood zidovudine levels at times after adm

presence of fat malabsorption
Time since administration of zidovu

0 15 30 45 60 90 120
Malabsorption patients
0.08 13.15 5.70 3.22 2.69 1.91 1.72
0.08 0.08 0.14 2.10 6.37 4.89 2.11
0.08 0.08 3.29 3.47 1.42 1.61 1.41
0.08 0.08 1.33 1.71 3.30 1.81 1.16
0.08 6.69 8.27 5.02 3.98 1.90 1.24
0.08 4.28 4.92 1.22 1.17 0.88 0.34
0.08 0.13 9.29 6.03 3.65 2.32 1.25
0.08 0.64 1.19 1.65 2.37 2.07 2.54
0.08 2.39 3.53 6.28 2.61 2.29 2.23
Normal absorption patients:
0.08 3.72 16.02 8.17 5.21 4.84 2.12
0.08 6.72 5.48 4.84 2.30 1.95 1.46

0.08 9.98 7.28 3.46 2.42 1.69 0.70
0.08 1.12 7.27 3.77 2.97 1.78 1.27
0.08 13.37 17.61 3.90 5.53 7.17 5.16
There is no simple answer to the question: ‘how large is a large sample?’.

We should be reasonably safe with inferences about means if the sample
is greater than 100 for a single sample, or if both samples are greater
than 50 for two samples. The application of statistical methods is a matter
of judgement as well as knowledge.
10.7 * Serial data

Table 10.5 shows levels of zidovudine (AZT) in the blood of AIDS patients
at several times after administration of the drug, for patients with normal
fat absorption or fat malabsorption. A line graph of these data was shown
in Figure 5.6. One common approach to such data is to carry out a two-
sample t test at each time separately, and researchers often ask at what
time the difference becomes significant. This is a misleading question, as
significance is a property of the sample rather than the population. The
difference at 15 min may not be significant because the sample is small
and the difference to be detected is small, not because there is no
difference in the population. Further, if we do this for each time point we
are carrying out multiple significance tests (§9.10) and each test only
uses a small part of the data so we are losing power (§9.9). It is better to
ask whether there is any evidence of a difference between the response
of normal and malabsorption subjects over the whole period of
observation.
The simplest approach is to reduce the data for a subject to one number.
We can use the highest value attained by the subject, the time at which
this peak value was reached, or the area under the curve. The first two
are self-explanatory.
The area under the curve or (AUC) is found by drawing a line through
all the points and finding the area between it and the horizontal axis. The
‘curve’ is ususally formed by a series of straight lines found by joining all
the points for the subject, and Figure 10.7 shows this for the first subject
in Table 10.5. The area under the curve can be calculated by taking each
straight line segment and calculating the area under this. This is the base
multiplied by the average of the two vertical heights. We calculate this for
each line segment, i.e. between each pair of adjacent time points, and
add. Thus for the first subject we get (15 - 0) × (0.08 + 13.15)/2 + (30 -
15) × (13.15 + 5.70)/2 + … + (360 - 300) × (0.43 + 0.32)/2 = 667.425.
This can be done fairly easily by most statistical computer packages. The
area for each subject is shown in Table 10.6.
Fig. 10.7. Calculation of the area under the curve for one subject
Table 10.6. Area under the curve for data of

Table 10.5
Malabsorption patients Normal patients
667.425 256.275 919.875
569.625 527.475 599.850
306.000 388.800 499.500
298.200 505.875 472.875
617.850 1377.975
Fig. 10.8. Normal plots for area under the curve and log area for the
data of Table 10.5
10.8 * Comparing two variances by the F test

We can test the null hypothesis that two population variances are equal
using the F distribution. Provided the data are from a Normal distribution,
the ratio of two independent estimates of the same variance will follow a
F distribution (§7A), the degrees of freedom being the degrees of
freedom of the two estimates. The F distribution is defined as that of the
ratio of two independent Chi-squared variables divided by their degrees
of freedom:
where m and n are the degrees of freedom (§7A). For Normal data the
distribution of a sample variance s2 from n observations is that of σ2χ2n/(n
- 1) and when we divide one estimate of variance by another to give the F
ratio, the σ2 cancels out. Like other distributions derived from the Normal,
the F distribution cannot be integrated and so we must use a table.
Because it has two degrees of freedom, the table is cumbersome,
covering several pages, and I shall omit it. Most F methods are done
using computer programs which calculate the probability directly. The
table is usually only given as the upper percentage points.
To test the null hypothesis, we divide the larger variance by the smaller.
For the skinfold data of §10.4, the variances are 5.860 with 19 degrees of
freedom for the Crohn's patients and 3.860 with 8 degrees of freedom for
the coeliacs, giving F = 5.860/3.860 = 1.52. The probability of this being
exceeded by the F distribution with 19 and 8 degrees of freedom is 0.3,
the 5% point of the distribution being 3.16, so there is no evidence from
these data that the variance of skinfold differs between patients with
Crohn's disease and coeliac disease.
Several variances can be compared by Bartlett's test or the Levene test

(see Armitage and Berry 1994, Snedecor and Cochran 1980).
Table 10.7. Mannitol and lactulose gut permeability tes

and controls
HIV % HIV
Diarrhoea %lactulose
status Mannitol status
AIDS Yes 14.9 1.17 ARC
AIDS Yes 16.82 0.367 HIV+

AIDS No 13.95 0.6 HIV-
AIDS No 12.455 0.4 HIV-
AIDS No 10.45 0.18 HIV-
AIDS No 8.36 0.189 HIV-
AIDS No 7.423 0.175 HIV-
AIDS No 2.657 0.039 HIV-
AIDS No 19.95 1.43 HIV-
AIDS No 12.59 0.25 HIV-

AIDS No 15.22 0.29 HIV-
AIDS No 17.71 0.47 HIV-
AIDS Yes 7.256 0.252 HIV-
AIDS No 17.75 0.47 HIV-
ARC Yes 7.42 0.21 HIV-
ARC Yes 9.174 0.399 HIV-
ARC Yes 9.77 0.215 HIV-
ARC No 22.03 0.651
10.9 * Comparing several means using analysis

of variance
Consider the data of Table 10.7. These are measures of gut permeability
obtained from four groups of subjects, diagnosed with AIDS, AIDS
related com-plex (ARC), asymptomatic HIV positive, and HIV negative
controls. We want to investigate the differences between the groups.
One approach would be to use the t test to compare each pair of groups.
This has disadvantages. First, there are many comparisons, m(m - 1)/2
where m is the number of groups. The more groups we have, the more
likely it is that two of them will be far enough apart to produce a
‘significant’ difference when the null hypothesis is true and the population
means are the same (§9.10). Second, when groups are small, there may
not be many degrees of freedom for the estimate of variance. If we can
use all the data to estimate variance we will have more
degrees of freedom and hence a more powerful comparison. We can do

this by analysis of variance, which compares the variation between the
groups to the variation within the groups.
Table 10.8. Some artificial data to illustrate

how analysis of variance works
Group 1 Group 2 Group 3 Group 4
6 4 7 3
7 5 9 5
8 6 10 6
8 6 11 6
9 6 11 6
11 8 13 8
Mean 8.167 5.833 10.167 5.667
To illustrate how the analysis of variance, or anova, works, I shall use

some artificial data, as set out in Table 10.8. In practice, equal numbers
in each group are unusual in medical applications. We start by estimating
the common variance within the groups, just as we do in a two-sample t
test (§10.3). We find the sum of squares about the group mean for each
group and add them. We call this the within groups sum of squares.
For Table 10.8 this gives 57.833. For each group we estimate the mean
from the data, so we have estimated 4 parameters and have 24 - 4 = 20
degrees of freedom. In general, for m groups of size n each we have nm
- m = m(n - 1) degrees of freedom. This gives us an estimate of variance
of
This is the within groups variance or residual variance. There is an

assumption here. For a common variance, we assume that the variances
are the same in the four populations represented by the four groups.
We can also find an estimate of variance from the group means. The
variance of the four group means is 4.562. If there were no difference
between the means in the population from which the sample comes, this
variance would be the variance of the sampling distribution of the mean
of n observations, which is s2/n, the square of the standard error (§8.2).
Thus n times this variance should be equal to the within groups variance.
For the example, this is 4.562 × 6 = 27.375. which is much greater than
the 2.892 found within the groups. We express this by the ratio of one
variance estimate to the other, between groups over within groups, which
we call the variance ratio or F ratio. If the null hypothesis is true and if the
observations are from a Normal distribution with uniform variance, this
ratio follows a known distribution, the F distribution with m - 1 and n - 1
degrees of freedom (§10.8).
For the example we would have 3 and 20 degrees of freedom and
If the null hypothesis were true, the expected value of this ratio would be
1.0.
A large value gives us evidence of a difference between the means in the

four populations. For the example we have a large value of 9.47 and the
probability of getting a value as big as this if the null hypothesis were true
would be 0.000 4. Thus there is a significant difference between the four
groups.
Table 10.9. One-way analysis of variance for the data of

10.8
Source Degrees
Sum of Mean Variance
of of
squares square ratio (F)
variation freedom
Total 23 139.958
Between 3 82.125 27.375 9.47

groups
Within 20 57.833 2.892

groups
Table 10.10. One-way analysis of variance for the manni
Source Degrees
of of
squares square ratio (F
variation freedom
Total 58 1559.036
Between 3 49.012 16.337 0.6

groups
Residual 55 1510.024 27.455
We can set these calculations out in an analysis of variance table, as

shown in Table 10.9. The sum of squares in the ‘between groups’ row is
the sum of squares of the group means times n. We call this the between
groups sum of squares. Notice that in the ‘degrees of freedom’ and
‘sum of squares’ columns the ‘within groups’ and ‘between groups’ rows
add up to the total. The within groups sum of squares is also called the
residual sum of squares, because it is what is left when the group effect
is removed, or the error sum of squares, because it measures the
random variation or error remaining when all systematic effects have
been removed.
The sum of squares of the whole data, ignoring the groups is called the
total sum of squares. It is the sum of the between groups and within
groups sum of squares.
Returning to the mannitol data, as so often happens the groups are of
unequal size. The calculation of the between groups sum of squares
becomes more complicated and we usually do it by subtracting the within
groups sum of squares from the total sum of squares. Otherwise, the
table is the same, as shown in Table 10.10. As these calculations are
usually done by computer the extra complexity in calculation does not
matter. Here there is no significant difference between the groups.
If we have only two groups, one-way analysis of variance is another way
of doing a two-sample t test. For example, the analysis of variance table
for the comparison of average capillary density (§10.3) is shown in Table
10.11. The probability is the same and the F ratio, 25.78, is the square of
the t statistic, 5.08. The residual mean square is the common variance of
the t test.
Table 10.11. One-way analysis of variance for the comp

mean capillary density between ulcerated patients and
Table 10.2
Source Degrees
of of
squares square ratio (
variation freedom
Total 41 3506.57
Between 1 1374.114 1374.114 25.78

groups
Residual 40 2132.458 53.311

Fig. 10.9. Plots of the mannitol data, showing that the assumptions
of Normal distribution and homoscedasticity are reasonable
10.10 * Assumptions of the analysis of variance

There are two assumptions for analysis of variance: that data come from
Normal distributions within the groups and that the variances of these
distributions are the same. The technical term for uniformity of variance is
homoscedasticity; lack of uniformity is heteroscedasticity.
Heteroscedasticity can affect analyses of variance a lot and we try to
guard against it.
We can examine these assumptions graphically. For mannitol (Figure
10.9) the scatter plot for the groups shows that the spread of data in each
group is similar, suggesting that the assumption of uniform variance is
met, the histogram looks Normal and Normal plot looks straight. This is
not the case for the lactulose data, as Figure 10.10 shows. The variances
are not uniform and the histogram and Normal plot suggest positive
skewness. As is often the case, the group with the highest mean, AIDS,
has the greatest spread. The square root transformation of the lactulose
fits better, giving a good Normal distribution although the variability is not
uniform. The log transform over-compensates for skewness, by
producing skewness in the opposite direction, though the variances
appear uniform. Either the square root or the logarithmic transformation
would be better than the raw data. I picked the square root because the
distribution looked better. Table 10.12 shows the analysis of variance for
square root transformed lactulose.
There are also significance tests which we can apply for Normal
distribution and homoscedasticity. I shall omit the details.
10.11 * Comparison of means after analysis of

variance
Concluding from Tables 10.9 and 10.12 that there is a significant
difference between the means is rather unsatisfactory. We want to know
which means differ
from which. There are a number of ways of doing this, called multiple
comparisons procedures. These are mostly designed to give only one
type I error (§9.3) per 20 analyses when the null hypothesis is true, as
opposed to doing t tests for each pair of groups, which gives one error
per 20 comparisons when the null hypothesis is true. I shall not go into
details, but look at a couple of examples. There are several tests which
can be used when the numbers in each group are the same, Tukey's
Honestly Significant Difference, the Newman-Keuls sequential procedure
(both called Studentized range tests), Duncan's multiple range test, etc.
The one you use will depend on which computer program you have. The
results of the Newman-Keuls sequential procedure for the data of Table
10.8 are shown in Table 10.13. Group 1 is significantly different from
groups 2 and 4, and group 3 from groups 2 and 4. At the 1% level, the
only significant differences are between group 3 and groups 2 and 4.
Fig. 10.10. Plots of the lactulose data on the natural scale and after
square root and log transformation
Table 10.12. One-way analysis of variance for the squa

transformed lactulose data of Table 10.7
Source Degrees
of of
variation freedom
Total 58 3.25441
HIV 3 0.42870 0.14290 2.78
status
Residual 55 2.82571 0.05138
For unequal-sized groups, the choice of multiple comparison procedures

is
more limited, Gabriel's test can be used with unequal-sized groups. For
the root transformed lactulose data, the results of Gabriel's test are
shown in Table 10.14. This shows that the AIDS subjects are significantly
different from the asymptomatic HIV+ patients and from the HIV-controls.
For the mannitol data, most multiple comparison procedures will give no
significant differences because they are designed to give only one type I
error per analysis of variance. When the F test is not significant, no group
comparisons will be either.
Table 10.13. The Newman-Keuls test for the

data of Table 10.8
0.05 level 0.01 level

Group Group Group Group
1 2 3 1 2 3
2 S 2 N
3 N S 3 N S
4 S N S 4 N N S
S = significant, N = not significant.
Table 10.14. Gabriel's test for the root transformed

lactulose data
0.05 level 0.01 level
Group Group Group Group

AIDS ARC HIV+ AIDS ARC
ARC N ARC N
HIV+ S N HIV+ N N
HIV- S N N HIV- N N
S = significant, N = not significant.
10.12 * Random effects in analysis of variance

Although the technique is called analysis of variance, in §10-9-11 we
have been using it for the comparison of means. In this section we shall
look at another application, where we shall indeed use anova to look at
variances. When we estimate and compare the means of groups
representing different diagnoses, different treatments, etc., we call these
fixed effects. In other applications, groups are members of a random
sample from a larger population and, rather than estimate the mean of
each group, we estimate the variance between them. We called these
groups random effects.
Consider Table 10.15, which shows repeated measurements of pulse
rate on a group of medical students. Each measurement was made by a
different observer. Observations made repeatedly under the same
circumstances are called replicates and here we have two replicates per
subject. We can do a one way analysis of variance on these data, with
subject as the grouping factor (Table 10.16).
The test of significance in Table 10.16 is redundant, because we know
each pair of measurements is from a different person, and the null
hypothesis that all pairs are from the same population is clearly false.
What we can use this anova
for is to estimate some variances. There are two different variances in the
data. One is between measurements on the same person, the within-
subject variance which we shall denote by σ2w. In this example the
within subject variance is the measurement error, and we shall assume it
is the same for everyone. The other is the variance between the subjects'
true or average pulse rates, about which the individual measurements for
a subject are distributed. This is the average of all possible
measurements for that subject, not the average of the two measurements
we actually have. This variance is the between-subjects variance and
we shall denote it by σ2b. A single measurement observed from a single
individual is the sum of the subject's true pulse rate and the
measurement error. Such measurements therefore have variance σ2b +
σ2w. We can estimate both these variances from the anova table.
Table 10.15. Paired measurements of 30 second pulse i

45 medical students
Pulse A Pulse A
Subject Subject Subject
B B
1 46 42 16 34 36 31
2 50 42 17 30 36 32
3 39 37 18 35 45 33
4 40 54 19 32 34 34
5 41 46 20 44 46 35
6 35 35 21 39 42 36
7 31 44 22 34 37 37
8 43 35 23 36 38 38
9 47 45 24 33 34 39
10 48 36 25 34 35 40
11 32 46 26 51 48 41
12 36 34 27 31 30 42
13 37 30 28 30 31 43
14 34 36 29 42 43 44
15 38 36 30 39 35 45
Table 10.16. One-way analysis of variance for the 30 se

pulse data of Table 10.15
Source Degrees
of of
variation freedom
Total 89 3.054.99
Between 44 2408.49 54.74 3.81

subjects
Within 45 646.50 14.37

subjects
For the simple example of the same number of replicates m on each of n

subjects, the estimation of the variances is quite simple. We estimate
σ2w, directly from the mean square within subjects, MSw, giving an
estimate s2w. We can show (although I shall omit it) that the mean square
between subjects, MSb, is an estimate of mσ2b + σ2w. The variance ratio,
F = MSb/MSw, will be expected to be 1.0 if σ2b = 0, i.e. if the null
hypothesis that all subjects are the same is true. We can estimate σ2b by
s2b = (MSb - MSw)/m.
For the example, s2w = 14.37 and s2b = (54.74 - 14.37)/2 = 20.19. Thus
the
variability between measurements by different observers on the same

subject is not much less than the variability between the underlying pulse
rate between different subjects. The measurement (by these untrained
and inexperience observers) does not tell us much about the subjects.
We shall see a practical application in the study of measurement error
and observer variation in §15.2, and consider another aspect of this
analysis, intraclass correlation, in §11.13.
Table 10.17. Number of X-ray requests conforming to t

each practice in the intervention and controls groups
1994)
Intervention group
Number of Number of
Percentage
requests requests
Total Conforming conforming Total
20 20 100 7
7 7 100 37
16 15 94 38
31 28 90 28
20 18 90 20
24 21 88 19
7 6 86 9
6 5 83 25
30 25 83 120
66 53 80 89
5 4 80 22
43 33 77 76
43 32 74 21
23 16 70 127
64 44 69 22
6 4 67 34
18 10 56 10
Total 429 341 704
Mean 81.6
SD 11.9
If we have different numbers of replicates per subject or other factors to

consider (e.g. if each observer made two repeated measurements) the
analysis becomes fiendishly complicated (see Searle et al. 1992, if you
must). These estimates of variance deserve confidence intervals like any
other estimate, but these are even more fiendishly complicated, as
Burdick and Graybill (1992) convincingly demonstrate. I would
recommend you consult a statistician experienced in these matters, if you
can find one.
10.13 * Units of analysis and cluster-randomized

trials
A cluster-randomized study (§2.11) is one where a group of subjects,
such as the patients in a hospital ward or a general practice list, are
randomized to the same treatment together. The treatment might be
applied to patient directly, such as an offer of breast cancer screening to
all eligible women in a district, or be applied to the care provider, such as
treatment guidelines given to the GP. The design of the study must be
taken into account in the analysis.
Table 10.17 shows an example (Oakeshott et al. 1994, Kerry and Bland
1998).
Fig. 10.11. Scatter plots and Normal plots for the data of Table
10.17, showing the effect of an arcsine square root transformation
In this study guidelines as to appropriate referral for X-ray were given to

GPs in 17 practices and another 17 practices served as controls. We
could say we have 341 out of 429 appropriate referrals in the treated
group and 509 out of 704 in the control group and compare these
proportions as in §8.6 and §9.8. This would be wrong, because to follow
a Binomial distribution, all the referrals must be independent (§6.4). They
are not, as the individual GP may have a profound effect on the decision
to refer. Even where the practitioner is not directly involved, members of
a cluster may be more similar to one another then they are to members
of another cluster and so not be independent. Ignoring the clustering may
result in confidence intervals which are too narrow and P values which
are too small, producing spurious significant differences.
The easiest way to analyse the data from such studies is to make the
experimental unit, that which is randomized (§2.11), the unit of analysis.
We can construct a summary statistic for each cluster and then analyse
these summary values. The idea is similar to the analysis of repeated
measurements on the same subject, where we construct a single
summary statistic over the times for each individual (§10.7). For Table
10.17, the practice's percentage of referrals which are appropriate is the
summary statistic. The mean percentages in the two groups can then be
compared by the two-sample t method. The observed difference is 81.6 –
73.6 = 8.0 and the standard error of the difference is 4.3. There are 32
degrees of freedom and, from Table 10.1, the 5% point of the t
distribution is 2.04. This gives a 95% confidence interval for the treatment
difference of
8.0 ± 2.037 × 4.3, or -1 to 17 percentage points. For the test of

significance, the test statistic is 8.0/4.3 = 1.86, P = 0.07.
In this example, each observation is a Binomial proportion, so we could
consider an arcsine square root transformation of the proportions (§10.4).
As Figure 10.11 shows, if anything the transformation makes the fit to the
Normal distribution worse. This is reflected in a larger P value, giving P =
0.10.
There is a widely varying number of referrals, between practices, which
must reflect the list size and number of GPs in the practice. We can take
this into account with an analysis which weights each observation by the
numbers of referrals. Bland and Kerry (1998) give details.
Appendices
10A Appendix: The ratio mean/standard
error
As if by magic, we have our sample mean over its standard error. I shall
not bother to go into this detail for the other similar ratios which we shall
encounter. Any quantity which follows a Normal distribution with mean
zero (such as [x with bar above] - µ), divided by its standard error, will
follow a t distribution provided the standard error is based on one sum of
squares and hence is related to the Chi-squared distribution.

50. The paired t test is:
(a) impractical for large samples;
(b) useful for the analysis of qualitative data;
(c) suitable for very small samples;
(d) used for independent samples;
(e) based on the Normal distribution.
View Answer
51. Which of the following conditions must be met for a valid t test
between the means of two samples:
(a) the numbers of observations must be the same in the two
groups;
(b) the standard deviations must be approximately the same in the
two groups;
(c) the means must be approximately equal in the two groups;
(d) the observations must be from approximately Normal
distributions;
(e) the samples must be small.
View Answer
52. In a two-sample clinical trial, one of the outcome measures was

highly skewed. To test the difference between the levels of this
measure in the two groups of patients, possible approaches
include:
(a) a standard t test using the observations;
(b) a Normal approximation if the sample is large;
(c) tranaforming the data to a Normal distribution and using a t
test;
(d) a sign test;
(e) the standard error of the difference between two proportions.
View Answer
53. In the two-sample t test, deviation from the Normal distribution

by the data may seriously affect the validity of the test if:
(a) the sample sizes are equal;
(b) the distribution followed by the data is highly skewed;
(c) one sample is much larger than the other;
(d) both samples are large;
(e) the data deviate from a Normal distribution because the
measurement unit is large and only a few values are possible.
View Answer
Table 10.18. Semen analyses for successful and

unsuccessful sperm donors (Paraskevaides et al.
1991)
Unsuccessful
Successful donors
donors
n Mean (sd) n Mean (sd)
Volume 17 3.14 (1.28) 19 2.91

(ml)
Semen 18 146.4 (95.7) 19 124.8

count
(106/ml)
% Motility 17 60.7 (9.7) 19 58.5
% 13 22.8 (8.4) 16 20.3

Abnormal
morphology
All differences not significant, t test.
54. Table 10.18 shows a comparison of successful (i.e. fertile) and

unsuccessful artificial insemination donors. The authors concluded
that ‘Conventional semen analysis may be too insensitive an
indicator of high fertility [in AID]’:
(a) the table would be more informative if P values were given;
(b) the t test is important to the conclusion given:
(c) it is likely that semen count follows a Normal distribution;
(d) if the null hypothesis were true, the sampling distribution of the
t test statistic for semen count would approximate to a t
distribution;
(e) if the null hypothesis were false, the power of the t test for
semen count could be increased by a log transformation.
View Answer
55. If we take samples of size n from a Normal distribution and

calculate the sample mean [x with bar above] and variance s2:
(a) samples with large values of [x with bar above] will tend to
have large s2;
(b) the sampling distribution of [x with bar above] will be Normal;
(c) the sampling distribution of s2 will be related to the Chi-squared
distribution with n - 1 degrees of freedom;
(e) the sampling distribution of s will be approximately Normal if n

> 20.
View Answer
56. In the one-way analysis of variance table for the comparison of

three groups:
(a) the group mean square + the error mean square = the total
mean square;
(b) there are two degrees of freedom for groups;
(c) the group sum of squares + the error sum of squares = the total
sum of squares;
(d) the numbers in each group must be equal;
(e) the group degrees of freedom + the error degrees of freedom =
the total degrees of freedom.
View Answer
10E Exercise: The paired t method

Table 10.19 shows the total static compliance of the respiratory system
and the arterial oxygen tension (pa(O2)) in 16 patients in intensive care
(Al-Saady, personal communication). The patients' breathing was
assisted by a respirator
and the question was whether their respiration could be improved by

varying the characteristics of the air flow. Table 10.19 compares a
constant inspiratory flow waveform with a decelerating inspiratory flow
waveform. We shall examine the effect of waveform on compliance.
Table 10.19. pa(O2) and compliance for two inspiratory

flow waveforms
Compliance
Patient pa(O2) (kPa)
(ml/cmH
Waveform Waveform
Constant Decelerating Constant Deceleratin
1 9.1 10.8 65.4 72.9
2 5.6 5.9 73.7 94.4
3 6.7 7.2 37.4 43.3
4 8.1 7.9 26.3 29.0
5 16.2 17.0 65.0 66.4
6 11.5 11.6 35.2 36.4
7 7.9 8.4 24.7 27.7
8 7.2 10.0 23.0 27.5
9 17.7 22.3 133.2 178.2
10 10.5 11.1 38.4 39.3
11 9.5 11.1 29.2 31.8

12 13.7 11.7 28.3 26.9
13 9.7 9.0 46.6 45.0
14 10.5 9.9 61.5 58.2
15 6.9 6.3 25.7 25.7
16 18.1 13.9 48.7 42.3
1. Calculate the changes in compliance. Find a stem and leaf plot

(hint: you will need both a zero and a minus zero row).
View Answer
2. As a check on the validity of the t method, plot the difference

against the subject's mean compliance. Do they appear to be
related?
View Answer
3. Calculate the mean, variance, standard deviation and standard

error of the mean for the compliance differences.
View Answer
4. Even though the compliance differences are far from a Normal

distribution, calculate the 95% confidence interval using the t
distribution. We will compare this with that for transformed data.
View Answer
5. Find the logarithms of the compliance and repeat steps 1 to 3. Do

the assumptions of the t distribution method apply more closely?
View Answer
6. Calculate the 95% confidence interval for the log difference and
transform back to the original scale. What does this mean and how
does it compare to that based on the untransformed data?
View Answer
7. What can be concluded about the effect of inspiratory waveform

on static compliance in intensive care patients?
View Answer
> Table of Contents > 11 - Regression and correlation
11
Regression and correlation
11.1 Scatter diagrams

In this chapter I shall look at methods of analysing the relationship
between two quantitative variables. Consider Table 11.1, which shows
data collected by a group of medical students in a physiology class.
Inspection of the data suggests that there may be some relationship
between FEV1 and height. Before trying to quantify this relationship, we
can plot the data and get an idea of its nature. The usual first plot is a
scatter diagram, §5.6. Which variable we choose for which axis depends
on our ideas as to the underlying relationship between them, as
discussed below. Figure 11.1 shows the scatter diagram for FEV1 and
height.
Inspection of Figure 11.1 suggests that FEVl increases with height. The
next step is to try and draw a line which best represents the relationship.
The simplest line is a straight one; I shall consider more complicated
relationships in Chapter 17.
The equation of a straight line relationship between variables x and y is y
= a + bx, where a and b are constants. The first, a, is called the
intercept. It is the value of y when x is 0. The second, b, is called the
slope or gradient of the line. It is the increase in y corresponding to an
increase of one unit in x. Their geometrical meaning is shown in Figure
11.2. We can find the values of a and b which best fit the data by
regression analysis.
11.2 Regression
Regression is a method of estimating the numerical relationship
between
variables. For example, we would like to know what is the mean or

expected FEV1 for students of a given height, and what increase in FEV1
is associated with a unit increase in height.
Table 11.1. FEV1 and height for 20 male medical

students
Height FEV1 Height FEV1 Height FEV1

(cm) (litres) (cm) (litres) (cm) (litres)
164.0 3.54 172.0 3.78 178.0 2.98
167.0 3.54 174.0 4.32 180.7 4.80
170.4 3.19 176.0 3.75 181.0 3.96
171.2 2.85 177.0 3.09 183.1 4.78
171.2 3.42 177.0 4.05 183.6 4.56
171.3 3.20 177.0 5.43 183.7 4.68
172.0 3.60 177.4 3.60

Fig. 11.1. Scatter diagram showing the relationship between FEV1
and height for a group of male medical students
Fig. 11.2. Coefficients of a straight line
The name ‘regression’ is due to Galton (1886), who developed the

technique to investigate the relationship between the heights of children
and of their parents. He observed that if we choose a group of parents of
a given height, the mean height of their children will be closer to the
mean height of the population than is the given height. In other words, tall
parents tend to be taller than their children, short parents tend to be
shorter. Galton termed this phenomenon ‘regression towards mediocrity’,
meaning ‘going back towards the average’. It is now called regression
towards the mean (§11.4). The method used to investigate it was called
regression analysis and the name has stuck. However,
in Galton's terminology there was ‘no regression’ if the relationship

between the variables was such that one predicted the other exactly; in
modern terminology there is no regression if the variables are not related
at all.
In regression problems we are interested in how well one variable can be
used to predict another. In the case of FEV1 and height, for example, we
are concerned with estimating the mean FEV1 for a given height rather
than mean height for given FEV1. We have two kinds of variables: the
outcome variable which we are trying to predict, in this case FEV1, and
the predictor or explanatory variable, in this case height. The predictor
variable is often called the independent variable and the outcome
variable is called the dependent variable. However, these terms have
other meanings in probability (§6.2), so I shall not use them. If we denote
the predictor variable by X and the outcome by Y, the relationship
between them may be written as
where a and b are constants and E is a random variable with mean 0,

called the error, which represents that part of the variability of Y which is
not explained by the relationship with X. If the mean of E were not zero,
we could make it so by changing a. We assume that E is independent of
X.
11.3 The method of least squares
If the points all lay along a line and there was no random variation, it
would be easy to draw a line on the scatter diagram. In Figure 11.1 this is
not the case. There are many possible values of a and b which could
represent the data and we need a criterion for choosing the best line.
Figure 11.3 shows the deviation of a point from the line, the distance from
the point to the line in the Y direction, The line will fit the data well if the
deviations from it are small, and will fit badly if they are large. These
deviations represent the error E, that part of the variable Y not explained
by X. One solution to the problem of finding the best line is to choose that
which leaves the minimum amount of the variability of Y unexplained, by
making the variance of E a minimum. This will be achieved by making the
sum of squares of the deviations about the line a minimum. This is called
the method of least squares and the line found is the least squares line.
The method of least squares is the best method if the deviations from the
line are Normally distributed with uniform variance along the line. This is
likely to be the case, as the regression tends to remove from Y the
variability between subjects and leave the measurement error, which is
likely to be Normal. I shall deal with deviations from this assumption in
§11.8.
Many users of statistics are puzzled by the minimization of variation in
one direction only. Usually both variables are measured with some error
and yet we seem to ignore the error in X. Why not minimize the
perpendicular distances to the line rather than the vertical? There are two
reasons for this. First, we are finding the beat prediction of Y from the
observed values of X, not from the
‘true’ values of X. The measurement error in both variables is one of the

causes of deviations from the line, and is included in these deviations
measured in the Y direction. Second, the line found in this way depends
on the units in which the variables are measured. For the data of Table
11.1 the line found by this method is
FEV1 (litre) = -9.33 + 0.075 × height (cm)
If we measure height in metres instead of centimetres, we get
FEV1 (litre) = -34.70 + 22.0 × height (m)
Thus by this method the predicted FEV1 for a student of height 170cm is
3.42 litres, but for a student of height 1.70m it is 2.70 litres. This is clearly
unsatisfactory and we will not consider this approach further.
Fig. 11.3. Deviations from the line in the y direction
Returning to Figure 11.3, the equation of the line which minimizes the
sum of squared deviations from the line in the outcome variable is found
quite easily (§11A). The solution is:
We then find the intercept a by

The equation Y = a + bX is called the regression equation of Y on X, Y
being the outcome variable and X the predictor. The gradient, b, is also
called the regression coefficient. We shall calculate it for the data of
Table 11.1. We have
We do not need the sum of squares for Y yet, but we shall later.
Hence the regression equation of FEV1 on height is

FEV = -9.19 + 0.0744 × height
Figure 11.4 shows the line drawn on the scatter diagram.
The coefficients a and b have dimensions, depending on those of X and
Y. If we change the units in which X and Y are measured we also change
a and b, but we do not change the line. For example, if height is
measured in metres we divide the xi by 100 and we find that b is
multiplied by 100 to give b = 7.4389 litres/m. The line is
FEV1 (litres) = -9.19 + 7.44 × height (m)
This is exactly the same line on the scatter diagram.
Fig. 11.4. The regression of FEV1 on height

Fig. 11.5. The two regression lines for the data of Tables 11.1 and
10.15
11.4 * The regression of X on Y

What happens if we change our choice of outcome and predictor
variables? The regression equation of height on FEVl is
height = 158 + 4.54 × FEV1
This is not the same line as the regression of FEV1 on height. For if we
rearrange this equation by dividing each side by 4.54 we get
FEVl = -34.8 + 0.220 × height
The slope of the regression of height on FEV1 is greater than that of
FEV1 on height (Figure 11.5). In general, the slope of the regression of X
on Y is greater than that of Y on X, when X is the horizontal axis. Only if
all the points lie exactly on a straight line are the two equations the same.
Figure 11.5 also shows the two 30 second pulse measurements of Table
10.15, with the lines representing the regression of the second
measurement on the
first and the first measurement on the second. The regression equations
are 2nd pulse = 17.3 + 0.572 × 1st pulse and 1st pulse = 14.9 + 0.598 ×
2nd pulse. Each regression coefficient is less than one. This means that
for subjects with any given first pulse measurement, the predicted second
pulse measurement will be closer to the mean than the first
measurement, and for any given second pulse measurement, the
predicted first measurement will be closer to the mean than the second
measurement. This is regression towards the mean (§11.2). Regression
towards the mean is a purely statistical phenomenon, produced by the
selection of the given value of the predictor and the imperfect relationship
between the variables. Regression towards the mean may manifest itself
in many ways. For example, suppose we measure the blood pressure of
an unselected group of people and then select subjects with high blood
pressure, e.g. diastolic >95 mm Hg. If we then measure the selected
group again, the mean diastolic pressure for the selected group will be
less on the second occasion than on the first, without any intervention or
treatment. The apparent fall is caused by the initial selection.
11.5 The standard error of the regression

coefficient
In any estimation procedure, we want to know how reliable our estimates
are. We do this by finding their standard errors and hence confidence
intervals. We can also test hypotheses about the coefficients, for
example, the null hypothesis that in the population the slope is zero and
there is no linear relationship. The details are given in §11C. We first find
the sum of squares of the deviations from the line, that is, the difference
between the observed yi and the values predicted by the regression line.
This is
In order to estimate the variance we need the degrees of freedom with

which to divide the sum of squares. We have estimated not one
parameter from the data, as for the sum of squares about the mean
(§4.6), but two, a and b. We lose two degrees of freedom, leaving us with
n - 2. Hence the variance of Y about the line, called the residual
variance, is
If we are to estimate the variation about the line, we must assume that it
is the same all the way along the line, i.e. that the variance is uniform.
This is the same as for the two-sample t method (§10.3) and analysis of
variance (§10.9). For the
FEV1 data the sum of squares due to the regression is 0.074 3892 ×
576.352 = 3.18937 and the sum of squares about the regression is
9.43868 - 3.18937 = 6.24931. There are 20 - 2 = 18 degrees of freedom,
so the variance about the regression is s2 = 6.2493/18 = 0.34718. The
standard error of b is given by
We have already assumed that the error E is Normally distributed, so b

must be, too. The standard error is based on a single sum of squares, so
b/SE(b) is an observation from the t distribution with n - 2 degrees of
freedom (§10.1). We can find a 95% confidence interval for b by taking t
standard errors on either side of the estimate. For the example, we have
18 degrees of freedom. From Table 10.1, the 5% point of the t distribution
is 2.10. so the 95% confidence interval for b is 0.074389 - 2.10 × 0.02454
to 0.074389 + 2.10 × 0.02454 or 0.02 to 0.13 litres/cm. We can see that
FEV1 and height are related, though the slope is not very well estimated.
We can also test the null hypothesis that, in the population, the slope = 0
against the alternative that the slope is not equal to 0, a relationship in
either direction. The test statistic is b/SE(b) and if the null hypothesis is
true this will be from a t distribution with n - 2 degrees of freedom. For the
example,
From Table 10.1 this has two-tailed probability of less than 0.01. The
computer tells us that the probability is about 0.007. Hence the data are
inconsistent with the null hypothesis and the data provide fairly good
evidence that a relationship exists. If the sample were much larger, we
could dispense with the t distribution and use the Standard Normal
distribution in its place.
11.6 * Using the regression line for prediction

We can use the regression equation to predict the mean or expected Y
for any given value of X. This is called the regression estimate of Y. We
can use this to say whether any individual has an observed Y greater or
less than would be expected given X. For example, the predicted FEVl
for students with height 177cm is -9.19 + 0.0744 × 177 = 3.98 litres.
Three subjects had height 177cm. The first had observed FEVl of 5.43
litres, 1.45 litres above that expected. The second had a rather low FEVl
of 3.09 litres, 0.89 litres below expectation, while the third with an FEVl of
4.05 litres was very close to that predicted. We can use this clinically to
adjust a measured lung function for height and thus get a better idea of
the patient's status. We would, of course, use a much larger sample to
establish a precise estimate of the regression equation. We can also use
a variant of the method (§17.1) to adjust FEV1 for height in comparing
different groups, where we can both remove variation in FEV1 due to
variation in height
and allow for differences in mean height between the groups. We may
wish to do this to compare patients with respiratory disease on different
therapies, or to compare subjects exposed to different environmental
factors, such as air pollution, cigarette smoking, etc.
Fig. 11.6. Confidence intervals for the regression estimate
As with all sample estimates, the regression estimate is subject to

sampling variation. We estimate its precision by standard error and
confidence interval in the usual way. The standard error of the expected
Y for an observed value x is
We need not go into the algebraic details of this. It is very similar to that
in §11C. For x = 177 we have
This gives a 95% confidence interval of 3.98 - 2.10 × 0.138 to 3.98 + 2.10
× 0.138 giving from 3.69 to 4.27 litres. Here 3.98 is the estimate and 2.10
is the 5% point of the t distribution with n - 2 = 18 degrees of freedom.
The standard error is a minimum at x = [x with bar above], and increases
as we move away from [x with bar above] in either direction. It can be
useful to plot the standard error and 95% confidence interval about the
line on the scatter diagram. Figure 11.6 shows this for the FEV1 data.
Notice that the lines diverge considerably as we reach the extremes of
the data. It is very dangerous to extrapolate beyond the data. Not only do
the standard errors become very wide, but we often have no reason to
suppose that the straight line relationship would persist.
The intercept a, the predicted value of Y when X = 0, is a special case of
this. Clearly, we cannot actually have a medical student of height zero
and with FEV1 of -9.19 litres. Figure 11.6 also shows the confidence
interval for the regression estimate with a much smaller scale, to show
the intercept. The confidence interval is very wide at height = 0, and this
does not take account of
any breakdown in linearity.

We may wish to use the value of X for a subject to estimate that subject's
individual value of Y, rather than the mean for all subjects with this X. The
estimate is the same as the regression estimate, but the standard error is
much greater:
For a student with a height of 177cm. the predicted FEVl is 3.98 litres,
with standard error 0.61 litres. Figure 11.7 shows the precision of the
prediction of a further observation. As we might expect, the 95%
confidence intervals include all but one of the 20 observations. This is
only going to be a useful prediction when the residual variance s2 is
small.
We can also use the regression equation of Y on X to predict X from Y.
This is much less accurate than predicting Y from X. The standard errors
are
For example, if we use the regression of height on FEV1 (Figure 11.5) to

predict the FEV1 of an individual student with height 177cm, we get a
prediction of 4.21 litres, with standard error 1.05 litres. This is almost
twice the standard error obtained from the regression of FEV1 on height,
0.61. Only if there is no possibility of deviations in X fulfilling the
assumptions of Normal distribution and uniform variance, and so no way
of fitting X = a + bY, should we consider predicting X from the regression
of Y on X. This might happen if X is fixed in advance, e.g. the dose of a
drug.
11.7 * Analysis of residuals

It is often very useful to examine the residuals, the differences between
the observed and predicted Y. This is best done graphically. We can
assess the assumption of a Normal distribution by looking at the
histogram or Normal plot (§7.5). Figure 11.8 shows these for the FEVl
data. The fit is quite good.
Figure 11.9 shows a plot of residuals against the predictor variable. This
plot enables us to examine deviations from linearity. For example, if the
true relationship were quadratic, so that Y increases more and more
rapidly as X increases, we should see that the residuals are related to X.
Large and small X would tend to have positive residuals whereas central
values would have negative residuals. Figure 11.9 shows no relationship
between the residuals and height, and the linear model seems to be an
adequate fit to the data.
Fig. 11.7. Confidence interval for a further observation

Fig. 11.8. Distribution of residuals for the FEV1 data
Fig. 11.9. Residuals against height for the FEV1 data

Fig. 11.10. Data which do not meet the conditions of the method of
least squares, before and after log transformation
Figure 11.9 shows something else, however. One point stands out as
having a rather larger residual than the others. This may be an outlier, a
point which may well come from a different population. It is often difficult
to know what to do with such data. At least we have been warned to
double check this point for transcription errors. It is all too easy to
transpose adjoining digits when transferring data from one medium to
another. This may have been the case here, as an FEV1 of 4.53, rather
than the 5.43 recorded, would have been more in line with the rest of the
data. If this happened at the point of recording, there is not much we can
do about it. We could try to measure the subject again, or exclude him
and see whether this makes any difference. I think that, on the whole, we
should work with all the data unless there are very good reasons for not
doing so. I have retained this case here.
11.8 * Deviations from assumptions in

regression
Both the appropriateness of the method of least squares and the use of
the t distribution for confidence intervals and tests of significance depend
on the assumption that the residuals are from a Normal distribution with
uniform variance. This assumption is easily met, for the same reasons
that it is in the paired t test (§10.2). The removal of the variation due to X
tends to remove some of the variation between individuals, leaving the
measurement error. Problems can arise, however, and it is always a good
idea to plot the original scatter diagram and the residuals to check that
there are no gross departures from the assumptions of the method. Not
only does this help preserve the validity of the statistical method used,
but it may also help us learn more about the structure of the data.
Figure 11.10 shows the relationship between gestational age and cord
blood levels of AVP, the antidiuretic hormone, in a sample of male
foetuses. The variability of the outcome variable AVP depends on the
actual value of the variable, being larger for large values of AVP. The
assumptions of the method of least squares do not apply. However, we
can use a transformation as we did for the comparison of means in
§10.4. Figure 11.10 also shows the data after AVP has been log
transformed, together with the least squares line.
As in §10.4, the transformation is found by trial and error. The log

transformation enables us to interpret the regression coefficient in a way
which other transformations do not. I used logs to base 10 for this
transformation and got the following regression equation:
log10(AVP) = -0.651 253 + 0.011 771 × gestational age
This means that for every one day increase in gestational age,
log10(AVP) increases by 0.011771. Adding 0.011771 to log10(AVP)
multiplies AVP by 100.011771 = 1.027 the antilog of 0.011771. We can
antilog the confidence limits for the slope to give the confidence interval
for this factor.
It may be more convenient to report the increase per week or per month.
These would be factors of 100.011 771 × 7 = 1.209 or 100.011771 × 30 = 2.255
respectively. When the data are a random sample, it is often convenient
to quote the slope calculated from logs as the effect of a difference of one
standard deviation of the predictor. For gestational age the standard
deviation is 61.16104 days, so the effect of a change of one SD is to
multiple AVP by 100.011771×61.16104 = 5.247, so a difference of one
standard deviation is associated with a fivefold increase in AVP. Another
approach is to look at the difference between two centiles, such as the
10th and the 90th. For gestational age these are 98 and 273 days, so the
effect on AVP would be to multiply it by 100.011771 × (273–98) = 114.796.
Thus the difference over this intercentile range is to raise AVP 115-fold.
11.9 Correlation
The regression method tells us something about the nature of the
relationship between two variables, how one changes with the other, but
it does not tell us how close that relationship is. To do this we need a
different coefficient, the correlation coefficient. The correlation coefficient
is based on the sum of products about the mean of the two variables, so I
shall start by considering the properties of the sum of products and why it
is a good indicator of the closeness of the relationship.
Figure 11.11 shows the scatter diagram of Figure 11.1 with two new axes
drawn through the mean point. The distances of the points from these
axes represent the deviations from the mean. In the top right section of
Figure 11.11, the deviations from the mean of both variables, FEV1 and
height, are positive. Hence, their products will be positive. In the bottom
left section, the deviations from the mean of the two variables will both be
negative. Again, their product will be positive. In the top left section of
Figure 11.11, the deviations of FEV1 from its mean will be positive, and
the deviation of height from its mean will be negative. The product of
these will be negative. In the bottom right section, the product will again
be negative. So in Figure 11.11 nearly all these products will be positive,
and their sum will be positive. We say that there is a positive correlation
between the two variables; as one increases so does the other. If one
variable decreased as the other increased, we would have a scatter
diagram where most of the points lay in the top left and bottom right
sections. In this
case the sum of the products would be negative and there would be a
negative correlation between the variables. When the two variables are
not related, we have a scatter diagram with roughly the same number of
points in each of the sections. In this case, there are as many positive as
negative products, and the sum is zero. There is zero correlation or no
correlation. The variables are said to be uncorrelated.
Fig. 11.11. Scatter diagram with axes through the mean point
The value of the sum of products depends on the units in which the two
variables are measured. We can find a dimensionless coefficient if we
divide the sum of products by the square roots of the sums of squares of
X and Y. This gives us the product moment correlation coefficient, or
the correlation coefficient for short, usually denoted by r.
If the n pairs of observations are denoted by (xi,yi), then r is given by
For the FEV1 and height we have
The effect of dividing the sum of products by the root sum of squares of
deviations of each variable is to make the correlation coefficient lie
between -1.0 and +1.0. When all the points lie exactly on a straight line
such that Y increases as X increases, r = 1. This can be shown by putting
a + bxi in place of yi in the equation for r; everything cancels out leaving r
= 1. When all the points lie exactly on a straight line with negative slope, r
= -1. When there is no relationship at all, r = 0, because the sum of
products is zero. The correlation coefficient describes the closeness of
the linear relationship between two variables. It does not matter which
variable we take to be Y and which to be X. There is no choice of
predictor and outcome variable, as there is in regression.
Fig. 11.12. Data where the correlation coefficient may be misleading
The correlation coefficient measures how close the points are to a

straight line. Even if there is a perfect mathematical relationship between
X and Y, the correlation coefficient will not be exactly 1 unless this is of
the form y = a + bx. For example, Figure 11.12 shows two variables
which are perfectly related but have r = 0.86. Figure 11.12 also shows
two variables which are clearly related but have zero correlation, because
the relationship is not linear. This shows again the importance of plotting
the data and not relying on summary statistics such as the correlation
coefficient only. In practice, relationships like those of Figures 11.12 are
rare in medical data, although the possibility is always there. More often,
there is so much random variation that it is not easy to discern any
relationship at all.
The correlation coefficient r is related to the regression coefficient b in a
simple way. If Y = a + bX is the regression of y on X, and X = a′ + b′Y is
the regression of X on Y, then r2 = bb′. This arises from the formulae for r
and b. For the FEV1 data, b = 0.074389 and b′ = 4.5424, so bb′ =
0.074389 × 4.5424 = 0.33790, the square root of which is 0.581 29, the
correlation coefficient. We also have
This is the proportion of variability explained, described in §11.5.
Table 11.2. Two-sided 5% and 1% points of the

distribution of the correlation coefficient, r, under the
null hypothesis
n 5% 1% n 5% 1% n 5%
3 1.00 1.00 16 0.50 0.62 29 0.37
4 0.95 0.99 17 0.48 0.61 30 0.36
5 0.88 0.96 18 0.47 0.59 40 0.31
6 0.81 0.92 19 0.46 0.58 50 0.28
7 0.75 0.87 20 0.44 0.56 60 0.25
8 0.71 0.83 21 0.43 0.55 70 0.24
9 0.67 0.80 22 0.42 0.54 80 0.22
10 0.63 0.77 23 0.41 0.53 90 0.21
11 0.60 0.74 24 0.40 0.52 100 0.20

12 0.58 0.71 25 0.40 0.51 200 0.14
13 0.55 0.68 26 0.39 0.50 500 0.09
14 0.53 0.66 27 0.38 0.49 1000 0.06
15 0.51 0.64 28 0.37 0.48
n = Number of observations.
11.10 Significance test and confidence interval

for r
Testing the null hypothesis that r = 0 in the population, i.e. that there is no
linear relationship, is simple. The test is numerically equivalent to testing
the null hypothesis that b = 0, and the test is valid provided at least one
of the variables is from a Normal distribution. This condition is effectively
the same as that for testing b, where the residuals in the Y direction must
be Normal, If b = 0, the residuals in the Y direction are simply the
deviations from the mean, and these will only be Normally distributed if Y
is. If the condition is not met, we can use a transformation (§11.8), or one
of the rank correlation methods (§12.4-5).
Because the correlation coefficient does not depend on the means or
variances of the observations, the distribution of the sample correlation
coefficient when the population coefficient is zero is easy to tabulate.
Table 11.2 shows the correlation coefficient at the 5% and 1% level of
significance. For the example we have r = 0.58 from 20 observations.
The 1% point for 20 observations is 0.56, so we have P < 0.01, and the
correlation is unlikely to have arisen if there were no linear relationship in
the population. Note that the values of r which can arise by chance with
small samples are quite high. With 10 points r would have to be greater
than 0.63 to be significant. On the other hand with 1 000 points very
small values of r, as low as 0.06, will be significant.
Finding a confidence interval for the correlation coefficient is more
difficult.
Even when X and Y are both Normally distributed, r does not itself
approach a Normal distribution until the sample size is in the thousands.
Furthermore, its distribution is rather sensitive to deviations from the
Normal in X and Y. However, if both variables are from Normal
distributions, Fisher's z transformation gives a Normally distributed
variable whose mean and variance are known in terms of the population
correlation coefficient which we wish to estimate. From this a confidence
interval can be found. Fisher's z transformation is
which follows a Normal distribution with mean
so for the lower limit we have
and for the upper limit

and the 95% confidence interval is 0.18 to 0.81. This is very wide,
reflecting the sampling variation which the correlation coefficient has for
small samples. Correlation coefficients must be treated with some
caution when derived from small samples.
The ease of the significance test compared to the relative complexity of
the confidence interval calculation has meant that in the past a
significance test was usually given for the correlation coefficient. The
increasing availability of computers with well-written statistical packages
should lead to correlation coefficients appearing with confidence intervals
in the future.
Table 11.3. Simulated data showing 10 pairs of measure

of two independent variables for four subjects
Subject 1 Subject 2 Subject 3
x y x y x y
47 51 49 52 51 46
46 53 50 56 46 48
50 57 42 46 46 47
52 54 48 52 45 55
46 55 60 53 52 49
36 53 47 49 54 61
47 54 51 52 48 53
46 57 57 50 47 48
36 61 49 50 47 50
44 57 49 49 54 44
Means 45.0 55.2 50.2 50.9 49.0 50.1
r = -0.33 r = 0.49 r = 0.06
P = 0.35 P = 0.15 P = 0.86
11.11 Uses of the correlation coefficient

The correlation coefficient has several uses. Using Table 11.2, it provides
a simple test of the null hypothesis that the variables are not linearly
related, with less calculation than the regression method. It is also useful
as a summary statistic for the strength of relationship between two
variables. This is of great value when we are considering the
interrelationships between a large number of variables. We can set up a
square array of the correlations of each pair of variables, called the
correlation matrix. Examination of the correlation matrix can be very
instructive, but we must bear in mind the possibility of non-linear
relationships. There is no substitute for plotting the data. The correlation
matrix also provides the starting point for a number of methods for
dealing with a large number of variables simultaneously.
Of course, for the reasons discussed in Chapter 3, the fact that two
variables are correlated does not mean that one causes the other.
11.12 * Using repeated observations

In clinical research we are often able to take several measurements on
the same patient. We may want to investigate the relationship between
two variables, and take pairs of readings with several pairs from each of
several patients. The analysis of such data is quite complex. This is
because the variability of measurements made on different subjects is
usually much greater than the variability between measurements on the
same subject, and we must take these two kinds of variability into
account. What we must not do is to put all the data together, as if they
were one sample.
Consider the simulated data of Table 11.3. The data were generated from
random numbers, and there is no relationship between X and Y at all.
First values of X and Y were generated for each ‘subject’, then a further
random number was added to make the individual ‘observation’. For each
subject separately,
there was no significant correlation between X and Y. For the subject

means, the correlation coefficient was r = 0.77, P = 0.23. However, if we
put all 40 observations together we get r = 0.53, P = 0.0004. Even though
the coefficient is smaller than that between subject means, because it is
based on 40 pairs of observations rather than 4 it becomes significant.
The data are plotted in Figure 11.13, with three other simulations. As the
null hypothesis is always true in these simulated data, the population
correlations for each ‘subject’ and for the means are zero. Because the
numbers of observations are small, the sample correlations vary greatly.
As Table 11.2 shows, large correlation coefficients can arise by chance in
small samples. However, the overall correlation is ‘significant’ in three of
the four simulations, though in different directions.
Fig. 11.13. Simulations of 10 pairs of observations on four subjects
We only have four subjects and only four points. By using the repeated
data, we are not increasing the number of subjects, but the statistical
calculation is done as if we have, and so the number of degrees of
freedom for the significance test is incorrectly increased and a spurious
significant correlation produced.
There are two simple ways to approach this type of data, and which is
chosen depends on the question being asked. If we want to know
whether subjects with a high value of X tend to have a high value of Y
also, we use the subject means and find the correlation between them. If
we have different numbers of observations for each subject, we can use
a weighted analysis, weighted by the number of observations for the
subject. If we want to know whether changes in one variable in the same
subject are parallelled by changes in the other, we need to use multiple
regression, taking subjects out as a factor (§17.1, §17.6). In either
case, we should not mix observations from different subjects

indiscriminately.
Fig. 11.14. Scatter plots of the 30 second pulse data as in Table
10.15 and with half the pairs of observations reversed
11.13 * Intraclass correlation

Sometimes we have pairs of observations where there is no obvious
choice of X and Y. The data of Table 10.15 are a good example. Each
subject has two measurements made by different observers, different
pairs of observers being used for each subject. The choice of X and Y is
arbitrary Figure 11.14 shows the data as in Table 10.15 and with half the
pairs arbitrarily reversed. The scatter plots look a little different and there
is no good reason to choose one against the other. The correlation
coefficients are a little different too: for the original order r = 0.584 8 and
for the second order r = 0.580 4. These are very similar, of course, but
which should we use?
It would be nice to have an average correlation coefficient across all the
245 possible orderings. This is provided by the intraclass correlation
coefficient or ICC. This can be found from the estimates of within
subject variance, s2w, and between subjects variance, s2b, found from the
analysis of variance in §10.12. We have:
For the example, s2w = 14.37 and s2b = 20.19 (§10.12). hence
The ICC was originally developed for applications such as correlation

between variables measured in pairs of twins (which twin is X and which
is Y?). We do not have to have pairs of measurements to use the ICC. It
works just as well for triplets or for any number of observations within the
groups, not necessarily all the same.
Although not used nearly as often as the product moment correlation
coefficient, the ICC has some important applications. One is in the study
of measurement error and observer variation (§15.2), where if
measurements are true
replicates the order in which they were made is not important. Another is
in the design of cluster-randomized trials where the group is the cluster
and may have hundreds of observations within it (§18.8).
Appendices
11A Appendix: The least squares
estimates
This section requires knowledge of calculus. We want to find a and b so

that the sum of squares about the line y = a + bx is a minimum. We
therefore want to minimize Σ(yi - a - bxi)2. This will have a minimum when
the partial differentials with respect to a and b are both zero.
Subtracting this from the second equation we get
This gives us
11B Appendix: Variance about the

regression line
11C Appendix: The standard error of b
To find the standard error of b, we must bear in mind that in our

regression model all the random variation is in Y. We first rewrite the sum
of products:
The variance of a constant times a random variable is the square of the

constant times the variance of the random variable (§6.6). The xi are
constants, not random variables, so
VAR(yi) is the same for all yi, say VAR(yi) = s2. Hence
The standard error of b is the square root of this.

57. In Figure 11.15(a):
(a) predictor and outcome are independent;
(b) predictor and outcome are uncorrelated;
(c) the correlation between predictor and outcome is less than 1;
(d) predictor and outcome are perfectly related;
(e) the relationship is best estimated by simple linear regression.
View Answer
58. In Figure 11.15(b):

(a) predictor and outcome are independent random variables;
(b) the correlation between predictor and outcome is close to zero;
(c) outcome increases as predictor increases;
(d) predictor and outcome are linearly related;
(e) the relationship could be made linear by a logarithmic
transformation of the outcome.
View Answer
Fig. 11.15. Scatter diagrams
59. A simple linear regression equation:

(a) describes a line which goes through the origin;
(b) describes a line with zero slope;
(c) is not affected by changes of scale;
(d) describes a line which goes through the mean point;
(e) is affected by the choice of dependent variable.
View Answer
60. If the t distribution is used to find a confidence interval for the

slope of a regression line:
(a) deviations from the line in the independent variable must follow
a Normal distribution;
(b) deviations from the line in the dependent variable must follow a
Normal distribution;
(c) the variance about the line is assumed to be the same
throughout the range of the predictor variable;
(d) the y variable must be log transformed;
(e) all the points must lie on the line.
View Answer
61. The product moment correlation coefficient, r:

(a) must lie between -1 and +1;
(b) can only have a valid significance test carried out when at least
one of the variables is from a Normal distribution;
(c) is 0.5 when there is no relationship;
(d) depends on the choice of dependent variable;
(e) measures the magnitude of the change in one variable
associated with a change in the other.
View Answer
11E Exercise: Comparing two regression lines

Table 11.4 and Figure 11.16 show the PEFR and heights of samples of
male and female medical students. Table 11.5 shows the sums of
squares and products for these data.
1. Estimate the slopes of the regression lines for females and males.
View Answer
2. Estimate the standard errors of the slopes.

View Answer
3. Find the standard error for the difference between the slopes,
which are independent. Calculate a 95% confidence interval for the
difference.
View Answer
4. Use the standard error to test the null hypothesis that the slopes
are the same in the population from which these data come.
View Answer
Fig. 11.16. PEFR and height for female and male medical students
Table 11.4. Height and PEFR in a sample of me
Females
Ht PEFR Ht PEFR Ht PEFR Ht PEFR
155 450 163 428 168 480 164 540
155 475 163 548 168 595 167 470

155 503 164 485 169 510 167 530
158 440 165 485 170 455 167 598
160 360 166 430 171 430 168 510
161 383 166 440 171 537 168 560
161 461 166 485 172 442 170 510
161 470 166 510 172 463 170 547
161 470 167 415 172 490 170 553
161 475 167 455 174 540 170 560
161 480 167 470 174 540 171 460
162 450 167 500 176 535 171 473
162 475 168 430 177 513 171 550
162 550 168 440 181 522 171 575
163 370 172 480
172 550
172 620
174 550
174 550
174 616
Table 11.5. Summary statistics for height and

PEFR in a sample of medical students
Females Males
Number 43 58
Sum of squares, height 1469.9 2292.0
Sum of squares, PEFR 101124.8 226994.1
Sum of products about 4220.1 9048.2

mean
> Table of Contents > 12 - Methods based on rank order
12
Methods based on rank order
12.1 *Non-parametric methods

In Chapters 10 and 11 I described a number of methods of analysis
which relied on the assumption that the data came from a Normal
distribution. To be more precise, we could say the data come from one of
the Normal family of distributions, the particular Normal distribution
involved being defined by its mean and standard deviation, the
parameters of the distribution. These methods are called parametric
because we estimate the parameters of the underlying Normal
distribution. Methods which do not assume a particular family of
distributions for the data are said to be non-parametric. In this and the
next chapter I shall consider some non-parametric tests of significance.
There are many others, but these will illustrate the general principle. We
have already met one non-parametric test, the sign test (§9.2). The large
sample Normal test could also be regarded as non-parametric.
It is useful to distinguish between three types of measurements scales.
On an interval scale, the size of the difference between two values on
the scale has a consistent meaning. For example, the difference in
temperature between 1°C and 2°C is the same as the difference between
31°C and 32°C. On an ordinal scale, observations are ordered, but
differences may not have a meaning. For example, anxiety is often
measured using sets of questions, the number of positive answers giving
the anxiety scale. A set of 36 questions would give a scale from 0 to 36.
The difference in anxiety between scores of 1 and 2 is not necessarily the
same as the difference between scores 31 and 32. On a nominal scale,
we have a qualitative or categorical variable, where individuals are
grouped but not necessarily ordered. Eye colour is a good example.
When categories are ordered, we can treat the scale as either ordered or
nominal, as appropriate.
All the methods of Chapters 10 and 11 apply to interval data, being based
on differences of observations from the mean. Most of the methods in
this chapter apply to ordinal data. Any interval scale which does not meet
the requirements of Chapters 10 and 11 may be treated as ordinal, since
it is, of course, ordered. This is the more common application in medical
work.
General texts such as Armitage and Berry (1994), Snedecor and
Cochran (1980) and Colton (1974) tend not to go into a lot of detail about
rank and related methods, and more specialized books are needed
(Siegel 1956, Conover 1980).
12.2 *The Mann-Whitney U test

This is the non-parametric analogue of the two-sample t test (§10.3). It
works like this. Consider the following artificial data showing observations
of a variable in two independent groups, A and B:
A 7 4 9 17
B 11 6 21 14
We want to know whether there is any evidence that A and B are drawn
from populations with different levels of the variable. The null hypothesis
is that there is no tendency for members of one population to exceed
members of the other. The alternative is that there is such a tendency, in
one direction or the other. First we arrange the observations in ascending
order, i.e. we rank them:
4 6 7 9 11 14 17 21
A B A A B B A B
We now choose one group, say A. For each A, we count how many Bs
precede it. For the first A, 4, no Bs precede. For the second, 7, one B
precedes, for the third A, 9, one B, for the fourth, 17, three Bs. We add
these numbers of preceding Bs together to give U = 0 + 1 + 1 + 3 = 5.
Now, if U is very small, nearly all the As are less than nearly all the Bs. If
U is large, nearly all As are greater than nearly all Bs. Moderate values of
U mean that As and Bs are mixed. The minimum U is 0, when all Bs
exceed all As, and maximum U is n1 × n2 when all As exceed all Bs. The
magnitude of U has a meaning, because U/n1n2 is an estimate of the
probability that an observation drawn at random from population A would
exceed an observation drawn at random from population B.
There is another possible U, which we will call U′, obtained by counting
the number of As before each B, rather than the number of Bs before
each A. This would be 1 + 3 + 3 + 4 = 11. The two possible values of U
and U′ are related by U + U′ = n1n2. So we subtract U′ from n1n2 to give 4
× 4 - 11 = 5.
If we know the distribution of U under the null hypothesis that the
samples come from the same population, we can say with what
probability these data could have arisen if there were no difference. We
can carry out the test of significance. The distribution of U under the null
hypothesis can be found easily. The two sets of four observations can be
arranged in 70 different ways, from AAAABBBB to BBBBAAAA (8!/4!4! =
70, §6A). Under the null hypothesis these arrangements are all equally
likely and, hence, have probability 1/70. Each has its value of U, from 0 to
16, and by counting the number of arrangements which give each value
of U we can find the probability of that value. For example, U = 0 only
arises from the order AAAABBBB and so has probability 1/70 = 0.014. U
= 1 only arises from AAABABBB and so has probability 1/70 = 0.014
also. U = 2 can arise in two ways: AAABBABB and AABAABBB. It has
probability 2/70 = 0.029. The full set of probabilities is shown in Table
12.1.
We apply this to the example. For groups A and B, U = 5 and the
probability of this is 0.071. As we did for the sign test (§9.2) we consider
the probability of more extreme values of U, U = 5 or less, which is 0.071
+ 0.071 + 0.043 + 0.029 + 0.014 + 0.014 = 0.242.
This gives a one sided test. For a two-sided test, we must consider the
probabilities of a difference as extreme in the opposite direction. We can
see from Table 12.1 that the distribution of U is symmetrical, so the
probability of an equally extreme value in the opposite direction is also
0.242, hence the two-sided probability is 0.242 + 0.242 = 0.484. Thus the
observed difference would have been quite probable if the null
hypothesis were true and the two samples could have come from the
same population.
Table 12.1. Distribution of the Mann-Whitney U

statistic, for two samples of size 4
U Probability U Probability U Probability
0 0.014 6 0.100 12 0.071
1 0.014 7 0.100 13 0.043
2 0.029 8 0.114 14 0.029
3 0.043 9 0.100 15 0.014
4 0.071 10 0.100 16 0.014
5 0.071 11 0.071
Table 12.2. Two-sided 5% points for the distribution of t

n1
2 3 4 5 6 7 8 9 10
2 - - - - - - 0 0 0
3 - - - 0 1 1 2 2 3
4 - - 0 1 2 3 4 4 5
5 - 0 1 2 3 5 6 7 8
6 - 1 2 3 5 6 8 10 11
7 - 1 3 5 6 8 10 12 14
8 0 2 4 6 8 10 13 15 17
9 0 2 4 7 10 12 15 17 20
10 0 3 5 8 11 14 17 20 23
11 0 3 6 9 13 16 19 23 26
12 1 4 7 11 14 18 22 26 29
13 1 4 8 12 16 20 24 28 33
14 1 5 9 13 17 22 26 31 36
15 1 5 10 14 19 24 29 34 39
16 1 6 11 15 21 26 31 37 42
17 2 6 11 17 22 28 34 39 45
18 2 7 12 18 24 30 36 42 48
19 2 7 13 19 25 32 38 45 52
20 2 8 13 20 27 34 41 48 55
If U is less than or equal to the tabulated value the difference
In practice, there is no need to carry out the summation of probabilities

described above, as these are already tabulated. Table 12.2 shows the
5% points of U for each combination of sample sizes n1 and n2 up to 20.
For our groups A and B, U = 5. we find the n2 = 4 column and the n1 = 4
row. From this we see that the 5% point for U is 0, and so U = 5 is not
significant. If we had calculated the larger of the two values of U, 11, we
can use Table 12.2 by finding the lower value, n1n2 - U = 16 - 11 = 5.
Table 12.3. Biceps skinfold thickness (mm) in
two groups of patients
Crohn's Disease Coeliac Disease
1.8 2.8 4.2 6.2 1.8 3.8
2.2 3.2 4.4 6.6 2.0 4.2
2.4 3.6 4.8 7.0 2.0 5.4
2.5 3.8 5.6 10.0 2.0 7.6
2.8 4.0 6.0 10.4 3.0
We can now turn to the practical analysis of some real data. Consider the
biceps skinfold thickness data of Table 10.4, reproduced as Table 12.3.
We will analyse these using the Mann-Whitney U test. Denote the
Crohn's disease group by A and the coeliac group by B. The joint order is
as follows:
Let us count the As before each B. Immediately we have a problem. The

first A and the first B have the same value. Does the first A come before
the first B or after it? We resolve this dilemma by counting one half for the
tied A. The ties between the second, third and fourth Bs do not matter, as
we can count the number of As before each without difficulty. We have for
the U statistic:
U = 0.5 + 1 + 1 + 1 + 6 + 8.5 + 10.5 + 13 + 18 = 59.5
This is the lower value, since n1n2 = 9 × 20 = 180 and so the middle
value is 90. We can therefore refer U to Table 12.2. The critical value at
the 5% level for groups size 9 and 20 is 48, which our value exceeds.
Hence the difference is not significant at the 5% level and the data are
consistent with the null hypothesis that there is no tendency for members
of one population to exceed members of the other. This is the same as
the result of the t test of §10.4.
For larger values of n1 and n2 calculation of U can be rather tedious. A
simple formula for U can be found using the ranks. The rank of the lowest
observation is 1, of the next is 2, and so on. If a number of observations
are tied, each having the same value and hence the same rank, we give
each the average of the ranks they would have were they ordered. For
example, in the skinfold data the first two observations are each 1.8.
They each receive rank (1 + 2)/2 = 1.5. The third, fourth and fifth are tied
at 2.0, giving each of them rank (3 + 4 + 5)/3 = 4. The sixth, 2.2, is not
tied and so has rank 6. The ranks for the skinfold data are as follows:
skinfold 1.8 1.8 2.0 2.0 2.0 2.2 2.4 2.5 2.8 2.8
group A B B B B A A A A A
rank 1.5 1.5 4 4 4 6 7 8 9.5 9.5
r1 r2 r3 r4
skinfold 3.0 3.2 3.6 3.8 3.8 4.0 4.2 4.2 4.4 4.8
group B A A A B A A B A A
rank 11 12 13 14.5 14.5 16 17.5 17.5 19 20
r5 r6 r7
skinfold 5.4 5.6 6.0 6.2 6.6 7.0 7.6 10.0 10.4
group B A A A A A B A A
rank 21 22 23 24 25 26 27 28 29
r8 r9
We denote the ranks of the B group by r1,r2,…,rn1. The number of As
preceding the first B must be r1 - 1, since there are no Bs before it and it
is the r1th observation. The number of As preceding the second B is r2 -
2, since it is the r2th observation, and one preceding observation is a B.
Similarly, the number preceding the third B is r3 - 3, and the number
preceding the ith B is ri - i. Hence we have:
That is, we add together the ranks of all the n1 observations, subtract
n1(n1 + 1)/2 and we have U. For the example, we have
as before. This formula is sometimes written

But this is simply based on the other group, since U + U′ = n1n2. For
testing we use the smaller value, as before.
is an observation from a Standard Normal distribution. For the example,

n1 = 9 and n2 = 20. we have
From Table 7.1 this gives two-sided probability = 0.15, similar to that
found by the two sample t test (§10.3).
Neither Table 12.2 nor the above formula for the standard deviation of U
take ties into account; both assume the data can be fully ranked. Their
use for data with ties is an approximation. For small samples we must
accept this. For the Normal approximation, ties can be allowed for using
the following formula for the standard deviation of U when the null
hypothesis is true:
The Mann-Whitney U test is a non-parametric analogue of the two
sample t test. The advantage over the t test is that the only assumption
about the distribution of the data is that the observations can be ranked,
whereas for the t test we must assume the data are from Normal
distributions with uniform variance. There are disadvantages. For data
which are Normally distributed, the U test is less powerful than the t test,
i.e. the t test, when valid, can detect
smaller differences for given sample size. The U test is almost as

powerful for moderate and large sample sizes, and this difference is
important only for small samples. For very small samples, e.g. two
groups of three observations, the test is useless as all possible values of
U have probabilities above 0.05 (Table 12.2). The U test is primarily a test
of significance. The t method also enables us to estimate the size of the
difference and gives a confidence interval. Although as noted above
U/n1n2 has an interpretation, we cannot, so far as I know, find a
confidence interval for it.
Table 12.4. Frequency distributions of number

of nodes involved in breast cancers detected
at screening and detected in the intervals
between screens (data of Mohammed Raja)
Screening cancers Interval cancers

Nodes Freqency Nodes Frequency
0 291 0 66
1 43 1 22
2 16 2 7
3 20 3 7
4 13 4 2
5 3 5 4
6 1 6 4
7 4 7 3
8 3 8 3
9 1 9 2
10 1 10 2
11 2 12 2
12 1 13 1
15 1 15 1
16 1 16 1
17 2 20 1
18 2
20 1
27 1
33 1
Total 408 128
Mean 1.21 2.19
Median 0 0
75%ile 1 3
The null hypothesis of the Mann–Whitney test is sometimes presented as

being that the populations have the same median. There is even a
confidence interval for the difference between two medians based on the
Mann–Whitney test (Campbell and Gardner 1989). This is surprising, as
the medians are not involved in the calculation. Furthermore, we can
have two groups which are significantly different using the Mann–Whitney
U test yet have the same median. Table 12.4
shows an example. The majority of observations in both groups are zero,
so transformation to the Normal is impossible. Although the samples are
quite large, the distribution is so skew that a rank method, appropriately
adjusted for ties, may be safer than the method of §9.7. The Mann–
Whitney U test was highly significant, yet the medians are both zero. As
the medians were equal, I suggested the 75th percentile as a measure of
location for the distributions.
The reason for these two different views of the Mann–Whitney U test lies
in the assumptions we make about the distributions in the two
populations. If we make no assumptions, we can test the null hypothesis:
that the probability that a member of the first population drawn at random
will exceed a member of the second population drawn at random is one
half. Some people choose to make an assumption about the distributions:
that they have the same shape and differ only in location (mean or
median). If this assumption is true, then if the distributions are different
the medians must be different. The means must differ by the same
amount. It is a very strong assumption. For example, if it is true then the
variances must be the same in the two populations. For the reasons
given in §10.5 and §7A, it is unlikely that we could get this if the
distributions were not Normal. Under this assumption the Mann–Whitney
U test will rarely be valid if the two sample t test is not valid also.
There are other non-parametric tests which test the same or similar null
hypotheses. Two of these, the Wilcoxon two sample test and the Kendall
Tau test, are different versions of the Mann–Whitney U test which were
developed around the same time and later shown to be identical. These
names are sometimes used interchangeably. The test statistics and
tables are not the same, and the user must be very careful that the
calculation of the test statistic being used corresponds to the table to
which it is referred. Another difficulty with tables is that some are drawn
so that for a significant difference U must be less than or equal to the
tabulated value (as in Table 12.2), for others U must be strictly less than
the tabulated value.
For more than two groups, the rank analogue of one-way analysis of
variance (§10.9) is the Kruskal–Wallis test, see Conover (1980) and
Siegel (1956). Conover (1980) also describes a multiple comparison test
for the pairs of groups, similar to those described in §10.11.
12.3 *The Wilcoxon matched pairs test
This test is an analogue of the paired t test. We have a sample measured
under two conditions and the null hypothesis is that there is no tendency
for the outcome on one condition to be higher or lower than the other.
The alternative hypothesis is that the outcome on one condition tends to
be higher or lower than the other. As the test is based on the magnitude
of the differences, the data must be interval.
Consider the data of Table 12.5, previously discussed in §2.6 and §9.2,
where we used the sign test for the analysis. In the sign test, we have
ignored the magnitude of differences, and only considered their signs. If
we can use information
about the magnitude, we would hope to have a more powerful test.

Clearly, we must have interval data to do this. To avoid making
assumptions about the distribution of the differences, we use their rank
order in a similar manner to the Mann–Whitney U test.
Table 12.5. Results of a trial of pronethalol for the preve

of angina pectoris (Pritchard et al. 1963), in rank orde
differences
Number of attacks Difference Rank of differenc

while on placebo –
Placebo Pronethalol pronethalol All Positive
2 0 2 1.5 1.5
17 15 2 1.5 1.5
3 0 3 3 3
7 2 5 4 4
8 1 7 6 6
14 7 7 6 6
23 16 7 6 6
34 25 9 8 8
79 65 14 9 9
60 41 19 10 10
323 348 -25 11
71 29 42 12 12
Sum of 67
ranks
First, we rank the differences by their absolute values, i.e. ignoring the
sign. As in §12.2, tied observations are given the average of their ranks.
We now sum the ranks of the positive differences, 67, and the ranks of
the negative differences, 11 (Table 12.5). If the null hypothesis were true
and there was no difference, we would expect the rank sums for positive
and negative differences to be about the same, equal to 39 (their
average). The test statistic is the lesser of these sums, T. The smaller T
is, the lower the probability of the data arising by chance.
The distribution of T when the null hypothesis is true can be found by
enumerating all the possibilities, as described for the Mann–Whitney U
statistic. Table 12.6 gives the 5% and 1% points for this distribution, for
sample size n up to 25. For the example, n = 12 and so the difference
would be significant at the 5% level if T were less than or equal to 14. We
have T = 11, so the data are not consistent with the null hypothesis. The
data support the view that there is a real tendency for patients to have
fewer attacks while on the active treatment.
From Table 12.6, we can see that the probability that T ≤ 11 lies between
0.05 and 0.01. This is greater than the probability given by the sign test,
which was 0.006 (§9.2). Usually we would expect greater power, and
hence lower probabilities when the null hypothesis is false, when we use
more of the information. In this case, the greater probability reflects the
fact that the one negative difference, -25, is large. Examination of the
original data shows that this individual had very large numbers of attacks
on both treatments, and it seems possible that he may belong to a
different population from the other eleven.
Like Table 12.2, Table 12.6 is based on the assumption that the
differences can be fully ranked and there are no ties. Ties may occur in
two ways in this
test. Firstly, ties may occur in the ranking sense. In the example we had
two differences of +2 and three of +7. These were ranked equally: 1.5
and 1.5. and 6, 6 and 6. When ties are present between negative and
positive differences, Table 12.6 only approximates to the distribution of T.

distribution of T (lower value) in the Wilcoxon
one-sample test
Probability Probability
that T ≤ the that T ≤ the
Sample tabulated Sample tabulated
size n value size n value
5% 1% 5% 1%
5 - - 16 30 19
6 1 - 17 35 23
7 2 - 18 40 28
8 4 0 19 46 32
9 6 2 20 52 37
10 8 3 21 59 43
11 11 5 22 66 49
12 14 7 23 73 55
13 17 10 24 81 61
14 21 13 25 90 68
15 25 16
Ties may also occur between the paired observations, where the
observed difference is zero. In the same way as for the sign test, we omit
zero differences (§9.2). Table 12.6 is used with n as the number of non-
zero differences only, not the toal number of differences. This seems odd,
in that a lot of zero differences would appear to support the null
hypothesis. For example, if in Table 12.5 we had another dozen patients
with zero differences, the calculation and conclusion would be the same.
However, the mean difference would be smaller and the Wilcoxon test
tells us nothing about the size of the difference, only its existence. This
illustrates the danger of allowing significance tests to outweigh all other
ways of looking at the data.
is from a Standard Normal distribution if the null hypothesis is true. For

the example of Table 12.5, we have:
From Table 7.1 this gives a two-tailed probability of 0.028, similar to that
obtained from Table 12.6.
We have three possible tests for paired data, the Wilcoxon, sign and
paired t methods. If the differences are Normally distributed, the t test is
the most powerful test. The Wilcoxon test is almost as powerful, however,
and in practice the difference is not great except for small samples. Like
the Mann–Whitney U test, the Wilcoxon is useless for very small
samples. The sign test is similar in power to the Wilcoxon for very small
samples, but as the sample size increases the Wilcoxon test becomes
much more powerful. This might be expected since the Wilcoxon test
uses more of the information. The Wilcoxon test uses the magnitude of
the differences, and hence requires interval data. This means that, as for
t methods, we will get different results if we transform the data. For truly
ordinal data we should use the sign test. The paired t method also gives
a confidence interval for the difference. The Wilcoxon test is purely a test
of significance, but a confidence interval for the median difference can be
found using the Binomial method described in §8.9.
12.4 * Spearman's rank correlation coefficient, ρ

We noted in Chapter 11 the sensitivity to assumptions of Normality of the
product moment correlation coefficient, r. This led to the development of
non-parametric approaches based on ranks. Spearman's approach was
direct. First we rank the observations, then calculate the product moment
correlation of the ranks, rather than of the observations themselves. The
resulting statistic has a distribution which does not depend on the
distribution of the original variables. It is usually denoted by the Greek
letter ρ, pronounced ‘rho’, or by rs.
Table 12.7 shows data from a study of the geographical distribution of a
tumour, Kaposi's sarcoma, in mainland Tanzania. The incidence rates
were calculated from cancer registry data and there was considerable
doubt that all cases were notified. The degree of reporting of cases may
have been related to population density or availability of health services.
In addition, incidence was closely related to age and sex (where
recorded) and so could be related to the age and sex distribution in the
region. To check that none of these were producing artefacts in the
geographical distribution, I calculated the rank correlation of disease
incidence with each of the possible explanatory variables. Table 12.7
shows the relationship of incidence to the percentage of the population
living within 10km of a health centre. Figure 12.1 shows the scatter
diagram of these data. The percentage within 10km of a health centre is
very highly skewed, whereas the disease incidence appears somewhat
bimodal. The assumption of the product moment correlation do not
appear to be met, so rank correlation was preferred.
Table 12.7. Incidence of Kaposi's sarcoma and access

population to health centres for each region of mainla
Tanzania (Bland et al. 1977)
Percent Rank order

Incidence population
per within
Region
million 10km of Incidence
per year health
centre
Coast 1.28 4.0 1
Shinyanga 1.66 9.0 2
Mbeya 2.06 6.7 3
Tabora 2.37 1.8 4
Arusha 2.46 13.7 5
Dodoma 2.60 11.1 6

Kigoma 4.22 9.2 7
Mara 4.29 4.4 8
Tanga 4.54 23.0 9
Singida 6.17 10.8 10
Morogoro 6.33 11.7 11
Mtwara 6.40 14.8 12
Westlake 6.60 12.5 13
Kilimanjaro 6.65 57.3 14
Ruvuma 7.21 6.6 15
Iringa 8.46 2.6 16
Mwanza 8.54 20.7 17

Fig. 12.1. Incidence of Kaposi's sarcoma per million per year and
percentage of population within 10km of a health centre, for 17
regions of mainland Tanzania
The calculation of Spearman's ρ proceeds as follows. The ranks for the

two variables are found (Table 12.7). We apply the formula for the
product moment correlation (§11.9) to these ranks. We define:

distribution of Spearman's ρ
Probability that ρ is as far or further

Sample
from 0 than the tabulated value
size n
5% 1%
4 - -
5 1.00 -
6 0.89 1.00
7 0.82 0.96
8 0.79 0.93
9 0.70 0.83
10 0.68 0.81
We have ignored the problem of ties in the above. We treat observations
with the same value as described in §12.2. We give them the average of
the ranks they would have if they were separable and apply the rank
correlation formula as described above. In this case the distribution of
Table 12.8 is only approximate.
There are several ways of calculating this coefficient, resulting in
formulae which appear quite different, though they give the same result
(see Siegel 1956).
12.5 * Kendall's rank correlation coefficient, τ

Spearman's rank correlation is quite satisfactory for testing the null
hypothesis of no relationship, but is difficult to interpret as a
measurement of the strength of the relationship. Kendall developed a
different rank correlation coefficient. Kendall's τ, which has some
advantages over Spearman's. (The Greek letter τ is pronounced ‘tau’.) It
is rather more tedious to calculate than Spearman's, but in the computer
age this hardly matters. For each pair of subjects we
observe whether the subjects are ordered in the same way by the two
variables, a concordant pair, ordered in opposite ways, a discordant
pair, or equal for one of the variables and so not ordered at all, a tied
pair. Kendall's τ is the proportion of concordant pairs minus the
proportion of discordant pairs. τ will be one if the rankings are identical,
as all pairs will be ordered in the same way, and minus one if the
rankings are exactly opposite, as all pairs will be ordered in the opposite
way.
We shall denote the number of concordant pairs (ordered the same way)
by nc, the number of discordant pairs (ordered in opposite ways) by nd,
and the difference, nc - nd, by S. There are n(n - 1)/2 pairs altogether, so
When there are no ties, nc + nd = n(n - 1)/2.

The simplest way to calculate nc is to order the observations by one of
the variables, as in Table 12.7 which is ordered by disease incidence.
Now consider the second ranking (% population within 10 km of a health
centre). The first region, Coast, has 14 regions below it which have
greater rank, so the pairs formed by the first region and these will be in
the correct order. There are 2 regions below it which have lower rank, so
the pairs formed by the first region and these will be in the opposite order.
The second region, Shinyanga, has 10 regions below it with greater rank
and so contributes 10 further pairs in the correct order. Note that the pair
‘Coast and Shinyanga’ has already been counted. There are 5 pairs in
opposite order. The third region, Mbeya, has 10 regions below it in the
same order and 4 in opposite orders, and so on. We add these numbers
to get nc and nd:
nc = 14 + 10 + 10 + 13 + 4 + 6 + 7 + 8 + 1 + 5 + 4 + 2 + 2 + 0 + 1 + 1 + 0
= 88
nd = 2 + 5 + 4 + 0 + 8 + 5 + 3 + 1 + 7 + 2 + 2 + 3 + 2 + 3 + 1 + 0 + 0 = 48
The number of pairs is n(n - 1)/2 = 17 × 16/2 = 136. Because there are
no ties, we could also calculate nd by nd = n(n - 1)/2 - nc = 136 - 88 = 48.
S = nc - nd = 88 - 48 = 40. Hence τ = S/ (n(n - 1)/2) = 40/136 = 0.29.
When there are ties, τ cannot be one. However, we could have perfect
correlation if the ties were between the same subjects for both variables.
To allow for this, we use a different version of τ, τb. Consider the
denominator. There are n(n - 1)/2 possible pairs. If there are t individuals
tied at a particular rank for variable X, no pairs from these t individuals
contribute to S. There are t(t - 1)/2 such pairs. If we consider all the
groups of tied individuals we have Σ t(t - 1)/2 pairs which do not
contribute to S, summing over all groups of tied ranks. Hence the total
number of pairs which can contribute to S is n(n - 1) - Σ t(t - 1)/2, and S
cannot be greater than n(n - 1)/2 - Σ t(t - 1)/2. The size of S is also limited
by ties in the second ranking. If we denote the number of individuals
with the same value of Y by u, then the number of pairs which can
contribute to S is n(n - 1)/2 - Σ u(u - 1)/2. We now define τb by
Note that if there are no ties, Σt (t -1)/2 = 0 = Σ. When the rankings are
identical τb = 1, no matter how many ties there are. Kendall (1970) also
discusses two other ways of dealing with ties, obtaining coefficients τa
and τc, but their use is restricted.
We often want to test the null hypothesis that there is no relationship
between the two variables in the population from which our sample was
drawn. As usual, we are concerned with the probability of S being as or
more extreme (i.e. far from zero) than the observed value. Table 12.9
was calculated in the same way as Tables 12.1 and 12.2. It shows the
probability of being as extreme as the observed value of S for n up to 10.
For convenience, S is tabulated rather than τ. When ties are present this
is only an approximation.
When the sample size is greater than 10, S has an approximately Normal
distribution under the null hypothesis, with mean zero. If there are no ties,
the variance is
When there are ties, the variance formula is very complicated (Kendall
1970). I shall omit it, as in practice these calculations will be done using
computers anyway. If there are not many ties it will not make much
difference if the simple form is used.
For the example, S = 40, n = 17 and there are no ties, so the Standard
Normal variate is
From Table 7.1 of the Normal distribution we find that the two-sided
probability of a value as extreme as this is 0.06 × 2 = 0.12, which is very
similar to that found using Spearman's ρ. The product moment
correlation, r, gives r = 0.30, P = 0.24, but of course the non-Normal
distributions of the variables make this P invalid.
Why have two different rank correlation coefficients? Spearman's ρ is
older than Kendall's τ, and can be thought of as a simple analogue of the
product moment correlation coefficient, Pearson's r. τ is a part of a more
general and consistent system of ranking methods, and has a direct
interpretation, as the difference between the proportions of concordant
and discordant pairs. In general,
the numerical value of ρ is greater than that of τ. It is not possible to

calculate τ from ρ or ρ from τ, they measure different sorts of correlation.
ρ gives more weight to reversals of order when data are far apart in rank
than when there is a reversal close together in rank, τ does not. However
in terms of tests of significance both have the same power to reject a
false null hypothesis, so for this purpose it does not matter which is used.

distribution of S for Kendall's τ
Probability that S is as far or further

Sample from the expected than the
size n tabulated value
5% 1%
4 - -
5 10 -
6 13 15
7 15 19
8 18 22
9 20 26
10 23 29
12.6 * Continuity corrections

In this chapter, when samples were large we have used a continuous
distribution, the Normal, to approximate to a discrete distribution. U, T or
S. For example, Figure 12.2 shows the distribution of the Mann—Whitney
U statistic for n1 = 4, n2 = 4 (Table 12.1) with the corresponding Normal
curve. From the exact distribution, the probability that U < 2 is 0.014 +
0.014 + 0.029 = 0.057. The corresponding Standard Normal deviate is
This has a probability of 0.048, interpolating in Table 7.1. This is smaller

than the exact probability. The disparity arises because the continuous
distribution gives probability to values other than the integers 0, 1, 2, etc.
The estimated probability for U = 2 can be found by the area under the
curve between U = 1.5 and U = 2.5. The corresponding Normal deviates
are -1.876 and -1.588, which have probabilities from Table 7.1 of 0.030
and 0.056. This gives the estimated probability for U = 2 to be 0.056 -
0.030 = 0.026, which compares quite well with the exact figure of 0.029.
Thus to estimate the probability that U < 2, we estimate the area below U
= 1.5, not below U = 2. This gives us a Standard Normal deviate of
-1.588, as already noted, and hence a probability of 0.056. This
corresponds remarkably well with the exact probability of 0.057,
especially when we consider how small n1 and n2 are.
We will get a better approximation from our Standard Normal deviate if
we make U closer to its expected value by 1/2. In general, we get a better
fit if we
make the observed value of the statistic closer to its expected value by
half of the interval between adjacent discrete values. This is a continuity
correction.
Fig. 12.2. Distribution of the Mann-Whitney U statistic, n1 = 4, n2 = 4,

when the null hypothesis is true, with the corresponding Normal
distribution and area estimating PROB(U = 2)
For S, the interval between adjacent values is 2, not 1, for S = nc - nd =
2nc - n(n - 1)/2, and nc is an integer. A change of one unit in nc produces
a change of two units in S. The continuity correction is therefore half of 2,
which is 1. We make S closer to the expected value of 0 by 1 before
applying the Normal approximation. For the Kaposi's sarcoma data, we
had S = 40, with n = 17. Using the continuity correction gives
This gives a two-sided probability of 0.066 × 2 = 0.13, slightly greater

than the uncorrected value of 0.12.
Continuity corrections are important for small samples; for large samples
they are negligible. We shall meet another in Chapter 13.
12.7 * Parametric or non-parametric methods?

For many statistical problems there are several possible solutions, just as
for many diseases there are several treatments, similar perhaps in their
overall efficacy but displaying variation in their side effects, in their
interactions with other diseases or treatments and in their suitability for
different types of patients. There is often no one right treatment, but
rather treatment is decided on the presciber's judgement of these effects,
past experience and plain prejudice. Many problems in statistical analysis
are like this. In comparing the means of two small groups, for instance,
we could use a t test, a t test with a transformation, a Mann-Whitney U
test, or one of several others. Our choice
of method depends on the plausibility of Normal assumptions, the

importance of obtaining a confidence interval, the ease of calculation,
and so on. It depends on plain prejudice, too. Some users of statistical
methods are very concerned about the implications of Normal
assumptions and will advocate non-parametric methods wherever
possible, while others are too careless of the errors that may be
introduced when assumptions are not met.
I sometimes meet people who tell me that they have used non-parametric
methods throughout their analysis as if this is some kind of badge of
statistical purity. It is nothing of the kind. It may mean that their
significance tests have less power than they might have, and that results
are left as ‘not significant’ when, for example, a confidence interval for a
difference might be more informative.
On the other hand, such methods are very useful when the necessary
assumptions of the t distribution method cannot be made, and it would be
equally wrong to eschew their use. Rather, we should choose the method
most suited to the problem, bearing in mind both the assumptions we are
making and what we really want to know. We shall say more about what
method to use when in Chapter 14.
There is a common misconception that when the number of observations
is very small, usually said to be less than six, Normal distribution
methods such as t tests and regression must not be used and that rank
methods should be used instead. I have never seen any argument put
forward in support of this, but inspection of Tables 12.2, 12.6, 12.8, and
12.9 will show that it is nonsense. For such small samples rank tests
cannot produce any significance at the usual 5% level. Should one need
statistical analysis of such small samples, Normal methods are required.
12M * Multiple choice questions 62 to 66

62. For comparing the responses to a new treatment of a group of
patients with the responses of a control group to a standard
treatment, possible approaches include:
(a) the two-sample t method;
(b) the sign test;
(c) the Mann-Whitney U test;
(d) the Wilcoxon matched pairs test;
(e) rank correlation between responses to the treatments.
View Answer
63. Suitable methods for truly ordinal data include:
(a) the sign test;
(b) the Mann-Whitney U test;
(c) the Wilcoxon matched pairs test;
(d) the two sample t method;
(e) Kendall's rank correlation coefficient.
View Answer
64. Kendall's rank correlation coefficient between two variables:

(a) depends on which variable is regarded as the predictor;
(b) is zero when there is no relationship;
(c) cannot have a valid significance test when there are tied
observations;
(d) must lie between -1 and +1;
(e) is not affected by a log transformation of the variables.
View Answer
65. Tests of significance based on ranks:

(a) are always to be preferred to methods which assume the data
to be Normally distributed;
(b) are less powerful than methods based on the Normal
distribution when data are Normally distributed;
(c) enable confidence intervals to be estimated easily;
(d) require no assumptions about the data;
(e) are often to be preferred when data cannot be assumed to
follow any particular distribution.
View Answer
66. Ten men with angina were given an active drug and a placebo on
alternate days in random order. Patients were tested using the time
in minutes for which they could exercise until angina or fatigue
stopped them. The existence of an active drug effect could be
examined by:
(a) paired t test;
(b) Mann-Whitney U test;
(c) sign test;
(d) Wilcoxon matched pairs test;
(e) Spearman's ρ.
View Answer
12E * Exercise: Application of rank methods

In this exercise we shall analyse the respiratory compliance data of §10E
using non-parametric methods.
1. For the data of Table 10.19, use the sign test to test the null
hypothesis that changing the waveform has no effect on static
compliance.
View Answer
2. Test the same null hypothesis using a test based on ranks.

View Answer
3. Repeat step 1 using log transformed compliance. Does the

transformation make any difference?
View Answer
4. Repeat step 2 using log compliance. Why do you get a different

answer?
View Answer
5. What do you conclude about the effect of waveform from the non-
parametric tests?
View Answer
6. How do the conclusions of the parametric and non-parametric

approaches differ?
View Answer
> Table of Contents > 13 - The analysis of cross-tabulations
13
The analysis of cross-tabulations
13.1 The chi-squared test for association

Table 13.1 shows for a sample of mothers the relationship between
housing tenure and whether they had a preterm delivery. This kind of
cross-tabulation of frequencies is also called a contingency table or
cross-classification. Each entry in the table is a frequency, the number
of individuals having these characteristics (§4.1). It can be quite difficult
to measure the strength of the association between two qualitative
variables like these, but it is easy to test the null hypothesis that there is
no relationship or association between the two variables. If the sample is
large, we do this by a chi-squared test.
The chi-squared test for association in a contingency table works like
this. The null hypothesis is that there is no association between the two
variables, the alternative being that there is an association of any kind.
We find for each cell of the table the frequency which we would expect if
the null hypothesis were true. To do this we use the row and column
totals, so we are finding the expected frequencies for tables with these
totals, called the marginal totals.
There are 1443 women, of whom 899 were owner occupiers, a proportion
899/1443. If there were no relationship between time of delivery and
housing tenure, we would expect each column of the table to have the
same proportion, 899/1443, of its members in the first row. Thus the 99
patients in the first column would be expected to have 99 × 899/1443 =
61.7 in the first row. By ‘expected’ we mean the average frequency we
would get in the long run. We could not actually observe 61.7 subjects.
The 1344 patients in the second column would be expected to have 1344
× 899/1443 = 837.3 in the first row. The sum of these two expected
frequencies is 899, the row total. Similarly, there are 258 patients in the
second row and so we would expect 99 × 258/1443 = 17.7 in
the second row, first column and 1344 × 258/1443 = 240.3 in the second
row, second column. We calculate the expected frequency for each row
and column combination, or cell. The 10 cells of Table 13.1 give us the
expected frequencies shown in Table 13.2. Notice that the row and
column totals are the same as in Table 13.1. In general, the expected
frequency for a cell of the contingency table is found by
It does not matter which variable is the row and which the column.
Table 13.1. Contingency table showing time of

delivery by housing tenure
Housing tenure Preterm Term Total
Owner–occupier 50 849 899
Council tenant 29 229 258
Private tenant 11 164 175
Lives with parents 6 66 72
Other 3 36 39
Total 99 1344 1443
We now compare the observed and expected frequencies. If the two

variables are not associated, the observed and expected frequencies
should be close together, any discrepancy being due to random variation.
We need a test statistic which measures this. The differences between
observed and expected frequencies are a good place to start. We cannot
simply sum them as the sum would be zero, both observed and expected
frequencies having the same grand total, 1443. We can resolve this as
we resolved a similar problem with differences from the mean (§4.7), by
squaring the differences. The size of the difference will also depend in
some way on the number of patients. When the row and column totals
are small, the difference between observed and expected is forced to be
small. It turns out, for reasons discussed in §13A, that the best statistic is
This is often written as
For Table 13.1 this is
As will be explained in §13A, the distribution of this test statistic when the
null hypothesis is true and the sample is large enough is the Chi-squared
distribution (§7A) with (r - 1)(c - 1) degrees of freedom, where r is the
number of rows and
c is the number of columns. I shall discuss what is meant by ‘large
enough’ in §13.3. We are treating the row and column totals as fixed and
only considering the distribution of tables with these totals. The test is
said to be conditional on these totals. We can prove that we lose very
little information by doing this and we get a simple test.
Table 13.2. Expected frequencies under the

null hypothesis for Table 13.1
Housing tenure Preterm Term Total
Owner–occupier 61.7 837.3 899
Council tenant 17.7 240.3 258
Private tenant 12.0 163.0 175
Lives with parents 4.9 67.1 72
Other 2.7 36.3 39
Total 99 1344 1443

Fig. 13.1. Percentage point of the Chi-squared distribution
For Table 13.1 we have (5 - 1) × (2 - 1) = 4 degrees of freedom. Table

13.3 shows some percentage points of the Chi-squared distribution for
selected degrees of freedom. These are the upper percentage points, as
shown in Figure 13.1. We see that for 4 degrees of freedom the 5% point
is 9.49 and 1% point is 13.28, so our observed value of 10.5 has
probability between 1% and 5%, or 0.01 and 0.05. If we use a computer
program which prints out the actual probability, we find P = 0.03. The
data are not consistent with the null hypothesis and we can conclude that
there is good evidence of a relationship between housing tenure and time
of delivery.
The chi-squared statistic is not an index of the strength of the
association. If we double the frequencies in Table 13.1, this will double
chi-squared, but the strength of the association is unchanged. Note that
we can only use the chi-squared test when the numbers in the cells are
frequencies, not when they are percentages, proportions or
measurements.
Table 13.3. Percentage points of the Chi-

squared distribution
Degrees Probability that the tabulated

of value is exceeded (Figure 13.1)
freedom 10% 5% 1% 0.1%
1 2.71 3.84 6.63 10.83
2 4.61 5.99 9.21 13.82
3 6.25 7.81 11.34 16.27
4 7.78 9.49 13.28 18.47
5 9.24 11.07 15.09 20.52
6 10.64 12.59 16.81 22.46
7 12.02 14.07 18.48 24.32
8 13.36 15.51 20.09 26.13
9 14.68 16.92 21.67 27.88
10 15.99 18.31 23.21 29.59

11 17.28 19.68 24.73 31.26
12 18.55 21.03 26.22 32.91
13 19.81 22.36 27.69 34.53
14 21.06 23.68 29.14 36.12
15 22.31 25.00 30.58 37.70
16 23.54 26.30 32.00 39.25
17 24.77 27.59 33.41 40.79
18 25.99 28.87 34.81 42.31
19 27.20 30.14 36.19 43.82
20 28.41 31.41 37.57 45.32
Table 13.4. Cough during the day or at night at

age 14 for children with and without a history
of bronchitis before age 5 (Holland et al. 1978)
Bronchitis No Bronchitis Total

Cough 26 44 70
No cough 247 1002 1249
Total 273 1046 1319
13.2 Tests for 2 by 2 tables

Consider the data on cough symptom and history of bronchitis discussed
in §9.8. We had 273 children with a history of bronchitis of whom 26 were
reported to have day or night cough, and 1046 children without history of
bronchitis, of whom 44 were reported to have day or night cough. We can
set these data out as a contingency table, as in Table 13.4. We can also
use the chi-squared test to test the null hypothesis of no association
between cough and history. The expected values are shown in Table
13.5. The test statistic is
We have r = 2 rows and c = 2 columns, so there are (r - 1)(c - 1) = (2 - 1)

× (2-1)=1 degree of freedom. We see from Table 13.3 that the 5% point is
3.84, and the 1% point is 6.63, so we have observed something very
unlikely if the null hypothesis were true. Hence we reject the null
hypothesis of no association and conclude that there is a relationship
between present cough and history of bronchitis.
Table 13.5. Expected frequencies for Table

13.4
No
Bronchitis Total
bronchitis
Cough 14.49 55.51 70.00
No 258.51 990.49 1249.00

cough
Total 273.00 1046.00 1319.00
Now the null hypothesis ‘no association between cough and bronchitis’ is
the same as the null hypothesis ‘no difference between the proportions
with cough in the bronchitis and no bronchitis groups’. If there were a
difference, the variables would be associated. Thus we have tested the
same null hypothesis in two different ways. In fact these tests are exactly
equivalent. If we take the Normal deviate from §9.8, which was 3.49, and
square it, we get 12.2, the chi-squared value. The method of §9.8 and
§8.6 has the advantage that it can also give us a confidence interval for
the size of the difference, which the chi-squared method does not. Note
that the chi-squared test corresponds to the two-sided z test, even
though only the upper tail of the chi-squared distribution is used.
13.3 The chi-squared test for small samples

When the null hypothesis is true, the test statistic Σ (O - E)2/E, which we
can call the chi-squared statistic, follows the Chi-squared distribution
provided the expected values are large enough. This is a large sample
test, like those of §9.7 and §9.8. The smaller the expected values
become, the more dubious will be the test.
The conventional criterion for the test to be valid is usually attributed to
the great statistician W.G. Cochran. The rule is this: the chi-squared test
is valid if at least 80% of the expected frequencies exceed 5 and all the
expected frequencies exceed 1. We can see that Table 13.2 satisfies this
requirement, since only 2 out of 10 expected frequencies, 20%, are less
than 5 and none is less than 1. Note that this condition applies to the
expected frequencies, not the observed frequencies. It is quite
acceptable for an observed frequency to be 0, provided the expected
frequencies meet the criterion.
This criterion is open to question. Simulation studies appear to suggest
that the condition may be too conservative and that the chi-squared
approximation works for smaller expected values, especially for larger
numbers of rows and columns. At the time of writing the analysis of
tables based on small sample sizes, particularly 2 by 2 tables, is the
subject of hot dispute among statisticians. As yet, no-one has succeeded
in devising a better rule than Cochran's, so I would recommend keeping
to it until the theoretical questions are resolved. Any
chi-squared test which does not satisfy the criterion is always open to the
charge that its validity is in doubt.
Table 13.6. Observed and expected frequencies of cate

radiological appearance at six months as compare
appearance on admission in the MRC streptomycin tria
with an initial temperature of 100–100.9°F
Streptomycin Control
Radiological
assessment Observed Expected Observed
Improvement 13 8.4 5
Deterioration 2 4.2 7
Death 0 2.3 5
Total 15 15 17
If the criterion is not satisfied we can usually combine or delete rows and
columns to give bigger expected values. Of course, this cannot be done
for 2 by 2 tables, which we consider in more detail below. For example,
Table 13.6 shows data from the MRC streptomycin trial (§2.2), the results
of radiological assessment for a subgroup of patients defined by a
prognostic variable. We want to know whether there is evidence of a
streptomycin effect within this subgroup, so we want to test the null
hypothesis of no effect using a chi-squared test. There are 4 out of 6
expected values less than 5, so the test on this table would not be valid.
We can combine the rows so as to raise the expected values. Since the
small expected frequencies are in the ‘deterioration’ and ‘death’ rows, it
makes sense to combine these to give a ‘deterioration or death’ row. The
expected values are then all greater than 5 and we can do the chi-
squared test with 1 degree of freedom. This editing must be done with
regard to the meaning of the various categories. In Table 13.6, there
would be no point in combining rows 1 and 3 to give a new category of
‘considerable improvement or death’ to be compared to the remainder, as
the comparison would be absurd. The new table is shown in Table 13.7.
We have
Under the null hypothesis this is from a Chi-squared distribution with one
degree of freedom, and from Table 13.3 we can see that the probability of
getting a value as extreme as 10.8 is less than 1%. We have data
inconsistent with the null hypothesis and we can conclude that the
evidence suggests a treatment effect in this subgroup.
If the table does not meet the criterion even after reduction to a 2 by 2
table, we can apply either a continuity correction to improve the
approximation to the Chi-squared distribution (§13.5), or an exact test
based on a discrete distribution (§13.4).
Table 13.7. Reduction of Table 13.6 to a 2 by 2 tab
Radiological Streptomycin Control

assessment Observed Expected Observed
Improvement 13 8.4 5
Deterioration 2 6.6 12
or death
Total 15 15.0 17
Table 13.8. Artificial data to illustrate Fisher's

exact test
Survived Died Total
Treatment A 3 1 4
Treatment B 2 2 4
Total 5 3 8
13.4 Fisher's exact test

The chi-squared test described in §13.1 is a large sample test. When the
sample is not large and expected values are less than 5, we can turn to
an exact distribution like that for the Mann–Whitney U statistic (§12.2).
This method is called Fisher's exact test.
The exact probability distribution for the table can only be found when the
row and column totals are given. Just as with the large sample chi-
squared test, we restrict our attention to tables with these totals. This
difficulty has led to much controversy about the use of this test. I shall
show how the test works, then discuss its applicability.
Consider the following artificial example. In an experiment, we randomly
allocate 4 patients to treatment A and 4 to treatment B, and get the
outcome shown in Table 13.8. We want to know the probability of so
large a difference in mortality between the two groups if the treatments
have the same effect (the null hypothesis). We could have randomized
the subjects into two groups in many ways, but if the null hypothesis is
true the same three would have died. The row and column totals would
therefore be the same for all these possible allocations. If we keep the
row and column totals constant, there are only 4 possible tables, shown
in Table 13.9. These tables are found by putting the values 0, 1, 2, 3 in
the ‘Died in group A’ cell. Any other values would make the D total
greater than 3.
Now, let us label our subjects a to h. The survivors we will call a to e, and
the deaths f to h. How many ways can these patients be arranged in two
groups of 4 to give tables i, ii, iii and iv? Table i can arise in 5 ways.
Patients f, g, and h would have to be in group B, to give 3 deaths, and the
remaining member of B could be a, b, c, d or e. Table ii can arise in 30
ways. The 3 survivors in group A can be abc, abd, abe, acd, ace, ade,
bcd, bce, bde, cde, 10 ways. The death in A can be f, g or h, 3 ways.
Hence the group can be made up in 10 × 3 = 30 ways. Table iii is the
same as table ii, with A and B reversed, so arises in 30 ways. Table iv is
the same as table i with A and B reversed, so arises in 5 ways.
Hence we can arrange the 8 patients into 2 groups of 4 in 5 + 30 + 30 + 5

= 70 ways. Now, the probability of any one arrangement arising by
chance is 1/70, since they are all equally likely if the null hypothesis is
true. Table i arises from 5 of the 70 arrangements, so had probability 5/70
= 0.071. Table ii arises from 30 out of 70 arrangements, so has
probability 30/70 = 0.429. Similarly, Table iii has probability 30/70 =
0.429, and Table iv has probability 5/70 = 0.071.
Table 13.9. Possible tables for the totals of

Table 13.8
i. S D T
A 4 0 4
B 1 3 4
T 5 3 8
S D T
A 3 1 4
ii
B 2 2 4
T 5 3 8
S D T
A 2 2 4
iii.
B 3 1 4
T 5 3 8
S D T
A 1 3 4
iv.
B 4 0 4
T 5 3 8
Hence, under the null hypothesis that there is no association between

treatment and survival, Table ii, which we observed, has a probability of
0.429. It could easily have arisen by chance and so it is consistent with
the null hypothesis. As in §9.2, we must also consider tables more
extreme than the observed. In this case, there is one more extreme table
in the direction of the observed difference, Table i. In the direction of the
observed difference, the probability of the observed table or a more
extreme one is 0.071 + 0.429 = 0.5. This is the P value for a one-sided
test (§9.5).
Fisher's exact test is essentially one sided. It is not clear what the
corresponding deviations in the other direction would be, especially when
all the marginal totals are different. This is because in that case the
distribution is asymmetrical, unlike those of §12.2–5. One solution is to
double the one-sided probability to get a two-sided test when this is
required. I follow Armitage and Berry (1994) in preferring this option.
Another solution is to calculate probabilities for every possible table and
sum all probabilities less than or equal to the probability for the observed
table to give the P value. This may give a smaller P value than the
doubling method.
There is no need to enumerate all the possible tables, as above. The
probability can be found from a simple formula (§13B). The probability of
observing a set of frequencies f11, f12, f21, f22, when the row and column
totals are r1, r2, c1, and c2 and the grand total is n, is
(See §6A for the meaning of n!.) We can calculate this for each possible
table so find the probability for the observed table and each more
extreme one. For the example:
giving a total of 0.50 as before.

Unlike the exact distributions for the rank statistics, this distribution is
fairly easy to calculate but difficult to tabulate. A good table of this
distribution required a small book (Finney et al. 1963).
We can apply this test to Table 13.7. The 2 by 2 tables to be tested and
their probabilities are:
The total one-sided probability is 0.0014553, which doubled for a two-
sided test gives 0.0029. The method using all smaller probabilities gives
P = 0.00159. Either is larger than the probability for the X2 value of 10.6,
which is 0.0011.
Fisher's exact test was originally devised for the 2×2 table and only used
when the expected frequencies were small. This was because for larger
numbers and larger tables the calculations were impractical. With
computers things have changed, and Fisher's exact test can be done for
any 2×2 table. Some programs will also calculate Fisher's exact test for
larger tables as the number of rows and columns increases, the number
of possible tables increases very rapidly and it becomes impracticable to
calculate and store the probability for each one. There are specialist
programs such as StatExact which create a random sample of the
possible tables and use them to estimate a distribution of probabilities
whose tail area is then found. Methods which sample the possibilities in
this way are (rather endearingly) called Monte Carlo methods.
13.5 Yates' continuity correction for the 2 by 2

table
The discrepancy in probabilities between the chi-squared test and
Fisher's exact test arises because we are estimating the discrete
distribution of the test statistic by the continuous Chi-squared distribution.
A continuity correction like those of §12.6, called Yates' correction, can
be used to improve the fit. The observed frequencies change in units of
one, so we make them closer to their expected values by one half. Hence
the formula for the corrected chi-squared statistic for a 2 by 2 table is
This has probability 0.0037, which is closer to the exact probability,
though there is still a considerable discrepancy. At such extremely low
values any approximate probability model such as this is liable to break
down. In the critical area between 0.10 and 0.01, the continuity correction
usually gives a very good fit to the exact probability. As Fisher's exact test
is now so easy to do, Yates' correction may soon disappear.
13.6 * The validity of Fisher's and Yates'

methods
There has been much dispute among statisticians about the validity of
the exact test and the continuity correction which approximates to it.
Among the more argumentative of the founding fathers of statistical
inference, such as Fisher and Neyman, this was quite acrimonious. The
problem is still unresolved, and generating almost as much heat as light.
Note that although both are 2 by 2 tables, Tables 13.4 and 13.7 arose in
different ways. In Table 13.7, the column totals were fixed by the design
of the experiment and only the row totals are from a random variable. In
Table 13.4 neither row nor column totals were set in advance. Both are
from the Binomial distribution, depending on the incidence of bronchitis
and prevalence of chronic cough in the population. There is a third
possibility, that both the row and column totals are fixed. This is rare in
practice, but it can be achieved by the following experimental design. We
want to know whether a subject can distinguish an active treatment from
a placebo. We present him with 10 tablets, 5 of each, and ask him to sort
the tablets into the 5 active and 5 placebo. This would give a 2 by 2 table,
subject's choice versus truth, in which all row and column totals are
preset to 5. There are several variations on these types of table, too. It
can be shown that the same chi-squared test applies to all these cases
when samples are large. When samples are small, this is not necessarily
so. A discussion of the problem is well beyond the scope of this book. For
some of these cases. Fisher's exact test and Yates' correction may be
conservative, that
is, give rather larger probabilities than they should, though this is a matter
of debate. My own opinion is that Yates' correction and Fisher's exact test
should be used. If we must err, it seems better to err on the side of
caution.
Table 13.10. The 2 by 2 table in symbolic

notation
Total
a b a + b
c d c + d
Total a + c b + d a + b + c + d
13.7 Odds and odds ratios

If the probability of an event is p then the odds of that event is o = p/(1 -
p). The probability that a coin shows a head is 0.5, the odds is 0.5/(1 -
0.5) = 1. Note that ‘odds’ is a singular word, not the plural of ‘odd’. The
odds has advantages for some types of analysis, as it is not constrained
to lie between 0 and 1, but can take any value from zero to infinity. We
often use the logarithm to the base e of the odds, the log odds or logit:
This can vary from minus infinity to plus infinity and thus is very useful in
fitting regression type models (§17.8). The logit is zero when p = 1/2 and
the logit of 1 - p is minus the logit of p:
Consider Table 13.4. The probability of cough for children with a history
of bronchitis is 26/273 = 0.09524. The odds of cough for children with a
history of bronchitis is 26/247 = 0.10526. The probability of cough for
children without a history of bronchitis is 44/1046 = 0.04207. The odds of
cough for children without a history of bronchitis is 44/1002 = 0.04391.
One way to compare children with and without bronchitis is to find the
ratio of the proportions of children with cough in the two groups (the
relative risk, §8.6). Another is to find the odds ratio, the ratio of the odds
of cough in children with bronchitis and children without bronchitis. This is
(26/247)/(44/1002) = 0.10526/0.04391 = 2.39718. Thus the odds of
cough in children with a history of bronchitis is 2.397 18 times the odds of
cough in children without a history of bronchitis.
If we denote the frequencies in the table by a,b,c. and d, as in Table
13.10, the odds ratio is given by
This is symmetrical; we get the same thing by
We can estimate the standard error and confidence interval using the log
of the odds ratio (§13C). The standard error of the log odds ratio is:
Hence we can find the 95% confidence interval. For Table 13.4, the log
odds ratio is loge (2.39718) = 0.87429, with standard error
Provided the sample is large enough, we can assume that the log odds
ratio comes from a Normal distribution and hence the approximate 95%
confidence interval is
0.87429 - 1.96 × 0.25736 to 0.87429 + 1.96 × 0.25736 = 0.36986 to
1.37872
To get a confidence interval for the odds ratio itself we must antilog:
The odds ratio can be used to estimate the relative risk in a case-control
study. The calculation of relative risk in §8.6 depended on the fact that we
could estimate the risks. We could do this because we had a prospective
study and so knew how many of the risk group developed the symptom.
This cannot be done if we start with the outcome, in this case cough at
age 14, and try to work back to the risk factor, bronchitis, as in a case–
control study.
Table 13.11 shows data from a case–control study of smoking and lung
cancer (see §3.8). We start with a group of cases, patients with lung
cancer and a group of controls, here hospital patients without cancer. We
cannot calculate risks (the column totals would be meaningless and have
been omitted), but we can still estimate the relative risk.
Suppose the prevalence of lung cancer is p, a small number, and the
table is as Table 13.10. Then we can estimate the probability of both
having lung cancer and being a smoker by pa/(a + b), because a/(a + b)
is the conditional probability of smoking in lung cancer patients (§6.8).
Similarly, the probability of being a smoker without lung cancer is (1-
p)c/(c + d). The probability of being a smoker is therefore pa/(a + b) + (1 -
p)c/(c + d), the probability of being a smoker with lung cancer plus the
probability of being a smoker without lung cancer. Because p is much
smaller than 1 - p, the first term can be ignored and
the probability of being a smoker is approximately (1 - p)c/(c + d). The

risk of lung cancer for smokers is found by dividing the probability of
being a smoker with lung cancer by the probability being a smoker:
Table 13.11. Smokers and non-smokers among

male cancer patients and controls (Doll and
Hill 1950)
Smokers Non-smokers Total
Lung cancer 647 2 649
Controls 622 27 649
Similarly, the probability of both being a non-smoker and having lung

cancer is pb/(a + b) and the probability of being a non-smoker without
lung cancer is (1 - p)d/(c + d). The probability of being a non-smoker is
therefore pb/(a + b) + (1 - p)d/(c + d), and since p is much smaller than 1
- p, the first term can be ignored and the probability of being a non-
smoker is approximately (1 - p)d/(c + d). This gives a risk of lung cancer
among non-smokers of approximately
The relative risk of lung cancer for smokers is thus, approximately,

This is, of course, the odds ratio. Thus for case control studies the
relative risk is approximated by the odds ratio.
For Table 13.11 we have
Thus the risk of lung cancer in smokers is about 14 times that of non-
smokers. This is a surprising result from a table with so few non-smokers,
but a direct estimate from the cohort study (Table 3.1) is 0.90/0.07 = 12.9,
which is very similar. The log odds ratio is 2.64210 and its standard error
is
Hence the approximate 95% confidence interval is
Table 13.12. Cough during the day or at night and cigar

12-year-old boys (Bland et al. 1978)
Boy's smoking

Non-smoker Occasional Regular
Cough 266 20.4% 395 28.8% 80
No 1037 79.6% 977 71.2% 92

cough
Total 1303 100.0% 1372 100.0% 172

To get a confidence interval for the odds ratio itself we must antilog:
The very wide confidence interval is because the numbers of non-

smokers, particularly for lung cancer cases, are so small.
13.8 * The chi-squared test for trend

Consider the data of Table 13.12. Using the chi-squared test described in
§13.1, we can test the null hypothesis that there is no relationship
between reported cough and smoking against the alternative that there is
a relationship of some sort. The chi-squared statistic is 64.25, with 2
degrees of freedom, P < 0.001. The data are not consistent with the null
hypothesis.
Now, we would have got the same value of chi-squared whatever the
order of the columns. The test ignores the natural ordering of the
columns, but we might expect that if there were a relationship between
reported cough and smoking, the prevalence of cough would be greater
for greater amounts of smoking. In other words, we look for a trend in
cough prevalence from one end of the table to the other. We can test for
this using the chi-squared test for trend.
First, we define two random variables. X and Y, whose values depend on
the categories of the row and column variables. For example, we could
put X=1 for non-smokers, X = 2 for occasional smokers and X = 3 for
regular smokers, and put Y = 1 for ‘cough’ and Y = 2 for ‘no cough’. Then
for a non-smoker who coughs, the value of X is 1 and the value of Y is 1.
Both X and Y may have more than two categories, provided both are
ordered. If there are n individuals, we have n pairs of observations (xi, yi).
If there is a linear trend across the table, there will be linear regression of
Y on X which has non-zero slope. We fit the usual least squares
regression line, Y = a + bX, where
and where s2 is the estimated variance of Y. In simple linear regression,

as described in Chapter 11, we are usually concerned with estimating b
and making statements about its precision. Here we are only going to test
the null hypothesis that in the population b=0. Under the null hypothesis,
the variance about the line is equal to the total variance of Y, since the
line has zero slope. We use the
estimate
(We use n as the denominator, not n - 1, because the test is conditional

on the row and column totals as described in §13A. There is a good
reason for it, but it is not worth going into here.) As in §11.5, the standard
error of b is
For practical calculations we use the alternative forms of the sums of

squares and products:
Note that it does not matter which variable is X and which is Y. The sums
of squares and products are easy to work out. For example, for the
column variable, X, we have 1303 individuals with X=1, 1372 with X=2
and 172 with X=3. For our data we have
Similarly, Σy2i = 9165 and Σyi = 4953;
= 59.47
If the null hypothesis is true, χ2i is an observation from the Chi-squared
distribution with 1 degree of freedom. The value 59.47 is highly unlikely
from this distribution and the trend is significant.
There are several points to note about this method. The choice of values
for X and Y is arbitrary. By putting X = 1, 2 or 3 we assumed that the
difference between non-smokers and occasional smokers is the same as
that between occasional smokers and smokers. This need not be so and
a different choice of X would give a different chi-squared for trend
statistic. The choice is not critical, however. For example, putting X = 1, 2
or 4, so making regular smokers more different from occasional smokers
than occasional smokers are from non-smokers, we get x2 for trend to be
64.22. The fit to the data is rather better, but the conclusions are
unchanged.
The trend may be significant even if the overall contingency table chi-
squared is not. This is because the test for trend has greater power for
detecting trends than has the ordinary chi-squared test. On the other
hand, if we had an association where those who were occasional
smokers had far more symptoms than either non-smokers or regular
smokers, the trend test would not detect it. If the hypothesis we wish to
test involves the order of the categories, we should use the trend test, if it
does not we should use the contingency table test of §13.1. Note that the
trend test statistic is always less than the overall chi-squared statistic.
The distribution of the trend chi-squared statistic depends on a large
sample regression model, not on the theory given in §13A. The table
does not have to meet Cochran's rule (§13.3) for the trend test to be
valid. As long as there are at least 30 observations the approximation
should be valid.
Some computer programs offer a slightly different test, the Mantel–
Haenzsel trend test (not to be confused with the Mantel–Haenzsel
method for combining 2 by 2 tables, §17.11). This is almost identical to
the method described here. As an alternative to the chi-squared test for
trend, we could calculate Kendall's rank correlation coefficient, τb,
between X and Y (§12.5). For Table 13.12 we get τb = -0.136 with
standard error 0.018. We get a χ21 statistic by (τb/SE(τb))2 = 57.09. This is
very similar to the X2 for trend value 59.47.
13.9 * Methods for matched samples

The chi-squared test described above enables us, among other things, to
test the null hypothesis that binomial proportions estimated from two
independent samples are the same. We can do this for the one sample or
matched sample problem also. For example, Holland et al. (1978)
obtained respiratory symptom questionnaires for 1319 Kent
schoolchildren at ages 12 and 14. One question we asked was whether
the prevalence of reported symptoms was different at the two ages. At
age 12, 356 (27%) children were reported to have had severe colds in
the past 12 months compared to 468 (35%) at age 14. Was there
evidence of a real increase? Just as in the one sample or paired t test
(§10.2) we would hope
to improve our analysis by taking into account the fact that this is the
same sample. We might expect, for instance, that symptoms on the two
occasions will be related.
Table 13.13. Severe colds reported at two ages

for Kent schoolchildren (Holland et al. 1978)
Severe colds at
Severe colds at age 14 Total
age 12
Yes No
Yes 212 144 356
No 256 707 963
Total 468 851 1319
The method which enables us to do this is McNemar's test, another

version of the sign test. We need to know that 212 children were reported
to have colds on both occasions. 144 to have colds at 12 but not at
14,256 to have colds at 14 but not at 12 and 707 to have colds at neither
age. Table 13.13 shows the data in tabular form.
The null hypothesis is that the proportions saying yes on the first and
second occasions are the same, the alternative being that one exceeds
the other. This is a hypothesis about the row and column totals, quite
different from that for the contingency table chi-squared test. If the null
hypothesis were true we would expect the frequencies for ‘yes, no’ and
‘no, yes’ to be equal. In other words, as many should go up as down.
(Compare this with the sign test, §9.2.) If we denote these frequencies by
fyn and fny, then the expected frequencies will be (fyn + fny)/2. We get the
test statistic:
which follows a Chi-squared distribution provided the expected values

are large enough. There are two observed frequencies and one
constraint, that the sum of the observed frequencies = the sum of the
expected frequencies. Hence there is one degree of freedom. Like the
chi-squared test (§13.1) and Fisher's exact test (§13.4), we assume a
total to be fixed. In this case it is fyn + fny, not the row and column totals,
which are what we are testing. The test statistic can be simplified
considerably, to:
For Table 13.13, we have
This can be referred to Table 13.3 with one degree of freedom and is
clearly highly significant. There was a difference between the two ages.
As there was no change in any of the other symptoms studied, we
thought that this was possibly due to an epidemic of upper respiratory
tract infection just before the second questionnaire.
There is a continuity correction, again due to Yates. If the observed
frequency fyn increases by 1, fny decreases by 1 and fyn - fny increases by
2. Thus half the difference between adjacent possible values is 1 and we
make the observed difference nearer to the expected difference (zero) by
1. Thus the continuity corrected test statistic is
where |fyn - fny| is the absolute value, without sign. For Table 13.13:
There is very little difference because the expected values are so large
but if the expected values are small, say less than 20, the correction is
advisable. For small samples, we can also take fny as an observation
from the Binomial distribution with p = ½ and n = fyn + fny and proceed as
for the sign test (§9.2).
We can find a confidence interval for the difference between the
proportions. The estimated difference is p1 - p2 = (fyn - fyn)/n. We
rearrange this:
We can treat the fyn as an observation from a Binomial distribution with

parameter n = fyn + fny, which, of course, we are treating as fixed. (I am
using n here to mean the parameter of the Binomial distribution as in
§6.4, not to mean the total sample size.) We find a confidence interval for
fyn/(fyn + fny) using either the z method of §8.4 or the exact method of
§8.8. We then multiply these limits by 2, subtract 1 and multiply by (fyn +
fny)/n.
For the example, the estimated difference is (144 - 256)/1319 = -0.085.
For the confidence interval, fyn + fny = 400 and fyn = 144. The 95%
confidence interval for fyn/(fyn + fny) is 0.313 to 0.407 by the large sample
method. Hence the confidence interval for p1 - p2 is (2 × 0.313 - 1) ×
400/1319 = -0.113 to (2 × 0.407 - 1) × 400/1319 = -0.056. We estimate
that the proportion of colds on the first occasion was less than that on the
second by between 0.06 and 0.11.
We may wish to compare the distribution of a variable with three or more
categories in matched samples. If the categories are ordered, like
smoking experience in Table 13.12, we are usually looking for a shift from
one end of the distribution to the other, and we can use the sign test
(§9.2), counting positives when smoking increased, negative when it
decreased, and zero if the category
was the same. When the categories are not ordered, as Table 13.1 there
is a test due to Stuart (1955), described by Maxwell (1970). The test is
difficult to do and the situation is very unusual, so I shall omit details. My
free program Clinstat will do it (§1.3).
Table 13.14. Parity of 125 women attending
antenatal clinics at St. George's Hospital,
with the calculation of the chi-squared
goodness of fit test
We can also find an odds ratio for the matched table, called the
conditional odds ratio. Like McNemar's method, it uses the frequencies
in the off diagonal only. The estimate is very simple: fyn/fny. Thus for Table
13.13 the odds of having severe colds at age 12 is 144/256 = 0.56 times
that at age 14. This example is not very interesting, but the method is
particularly useful in matched case–control studies, where it provides an
estimate of the relative risk. A confidence interval is provided in the same
way as for the difference between proportions. We can estimate p =
fyn/(fyn + fny) and then the odds ratio is given by p/(1 - p). For the
example, p = 144/400 = 0.36 and turning p back to the odds ratio p/(1 - p)
= 0.36/(1 - 0.36) = 0.56 as before. The 95% confidence interval for p is
0.313 to 0.4071, as above. Hence the 95% confidence interval for the
conditional odds ratio is 0.31/(1 - 0.31) = 0.45 to 0.41/(1 - 0.41) = 0.69.
13.10 * The chi-squared goodness of fit test

Another use of the Chi-squared distribution is the goodness of fit test.
Here we test the null hypothesis that a frequency distribution follows
some theoretical distribution such as the Poisson or Normal. Table 13.14
shows a frequency distribution. We shall test the null hypothesis that it is
from a Poisson distribution, i.e. that conception is a random event among
fertile women.
First we estimate the parameter of the Poisson distribution, its mean, µ,
in this case 0.816. We then calculate the probability for each value of the
variable, using the Poisson formula of §6.7:
where r is the number of events. The probabilities are shown in Table

13.14. The probability that the variable exceeds five is found by
subtracting the probabilities for 0, 1, 2, 3, 4, and 5 from 1.0. We then
multiply these by the number of
observations, 125, to give the frequencies we would expect from 125

observations from a Poisson distribution with mepn 0.816.
Table 13.15. Time of onset of 554 strokes Wroe

et al. (1992)
Time Frequency Time Frequency
00.01– 21 12.01– 34
02.00 14.00
02.01– 16 14.01– 59
04.00 16.00
04.01– 22 16.01– 44
06.00 18.00
06.01– 104 18.01– 51
08.00 20.00
08.01– 95 20.01– 32
10.00 22.00
10.01– 66 22.01– 10
12.00 24.00
We now have a set of observed and expected frequencies and can

compute a chi-squared statistic in the usual way. We want all the
expected frequencies to be greater than 5 if possible. We achieve this
here by combining all the categories for parity greater than or equal to 3.
We then add (O - E)2/E for the categories to give a χ2 statistic. We now
find the degrees of freedom. This is the number of categories minus the
number of parameters fitted from the data (one in the example) minus
one. Thus we have 4-1-1 = 2 degrees of freedom. From Table 13.3 the
observed χ2 value of 2.99 has P > 0.10 and the deviation from the
Poisson distribution is clearly not significant.
The same test can be used for testing the fit of any distribution. For
example, Wroe et al. (1992) studied diurnal variation in onset of strokes.
Table 13.15 shows the frequency distribution of times of onset. If the null
hypothesis that there is no diurnal variation were true, the time at which
strokes occurred would follow a Uniform distribution (§7.2). The expected
frequency in each time interval would be the same. There were 554
cases altogether, so the expected frequency for each time is 554/12 =
46.167. We then work out (O - E)2/E for each interval and add to give the
chi-squared statistic, in this case equal to 218.8. There is only one
constraint, that the frequencies total 554, as no parameters have been
estimated. Hence if the null hypothesis were true we would have an
observation from the Chi-squared distribution with 12 - 1 = 11 degrees of
freedom. The calculated value of 218.8 is very unlikely, P < 0.001 from
Table 13.3, and the data are not consistent with the null hypothesis.
When we test the equality of a set of frequencies like this the test is also
called the Poisson heterogeneity test.
Appendices
13A Appendix: Why the chi-squared test
works
We noted some of the properties of the Chi-squared distribution in §7A.

In particular, it is the sum of the squares of a set of independent Standard
Normal variables, and if we look at a subset of values defined by
independent linear relationships between these variables we lose one
degree of freedom for each constraint. It is on these two properties that
the chi-squared test depends.
Suppose we did not have a fixed size to the birth study of Table 13.1, but
observed subjects as they delivered over a fixed time. Then the number
in
a given cell of the table would be from a Poisson distribution and the set
of Poisson variables corresponding to the cell frequency would be
independent of one another. Our table is one set of samples from these
Poisson distributions. However, we do not know the expected values of
these distributions under the null hypothesis; we only know their
expected values if the table has the row and column totals we observed.
We can only consider the subset of outcomes of these variables which
has the observed row and column totals. The test is said to be
conditional on these row and column totals.
Table 13.16. Symbolic representation of a 2 × 2

table
Total
f11 f12 r1
f21 f22 r2
Total c1 c2 n
The mean and variance of a Poisson variable are equal (§6.7). If the null
hypothesis is true, the means of these variables will be equal to the
expected frequency calculated in §13.1. Thus O, the observed cell
frequency, is from a Poisson distribution with mean E, the expected cell
frequency, and standard deviation √E. Provided E is large enough, this
Poisson distribution will be approximately Normal. Hence (O - E)/√E is
from a Normal distribution mean 0 and variance 1. Hence if we find
this is the sum of the squares of a set of Normally distributed random

variables with mean 0 and variance 1, and so is from a Chi-squared
distribution (§7A).
We will now find the degrees of freedom. Although the underlying
variables are independent, we are only considering a subset defined by
the row and column totals. Consider the table as in Table 13.16. Here, f11
to f22 are the observed frequencies, r1, r2 the row totals, c1, c2 the column
totals, and n the grand total. Denote the corresponding expected values
by e11 to e22. There are three linear constraints on the frequencies:
Any other constraint can be made up of these. For example, we must

have
This can be found by subtracting the second equation from the first. Each
of these linear constraints on f11 to f22 is also a linear constraint on (f11 -
e11)/√e11
to (f22 - e22)/√e22. This is because e11 is fixed and so (f11 - e11)/√e11 is a

linear function of f11. There are four observed frequencies and so four (O
- E)/√E variables, with three constraints. We lose one degree of freedom
for each constraint and so have 4 - 3 = 1 degree of freedom.
If we have r rows and c columns, then we have one constraint that the
sum of the frequencies is n. Each row must add up, but when we reach
the last row the constraint can be obtained by subtracting the first r - 1
rows from the grand total. The rows contribute only r - 1 further
constraints. Similarly the columns contribute c - 1 constraints. Hence,
there being rc frequencies, the degrees of freedom are
So we have degrees of freedom given by the number of rows minus one

times the number of columns minus one.
13B Appendix: The formula for Fisher's

exact test
The derivation of Fisher's formula is strictly for the algebraically minded.

Remember that the number of ways of choosing r things out of n things
(§6A) is n!/r!(n - r)!. Now, suppose we have a 2 by 2 table made up of n
as shown in Table 13.16. First, we ask how many ways n individuals can
be arranged to give marginal totals, r1, r2, c1 and c2. They can be
arranged in columns in n!/c1!c2! ways, since we are choosing c1 objects
out of n, and in rows n!/r1!r2! ways. (Remember n - c1 = c2 and n - r1 = r2.)
Hence they can be arranged in
ways. For example, the table with totals
can happen in
As we saw in §13.4, the columns can be arranged in 70 ways. Now we

ask, of these ways how many make up a particular table? We are now
dividing the n into four groups of sizes f11, f12, f21 and f12. We can choose
the first group in n!/f11!(n - f11)! ways, as before. We are now left with n -
f11 individuals, so we can choose f12 in (n - f11)!/f12!(n - f11 - f12)!. We are
now left with n - f11 - f12, and so we choose f21 in (n - f11 - f12)!/f21! ways.
This leaves n - f11 - f12 - f21, which is, of course, equal to f22 and so f22
can only be chosen in one way. Hence we have altogether:
because n - f11 - f12 -f12 = f22. So out of the
possible tables, the given tables arises in
ways. The probability of this table arising by chance is

13C Appendix: Standard error for the log
odds ratio
This is for the mathematical reader. We start with a general result

concerning log transformations. If X is a random variable with mean µ,
the approximate variance of loge(X) is given by
If an event happens a times and does not happen b times, the log odds is
loge(a/b) - loge(a) - loge(b). The frequencies a and b are from
independent Poisson distributions with means estimated by a and b
respectively. Hence their variances are estimated by 1/a and 1/b
respectively. The variance of the log odds is given by
The standard error of the log odds is thus given by
The log odds ratio is the difference between the log odds:
The variance of the log odds ratio is the sum of the variances of the log
odds and for table 2 we have
The standard error is the square root of this:

67. The standard chi-squared test for a 2 by 2 contingency table is
valid only if:
(a) all the expected frequencies are greater than five;
(b) both variables are continuous;
(c) at least one variable is from a Normal distribution;
(d) all the observed frequencies are greater than five;
(e) the sample is very large.
View Answer
68. In a chi-squared test for a 5 by 3 contingency table:

(a) variables must be quantitative;
(b) observed frequencies are compared to expected frequencies;
(c) there are 15 degrees of freedom;
(d) at least 12 cells must have expected values greater than five;
(e) all the observed values must be greater than one.
View Answer
Table 13.17. Cough first thing in the morning
in a group of schoolchildren, as reported by
the child and by the child's parents (Bland et
al. 1979)
Child's report
Parents' report Total
Yes No
Yes 29 104 133
No 172 5097 5269
Total 201 5201 5402
69. In Table 13.17:

(a) the association between reports by parents and children can
be tested by a chi-squared test;
(b) * the difference between symptom prevalence as reported by
children and parents can be tested by McNemar's test;
(c) * if McNemar's test is significant, the contingency chi-squared
test is not valid;
(d) the contingency chi-squared test has one degree of freedom;
(e) it would be important to use the continuity correction in the
contingency chi-squared test.
View Answer
70. Fisher's exact test for a contingency table:
(a) applies to 2 by 2 tables;
(b) usually gives a larger probability than the ordinary chi-squared
test;
(c) usually gives about the same probability as the chi-squared
test with Yates' continuity correction;
(d) is suitable when expected frequencies are small;
(e) is difficult to calculate when the expected frequencies are
large.
View Answer
71. When an odds ratio is calculated from a 2 by 2 table:

(a) the odds ratio is a measure of the strength of the relationship
between the row and column variables;
(b) if the order of the rows and the order of the columns is
reversed, the odds ratio will be unchanged;
(c) the ratio may take any positive value;
(d) the odds ratio will be changed to its reciprocal if the order of
the columns is changed;
(e) the odds ratio is the ratio of the proportions of observations in
the first row for the two columns.
View Answer
Table 13.18. Bird attacks on milk bottles

reported by cases of Campylobacter jejuni
infection and controls (Southern et al. 1990)
Number of days of Number of

week when attacks OR
took place Cases Controls
0 3 42 1
1–3 11 3 51
4–5 5 1 70
6–7 10 1 140
72. Table 13.18 appeared in the report of a case control study of

infection with Campylobacter jejuni (§3E):
(a) * a chi-squared test for trend could be used to test the null
hypothesis that risk of disease does not increase with the number
of bird attacks;
(b) ‘OR’ means the odds ratio;
(c) * a significant chi-squared test would show that risk of disease
increases with increasing numbers of bird attacks;
(d) ‘OR’ provides an estimate of the relative risk of Campylobacter
jejuni infection;
(e) * Kendall's rank correlation coefficient, τb, could be used to test
the null hypothesis that risk of disease does not increase with the
number of bird attacks.
View Answer
73. * McNemar's test could be used:

(a) to compare the numbers of cigarette smokers among cancer
cases and age and sex matched healthy controls;
(b) to examine the change in respiratory symptom prevalence in a
group of asthmatics from winter to summer;
(c) to look at the relationship between cigarette smoking and
respiratory symptoms in a group of asthmatics;
(d) to examine the change in PEFR in a group of asthmatics from
winter to summer;
(e) to compare the number of cigarette smokers among a group of
cancer cases and a random sample of the general population.
View Answer
13E Exercise: Admissions to hospital in a

heatwave
In this exercise we shall look at some data assembled to test the
hypothesis that there is a considerable increase in the number of
admissions to geriatric wards during heatwaves. Table 13.19 shows the
number of admissions to geriatric wards in a health district for each week
during the summers of 1982, which was cold, and 1983, which was hot.
Also shown are the average of the daily peak temperatures for each
week.
1. When do you think the heatwave began and ended?
View Answer
2. How many admissions were there during the heatwave and in the
correspond ing period of 1982? Would this be sufficient evidence to
conclude that heatwaves produce an increase in admissions?
View Answer
3. We can use the periods before and after the heatwave weeks as
controls for changes in other factors between the years. Divide the
years into three periods, before, during, and after the heatwave and
set up a two-way table showing numbers of admissions by period
and year.
View Answer
Table 13.19. Mean peak daily temperatures for each we

September of 1982 and 1983, with geriatric admissions
(Fish 1985)
Mean peak, Mean peak,

Admissions
°C
Week Week
1982 1983 1982 1983 1982
1 12.4 15.3 24 20 12 21.7
2 18.2 14.4 22 17 13 22.5
3 20.4 15.5 21 21 14 25.7
4 18.8 15.6 22 17 15 23.6
5 25.3 19.6 24 22 16 20.4
6 23.2 21.6 15 23 17 19.6
7 18.6 18.9 23 20 18 20.2

8 19.4 22.0 21 16 19 22.2
9 20.6 21.0 18 24 20 23.3
10 23.4 26.5 21 21 21 18.1
11 22.8 30.4 17 20 22 17.3
4. We can use this table to test for a heatwave effect. State the null
hypothesis and calculate the frequencies expected if the null
hypothesis were true.
View Answer
5. Test the null hypothesis. What conclusions can you draw?

View Answer
6. What other information could be used to test the relationship

between heat waves and geriatric admissions?
View Answer
> Table of Contents > 14 - Choosing the statistical method
14
Choosing the statistical method
14.1 * Method oriented and problem oriented

teaching
The choice of method of analysis for a problem depends on the
comparison to be made and the data to be used. In Chapters 8, 9, 10, 11,
12, and 13, statistical methods have been arranged largely by type of
data, large samples, Normal, ordinal, categorical, etc, rather than by type
of comparison. In this chapter we look at how the appropriate method is
chosen for the three most common problems in statistical inference:
comparison of two independent groups, for example, groups of patients
given different treatments;
comparison of the response of one group under different conditions, as
in a cross-over trial, or of matched pairs of subjects, as in some case–
control studies;
investigation of the relationship between two variables measured on
the same sample of subjects.
This chapter acts as a map of the methods described in Chapters 8, 9,
10, 11, 12, and 13. Subsequent chapters describe methods for special
problems in clinical medicine, population study, dealing with several
factors at once, and the choice of sample size.
As was discussed in §12.7, there are often several different approaches
to even a simple statistical problem. The methods described here and
recommended for particular types of question may not be the only
methods, and may not always be universally agreed as the best method.
Statisticians are at least as prone to disagree as clinicians. However,
these would usually be considered as valid and satisfactory methods for
the purposes for which they are suggested here. When there is more
than one valid approach to a problem, they will usually be found to give
similar answers.
14.2 * Types of data

The study design is one factor which determines the method of analysis,
the variable being analysed is another. We can classify variables into the
following types:
Ratio scales
The ratio of two quantities has a meaning, so we can say that one
observation is twice another. Human height is a ratio scale. Ratio scales
allow us to carry out power transformations like log or square root.
Interval scales
The interval or distance between points on the scale has precise
meaning, a change of one unit at one scale point is the same as a
change of one unit at another. For example, temperature in °C is an
interval scale, though not a ratio scale because the zero is arbitrary. We
can add and subtract on an interval scale. All ratio scales are also
interval scales. Interval scales allow us to calculate means and
variances, and to find standard errors and confidence intervals for these.
Ordinal scale
The scale enables us to order the subjects, from that with the lowest
value to that with the highest. Any ties which cannot be ordered are
assumed to be because the measurement is not sufficiently precise. A
typical example would be an anxiety score calculated from a
questionnaire. A person scoring 10 is more anxious than a person
scoring 8, but not necessarily higher by the same amount that a person
scoring 4 is higher than a person scoring 2.
Ordered nominal scale

We can group subjects into several categories, which have an order. For
example, we can ask patients if their condition is much improved,
improved a little, no change, a little worse, much worse.
Nominal scale
We can group subjects into categories which need not be ordered in any
way. Eye colour is measured on a nominal scale.
Dichotomous scales
Subjects are grouped into only two categories, for example: survived or
died. This is a special case of the nominal scale.
Clearly these classes are not mutually exclusive, and an interval scale is
also ordinal. Sometimes it is useful to apply methods appropriate to a
lower level of measurement, ignoring some of the information. The
combination of the type of comparsion and the scale of measurement
should direct us to the appropriate method.
14.3 * Comparing two groups

The methods used for comparing two groups are summarized in Table
14.1.
Interval data. For large samples, say more than 50 in each group,
confidence intervals for the mean can be found by the Normal
approximation (§8.5). For smaller samples. confidence intervals for the
mean can be found using the t distribution provided the data follow or can
be transformed to a Normal distribution (§10.3, §10.4). If not, a
significance test of the null hypothesis that the means are equal can be
carried out using the Mann–Whitney U test (§12.2). This can be useful
when the data are censored, that is, there are values too small or too
large to measure. This happens, for example, when concentrations are
too small to measure and labelled ‘not detectable’. Provided that data are
from Normal distributions, it is possible to compare the variances of the
groups using the F test (§10.8).
Ordinal data. The tendency for one group to exceed members of the
other is tested by the Mann–Whitney U test (§12.2).
Ordered nominal data. First the data is set out as a two way table, one
variable being group and the other the ordered nominal data. A chi-
squared test
(§13.1) will test the null hypothesis that there is no relationship between
group and variable, but takes no account of the ordering. This is done by
using the chi-squared test for trend, which takes the ordering into account
and provides a much more powerful test (§13.8).
Table 14.1. Methods for comparing two

samples
Size of
Type of data Method
sample
Interval Large, > 50 Normal

each sample distribution for
means (§8.5,
§9.7)
Small, < 50 Two-sample t

each method (§10.3)
sample, with
Normal
distribution
and uniform
variance
Small, < 50 Mann–Whitney

each U test (§12.2)
sample,
non-Normal
Ordinal Any Mann–Whitney

U test (§12.2)
Nominal, Large, n > Chi-squared for

ordered 30 trend (§13.8)
Nominal, not Large, most Chi-squared

ordered expected test (§13.1)
frequencies
> 5
Small, more Reduce number

than 20% of categories by
expected combining or
frequencies excluding as
< 5 appropriate
(§13.3)
Dichotomous Large, all Comparison of

expected two proportions
frequencies (§8.6, §9.8),
> 5 chi-squared test
(§13.1), odds
ratio (§13.7)
Small, at Chi-squared
least one test with Yates'
expected correction
frequency < (§13.5),
5 Fisher's exact
test (§13.4)
Nominal data. Set the data out as a two way table as described above.
The chi-squared test for a two way table is the appropriate test (§13.1).
The condition for validity of the test, that at least 80% of the expected
frequencies should be greater than 5, must be met by combining or
deleting categories as appropriate (§13.3). If the table reduces to a 2 by 2
table without the condition being met, use Fisher's exact test.
Dichotomous data. For large samples, either present the data as two
proportions and use the Normal approximation to find the confidence
interval for the difference (§8.6), or set the data up as a 2 by 2 table and
do a chi-squared test (§13.1). These are equivalent methods. An odds
ratio can also be calculated (§13.7). If the sample is small, the fit to the
Chi-squared distribution can be improved by using Yates' correction
(§13.5). Alternatively, use Fisher's exact test (§13.4).
Table 14.2. Methods for differences in one or

paired sample
Type of data Size of sample Method
Interval Large, > 100 Normal

distribution
(§8.3)
Small, < 100, Paired t

Normal method
differences (§10.2)
Small, < 100, Wilcoxon

non-Normal matched pairs
differences test (§12.3)
Ordinal Any Sign test

(§9.2)
Nominal, Any Sign test

ordered (§9.2)
Nominal Any Stuart test

(§13.9)
Dichotomous Any McNemar's

test (§13.9)
14.4 * One sample and paired samples
Methods of analysis for paired samples are summarized in Table 14.2.
Interval data. Inferences are on differences between the variable as
observed on the two conditions. For large samples, say n > 100, the
confidence interval for the mean difference is found using the Normal
approximation (§8.3). For small samples, provided the differences are
from a Normal distribution, use the paired t test (§10.2). This assumption
is often very reasonable, as most of the variation between individuals is
removed and random error is largely made up of measurement error.
Furthermore, the error is the result of two added measurement errors and
so tends to follow a Normal distribution anyway. If not, transformation of
the original data will often make differences Normal (§10.4). If no
assumption of a Normal distribution can be made, use the Wilcoxon
signed-rank matched-pairs test (§12.3).
It is rarely asked whether there is a difference in variability in paired data.
This can be tested by finding the differences between the two conditions
and their sum. Then if there is no change in variance the correlation
between difference and sum has expected value zero (Pitman's test).
This is not obvious but it is true.
Ordinal data. If the data do not form an interval scale, as noted in §14.2
the difference between conditions is not meaningful. However, we can
say what direction the difference is in, and this can be examined by the
sign test (§9.2).
Ordered nominal data. Use the sign test, with changes in one direction
being positive, in the other negative, no change as zero (§9.2).
Nominal data. With more than two categories, this is difficult. Use Stuart's
generalization to more than two categories of McNemar's test (§13.9).
Dichotomous data. Here we are comparing the proportions of individuals
in a given state under the two conditions. The appropriate test is
McNemar's test (§13.9).
14.5 * Relationship between two variables

The methods for studying relationships between variables are
summarized in Table 14.3. Relationships with dichotomous variables can
be studied as the difference between two groups (§14.3), the groups
being defined by the two states of the dichotomous variable.
Dichotomous data have been excluded from the text of this section, but
are included in Table 14.3.
Interval and interval data. Two methods are used: regression and
correlation. Regression (§11.2, §11.5) is usually preferred, as it gives
information about the nature of the relationship as well as about its
existence. Correlation (§11.9) measures the strength of the relationship.
For regression, residuals about the line must follow a Normal distribution
with uniform variance. For estimation, the correlation coefficient requires
an assumption that both variables follow a Normal distribution, but to test
the null hypothesis only one variable needs to follow a Normal
distribution. If neither variable can be assumed to follow a Normal
distribution or be transformed to it (§11.8), use rank correlation (§12.4,
§12.5).
Interval and ordinal data. Rank correlation coefficient (§12.4, §12.5).
Interval and ordered nominal data. This can be approached by rank
correlation, using Kendall's τ (§12.5) because it copes with the large
number of ties better than does Spearman's ρ, or by analysis of variance
as described for interval and nominal data. The latter requires an
assumption of Normal distribution and uniform variance for the interval
variable. These two approaches are not equivalent.
Interval and nominal data. If the interval scale follows a Normal
distribution, use one-way analysis of variance (§10.9). The assumption is
that within categories the interval variable is from Normal distributions
with uniform variance. If this assumption is not reasonable, use Kruskal–
Wallis analysis of variance by ranks (§12.2).
Ordinal and ordinal data. Use a rank correlation coefficient, Spearman's ρ
(§12.4) or Kendall's τ (§12.5). Both will give very similar answers for
testing the null hypothesis of no relationship in the absence of ties. For
data with many ties and for comparing the strengths of different
relationships, Kendall's τ is preferable.
Ordinal and ordered nominal data. Use Kendall's rank correlation
coefficient, τ (§12.5).
Ordinal and nominal data. Kruskal–Wallis one-way analysis of variance
by ranks (§12.2).
Ordered nominal and ordered nominal data. Use chi-squared for trend
(§13.8).
Ordered nominal and nominal data. Use the chi-squared test for a two-
way table (§13.1).
Nominal and nominal data. Use the chi-squared test for a two-way table
(§13.1), provided the expected values are large enough. Otherwise use
Yates' correction (§13.5) or Fisher's exact test (§13.4).
Table 14.3. Methods for relationships between variables
Interval,
Interval,
non- Ordinal
Normal
Normal
Interval Regression Regression Rank

Normal (§11.2) (§11.2) correlation
correlation Rank (§12.4,
(§11.9) correlation §12.5)
(§12.4,
§12.5)
Interval, non- Regression Rank Rank

Normal (§11.2) correlation correlation
rank (§12.4, (§12.4,
correlation §12.5) §12.5)
(§12.4,
§12.5)
Ordinal Rank Rank Rank

correlation correlation correlation
(§12.4, (§12.4, (§12.4,
§12.5) §12.5) §12.5)
Nominal, Kendall's Kendall's Kendall's

ordered rank rank rank
correlation correlation correlation
(§12.5) (§12.5) (§12.5)
Nominal Analysis of Kruskal– Kruskal–

variance Wallis test Wallis test
(§10.9) (§12.2) (§12.2)
Dichotomous t test Large Mann–

(§10.3) sample Whitney U
Normal Normal test (§12.2
test (§8.5, test (§8.5,
§9.7) §9.7)
Mann–
Whitney U
test (§12.2)
Nominal,
Nominal Dichotomous
ordered
Interval Rank Analysis of t test (§10.3
Normal correlation variance Normal test
(§12.4, (§10.9) (§8.5, §9.7
§12.5)
Interval, non- Kendall's Kruskal- Large

Normal rank Wallis test sample
correlation (§12.2) Normal test
(§12.5) (§8.5, §9.7
Mann–
Whitney U
test (§12.2
Ordinal Kendall's Kruskal- Mann-

rank Wallis test Whitney U
correlation (§12.2) test (§12.2
(§12.5)
Nominal, Chi- Chi- Chi-squared

ordered squared squared test for trend
test for test (§13.1) (§13.8)
trend
(§13.8)
Nominal Chi- Chi- Chi-squared

squared squared test (§13.1
test (§13.1) test (§13.1)
Dichotomous Chi- Chi- Chi-squared
squared squared test (§13.1
test for test (§13.1) §13.5)
trend Fisher's
(§13.8) exact test
(§13.4)

(* Each branch is either true or false)
74. The following variables have interval scales of measurement:
(a) height;
(b) presence or absence of asthma;
(c) Apgar score;
(d) age;
(e) Forced Expiratory Volume.
View Answer
75. The following methods may be used to investigate a relationship

between two continuous variables:
(a) paired t test;
(b) the correlation coefficient, r;
(c) simple linear regression;
(d) Kendall's τ;
(e) Spearman's ρ.
View Answer
76. When analysing nominal data the following statistical methods

may be used:
(a) simple linear regression;
(b) correlation coefficient, r;
(c) paired t test;
(d) Kendall's τ;
(e) chi-squared test.
View Answer
77. To compare levels of a continuous variable in two groups,

possible methods include:
(a) the Mann–Whitney U test;
(b) Fisher's exact test;
(c) a t test;
(d) Wilcoxon matched-pairs signed-rank test;
(e) the sign test.
View Answer
Table 14.4. Number of rejection episodes over

16 weeks following heart transplant in two
groups of patients
Episodes GroupA GroupB Total
0 10 8 18
1 15 6 21
2 4 0 4
3 3 0 3
Total patients 32 14 46
78. Table 14.4 shows the number of rejection episodes following

heart transplant in two groups of patients:
(a) the rejection rates in the two populations could be compared by
a Mann–Whitney U test;
(b) the rejection rates in the two populations could be compared by
a two-sample t test;
(c) the rejection rates in the two populations could be compared by
a chi-squared test for trend:
(d) the chi-squared test for a 4 by 2 table would not be valid;
(e) the hypothesis that the number of episodes follows a Poisson
distribution could be investigated using a chi-squared test for
goodness of fit.
View Answer
79. Twenty arthritis patients were given either a new analgesic or

aspirin on successive days in random order. The grip strength of
the patients was measured. Methods which could be used to
investigate the existence of a treatment effect include:
(a) Mann–Whitney U test;
(b) paired t method;
(c) sign test;
(d) Normal confidence interval for the mean difference;
(e) Wilcoxon matched-pairs signed-rank test.
View Answer
80. In a study of boxers, computer tomography revealed brain

atrophy in 3 of 6 professionals and 1 of 8 amateurs (Kaste et al.
1982). These groups could be compared using:
(a) Fisher's exact test;
(b) the chi-squared test;
(c) the chi-squared test with Yates' correction;
(d) * McNemar's test;
(e) the two-sample t test.
View Answer
Table 14.5. Gastric pH and urinary nitrite concentration

in 26 subjects (Hall and Northfield, private
communication)
pH Nitrite pH Nitrite pH Nitrite pH
1.72 1.64 2.64 2.33 5.29 50.6 5.77
1.93 7.13 2.73 52.0 5.31 43.9 5.86
1.94 12.1 2.94 6.53 5.50 35.2 5.90

2.03 15.7 4.07 22.7 5.55 83.8 5.91
2.11 0.19 4.91 17.8 5.59 52.5 6.03
2.17 1.48 4.94 55.6 5.59 81.8
2.17 9.36 5.18 0.0 5.17 21.9
14E * Exercise: Choosing a statistical method

1. In a cross-over trial to compare two appliances for ileostomy
patients, of 14 patients who received system A first, 5 expressed a
preference for A, 9 for system B and none had no preference. Of the
patients who received system B first, 7 preferred A, 5 preferred B
and 4 had no preference. How would you decide whether one
treatment was preferable? How would you decide whether the order
of treatment influenced the choice?
View Answer
2. Burr et al. (1976) tested a procedure to remove house-dust mites

from the bedding of adult asthmatics in attempt to improve
subjects' lung function, which they measured by PEFR. The trial
was a two period cross-over design, the control or placebo
treatment being thorough dust removal from the living room. The
means and standard errors for PEFR in the 32 subjects were:
active treatment: 335 litres/min, SE = 19.6 litres/min
placebo treatment: 329 litres/min, SE = 20.8 litres/min
differences within subjects: (treatment – placebo) 6.45 litres/min, SE
= 5.05 litres/min
How would you decide whether the treatment improves PEFR?
View Answer
3. In a trial of screening and treatment for mild hypertension

(Reader et al. 1980), 1138 patients completed the trial on active
treatment, with 9 deaths, and 1080 completed on placebo, with 19
deaths. A further 583 patients allocated to active treatment
withdrew, of whom 6 died, and 626 allocated to placebo withdrew, of
whom 16 died during the trial period. How would you decide
whether screening and treatment for mild hypertension reduces the
risk of dying?
View Answer
4. Table 14.5 shows the pH and nitrite concentrations in samples of

gastric fluid from 26 patients. A scatter diagram is shown in Figure
14.1. How would you assess the evidence of a relationship between
pH and nitrite concentration?
View Answer
5. The lung function of 79 children with a history of hospitalization

for whooping cough and 178 children without a history of whooping
cough, taken from the same school classes, was measured. The
mean transit time for the whooping cough cases was 0.49 seconds
(s.d. = 0.14 seconds) and for the controls 0.47 seconds (s.d. = 0.11
seconds), (Johnston et al. 1983). How could you analyse the
difference in lung function between children who had had whooping
cough and those who had not? Each case had two matched
controls. If you had all the data, how could you use this
information?
View Answer
Fig. 14.1. Gastric pH and urinary nitrite
Table 14.6. Visual acuity and results of a

contrast sensitivity vision test before and after
cataract surgery (Wilkins, personal
communication)
Contrast sensitivity
Visual acuity
Case test
Before After Before After
1 6/9 6/9 1.35 1.50
2 6/9 6/9 0.75 1.05

3 6/9 6/9 1.05 1.35
4 6/9 6/9 0.45 0.90
5 6/12 6/6 1.05 1.35
6 6/12 6/9 0.90 1.20
7 6/12 6/9 0.90 1.05
8 6/12 6/12 1.05 1.20
9 6/12 6/12 0.60 1.05
10 6/18 6/6 0.75 1.05
11 6/18 6/12 0.90 1.05
12 6/18 6/12 0.90 1.50
13 6/24 6/18 0.45 0.75
14 6/36 6/18 0.15 0.45
15 6/36 6/36 0.45 0.60
16 6/60 6/9 0.45 1.05

17 6/60 6/12 0.30 1.05
6. Table 14.6 shows some data from a pre- and post-treatment study
of cataract patients. The second number in the visual acuity score
represents the size of letter which can be read at a distance of six
metres, so high numbers represent poor vision. For the contrast
sensitivity test, which is a measurement, high numbers represent
good vision. What methods could be used to test the difference in
visual acuity and in the contrast sensitivity test pre- and post-
operation? What method could be used to investigate the
relationship between visual acuity and the contrast sensitivity test
post-operation?
View Answer
Table 14.7. Asthma or wheeze by maternal age

(Anderson et al. 1986)
Mother's age at
Asthma or wheeze child's birth
reported
15–19 20–29 30+
Never 261 4017 2146
Onset by age 7 103 984 487

Onset from 8 to 11 27 189 95
Onset from 12 to 16 20 157 67
Table 14.8. Colon transit time (hours) in groups of mo

immobile elderly patients (data of Dr Michael O'Co
Mobile patients Immobile patients
8.4 21.6 45.5 62.4 68.4 15.6 38.8 54.0
14.4 25.2 48.0 66.0 24.0 42.0 54.0
19.2 30.0 50.4 66.0 24.0 43.2 57.6
20.4 36.0 60.0 66.0 32.4 47.0 58.8
20.4 38.4 60.0 67.2 34.8 52.8 62.4
n1 = 21, [x with bar above]1 = n1 = 21, [x with bar abo

42.57, s1 = 20.58 49.63, s2 = 16.39
7. Table 14.7 shows the relationship between age of onset of asthma

in children and maternal age at the child's birth. How would you test
whether these were related? The children were all born in one week
in March, 1958. Apart from the possibility that young mothers in
general tend to have children prone to asthma, what other possible
explanations are there for this finding?
View Answer
8. In a study of thyroid hormone in premature babies, we wanted to

study the relationship of free T3 measured at several time points
over seven days with the number of days the babies remained
oxygen dependent. Some babies died, mostly within a few days of
birth, and some babies went home still oxygen dependent and were
not followed any longer by the researchers. How could you reduce
the series of T3 measurements on a baby to a single variable? How
could you test the relationship with time on oxygen?
View Answer
9. Table 14.8 shows colon transit times measured in a group of

elderly patients who were mobile and in a second group who were
unable to move independently. Figure 14.2 shows a scatter diagram
and histogram and Normal plot of residuals for these data. What two
statistical approaches could be used here? Which would you prefer
and why?
View Answer
Fig. 14.2. Scatter plot, histogram, and Normal plot for the colon
transit time data of Table 14.8
> Table of Contents > 15 - Clinical measurement
15
Clinical measurement
15.1 Making measurements

In this chapter we shall look at a number of problems associated with
clinical measurement. These include how precisely we can measure, how
different methods of measurement can be compared, how measurements
can be used in diagnosis and how to deal with incomplete measurements
of survival.
When we make a measurement, particularly a biological measurement,
the number we obtain is the result of several things: the true value of the
quantity we want to measure, biological variation, the measurement
instrument itself, the position of the subject, the skill, experience and
expectations of the observer, and even the relationship between observer
and subject. Some of these factors, such as the variation within the
subject, are outside the control of the observer. Others, such as position,
are not, and it is important to standardize these. One which is most under
our control is the precision with which we read scales and record the
result. When blood pressure is measured, for example, some observers
record to the nearest 5 mm Hg, others to the nearest 10 mm Hg. Some
observers may record diastolic pressure at Korotkov sound four, others at
five. Observers may think that as blood pressure is such a variable
quantity, errors in recording of this magnitude are unimportant. In the
monitoring of the individual patient, such lack of uniformity may make
apparent changes difficult to interpret. In research, imprecise
measurement can lead to problems in the analysis to loss of power.
How precisely should we record data? While this must depend to some
extent on the purpose for which the data are to be recorded, any data
which are to be subjected to statistical analysis should be recorded as
precisely as possible. A study can only be as good as the data, and data
are often very costly and time-consuming to collect. The precision to
which data are to be recorded and all other. procedures to be used in
measurement should be decided in advance and stated in the protocol,
the written statement of how the study is to be carried out. We should
bear in mind that the precision of recording depends on the number of
significant figures (§5.2) recorded, not the number of decimal places. The
observations 0.15 and 1.66 from Table 4.8, for example, are both
recorded to two decimal places, but 0.15 has two significant figures and
1.66 has three. The second observation is recorded more precisely. This
becomes very important when we come to analyse the data, for the data
of Table 4.8 have a
skew distribution which we wish to log transform. The greater imprecision

of recording at the lower end of the scale is magnified by the
transformation.
In measurement there is usually uncertainty in the last digit. Observers
will often have some values for this last digit which they record more
often than others. Many observers are more likely to record a terminal
zero than a nine or a one, for example. This is known as digit
preference. The tendency to read blood pressure to the nearest 5 or 10
mm Hg mentioned above is an example of this. Observer training and
awareness of the problem help to minimize digit preference, but if
possible readings should be taken to sufficient significant figures for the
last digit to be unimportant. Digit preference is particularly important
when differences in the last digit are of importance to the outcome, as it
might be in Table 15.1, where we are dealing with the difference between
two similar numbers. Because of this it is a mistake to have one
measurer take readings under one set of conditions and a second under
another, as their degree of digit preference may differ. It is also important
to agree the precision to which data are to be recorded and to ensure
that instruments have sufficiently fine scales for the job in hand.
15.2 * Repeatability and measurement error

I have already discussed some factors which may produce bias in
measurements (§2.7, §2.8, §3.6). I have not yet considered the natural
biological variability, in subject and in measurement method, which may
lead to measurement error. ‘Error’ comes from a Latin root meaning ‘to
wander’, and its use in statistics in closely related to this, as in §11.2, for
example. Thus error in measurement may include the natural continual
variation of a biological quantity, when a single observation will be used
to characterize the individual. For example, in the measurement of blood
pressure we are dealing with a quantity that varies continuously, not only
from heart beat to heart beat but from day to day, season to season, and
even with the sex of the measurer. The measurer, too, will show variation
in the perception of the Korotkov sound and reading of the manometer.
Because of this, most clinical measurements cannot be taken at face
value without some consideration being given to their error.
The quantification of measurement error is not difficult in principle. To do
it we need a set of replicate readings, obtained by measuring each
member of a sample of subjects more than once. We can then estimate
the standard deviation of repeated measurements on the same subject.
Table 15.1 shows some replicated measurements of peak expiratory flow
rate, made by the same observer (myself) with a Wright Peak Flow
Meter. For each subject, the measured PEFR varies from observation to
observation. This variation is the measurement error. We can quantify
measurement error in two ways: using the standard deviation for
repeated measurements on the same subject and by correlation.
Table 15.1. Pairs of readings made with a Wright

Peak Flow Meter on 17 healthy volunteers
PEFR PEFR
Subject (litres/min) Subject (litres/min)
First Second First Second
1 494 490 10 433 429
2 395 397 11 417 420
3 516 512 12 656 633
4 434 401 13 267 275
5 476 470 14 478 492
6 557 611 15 178 165
7 413 415 16 423 372
8 442 431 17 427 421
9 650 638
Table 15.2. Analysis of variance by subject for the PEFR

Table 15.1
Source Degrees Sum of Mean Variance
of of squares square ratio (
variation freedom
Total 33 445581.5
Between 16 441598.5 27599.9 117.8

subjects
Residual 17 3983.0 234.3

(within
subjects)
Table 15.3. Analysis of variance by subject for the log
transformed PEFR data of Table 15.1
Source Degrees
of of
squares square ratio (
variation freedom
Total 33 3.160104
Subjects 16 3.139249 0.196203 159.9

Residual 17 0.020855 0.001227
(within
subjects)
We should check to see whether the error does depend on the value of
the measurement, usually being larger for larger values. We can do this
by plotting a scatter diagram of the absolute value of the difference (i.e.
ignoring the sign) and the mean of the two observations (Figure 15.1).
For the PEFR data, there is no obvious relationship. We can check this
by calculating a correlation (§11.9) or rank correlation coefficient (§12.4,
§12.5). For Figure 15.1 we have τ = 0.17, P = 0.3, so there is little to
suggest that the measurement error is related to the size of the PEFR.
Hence the coefficient of variation is not as appropriate as the within
subjects standard deviation as a representation of the measurement
error. For most medical measurements, the standard deviation is either
independent of or proportional to the measurement and so one of these
two approaches can be used.
Fig. 15.1. Absolute difference versus sum for 17 pairs of Wright
Peak Flow Meter measurements
Measurement error may also be presented as the correlation coefficient

between pairs of readings. This is sometimes called the reliability of the
measurement, and is often used for psychological measurements using
questionnaire scales. However, the correlation depends on the amount of
variation between subjects. If we deliberately choose subjects to have a
wide spread of possible values, the correlation will be bigger than if we
take a random sample of subjects. Thus this method should only be used
if we have a representative sample of the subjects in whom we are
interested. The intra-class correlation coefficient (§11.13), which does not
take into account the order in which observations were taken and which
can be used with more than two observations per subject, is preferred for
this application. Applying the method of §11.13 to Table 15.1 we get ICC
= 0.98. ICC and sw are closely related, because ICC = 1-sw2/(sb2+sw2).
ICC therefore depends also on the variation between subjects, and thus
relates to the population of which the subjects can be considered a
random sample. Streiner and Norman (1996) give an interesting
discussion.
15.3 * Comparing two methods of measurement

In clinical measurement, most of the things we want to measure, hearts,
lungs, livers and so on, are deep within living bodies and out of reach.
This means that many of the methods we use to measure them are
indirect and we cannot be sure how closely they are related to what we
really want to know. When a new method of measurement is developed,
rather than compare its outcome to a set of known values we must often
compare it to another method just as indirect. This is a common type of
study, and one which is often badly done (Altman and Bland 1983, Bland
and Altman 1986).
Table 15.4 shows measurements of PEFR by two different methods, the
Wright meter data coming from Table 15.1. For simplicity, I shall use only
one measurement by each method here. We could make use of the
duplicate
data by using the average of each pair first, but this introduces an extra
stage in the calculation. Bland and Altman (1986) give details.
Table 15.4. Comparison of two methods of

measuring PEFR
PEFR
Subject (litres/min) Difference
number Wright Mini Wright - mini
meter meter
1 494 512 -18
2 395 430 -35
3 516 520 -4
4 434 428 6
5 476 500 -24
6 557 600 -43
7 413 364 49
8 442 380 62
9 650 658 -8
10 433 445 -12
11 417 432 -15
12 656 626 30
13 267 260 7
14 478 477 1
15 178 259 -81
16 423 350 73
17 427 451 -24
Total -36
Mean 2.1
S.d. 38.8
The first step in the analysis is to plot the data as a scatter diagram
(Figure 15.2). If we draw the line of equality, along which the two
measurements would be exactly equal, this gives us an idea of the extent
to which the two methods agree. This is not the best way of looking at
data of this type, because much of the graph is empty space and the
interesting information is clustered along the line. A better approach is to
plot the difference between the methods against the sum or average. The
sign of the difference is important, as there is a possibility that one
method may give higher values than the other and this may be related to
the true value we are trying to measure. This plot is also shown in Figure
15.2.
Two methods of measurement agree if the difference between
observations on the same subject using both methods is small enough
for us to use the methods interchangeably. How small this difference has
to be depends on the measurement and the use to which it is to be put. It
is a clinical, not a statistical, decision. We quantify the differences by
estimating the bias, which is the mean difference, and the limits within
which most differences will lie. We estimate these limits from the mean
and standard deviation of the differences. If we are to estimate these
quantities, we want them to be the same for high values and for low
values of the measurement. We can check this from the plot. There is no
clear evidence of a relationship between difference and mean in Figure
15.4, and we can check this by a test of significance using the correlation
coefficient. We get r = 0.19, P = 0.5.
The mean difference is close to zero, so there is little evidence of overall
bias.
We can find a confidence interval for the mean difference as described in

§10.2. The differences have a mean [d with bar above] = -2.1 litres/min,
and a standard deviation of 38.8. The standard error of the mean is thus
s/√n = 38.8/√17 = 9.41 litres/min and the corresponding value of t with 16
degrees of freedom is 2.12. The 95% confidence interval for the bias is
thus -2.1 ± 2.12 × 9.41 = -22 to +18 litres/min. Thus on the basis of these
data we could have a bias of as much as 22 litres/min, which could be
clinically important. The original comparison of these instruments used a
much larger sample and found that any bias was very small (Oldham et
al. 1979).
Fig. 15.2. PEFR measured by two different instruments, mini meter

versus Wright meter and difference versus mean of mini and Wright
meters
Fig. 15.3. Distribution of differences between PEFR measured by

two methods
The standard deviation of the differences between measurements made

by the two methods provides a good index of the comparability of the
methods. If we can estimate the mean and standard deviation reliably,
with small standard errors, we can then say that the difference between
methods will be at most two standard deviations on either side of the
mean for 95% of observations. These [d with bar above] ± 2s limits for
the difference are called the 95% limits of agreement. For the PEFR
data, the standard deviation of the differences is estimated to be 38.8
litres/min and the mean is -2 litres/min. Two standard deviations is
therefore 78 litres/min. The reading with the mini meter is expected to be
80 litres below to 76 litres above for most subjects. These limits are
shown as horizontal lines in Figure 15.4. The limits depend on the
assumption that the distribution of
the differences is approximately Normal, which can be checked by

histogram and Normal plot (§7.5) (Figure 15.3).
Fig. 15.4. Difference versus sum for PEFR measured by two

methods
On the basis of these data we would not conclude that the two methods
are comparable or that the mini meter could reliably replace the Wright
peak flow meter. As remarked in §10.2, this meter had received
considerable wear.
When there is a relationship between the difference and the mean, we
can try to remove it by a transformation. This is usually accomplished by
the logarithm, and leads to an interpretation of the limits similar to that
described in §15.2. Bland and Altman (1986, 1999) give details.
15.4 Sensitivity and specificity

One of the main reasons for making clinical measurements is to aid in
diagnosis. This may be to identify one of several possible diagnoses in a
patient, or to find people with a particular disease in an apparently
healthy population. The latter is known as screening. In either case the
measurement provides a test which enables us to classify subjects into
two groups, one group whom we think are likely to have disease in which
we are interested, and another group unlikely to have the disease. When
developing such a test, we need to compare the test result with a true
diagnosis. The test may be based on a continuous variable and the
disease indicated if it is above or below a given level, or it may be a
qualitative observation such as carcinoma in situ cells on a cervical
smear. In either case I shall call the test positive if it indicates the disease
and negative if not, and the disease positive if the disease is later
confirmed, negative if not.
How do we measure the effectiveness of the test? Table 15.5 shows
three artificial sets of test and disease data. We could take as an index of
test effectiveness the proportion giving the correct diagnosis from the
test. For Test 1 in the example it is 94%. Now consider Test 2, which
always gives a negative result. Test 2 will never detect any cases of the
disease. We are now right for 95% of the subjects! However, the first test
is useful, in that it detects some
cases of the disease, and the second is not, so this is clearly a poor
index.
Table 15.5. Some artificial test and diagnosis

data
Test 1 Test 2 Test 3

Disease Total
+ve -ve +ve -ve +ve -ve
Yes 4 1 0 5 2 3 5
No 5 90 0 95 0 95 95
Total 9 91 0 100 2 98 100
There is no one simple index which enables us to compare different tests

in all the ways we would like. This is because there are two things we
need to measure: how good the test is at finding disease positives, i.e.
those with the condition, and how good the test is at excluding disease
negatives, i.e. those who do not have the condition. The indices
conventionally employed to do this are:
In other words, the sensitivity is a proportion of disease positives who are

test positive, and the specificity is the proportion of disease negatives
who are test negatives. For our three tests these are:
Test 1 Test 2 Test 3
Sensitivity 0.80 0.00 0.40
Specificity 0.95 1.00 1.00

Test 2, of course, misses all the disease positives and finds all the
disease negatives, by saying all are negative. The difference between
Tests 1 and 3 is brought out by the greater sensitivity of 1 and the greater
specificity of 3. We are comparing tests in two dimensions. We can see
that Test 3 is better than Test 2, because its sensitivity is higher and
specificity the same. However, it is more difficult to see whether Test 3 is
better than Test 1. We must come to a judgement based on the relative
importance of sensitivity and specificity in the particular case.
Sensitivity and specificity are often multiplied by 100 to give percentages.
They are both binomial proportions, so their standard errors and
confidence intervals are found as described in §8.4 and §8.8. Because
the proportions are often near to 1.0, the large sample approach (§8.4)
may not be valid. The exact method using the Binomial probabilities
(§8.8) is preferable. Harper and Reeves (1999) point out that confidence
intervals are almost always omitted in studies of diagnostic tests reported
outside the major general medical journals, and recommend that they
should always be given. As the reader might expect, I agree with them!
The sample size required for the reliable estimation of sensitivity and
specificity can be calculated as described in §18.2.
Sometimes a test is based on a continuous variable. For example, Table
15.6 shows measurements of creatinekinase (CK) in patients with
unstable angina
and acute myocardial infarction. Figure 15.5(a) shows a scatter plot. We

wish to detect patients with AMI among patients who may have either
condition and this measurement is a potential test, AMI patients tending
to have high values. How do we choose the cut-off point? The lowest CK
in AMI patients is 90, so a cut-off below this will detect all AMI patients.
Using 80, for example, we would detect all AMI patients, sensitivity =
1.00, but would also only have 42% of angina patients below 80, so the
sensitivity = 0.42. We can alter the sensitivity and specificity by changing
the cut-off point. Raising the cut-off point will mean fewer cases will be
detected and so the sensitivity will be decreased. However, there will be
fewer false positives, positives on test but who do not in fact have the
disease, and the specificity will be increased. For example, if CK ≥ 100
were the criterion for AMI, sensitivity would be 0.96 and specificity 0.62.
There is a trade-off between sensitivity and specificity. It can be helpful to
plot sensitivity against specificity to examine this trade-off. This is called a
receiver operating characteristic or ROC curve. (The name comes
from telecommunications.)
We often plot sensitivity against 1–specificity, as in Figure 15.5(b). We

can see from Figure 15.5(b) that we can get both high sensitivity and high
specificity if we choose the right cut-off. With 1-specificity less than 0.1,
i.e. sensitivity greater than 0.9. we can get sensitivity greater than 0.9
also. In fact, a cut-off of 200 would give sensitivity = 0.93 and specificity =
0.91 in this sample. These estimates will be biased, because we are
estimating the cut-off and testing it in the same sample. We should check
the sensitivity and specificity of this cut-off in a different sample to be
sure.
Table 15.6. Creatinekinase in patients with unstable

angina and acute myocardial infarction (AMI) (data of
Frances Boa)
Unstable angina AMI
23 48 62 83 104 130 307 90 648
33 49 63 84 105 139 351 196 894
36 52 63 85 105 150 360 302 962
37 52 65 86 107 155 311 1015
37 52 65 88 108 157 325 1143
41 53 66 88 109 162 335 1458
41 54 67 88 111 176 347 1955

41 57 71 89 114 180 349 2139
42 57 72 91 116 188 363 2200
42 58 72 94 118 198 377 3044
43 58 73 94 121 226 390 7590
45 58 73 95 121 232 398 11138
47 60 75 97 122 257 545
48 60 80 100 126 257 577
48 60 80 103 130 297 629
Fig. 15.5. Scatter diagram and with ROC curve for the data of Table
15.6
The area under the ROC curve is often quoted (here it is 0.975 3). It
estimates the probability that a member of one population chosen at
random will exceed a member of the other population, in the same way
as does U/n1n2 in the Mann–Whitney U test (§12.2). It can be useful in
comparing different tests. In this study another blood test gave us an
area under the ROC curve = 0.9825, suggesting that the test may be
slightly better than CK.
We can also estimate the positive predictive value or PPV, the
probability that a subject who is test positive will be a true positive (i.e.
has the disease and is correctly classified), and the negative predictive
value or NPV, the probability that a subject who is test negative will be a
true negative (i.e. does not have the disease and is correctly classified).
These depend on the prevalence of the condition, Pprev, as well as the
sensitivity, Psens, and the specificity, pspec. If the sample is a single group
of people, we know the prevalence and can estimate PPV and NPV for
this population directly as simple proportions. If we started with a sample
of cases and a sample of controls, we do not know the prevalence, but
we can estimate PPV and NPV for a population with any given
prevalence. As described in §6.8, psens is the conditional probability of a
positive test given the disease, so the probability of being both test
positive and disease positive is psens × pprev. Similarly, the probability of
being both test negative and disease positive is (1 - pspec) × (1 - pprev).
The probability of being test positive is the sum of these (§6.2): psens ×
pprev + (1 - pspec) × (1 - pprev) and the PPV is
Similarly, the NPV is
In screening situations the prevalence is almost always small and the

PPV is low. Suppose we have a fairly sensitive and specific test, psens =
0.95 and pspec = 0.90, and the disease has prevalence pprev = 0.01 (1%).
Then
so only 8.8% of test positives would be true positives, but almost all test
negatives would be true negatives. Most screening tests are dealing with
much smaller prevalences than this, so most test positives are false
positives.
15.5 Normal range or reference interval

In §15.4 we were concerned with the diagnosis of particular diseases. In
this section we look at it the other way round and ask what values
measurements on normal, healthy people are likely to have. There are
difficulties in doing this. Who is ‘normal’ anyway? In the UK population
almost everyone has hard fatty deposits in their coronary arteries, which
result in death for many of them. Very few Africans have this; they die
from other causes. So it is normal in the UK to have an abnormality. We
usually say that normal people are the apparently healthy members of
the local population. We can draw a sample of these as described in
Chapter 3 and make the measurement on them.
The next problem is to estimate the set of values. If we use the range of
the observations, the difference between the two most extreme values,
we can be fairly confident that if we carry on sampling we will eventually
find observations outside it. and the range will get bigger and bigger
(§4.7). To avoid this we use a range between two quantiles (§4.7),
usually the 2.5 centile and the 97.5 centile, which is called the normal
range, 95% reference range or 95% reference interval. This leaves 5%
of normals outside the ‘normal range’, which is the set of values within
which 95% of measurements from apparently healthy individuals will lie.
A third difficulty comes from confusion between ‘normal’ as used in
medicine and ‘Normal distribution’ as used in statistics. This has led
some people to develop approaches which say that all data which do not
fit under a Normal curve are abnormal! Such methods are simply absurd,
there is no reason to suppose that all variables follow a Normal
distribution (§7.4, §7.5). The term ‘reference interval’, which is becoming
widely used, has the advantage of avoiding this confusion. However, the
most commonly used method of calculation rests on the assumption that
the variable follows a Normal distribution.
We have already seen that in general most observations fall within two
standard deviations of the mean, and that for a Normal distribution 95%
are within these limits with 2.5% below and 2.5% above. If we estimate
the mean and standard deviation of data from a Normal population we
can estimate the reference interval as [x with bar above] - 2s to [x with
bar above] + 2s.
Consider the FEV1 data of Table 4.5. We will estimate the reference
interval for FEV1 in male medical students. We have 57 observations,
mean 4.06 and standard deviation 0.67 litres. The reference interval is
thus from 2.7 to 5.4 litres. From Table 4.4 we see that in fact only one
student (2%) is outside these limits, although the sample is rather small.
Hence, provided Normal assumptions hold, the standard error of the limit
of the reference interval is
Compare the serum triglyceride measurements of Table 4.8. As already

noted (§4.4, §7.4). the data are highly skewed, and we cannot use the
Normal method directly. If we did, the lower limit would be 0.07, well
below any of the observations, and the upper limit would be 0.94, greater
than which are 5% of the observations. It is possible for such data to give
a negative lower limit.
Because of the obviously unsatisfactory nature of the Normal method for

some data, some authors have advocated the estimation of the
percentiles directly (§4.5), without any distributional assumptions. This is
an attractive idea. We want to know the point below which 2.5% of values
will fall. Let us simply rank the observations and find the point below
which 2.5% of the observations fall. For the 282 triglycerides, the 2.5 and
97.5 centiles are found as follows. For the 2.5 centile, we find i = q(n + 1)
= 0.025 × (282+1) = 7.08. The required quantile will be between the 7th
and 8th observation. The 7th is 0.21, the 8th is 0.22 so the 2.5 centile
would be estimated by 0.21 + (0.22 - 0.21) × (7.08 - 7) = 0.211. Similarly
the 97.5 centile is 1.039.
This approach gives an unbiassed estimate whatever the distribution.

The log transformed triglyceride would give exactly the same results.
Note that the Normal theory limits from the log transformed data are very
similar. We now look at the confidence interval. The 95% confidence
interval for the q quantile, here q being 0.025 or 0.975, estimated directly
from the data is found by the Binomial distribution method (§8.9). For the
triglyceride data, n = 282 and so for the lower limit, q = 0.025, we have
This gives j = 1.9 and k = 12.2, which we round up to j = 2 and k = 13. In

the triglyceride data the second observation, corresponding to j = 2, is
0.16 and the 13th is 0.26. Thus the 95% confidence interval for the lower
reference limit is 0.16 to 0.26. The corresponding calculation for q =
0.975 gives j = 270 and k = 281. The 270th observation is 0.96 and the
281st is 1.64, giving a 95% confidence interval for the upper reference
limit of 0.96 to 1.64. These are wider confidence intervals than those
found by the Normal method, those for the long tail particularly so. This
method of estimating percentiles in long tails is relatively imprecise.
15.6 * Survival data
We often have data which represent the time from some event to death,
such as time from diagnosis or from entry to a clinical trial, but survival
analysis does not have to be about death. In cancer studies we can use
survival analysis for the time to metastasis or to local recurrence of a
tumour, in a study of medical care we can use it to analyse the time to
readmission to hospital, in a study of breast-feeding we could look at the
age at which breast-feeding ceased or at which bottle feeding was first
introduced, and in a study of the treatment of infertility we can treat the
time from treatment to conception as survival data. We usually refer to
the terminal event, death, conception, etc., as the endpoint.
Problems arise in the measurement of survival because often we do not
know the exact survival times of all subjects. This is because some will
still be surviving when we want to analyse the data. When cases have
entered the study at different times, some of the recent entrants may be
surviving, but only have been observed for a short time. Their observed
survival time may be less than those cases admitted early in the study
and who have since died. The method of calculating survival curves
described below takes this into account. Observations which are known
only to be greater than some value are right censored, often shortened
to censored. (We get left censored data when the measurement
method cannot detect anything below some cut-off value, and
observations are recorded as ‘none detectable’. The rank methods in
Chapter 12 are useful for such data.)
Table 15.7. Survival time in years of patients

after diagnosis of parathyroid cancer
Alive Deaths
<1 <1
<1 2
1 6
1 6
4 7
5 9
6 9
8 11
10 14
10
17
Table 15.7 shows some survival data, for patients with parathyroid
cancer. The survival times are recorded in completed years. A patient
who survived for 6 years and then died can be taken as having lived for 6
years and then died in the seventh. In the first year from diagnosis. one
patient died, two patients were observed for only part of this year, and 17
survived into the next year. The subjects who have only been observed
for part of the year are censored, also called lost to follow-up or
withdrawn from follow-up. (These are rather misleading names, often
wrongly interpreted as meaning that these subjects have dropped out of
the study. This may be the case, but most of these subjects are simply
still alive and their further survival is unknown.) There is no information
about the survival of these subjects after the first year, because it has not
happened yet. These patients are only at risk of dying for part of the year
and we cannot say that 1 out of 20 died as they may yet contribute
another death in the first year. We can say that such patients will
contribute half a year of risk, on average, so the number of patient years
at risk in the first year is 18 (17 who survived and 1 who died) plus 2
halves for those withdrawn from follow-up, giving 19 altogether. We get
an estimate of the probability of dying in the first year of 1/19, and an
estimated probability of surviving of 1 - 1/19. We can do this for each
year until the limits of the data are reached. We thus trace the survival of
these patients estimating the probability of death or survival at each year
and the cumulative probability of survival to each year. This set of
probabilities is called a life table.
To carry out the calculation, we first set out for each year, x, the number
alive at the start, nx, the number withdrawn during the year, wx, and the
number at risk, rx, and the number dying, dx (Table 15.8). Thus in year 1
the number at the start is 20, the number withdrawn is 2, the number at
risk r1 = n1 - 1/2w1 = 20 - 1/2 × 2 = 19 and the number of deaths is 1. As
there were 2 withdrawals and 1 death the number at the start of year 2 is
17. For each year we calculate the probability of dying in that year for
patients who have reached the beginning of it, qx = dx/rx, and hence the
probability of surviving to the next year, px = 1 - qx. Finally we calculate
the cumulative survival probability.
Table 15.8. Life table calculation for parathyroid ca
Withdrawn Prob.
Number At
Year during Deaths of
at start risk
year death
x nx wx rx dx qx
1 20 2 19 1 0.0526
2 17 2 16 0 0
3 15 0 15 1 0.0667
4 14 0 14 0 0
5 14 1 13.5 0 0
6 13 1 12.5 0 0
7 12 1 11.5 2 0.1739
8 9 0 9 1 0.1111
9 8 1 7.5 0 0
10 7 0 7 2 0.2857
11 5 2 4 0 0
12 3 0 3 1 0.3333
13 2 0 2 0 0
14 2 0 2 0 0
15 2 0 2 1 0.5000
16 1 0 1 0 0
17 1 0 1 0 0
18 1 1 0.5 0 0
rx = nx - 1/2wx, qx = dx/rx, px = 1-qx, Px = pxPx-1.
For the first year, this is the probability of surviving that year, P1 = p1. For
the second year, it is the probability of surviving up to the start of the
second year, P1, times the probability of surviving that year, p2, to give P2
= p2P1. The probability of surviving for 3 years is similarly P3 = p3P2, and
so on. From this life table we can estimate the five year survival rate, a
useful measure of prognosis in cancer. For the parathyroid cancer, the
five year survival rate is 0.8842, or 88%. We can see that the prognosis
for this cancer is quite good. If we know the exact time of death or
withdrawal for each subject, then instead of using fixed time intervals we
use x as the exact time, with a row of the table for each time when either
an endpoint or a withdrawal occurs. Then rx = nx and we can omit the rx =
nx - 1/2wx step.
We can draw a graph of the cumulative survival probability, the survival
curve. This is usually drawn in steps, with abrupt changes in probability
(Figure 15.6). This convention emphasizes the relatively poor estimation
at the long survival end of the curve, where the small numbers at risk
produced large steps. When the exact times of death and censoring are
known, this is called a Kaplan-Meier survival curve. The times at which
observations are censored may be marked by small vertical lines above
the survival curve (Figure 15.7), and the number remaining at risk may be
written at suitable intervals below the time axis.
The standard error and confidence interval for the survival probabilities
can be found (see Armitage and Berry 1994). These are useful for
estimates such as five year survival rate. They do not provide a good
method for comparing
survival curves, as they do not include all the data, only using those up to
the chosen time. Survival curves start off together at 100% survival,
possibly diverge, but eventually come together at zero survival. Thus the
comparison would depend on the time chosen. Survival curves can be
compared by several significance tests, of which the best known is the
logrank test. This is a non-parametric test which makes use of the full
survival data without making any assumption about the shape of the
survival curve.
Fig. 15.6. Survival curve for parathyroid cancer patients

Table 15.9 shows the time to recurrence of gallstones following
dissolution by bile acid treatment or lithotrypsy. Here we shall compare
the two groups defined by having single or multiple gallstones, using the
logrank test. We shall look at the quantitative variables diameter of
gallstone and months to dissolve in §17.9. Figure 15.7 show the time to
recurrence for subjects with single primary gallstones and multiple
primary gallstones. The null hypothesis is that there is no difference in
recurrence-free survival time, the alternative that there is such a
difference. The calculation of the logrank test is set out in Table 15.10.
For each time at which a recurrence or a censoring occurred, we have
the numbers under observation in each group, n1 and n2, the number of
recurrences, d1 and d2 (d for death), and the number of censorings, w1
and w2 (w for withdrawal). For each time, we calculate the probability of
recurrence, pd = (d1 + d2)/(n1 + n2), which each subject would have if the
null hypothesis were true. For each group, we calculate the expected
number of recurrences, e1 = Pd × n1 and e2 = Pd × n2. We then calculate
the numbers at risk at the next time, n1 - d1 - w1 and n2 - d2 - w2. We do
this for each time. We then add the d1 and d2 columns to get the
observed numbers of recurrences, and the e1 and e2 columns to get the
numbers of recurrences expected if the null hypothesis were true.
We have observed frequencies of recurrence d1 and d2, and expected
frequencies e1, and e2. Of course, d1 + d2 = e1 + e2, so we only need to
calculate e1 as in Table 15.10. and hence e2 by subtraction. This only
works for two groups, however, and the method of Table 15.10 works for
any number of groups.
Table 15.9. Time to recurrence of gallstones follow

dissolution, whether previous gallstones were mult
maximum diameter of previous gallstones, and mon
previous gallstones took to dissolve
Time Rec. Mult. Diam. Dis. Time Rec. Mult.
3 No Yes 4 10 13 No No
3 No No 18 3 13 No No
4 No Yes 4 4 13 Yes Yes
5 No No 19 20 14 No Yes
6 Yes Yes 5 8 16 Yes Yes
6 Yes Yes 3 18 16 No No
6 No No 25 9 17 No Yes
6 No Yes 4 6 17 No Yes
6 Yes Yes 10 38 17 Yes No
6 Yes Yes 8 15 17 No Yes
6 No Yes 4 13 18 Yes No
8 Yes No 18 14 19 Yes Yes
8 No No 15 1 20 No No
9 No No 19 8 22 No No
11 No Yes 8 12 23 No No
11 No No 15 15 24 No No
11 Yes No 5 8 24 No Yes
11 No Yes 13 18 25 No No
11 Yes No 7 8 26 No No

12 Yes Yes 8 12 26 Yes No
12 Yes No 7 3 29 Yes No
12 Yes No 8 1 29 No Yes
12 No No 26 4 30 No No
13 No No 13 6 30 Yes Yes
31 No No 26 3 38 Yes Yes
31 No No 7 24 38 No No
32 No No 4 6 41 No No
32 No No 18 10 42 No No
33 No No 13 9 42 No Yes
34 No No 15 8 42 No Yes
34 No No 20 30 42 No Yes
34 No No 27 8 44 No Yes
35 No No 6 12 44 No Yes
36 No No 18 5 45 No No
36 No Yes 8 17 48 No No
37 No No 19 4 61 No No
Fig. 15.7. Gallstone-free survival after the dissolution of single and

multiple gall-stones
Table 15.10. Calculation for the logrank test
Time n1 d1 w1 n2 d2 w2 pd e1
3 65 0 1 79 0 2 0.000 0.000
4 64 0 0 77 0 1 0.000 0.000
5 64 0 1 76 0 0 0.000 0.000
6 63 0 1 76 5 5 0.036 2.266
7 62 0 0 66 2 1 0.016 0.969
8 62 1 1 63 2 2 0.024 1.488
9 60 0 1 59 1 1 0.008 0.504
10 59 0 0 57 1 0 0.009 0.509
11 59 2 1 56 1 5 0.026 1.539
12 56 2 2 50 3 3 0.047 2.642
13 52 0 4 44 1 1 0.010 0.542
14 48 0 2 42 0 1 0.000 0.000
16 46 0 1 41 2 0 0.023 1.057
17 45 1 1 39 0 3 0.012 0.536
18 43 1 0 36 1 1 0.025 1.089
19 42 0 1 34 1 1 0.013 0.553
20 41 0 3 32 0 0 0.000 0.000
21 38 0 0 32 0 3 0.000 0.000
22 38 0 2 29 0 0 0.000 0.000
23 36 0 1 29 0 0 0.000 0.000
24 35 0 2 29 1 1 0.016 0.547
25 33 0 2 27 1 0 0.017 0.550
26 31 1 1 26 0 1 0.018 0.544
28 29 1 1 25 0 0 0.019 0.537
29 27 1 1 25 1 1 0.038 1.038
30 25 0 1 23 2 1 0.042 1.042
31 24 0 2 20 0 1 0.000 0.000
32 22 0 2 19 1 1 0.024 0.537
33 20 0 1 17 0 0 0.000 0.000
34 19 0 3 17 0 1 0.000 0.000
35 16 0 1 16 0 0 0.000 0.000
36 15 0 2 16 0 3 0.000 0.000
37 13 0 1 13 0 3 0.000 0.000
38 12 0 2 10 1 0 0.045 0.545
40 10 0 1 9 0 0 0.000 0.000
41 9 0 2 9 0 0 0.000 0.000
42 7 0 1 9 0 3 0.000 0.000
43 6 1 0 6 0 0 0.083 0.500
44 5 0 0 4 0 2 0.000 0.000
45 5 0 1 4 0 0 0.000 0.000
47 4 0 0 4 0 1 0.000 0.000
48 4 0 2 3 0 0 0.000 0.000
53 2 0 0 3 0 1 0.000 0.000
60 2 1 0 2 0 0 0.250 0.500
61 1 0 1 2 0 0 0.000 0.000
65 0 0 0 2 0 1 0.000 0.000
70 0 0 0 1 0 1 0.000 0.000
Total 12 27 20.032
pd = (d1 + d2)/(n1 + n2), e1 = pdn1, e2 = pdn2.
We can test the null hypothesis that the risk of recurrence in any month is
equal for the two populations by a chi-squared test:
There is one constraint, that the two frequencies add to the sum of the
expected (i.e. the total number of recurrences), so we lose one degree of
freedom, giving 2-1 = 1 degree of freedom. From Table 13.3. this has a
probability of 0.01.
Some texts describe this test differently, saying that under the null
hypothesis d1 is from a Normal distribution with mean e1 and variance
e1e2/(e1 + e2). This is algebraically identical to the chi-squared method,
but only works for two groups.
The logrank test is non-parametric, because we make no assumptions
about either the distribution of survival time or any difference in
recurrence rates. It requires the survival or censoring times to be exact. A
similar method for grouped data as in Table 15.8 is given by Mantel
(1966).
The logrank test is a test of significance and, of course, an estimate of
the difference is preferable if we can get one. The logrank test calculation
can be used to give us one: the hazard ratio. This is the ratio of the risk
of death in group 1 to the risk of death in group 2. For this to make sense,
we have to assume that this ratio is the same at all times, otherwise there
could not be a single estimate. (Compare the paired t method, §10.2.)
The risk of death is the number of deaths divided by the population at
risk, but the population keeps changing due to censoring. However, the
populations at risk in the two groups are proportional to the numbers of
expected deaths, e1 and e2. We can thus calculate the hazard ratio by
For Table 15.10. we have
Thus we estimate the risk of recurrence with single stones to be 0.42

times the risk for multiple stones. The direct calculation of a confidence
interval for the hazard ratio is tedious and I shall omit it. Altman (1991)
gives details. It can also be done by Cox regression (§17.9).
15.7 * Computer aided diagnosis

Reference intervals (§15.5) are one area where statistical methods are
involved directly in diagnosis, computer aided diagnosis is another. The
‘aided’ is put in to persuade clinicians that the main purpose is not to do
them out of a job, but, naturally, they have their doubts. Computer aided
diagnosis is partly a statistical exercise. There are two types of computer
aided diagnosis: statistical methods, where diagnosis is based on a set of
data obtained from past cases, and
decision tree methods, which try to imitate the thought processes of an

expert in the field. We shall look briefly at each approach.
There are several methods of statistical computer aided diagnosis. One
uses discriminant analysis. In this we start with a set of data on
subjects whose diagnosis was subsequently confirmed, and calculate
one or more discriminant functions. A discriminant function has the form:
constant1 × variable1 + constant2 × variable2 +…+ constantk × variablek
The constants are calculated so that the values of the functions are as
similar as possible for members of the same group and as different as
possible for members of different groups. In the case of only two groups,
we have one discriminant function and all the subjects in one group will
have high values of the function and all subjects in the other will have low
values. For each new subject we evaluate the discriminant function and
use it to allocate the subject to a group or diagnosis. We can estimate the
probability of the subject falling in that group, and in any other. Many
forms of discriminant analysis have been developed to try and improve
this form of computer diagnosis, but it does not seem to make much
difference which is used. Logistic regression (§17.8) can also be used.
A different approach uses Bayesian analysis. This is based on Bayes'
theorem, a result about conditional probability (§6.8) which may be
stated in terms of the probability of diagnosis A being true if we have
observed data B, as:
If we have a large data set of known diagnoses and their associated
symptoms and signs, we can determine PROB(diagnosis A) easily. It is
simply the proportion of times A has been diagnosed. The problem of
finding the probability of a particular combination of symptoms and signs
is more difficult. If they are all independent, we can say that the
probability of a given symptom is the proportion of times it occurs, and
the probability of the symptom for each diagnosis is found in the same
way. The probability of any combination of symptoms can be found by
multiplying their individual probabilities together, as described in §6.2. In
practice the assumption that signs and symptoms are independent is
most unlikely to be met and a more complicated analysis would be
required to deal with this. However, some systems of computer aided
diagnosis have been found to work quite well with the simple approach.
Expert or knowledge-based systems work in a different way. Here the
knowledge of a human expert or group of experts in the field is converted
into a series of decision rules, e.g. ‘if the patient has high CK then the
patient has myocardial infarction, if not then on to the next decision’.
These systems can be modified by asking further experts to test the
system with cases from their own experience and to suggest further
decision rules if the program fails. They also have the advantage that the
program can ‘explain’ the reason for its ‘decision’ by listing the series of
steps which led to it. Most of Chapter 14 consists of rules of just
this type and could be turned into an expert system for statistical
analysis.
Although there have been some impressive achievements in the field of
computer diagnosis, it has to date made little progress towards
acceptance in routine medical practice. As computers become more
familiar to clinicians, more common in their surgeries and more powerful
in terms of data storage and processing speed, we may expect computer
aided diagnosis to become as well established as computer aided
statistical analysis is today.
15.8 * Number needed to treat

When a clinical trial has a dichotomous outcome measure, such as
survival or death, there are several ways in which we can express the
difference between the two treatments. These include the difference
between proportions of successes, ratio of proportions (risk ratio or
relative risk), and the odds ratio. The number needed to treat (NNT) is
the number of patients we would need to treat with the new treatment to
achieve one more success than we would on the old treatment (Laupacis
et al. 1988; Cook and Sackett 1995). It is the reciprocal of the difference
between the proportion of success on the new treatment and the
proportion on the old treatment. For example, in the MRC streptomycin
trial (Table 2.10) the survival rates after 6 months were 93% in
streptomycin group and 730.93 - 0.73 = 0.20 and the number needed to
treat to prevent one death over six months was 1/0.20 = 5. The smaller
the NNT, the more effective the treatment will be. The smallest possible
value for NNT is 1.0, when the proportions successful are 1.0 and 0.0.
This would mean that the new treatment was always effective and the old
treatment was never effective. The NNT cannot be zero. If the treatment
has no effect at all, the NNT will be infinite, because the difference in the
proportion of successes will be zero. If the treatment is harmful, so that
success rate is less than on the control treatment, the NNT will be
negative. The number is then called the number needed to harm
(NNH). This idea has caught on very quickly and has been widely used
and developed, for example as the number needed to screen (Rembold
1998).
The NNT is an estimate and should have a confidence interval. This is
apparently quite straightforward. We find the confidence interval for the
difference in the proportions, and the reciprocal of these limits are the
confidence limits for the NNT. For the MRC streptomycin trial the 95%
confidence interval for the difference is 0.0578 to 0.3352, reciprocals 17.3
and 3.0. Thus the 95% confidence interval for the NNT is 3 to 17.
This is deceptively simple. As Altman (1998) pointed out, there are
problems when the difference is not significant. The confidence interval
for the difference between proportions includes zero, so infinity is a
possible value for NNT, and negative values are also possible, i.e. the
treatment may harm. The confidence interval must allow for this.
For example, Henzi et al. (2000) calculated NNT for several studies,
including that of Lopez-Olaondo et al. (1996). This study compared
dexamethasone against placebo to prevent postoperative nausea and
vomiting. They observed
nausea in 5/25 patients on dexamethasone and 10/25 on placebo. Thus

the difference in proportions without nausea (success) is 0.80 - 0.60 =
0.20, 95% confidence interval -0.0479 to 0.4479 (§8.6). The number
needed to treat is the reciprocal of this difference, 1/0.20 = 5.0. The
reciprocals of the confidence limts are 1/(-0.0479) = -20.9 and 1/0.4479 =
2.2. But the confidence interval for the NNT is not -20.9 to 2.2. Zero,
which this includes, is not a possible value for the NNT. Since there may
be no treatment difference at all, zero difference between proportions, the
NNT may be infinite. In fact, the confidence interval for NNT is not the
values between -20.9 and 2.2, but the values outside this interval, i.e. 2.2
to infinity (number needed to achieve an extra success, NNT) and minus
infinity to -20.9 (number needed to achieve an extra failure, NNH). Thus
the NNT is estimated to be anything greater than 2.2, and the NNH to be
anything greater than 20.9. The confidence interval is in two parts, -∞ to
-20.9 and 2.2 to ∞. (‘∞’ is the symbol for infinity.) Henzi et al. (2000) quote
this confidence interval as 2.2 to -21, which they say the reader should
interpret as including infinity. Altman (1998) recommends ‘NNTH=21.9 to
∞ to NNTB 2.2’, where NNTH means ‘number needed to harm’ and
NNTB means ‘number needed to benefit’. I prefer ‘-∞ to -20.9, 2.2 to ∞’.
Here -∞ and ∞ each tell us that it does not matter which treatment is
used.
Two-part confidence intervals are not exactly intuitive and I think that the
problems of interpretation of NNT in negative trials limit its value to being
a supplementary description of trials results.

(Each answer is true or false)
81. * The repeatability or precision of measurements may be
measured by:
(a) the coefficient of variation of repeated measurements;
(b) the standard deviation of measurements between subjects;
(c) the standard deviation of the difference between pairs of
measurements;
(d) the standard deviation of repeated measurements within
subjects;
(e) the difference between the means of two sets of
measurements on the same set of subjects.
View Answer
82. The specificity of a test for a disease:

(a) has a standard error derived from the Binomial distribution;
(b) measures how well the test detects cases of the disease;
(c) measures how well the test excludes subjects without the
disease;
(d) measures how often a correct diagnosis is obtained from the
test;
(e) is all we need to tell us how good the test is.
View Answer
83. The level of an enzyme measured in blood is used as a

diagnostic test for a disease, the test being positive if the enzyme
concentration is above a critical value. The sensitivity of the
diagnostic test:
(a) is one minus the specificity;
(b) is a measure of how well the test detects cases of the disease;
(c) is the proportion of people with the disease who are positive on
the test;
(d) increases if the critical value is lowered;
(e) measures how well people without the disease are excluded.
View Answer
84. A 95% reference interval, 95% reference range, or normal range:
(a) may be calculated as two standard deviations on either side of
the mean;
(b) may be calculated directly from the frequency distribution;
(c) can only be calculated if the observations follow a Normal
distribution;
(d) gets wider as the sample size increases;
(e) may be calculated from the mean and its standard error.
View Answer
85. If the 95% reference interval for haematocrit in men is 43.2 to

49.2:
(a) any man with haematocrit outside these limits is abnormal;
(b) haematocrits outside these limits are proof of disease:
(c) a man with a haematocrit of 46 must be very healthy;
(d) a woman with a haematocrit of 48 has a haematocrit within
normal limits;
(e) a man with a haematocrit of 42 may be ill.
View Answer
86. * When a survival curve is calculated from censored survival

times:
(a) the estimated proportion surviving becomes less reliable as
survival time increases;
(b) individuals withdrawn during the first time interval are excluded
from the analysis;
(c) survival estimates depend on the assumption that survival
rates remain constant over the study period;
(d) it may be that the survival curve will not reach zero survival;
(e) the five year survival rate can be calculated even if some of the
subjects were identified less than five years ago.
View Answer
15E Exercise: A reference interval

In this exercise we shall estimate a reference interval. Mather et al.
(1979) measured plasma magnesium in 140 apparently healthy people,
to compare with a sample of diabetics. The normal sample was chosen
from blood donors and people attending day centres for the elderly in the
area of St. George's Hospital, to give 10 male and 10 female subjects in
each age decade from 15–24 to 75 years and over. Questionnaires were
used to exclude any subject with persistent
diarrhoea, excessive alcohol intake or who were on regular drug therapy

other than hypnotics and mild analgesics in the elderly. The distribution of
plasma magnesium is shown in Figure 15.8. The mean was 0.810
mmol/litre and the standard deviation 0.057 mmol/litre.
Fig. 15.8. Distribution of plasma magnesium in 140 apparently

healthy people
1. What do you think of the sampling method? Why use blood

donors and elderly people attending day centres?
View Answer
2. Why were some potential subjects excluded? Was this a good
idea? Why were certain drugs allowed for the elderly?
View Answer
3. Does plasma magnesium appear to follow a Normal distribution?

View Answer
4. What is the reference interval for plasma magnesium, using the

Normal distribution method?
View Answer
5. Find confidence intervals for the reference limits.

View Answer
6. Would it matter if mean plasma magnesium in normal people

increased with age? What method might be used to improve the
estimate of the reference interval in this case?
View Answer
> Table of Contents > 16 - Mortality statistics and population structure
16
Mortality statistics and population
structure
16.1 Mortality rates

Mortality statistics are one of our principal sources of information about
changing pattern of disease within a country and the differences in
disease tween countries. In most developed countries, any death must
be certified by-doctor, who records the cause, date and place of death
and some data about deceased. In Britain, these include the date of birth,
area of residence and known occupation. These death certificates form
the raw material from which mortality statistics are compiled by a national
bureau of censuses, in Britain the Office for National Statistics. The
numbers of deaths can be tabulated by cause, sex, age, types of
occupation, area of residence, and marital status. Table 5.1 shows one
such tabulation, of deaths by cause and sex.
For purposes of comparison we must relate the number of deaths to the
number in the population in which they occur. We have this information
fairly reliably at 10 year intervals from the decennial census of the
country. We can estimate the size and age and sex structure of the
population between censuses using registration of births and deaths.
Each birth or death is notified to an official registrar, and so we can keep
some track of changes in the population There are other, less well
documented changes taking place, such as immigration and emigration,
which mean that population size estimates between the census years are
only approximations. Some estimates, such as the numbers in different
occupations, are so unreliable that mortality data is only tabulated by
them for census years.
If we take the number of deaths over a given period of time and divide it
by the number in the population and the time period, we get a mortality
rate, the number of deaths per unit time per person. We usually take the
number of deaths over one calendar year, although when the number of
deaths is small we may take deaths over several years, to increase the
precision of the numerator. The number in the population is changing
continually, and we take as the denominator the estimated population at
the mid-point of the time period. Mortality rates are often very small
numbers, so we usually multiply them by a constant, such as 1000 or 100
000, to avoid strings of zeros after the decimal point.
When we are dealing with deaths in the whole population, irrespective of
age, the rate we obtain is called the crude mortality rate or crude
deathd rate.
The terms ‘death rate’ and ‘mortality rate’ are used interchangeably. We
calculate the crude mortality rate for a population as:
Table 16.1. Age-specific mortality rates and

age distribution in adult males, England and
Wales, 1901 and 1981
Age Age-specific death %Adult

group rate per 1000 per population in
(years) year age group
1901 1981 1901 1981
15–19 3.5 0.8 15.36 11.09

20–24 4.7 0.8 14.07 9.75
25–34 6.2 0.9 23.76 18.81
35–44 10.6 1.8 18.46 15.99
45–54 18.0 6.1 13.34 14.75
55–64 33.5 17.7 8.68 14.04
65–74 67.8 45.6 4.57 10.65
75–84 139.8 105.2 1.58 4.28
85+ 276.5 226.2 0.17 0.64
If the period is in years, this gives the crude mortality rate as deaths per
1000 population per year.
The crude mortality rate is so called because no allowance is made for
the age distribution of the population, and comparisons between
populations with different age structures. For example, in 1901 the crude
mortality rate among adult males (aged over 15 years) in England and
Wales was 15.7 per 1000 per year, and in 1981 it was 14.8 per 1000 per
year. It seems strange that with all the improvements in medicine,
housing and nutrition between these times there has been so little
improvement in the crude mortality rate. To see why we must look at the
age-specific mortality rates, the mortality rates within narrow age
groups. Age-specific mortality rates are usually calculated for one, five or
ten year age groups. In 1901 the age specific mortality rate for men aged
15 to 19 was 3.5 deaths per 1000 per year, whereas in 1981 it was only
0.8. As Table 16.1 shows, the age specific mortality rate in 1901 was
greater than that in 1981 for every age group. However in 1901 there was
a much greater proportion of the population in the younger age groups,
where mortality was low, than there was in 1981. Correspondingly, there
was a smaller proportion of the 1901 population than the 1981 population
in the higher mortality older age groups. Although mortality was lower at
any given age in 1981, the greater proportion of older people meant that
there were almost as many deaths as in 1901.
To eliminate the effects of different age structures in the populations
which we want to compare, we can look at the age-specific death rates.
But if we are comparing several populations, this is a rather cumbersome
procedure, and it is often more convenient to calculate a single summary
figure from the age-specific
rates. There are many ways of doing this, of which three are frequently
used: the direct and indirect methods of age standardization and the life
table.
Table 16.2. Calculation of the age standardized

mortality rate by the direct method
Standard Observed
Age
proportion in mortality
group a × i
age group rate per 1
(years)
(a) 000 (b)
15–19 0.1536 0.8 0.1229
20–24 0.1407 0.8 0.1126

25–34 0.2376 0.9 0.2138
35–44 0.1846 1.8 0.3323
45–54 0.1334 6.1 0.8137
55–64 0.0868 17.7 1.5364
65–74 0.0457 45.6 2.0839
75–84 0.0158 105.2 1.6622
85+ 0.0017 226.2 0.3845
Sum 7.2623
16.2 Age standardization using the direct

method
I shall describe the direct method first. We use a standard population
structure, i.e. a standard age distribution or set of proportions of people in
each age group. We then calculate the overall mortality rate which a
population with the standard age structure would have if it experienced
the age specific mortality rates of the observed population, the population
whose mortality rate is to be adjusted. We shall take the 1901 population
as the standard and calculate the mortality rate the 1981 population
would have experienced if the age distribution were the same as in 1901.
We do this by multiplying each 1981 age specific mortality rate by the
proportion in that age group in the standard 1901 population, and adding.
This then gives us an average mortality rate for the whole population, the
age-standardized mortality rate. For example, the 1981 mortality rate in
age group 15–19 was 0.8 per 1 000 per year and the proportion in the
standard population in this age group is 15.36% or 0.1536. The
contribution of this age group is 0.8 × 0.1536 = 0.1229. The calculation is
set out in Table 16.2.
If we used the population's own proportions in each age group in this
calculation we would get the crude mortality rate. Since 1901 has been
chosen as the standard population, its crude mortality rate of 15.7 is also
the age-standardized mortality rate. The age-standardized mortality rate
for 1981 was 7.3 per 1 000 men per year. We can see that there was a
much higher age-standardized mortality in 1901 than 1981, reflecting the
difference in age-specific mortality rates.
16.3 Age standardization by the indirect method

The direct method relies upon age-specific mortality rates for the
observed population. If we have very few deaths, these age-specific rates
will be very poorly estimated. This will be particularly so in the younger
age groups, where we may
even have no deaths at all. Such situations arise when considering

mortality due to particular conditions or in relatively small groups, such as
those defined by occupation. The indirect method of standardization is
used for such data. We calculate the number of deaths we would expect
in the observed population if it experienced the age-specific mortality
rates of a standard population. We then compare the expected number of
deaths with that actually observed.
Table 16.3. Age-specific mortality rates due to

cirrhosis of the liver and age distributions of
all men and medical practitioners, England
and Wales, 1971
Age Mortality per Number
Number
group million men of
of men
(years) per year doctors
15–24 5.859 3584320 1080
25–34 13.050 3065100 12860
35–44 46.937 2876170 11510
45–54 161.503 2965880 10330
55–64 271.358 2756510 7790
I shall take as an example the deaths due to cirrhosis of the liver among
male qualified medical practitioners in England and Wales, recorded
around the 1971 census. There were 14 deaths among 43570 doctors
aged below 65, a crude mortality rate of 14/43570 = 321 per million,
compared to 1423 out of 15247980 adult males (aged 15–64), or 93 per
million. The mortality among doctors appears high, but the medical
population may be older than the population of men as a whole, as it will
contain relatively few below the age of 25. Also the actual number of
deaths among doctors is small and any difference not explained by the
age effect may be due to chance. The indirect method enables us to test
this. Table 16.3 shows the age-specific mortality rates for cirrhosis of the
liver among all men aged 15 to 65, and the number of men estimated in
each ten-year-age group, for all men and for doctors. We can see that
the two age distributions do appear to be different.
The calculation of the expected number of deaths is similar to the direct
method, but different populations and rates are used. For each age
group, we take the number in the observed population, and multiply it by
the standard age specific mortality rate, which would be the probability of
dying if mortality in the observed population were the same as that in the
standard population. This gives us the number we would expect to die in
this age group in the observed population. We add these over the age
groups and obtain the expected number of deaths. The calculation is set
out in Table 16.4.
The expected number of deaths is 4.4965, which is considerably less
than the 14 observed. We usually express the result of the calculation as
the ratio of observed to expected deaths, called the standardized
mortality ratio or SMR. Thus the SMR for cirrhosis among doctors is
We usually multiply the SMR by 100 to get rid of the decimal point, and
report the SMR as 311. If we do not adjust for age at all, the ratio of the
crude death rates is 3.44, compared to the age adjusted figure of 3.11, so
the adjustment has made some, but not much, difference in this case.
Table 16.4. Calculation of the expected

number of deaths due to cirrhosis of the liver
among practitioners, using the indirect
method
Observed
Age Standard
population
group mortality a × b
number of
(years) rate (a)
doctors (b)
15–24 0.000005859 1080 0.0063

25–34 0.000013050 12860 0.1678
35–44 0.000046937 11510 0.5402
45–54 0.000161503 10330 1.6683
55–64 0.000271358 7790 2.1139
Total 4.4965
We can calculate a confidence interval for the SMR quite easily. Denote
the observed deaths by O and expected by E. It is reasonable to suppose
that the deaths are independent of one another and happening randomly
in time, so the observed number of deaths is from a Poisson distribution
(§6.7). The standard deviation of this Poisson distribution is the square
root of its mean and so can be estimated by the square root of the
observed deaths, √O. The expected number is calculated from a very
much larger sample and is so well estimated it can be treated as a
constant, so the standard deviation of 100 × O/E, which is the standard
error of the SMR, is estimated by 100 × √O/E. Provided the number of
deaths is large enough, say more than 10, an approximate 95%
confidence interval is given by
For the cirrhosis data the formula gives
The confidence interval clearly excludes 100 and the high mortality
cannot be ascribed to chance.
For small observed frequencies tables based on the exact probabilities of
the Poisson distribution are available (Pearson and Hartley 1970). The
calculations are easily done by computer and my free program Clinstat
(§1.3) does them. There is also an exact method for comparing two
SMRs, which Clinstat does. For the cirrhosis data the exact 95%
confidence interval is 170 to 522. This is
not quite the same as the large sample approximation. Better

approximations and exact methods of calculating confidence intervals are
described by Morris and Gardner (1989) and Breslow and Day (1987).
We can also test the null hypothesis that in the population the SMR =
100. If the null hypothesis is true, O is from a Poisson distribution with
mean E and hence standard deviation √E, provided the sample is large
enough, say E > 10. Then (O - E)/√E would be an observation from the
Standard Normal distribution if the null hypothesis were true. The sample
of doctors is too small for this test to be reliable, but if it were, we would
have (O - E)/√E = (14 - 4.4965)/√4.4965 = 4.48, P = 0.0001. Again, there
is an exact method. This gives P = 0.0005. As so often happens, large
sample methods become too liberal and give P values which are too
small when used with samples which are too small for the test to be valid.
The highly significant difference suggests that doctors are at increased
risk of death from cirrhosis of the liver, compared to employed men as a
whole. The news is not all bad for medical practitioners, however. Their
SMR for cancer of the trachea, bronchus and lung is only 32. Doctors
may drink, but they do not smoke!
16.4 Demographic life tables

We have already discussed a use of the life table technique for the
analysis of clinical survival data (§15.6). The life table was found by
following the survival of a group of subjects from some starting point to
death. In demography, which means the study of human populations,
this longitudinal method of analysis is impractical, because we could only
study people born more than 100 years ago. Demographic life tables are
generated in a different way, using a cross-sectional approach. Rather
than charting the progress of a group from birth to death, we start with
the present age-specific mortality rates. We then calculate what would
happen to a cohort of people from birth if these age-specific mortality
rates applied unchanged throughout their lives. We denote the probability
of dying between ages x and x + 1 years (the age-specific mortality rate
at age x) by qx. As in Table 15.8, the probability of surviving from age x to
x + 1 is px = 1 - qx. We now suppose that we have a cohort of size l0 at
age 0, i.e. at birth. l0 is usually 100000 or 10000. The number who would
still be alive after x years is lx. We can see that the number alive after x +
1 years is lx+1 = px × lx, so given all the px from x = 0 onwards we can
calculate the lx. The cumulative survival probability to age x is then Px =
lx/l0
Table 16.5 shows an extract from Life Table Number 11, 1950–52, for
England and Wales. With the exception of 1941, a life table like this has
been produced every 10 years since 1871, based on the decennial
census year. The life table is based on the census year because only
then do we have a good measure of the number of people at each age,
the denominator in the calculation of qx. A three year period is used to
increase the number of deaths for a year of age and so improve the
estimation of qx. Separate tables are produced for males and females
because the mortality of the two sexes is very different. Age specific
death rates are higher in males than females at every age. Between
census years life tables are still produced but are only published in an
abridged form, giving lx at five year intervals only after age five (Table
16.6).
Table 16.5. Extract from English Life Table

Number 11, 1950–52, Males
Expected Probability an Expected

Age
number individual dies life at
in
alive at between ages x age x
years
age x and x + 1 years
x lx qx ex
0 100000 0.03266 66.42
1 96734 0.00241 67.66
2 96501 0.00141 66.82
3 96395 0.00102 65.91
4 96267 0.00084 64.98
. . . .
. . . .
. . . .
100 23 0.44045 1.67
101 13 0.45072 1.62
102 7 0.46011 1.58
103 4 0.46864 1.53

104 2 0.47636 1.50
The final column in Tables 16.5 and 16.6 is the expected life,
expectation of life or life expectancy, ex. This is the average life still to
be lived by those reaching age x. We have already calculated this as the
expected value of the probability distribution of year of death (§6E). We
can do the calculation in a number of other ways. For example, if we add
lx+1, lx+2, lx+3, etc. we will get the total number of years to be lived,
because the lx+1 who survive to x + 1 will have added lx+1 years to the
total, the lx+2 of these who survive from x + 1 to x + 2 will add a further
lx+2 years, and so on. If we divide this sum by lx we get the average
number of whole years to be lived. If we then remember that people do
not die on their birthdays, but scattered throughout the year, we can add
half to allow for the average of half year lived in the year of death. We
thus get
i.e. summing the li from age x + 1 to the end of the life table.
If many people die in early life, with high age-specific death rates for
children, this has a great effect on expectation of life at birth. Table 16.7
shows expectation of life at selected ages from four English Life Tables
(Office for National Statistics 1997). In 1991, for example, expectation of
life at birth for males was 74 years, compared to only 40 years in 1841,
an improvement of 34 years. However expectation of life at age 45 in
1991 was 31 years compared to 23 years in 1841, an improvement of
only 8 years. At age 65, male expectation of life was 11
years in 1841 and 14 years in 1991, an even smaller change. Hence the
change in life expectancy at birth was due to changes in mortality in early
life, not late life.
Table 16.6. Abridged Life Table 1988–90,
England and Wales
Age Males Females
x lx ex lx ex
0 10000 73.0 10000 78.5
1 9904 72.7 9928 78.0
2 9898 71.7 9922 77.1
3 9893 70.8 9919 76.1
4 9890 69.8 9916 75.1
5 9888 68.8 9914 74.2
10 9877 63.9 9907 69.2
15 9866 58.9 9899 64.3
20 9832 54.1 9885 59.4
25 9790 49.3 9870 54.4

30 9749 44.5 9852 49.5
35 9702 39.7 9826 44.6
40 9638 35.0 9784 39.8
45 9542 30.3 9718 35.1
50 9375 25.8 9607 30.5
55 9097 21.5 9431 26.0
60 8624 17.5 9135 21.7
65 7836 14.0 8645 17.8
70 6689 11.0 7918 14.2
75 5177 8.4 6869 11.0
80 3451 6.4 5446 8.2
85 1852 4.9 3659 5.9
There is a common misconception that a life expectancy at birth of 40

years, as in 1841, meant that most people died about age 40. For
example (Rowe 1992):
Mothers have always provoked rage and resentment in their adult
daughters, while the adult daughters have always provoked anguish and
guilt in their mothers. In past centuries, however, such matched misery
did not last for long. Daughters could bury their rage and resentment
under a concern for duty while they cared for their mothers who, turning
40, rapidly aged, grew frail and died. Now mothers turning 40 are strong
and healthy, and only half way through their lives.
This is absurd. As Table 16.7 shows, since life expectancy was first
estimated women turning 40 have had average remaining lives of more
than 20 years. They did not rapidly age, grow frail, and die.
‘Expectation’ is used in its statistical sense of the average of a
distribution. It does not mean that each person can know when they will
die. From the most recent life table for England and Wales, for 1994–96
(Office for National Statistics 1998a), a man aged 53 (myself, for
example) has a life expectancy of 24 years. This is the average life time
which all men aged 53 years would have if the present age-specific
mortality rates do not change. (These should go down over time, putting
life-spans up.) About half of these men will have shorter lives and half
longer. If we could calculate life expectancies for men with different
combinations of risk factors, we might find that my life expectancy would

be decreased because I am short (so unfair I think) and fat and increased
because I do not smoke (like almost all medical statisticians) and am of
professional social class. However my expectation of life was adjusted, it
would remain an average, not a guaranteed figure for me.
Table 16.7. Life expectancy in 1841, 1901,

1951, and 1991, England and Wales
Age Sex Expectation of life in years

1841 1901 1951 1991
Birth Males 40 49 66 74
Females 42 52 72 79
15 yrs Males 43 47 54 59
Females 44 50 59 65
45 yrs Males 23 23 27 31
Females 24 26 31 36
65 yrs Males 11 11 12 14
Females 12 12 14 18
Life tables have a number of uses, both medical and non-medical.

Expectation of life provides a useful summary of mortality without the
need for a standard population. The table enables us to predict the future
size of and age structure of a population given its present state, called a
population projection. This can be very useful in predicting such things
as the future requirement for geriatric beds in a health district. Life tables
are also invaluable in non-medical applications, such as the calculation of
insurance premiums, pensions and annuities.
The main difficulty with prediction from a life table is finding a table which
applies to the populations under consideration. For the general
population of, say, a health district, the national life table will usually be
adequate, but for special populations this may not be the case. If we want
to predict the future need for care of an institutionalized population, such
as in a long stay psychiatric hospital or old peoples' home, the mortality
may be considerably greater than that in the general population.
Predictions based on the national life table can only be taken as a very
rough guide. If possible life tables calculated on that type of population
should be used.
16.5 Vital statistics

We have seen a number of occasions where ordinary words have been
given quite different meanings in statistics from those they have in
common speech; ‘Normal’ and ‘significant’ are good examples. ‘Vital
statistics’ is the opposite, a technical term which has acquired a
completely unrelated popular meaning. As far as the medical statistician
is concerned, vital statistics have nothing to do with the dimensions of
female bodies. They are the statistics relating to life and death: birth
rates, fertility rates, marriage rates and death rates. I have already
mentioned crude mortality rate, age-specific mortality rates, age-
standardized
mortality rate, standardized mortality ratio, and expectation of life. In this

section I shall define a number of other statistics which are often quoted
in the medical literature.
The infant mortality rate is the number of deaths under one year of age
divided by the number of live births, usually expressed as deaths per
1000 live births. The neonatal mortality rate is the same thing for
deaths in the first 4 weeks of life. The stillbirth rate is the number of
stillbirths divided by the total number of births, live and still. A stillbirth is a
child born dead after 28 weeks gestation. The perinatal mortality rate is
the number of stillbirths and deaths in the first week of life divided by the
total births, again usually presented per 1000 births. Infant and perinatal
mortality rates are regarded as particularly sensitive indicators of the
health status of the population. The maternal mortality rate is the
number of deaths of mothers ascribed to problems of pregnancy and
birth, divided by the total number of births. The birth rate is the number
of live births per year divided by the total population. The fertility rate is
the number of live births per year divided by the number of women of
childbearing age, taken as 15–44 years.
The attack rate for a disease is the proportion of people exposed to
infection who develop the disease. The case fatality rate is the
proportion of people with the disease who die from it. The prevalence of
a disease is the proportion of people who have it at one point in time. The
incidence is the number of new cases in one year divided by the number
at risk.
16.6 The population pyramid

The age distribution of a population can be presented as histogram,
using the methods of §4.3. However, because the mortality of males and
females is so different the age distributions for males and females are
also different. It is usual to present the age distributions for the two sexes
separately. Figure 16.1 shows the age distributions for the male and
female populations of England and Wales in 1901. Now, these
histograms have the same horizontal scale. The conventional way to
display them is with the age scale vertically and the frequency scale
horizontally as in Figure 16.2. The frequency scale has zero in the middle
and increases to the right for females and to the left for males. This is
called a population pyramid, because of its shape.
Figure 16.3 shows the population pyramid for England and Wales in
1991. The shape is quite different. Instead of a triangle we have an
irregular figure with almost vertical sides which begin to bend very
sharply inwards at about age 65. The post-war and 1960s baby booms
can be seen as bulges at ages 25–30 and 40–45. A major change in
population structure has taken place, with a vast increase in the
proportion of elderly. This has major implications for medicine, as the
care of the elderly has become a large proportion of the work of doctors,
nurses and their colleagues. It is interesting to see how this has come
about.
It is popularly supposed that people are now living much longer as a
result of modern medicine, which prevents deaths in middle life. This is
only partly true.
Fig. 16.1. Age distributions for the population of England and Wales,
by sex, 1901
Fig. 16.2. Population pyramid for England and Wales, 1901

Fig. 16.3. Population pyramid for England and Wales, 1991
As Table 16.7 shows, life expectancy at birth increased dramatically

between 1901 and 1991, but the increase in later life is much less. The
change is not an extension of every life by 25 years, which would be
seen at every age, but mainly a reduction in mortality in childhood and
early adulthood. Mortality in later life has changed relatively little. Now, a
big reduction in mortality in childhood would result in an increase in the
base part of the pyramid, as more children survived, unless there was a
corresponding fall in the number of babies being born. In the 19th
century, women were having many children and despite the high mortality
in childhood the number who survived into adulthood to have children of
their own exceeded that of their own parents. The population expanded
and this history is embodied in the 1901 population pyramid. In the 20th
century, infant mortality fell and people responded to this by having fewer
children. In 1841–45, the infant mortality rates were 148 per 1000 live
births, 138 in 1901–05, 10 in 1981–85 (OPCS 1992) and only 5.9 in 1997
(Office for National Statistics 1999). The birth rate was 32.2 per 1000
population per year in 1841–45, in 1901–05 it was 28.2, and in 1987–97
it was 13.5 (Office for National Statistics 1998b). The base of the pyramid
ceased to expand. As those who were in the base of the 1901 pyramid
grew older, the population in the top half of the pyramid increased. The
survivors of the 0–4 age group in the 1901 pyramid are the 90+ age
group in the 1991 pyramid. Had the birth rate not fallen, the population
would have continued to expand and we would have as great or greater a
proportion of young people in 1991 as we did in 1901, and a vastly larger
population. Thus the increase in the proportion of the elderly is not
primarily because adult lives have been extended, although this has a
small effect, but because fertility has declined. Life expectancy for the
elderly has changed relatively little. Most developed countries have
stable population pyramids like Figure 16.3 and those of most developing
countries have expanding pyramids like Figure 16.2.

87. Age-specific mortality rate:
(a) is a ratio of observed to expected deaths;
(b) can be used to compare mortality between different age
groups;
(c) is an age adjusted mortality rate;
(d) measures the number of deaths in a year;
(e) measures the age structure of the population.
View Answer
88. Expectation of life:

(a) is the number of years most people live;
(b) is a way of summarizing age-specific death rates:
(c) is the expected value of a particular probability distribution;
(d) varies with age:
(e) is derived from life tables.
View Answer
89. In a study of post-natal suicide (Appleby 1991), the SMR for

suicide among women who had just had a baby was 17 with a 95%
confidence interval 14 to 21 (all women = 100). For women who had
had a stillbirth, the SMR was 105 (95% confidence interval 31 to
277). We can conclude that:
(a) women who had just had a baby were less likely to commit
suicide than other women of the same age;
(b) women who had just had a still birth were less likely to commit
suicide than other women of the same age;
(c) women who had just had a live baby were less likely to commit
suicide than women of the same age who had had a stillbirth:
(d) it is possible that having a stillbirth increases the risk of suicide;
(e) suicidal women should have babies.
View Answer
90. In 1971, the SMR for cirrhosis of the liver for men was 773 for
publicans and innkeepers and 25 for window cleaners, both being
significantly different from 100 (Donnan and Haskey 1977). We can
conclude that:
(a) publicans are more than 7 times as likely as the average
person to die from cirrhosis of the liver;
(b) the high SMR for publicans may be because they tend to be
found in the older age groups;
(c) being a publican causes cirrhosis of the liver;
(d) window cleaning protects men from cirrhosis of the liver;
(e) window cleaners are at high risk of cirrhosis of the liver.
View Answer
91. The age and sex structure of a population may be described by:
(a) a life table;
(b) a correlation coefficient;
(c) a standardized mortality ratio;
(d) a population pyramid;
(e) a bar chart.
View Answer
92. The following statistics are adjusted to allow for the age
distribution of the population:
(a) age-standardized mortality rate;
(b) fertility rate;
(c) perinatal mortality rate;
(d) crude mortality rate;
(e) expectation of life at birth.
View Answer
16E Exercise: Deaths from volatile substance

abuse
Anderson et al. (1985) studied mortality associated with volatile
substance abuse (VSA), often called glue sniffing. In this study all known
deaths associated with VSA from 1971 to 1983 inclusive were collected,
using sources including three press cuttings agencies and a six-monthly
systematic survey of all coroners. Cases were also notified by the Office
of Population Censuses and Surveys for England and Wales and by the
Crown Office and procurators fiscal in Scotland.
Table 16.8 shows the age distribution of these deaths for Great Britain
and for Scotland alone, with the corresponding age distributions at the
1981 decennial census.
1. Calculate age-specific mortality rates for VSA per year and for the
whole period. What is unusual about these age-specific mortality
rates?
View Answer
2. Calculate the SMR for VSA deaths for Scotland.

View Answer
3. Calculate the 95% confidence interval for this SMR.

View Answer
4. Does the number of deaths in Scotland appear particularly high?

A part from a lot of glue sniffing, are there any other factors which
should be considered as possible explanations for this finding?
View Answer
Table 16.8. Volatile substance abuse mortality and

population size, Great Britain and Scotland. 1971–
83 (Anderson et al. 1985)
Age
group Great Britain Scotland
(years)
Population VSA Population

VSA deaths
(thousands) deaths (thousands)
0–9 0 6770 0 653

10–14 44 4271 13 425
15–19 150 4467 29 447
20–24 45 3959 9 394
25–29 15 3616 0 342
30–39 8 7408 0 0659
40–49 2 6055 0 574
50–59 7 6242 0 579
60+ 4 10769 0 962

> Table of Contents > 17 - Multifactorial methods
17
Multifactorial methods
17.1 * Multiple regression

In Chapters 10 and 11 we looked at methods of analysing the relationship
between a continuous outcome variable and a predictor. The predictor
could be quantitative, as in regression, or qualitative, as in one-way
analysis of variance. In this chapter we shall look at the extension of
these methods to more than one predictor variable, and describe related
methods for use when the outcome is dichotomous or censored survival
data. These methods are very difficult to do by hand and computer
programs are always used. I shall omit the formulae.
Table 17.1 shows the ages, heights and maximum voluntary contraction
of the quadriceps muscle (MVC) in a group of male alcoholics. The
outcome variable is MVC. Figure 17.1 shows the relationship between
MVC and height. We can
fit a regression line of the form MVC = a + b × height (§11.2–3). This

enables us to predict what the mean MVC would be for men of any given
height. But MVC varies with other things beside height. Figure 17.2
shows the relationship between MVC and age.
Table 17.1. Maximum voluntary contraction (MVC) of

quadriceps muscle, age and height, of 41 male
alcoholics (Hickish et al. 1989)
Age Height MVC Age Height MVC
(years) (cm) (newtons) (years) (cm) (newtons)
24 166 466 42 178
27 175 304 47 171
28 173 343 47 162
28 175 404 48 177
31 172 147 49 177
31 172 294 49 178
32 160 392 50 167
32 172 147 51 176
32 179 270 53 159
32 177 412 53 173
34 175 402 53 175
34 180 368 53 172

35 167 491 55 170
37 175 196 55 178
38 172 343 55 155
39 172 319 58 160
39 161 387 61 162
39 173 441 62 159
40 173 441 65 168
41 168 343 65 168
41 178 540
Fig. 17.1. Muscle strength (MVC) against height
Fig. 17.2. Muscle strength (MVC) against age
We can show the strengths of the linear relationships between all three
variables by their correlation matrix. This is a tabular display of the
correlation coefficients between each pair of variables, matrix being used
in its mathematical sense as a rectangular array of numbers. The
correlation matrix for the data of Table 17.1 is shown in Table 17.2. The
coefficients of the main diagonal are all 1.0, because they show the
correlation of the variable with itself, and the correlation matrix is
symmetrical about this diagonal. Because of this symmetry many
computer programs print only the part of the matrix below the diagonal.
Inspection of Table 17.2 shows that older men were shorter and weaker
than younger men. that taller men were stronger than shorter men, and
that the magnitudes of all three relationships was similar. Reference to
Table 11.2 with 41 - 2 = 39 degrees of freedom shows that all three
correlations are significant.
Table 17.2. Correlation matrix for the data of

Table 17.1
Age Height MVC
Age 1.000 -0.338 -0.417
Height -0.338 1.000 0.419
MVC -0.417 0.419 1.000
We could fit a regression line of the form MVC = a + b × age, from which
we could predict the mean MVC for any given age. However, MVC would
still vary with height. To investigate the effect of both age and height, we
can use multiple regression to fit a regression equation of the form
MVC = b0 + b1 × height + b2 × age
The coefficients are calculated by a least squares procedure, exactly the
same in principle as for simple regression. In practice, this is always done
using a computer program. For the data of Table 17.1, the multiple
regression equation is
MVC = -466 + 5.40 × height - 3.08 × age
From this, we would estimate the mean MVC of men with any given age
and height, in the population of which these are a sample.
There are a number of assumptions implicit here. One is that the
relationship between MVC and height is the same at each age, that is,
that there is no interaction between height and age. Another is that the
relationship between MVC and height is linear, that is of the form MVC =
a + b × height. Multiple regression analysis enables us to test both of
these assumptions.
Multiple regression is not limited to two predictor variables. We can have
any number, although the more variables we have the more difficult it
becomes to interpret the regression. We must, however, have more
points than variables, and as the degrees of freedom for the residual
variance are n - 1 - q if q variables are fitted, and this should be large
enough for satisfactory estimation of confidence intervals and tests of
significance. This will become clear after the next section.
17.2 * Significance tests and estimation in

multiple regression
As we saw in §11.5, the significance of a simple linear regression line can
be tested using the t distribution. We can carry out the same test using
analysis of variance. For the FEV1 and height data of Table 11.1 the
sums of squares and products were calculated in §11.3. The total sum of
squares for FEV1 is Syy = 9.438 68, with n - 1 = 19 degrees of freedom.
The sum of squares due to regression was calculated in §11.5 to be
3.189 37. The residual sum of squares, i.e. the sum of squares about the
regression line, is found by subtraction as 9.438 68 - 3.189 37 = 6.249
31, and this has n - 2 = 18 degrees of freedom. We
can now set up an analysis of variance table as described in §10.9,

shown in Table 17.3.
Table 17.3. Analysis of variance for the regression of F

height
Degrees
Source of Sum of Mean Variance
of
variation squares square ratio (F
freedom
Total 19 9.438
68
Due to 1 3.189 3.189 9.19

regression 37 37
Residual 18 6.249 0.347

(about 31 18
regression)
Table 17.4. Analysis of variance for the regression of M

height and age
Degrees
of
freedom
Total 40 503
344
Regression 2 131 65 6.72

495 748
Residual 38 371 9785

849
Note that the square root of the variance ratio is 3.03, the value of t found
in §11.5. The two tests are equivalent. Note also that the regression sum
of squares divided by the total sum of squares = 3.189 37/9.438 68 =
0.337 9 is the square of the correlation coefficient, r = 0.58 (§11.5,
§11.10). This ratio, sum of squares due to regression over total sum of
squares, is the proportion of the variability accounted for by the
regression. The percentage variability accounted for or explained by the
regression is 100 times this, i.e. 34%.
Returning to the MVC data, we can test the significance of the regression
of MVC on height and age together by analysis of variance. If we fit the
regression model in §17.1, the regression sum of squares has two
degrees of freedom, because we have fitted two regression coefficients.
The analysis of variance for the MVC regression is shown in Table 17.4.
The regression is significant; it is unlikely that this association could have
arisen by chance if the null hypothesis were true. The proportion of
variability accounted for, denoted by R2, is 131 495/503 344 = 0.26. The
square root of this is called the multiple correlation coefficient, R. R2 must
lie between 0 and 1, and as no meaning can be given to the direction of
correlation in the multivariate case, R is also taken as positive. The larger
R is, the more closely correlated with the outcome variable the set of
predictor variables are. When R = 1 the variables are perfectly correlated
in the sense that the outcome variable is a linear combination of the
others. When the outcome variable is not linearly related to any of the
predictor variables, R will be small, but not zero.
We may wish to know whether both or only one of our variables leads to
the association. To do this, we can calculate a standard error for each
regression coefficient (Table 17.5). This will be done automatically by the
regression program. We can use this to test each coefficient separately
by a t test. We can
also find a confidence interval for each, using t standard errors on either
side of the estimate. For the example, both age and height have P = 0.04
and we can conclude that both age and height are independently
associated with MVC.
Table 17.5. Coefficients for the regression of MVC on he

and age, with standard errors and confidence interva
Predictor Standard t
Coefficient P
variable error ratio
height 5.40 2.55 2.12 0.04
age -3.08 1.47 -2.10 0.04
intercept -465.63 460.33 -1.01 0.3

A difficulty arises when the predictor variables are correlated with one
another. This increases the standard error of the estimates, and variables
may have a multiple regression coefficient which is not significant despite
being related to the outcome variable. We can see that this will be so
most clearly by taking an extreme case. Suppose we try to fit
MVC = b0 + b1 × height + b2 × height
For the MVC data
MVC = -908 + 6.20 × height + 1.00 × height
is a regression equation which minimizes the residual sum of squares.
However, it is not unique, because
MVC = -908 + 5.20 × height + 2.00 × height
will do so too. The two equations give the same predicted MVC. There is
no unique solution, and so no regression equation can be fitted, even
though there is a clear relationship between MVC and height. When the
predictor variables are highly correlated the individual coefficients will be
poorly estimated and have large standard errors. Correlated predictor
variables may obscure the relationship of each with the outcome variable.
A different (and equivalent) way of testing the effects of two correlated
predictor variables separately is to proceed as follows. We fit three
models:
1. MVC on height and age, regression sum of squares = 131 495, d.f. = 2
2. MVC on height, regression sum of squares = 88 511, d.f. = 1
3. MVC on age, regression sum of squares = 87 471, d.f. = 1
Note that 88 511 + 87 471 = 175 982 is greater than 131 495. This is
because age and height are correlated. We then test the effect of height
if age is taken into account, referred to as the effect of height given age.
The regression sum of squares for height given age is the regression
sum of squares (age and height) minus regression sum of squares (age
only), which is 131 495 - 87 471 = 44 024. This has degrees of freedom =
2 - 1 = 1. Similarly, the effect of age allowing
for height, i.e. age given height, is tested by regression sum of squares
(age and height) minus regression sum of squares (height only) = 131
495-88 511 = 42 984, with degrees of freedom = 2 - 1 = 1. We can set all
this out in an analysis of variance table (Table 17.6). The third to sixth
rows of the table are indented for the source of variation, degrees of
freedom and sum of squares columns, to indicate that they are different
ways of looking at variation already accounted for in the second row. The
indented rows are not included when the degrees of freedom and sums
of squares are added to give the total. After adjustment for age there is
still evidence of a relationship between MVC and height, and after
adjustment for height there is still evidence of a relationship between
MVC and age. Note that the P values are the same as those found by a t
test for the regression coefficient. This approach is essential for
qualitative predictor variables with more than two categories (§17.6),
when several t statistics may be printed for the variable.
Table 17.6. Analysis of variance for the regression of M

height and age, showing adjusted sums of square
Degrees
of
freedom
Total 40 503 344

495 748
Age alone 1 87 87 8.94

471 471
Height 1 44 44 4.50
given age 024 024
Height 1 88 88 9.05
alone 511 511
Age 1 42 42 4.39
given 984 984
height
Residual 38 371 9 785

849
17.3 * Interaction in multiple regression

An interaction between two predictor variables arises when the effect of
one on the outcome depends on the value of the other. For example, tall
men may be stronger than short men when they are young, but the
difference may disappear as they age.
We can test for interaction as follows. We have fitted
An interaction may take two simple forms. As height increases, the effect
of age may increase so that the difference in MVC between young and
old tall men is greater than the difference between young and old short
men. Alternatively, as height increases, the effect of age may decrease.
More complex interactions are beyond the scope of this discussion. Now,
if we fit
MVC = b0 + b1 × height + b2 × age + b3 × height × age
for fixed height the effect of age is b2 + b3 × height. If there is no
interaction, the effect of age is the same at all heights, and b3 will be
zero. Of course, b3 will not
be exactly zero, but only within the limits of random variation. We can fit
such a model just as we fitted the first one. We get
Table 17.7. Analysis of variance for the interaction of h

and age
Degrees
of
freedom
Total 40 503
344

719 573
Height 2 131 65 8.09

and age 495 748
Height × 1 71 71 8.77
age 224 224
Residual 37 300 8 125

625
MVC = 4 661 - 24.7 × height - 112.8 × age + 0.650 × height × age

The regression is still significant, as we would expect. However, the
coefficients of height and age have changed; they have even changed
sign. The coefficient of height depends on age. The regression equation
can be written
MVC = 4 661 + (-24.7 + 0.650 × age) × height - 112.8 × age
The coefficient of height depends on age, the difference in strength
between short and tall subjects being greater for older subjects than for
younger.
The analysis of variance for this regression equation is shown in Table
17.7. The regression sum of squares is divided into two parts: that due to
age and height, and that due to the interaction term after the main effects
of age and height have been accounted for. The interaction row is the
difference between the regression row in Table 17.7, which has 3
degrees of freedom, and the regression row in Table 17.4, which has 2.
From this we see that the interaction is highly significant. The effects of
height and age on MVC are not additive. Another example of the
investigation of a possible interaction is given in §17.7.
17.4 * Polynomial regression

So far, we have assumed that all the regression relationships have been
linear, i.e. that we are dealing with straight lines. This is not necessarily
so. We may have data where the underlying relationship is a curve rather
than a straight line. Unless there is a theoretical reason for supposing
that a particular form of the equation, such as logarithmic or exponential,
is needed, we test for non-linearity by using a polynomial. Clearly, if we
can fit a relationship of the form
we can also fit one of the form
MVC = b0 + b1 × height + b2 × height2
Table 17.8. Analysis of variance for polynomial regress
MVC on height
Degrees
of
variation squares square ratio (F)
freedom
Total 40 503
344
Regression 2 89 103 44 4.09

552
Linear 1 88 88 7.03
522 522
Quadratic 1 581 581 0.05
Residual 38 414 12
241 584
to give a quadratic equation, and continue adding powers of height to

give equations which are cubic, quartic, etc.
Height and height squared are highly correlated, which can lead to
problems in estimation. To reduce the correlation, we can subtract a
number close to mean height from height before squaring. For the data of
Table 17.1, the correlation between height and height squared is 0.999 8.
Mean height is 170.7 cm, so 170 is a convenient number to subtract. The
correlation between height and height minus 170 squared is -0.44, so the
correlation has been reduced, though not eliminated. The regression
equation is
MVC = -961 + 7.49 × height + 0.092 × (height - 170)2
To test for non-linearity, we proceed as in §17.2. We fit two regression
equations, a linear and a quadratic. The non-linearity is then tested by
the difference between the sum of squares due to the quadratic equation
and the sum of squares due to the linear. The analysis of variance is
shown in Table 17.8. In this case the quadratic term is not significant, so
there is no evidence of non-linearity. Were the quadratic term significant,
we could fit a cubic equation and test the effect of the cubic term in the
same way. Polynomial regression of one variable can be combined with
ordinary linear regression of others to give regression equations of the
form
MVC = b0 + b1 × height + b2 × height2 + b3 × age
and so on. Royston and Altman (1994) have shown that quite complex
curves can be fitted with a small number of coefficients if we use log(x)
and powers -1, 0.5, 0.5, 1 and 2 in the regression equation.
17.5 * Assumptions of multiple regression

For the regression estimates to be optimal and the F tests valid, the
residuals (the difference between observed values of the dependent
variable and those predicted by the regression equation) should follow a
Normal distribution and have the same variance throughout the range.
We also assume that the relationships which we are modelling are linear.
These assumptions are the same as for simple linear regression (§11.8)
and can be checked graphically in the same way, using histograms,
Normal plots and scatter diagrams. If the assumptions of Normal
distribution and uniform variance are not met, we can use a

transformation as described in §10.4 and §11.8. Non-linearity can be
dealt with using polynomial regression.
Fig. 17.3. Histogram and Normal plot of residuals of MVC about

height and age
Fig. 17.4. Residuals against observed MVC, to check uniformity of

variance, and age, to check linearity
The regression equation of strength on height and age is MVC = -466 +

5.40 × height - 3.08 × age and the residuals are given by
residual = MVC - (-466 + 5.40 × height - 3.08 × age)
Figure 17.3 shows a histogram and a Normal plot of the residuals for the
MVC data. The distribution looks quite good. Figure 17.4 shows a plot of
residuals against MVC. The variability looks uniform. We can also check
the linearity by plotting residuals against the predictor variables. Figure
17.4 also shows the residual against age. There is an indication that
residual may be related to age. The possibility of a nonlinear relationship
can be checked by polynomial regression, which, in this case, does not
produce a quadratic term which approaches significance.
17.6 * Qualitative predictor variables

In §17.1 the predictor variables, height and age, were quantitative. In the
study from which these data come, we also recorded whether or not
subjects had
cirrhosis of the liver. Cirrhosis was recorded as ‘present’ or ‘absent’, so

the variable was dichotomous. It is easy to include such variables as
predictors in multiple regression. We create a variable which is 0 if the
characteristic is absent, 1 if present, and use this in the regression
equation just as we did height. The regression coefficient of this
dichotomous variable is the difference in the mean of the outcome
variable between subjects with the characteristic and subjects without. If
the coefficient in this example were negative, it would mean that subjects
with cirrhosis were not as strong as subjects without cirrhosis. In the
same way, we can use sex as a predictor variable by creating a variable
which is 0 for females and 1 for males. The coefficient then represents
the difference in mean between male and female. If we use only one,
dichotomous predictor variable in the equation, the regression is exactly
equivalent to a two-sample t test between the groups defined by the
variable (§10.3).
A predictor variable with more than two categories or classes is called a
class variable or a factor. We cannot simply use a class variable in the
regression equation, unless we can assume that the classes are ordered
in the same way as their codes, and that adjoining classes are in some
sense the same distance apart. For some variables, such as the
diagnosis data of Table 4.1 and the housing data of Table 13.1, this is
absurd. For others, such as the AIDS categories of Table 10.7, it is a very
strong assumption. What we do instead is to create a set of dichotomous
variables to represent the factor. For the AIDS data of Table 10.7, we can
create three variables:
hiv1 = 1 if subject has AIDS, 0 otherwise
hiv2 = 1 if subject has ARC, 0 otherwise
hiv3 = 1 if subject is HIV positive but has no symptoms, 0 otherwise
If the subject is HIV negative, all three variables are zero. hiv1, hiv2, and
hiv3 are called dummy variables. Some computer programs will
calculate the dummy variables automatically if the variable is declared to
be a factor, for others the user must define them. We put the three
dummy variables into the regression equation. This gives the equation:
mannitol = 11.4 - 0.066 × hiv1 - 2.56 × hiv2 - 1.69 × hiv3
Each coefficient is the difference in mannitol absorption between the
class represented by that variable and the class represented by all
dummy variables being zero, HIV negative, called the reference class.
The analysis of variance for this regression equation is shown in Table
17.9, and the F test shows that there is no significant relationship
between mannitol absorption and HIV status. The regression program
prints out standard errors and t tests for each dummy variable, but these t
tests should be ignored, because we cannot interpret one dummy
variable in isolation from the others.
Table 17.9. Analysis of variance for the regression of m

excretion on HIV status
Degrees
of
variation squares square ratio (
freedom
Total 58 1
559.035
Regression 3 49.011 16.337 0.60
Residual 55 1 27.455
510.024
Table 17.10. Two-way analysis of variance for mannitol e

with HIV status and diarrhoea as factors
Degrees
of
freedom
Total 58 1
559.035
Model 4 134.880 33.720 1.28
HIV 3 58.298 19.432 0.74
Diarrhoea 1 85.869 85.869 3.26
Residual 54 1 26.373
424.155
17.7 * Multi-way analysis of variance

A different approach to the analysis of multifactorial data is provided by
the direct calculation of analysis of variance. Table 17.9 is identical to the
one way analysis of variance for the same data in Table 10.8. We can
also produce analyses of variance for several factors at once. Table
17.10 shows the two-way analysis of variance for the mannitol data, the
factors being HIV status and presence or absence of diarrhoea. This
could be produced equally well by multiple regression with two
categorical predictor variables. If there were the same number of patients
with and without diarrhoea in each HIV group the factors would be
balanced. The model sum of squares would then be the sum of the sums
of squares for HIV and for diarrhoea, and these could be calculated very
simply from the total of the HIV groups and the diarrhoea groups. For
balanced data we can assess many categorical factors and their
interactions quite easily by manual calculation. See Armitage and Berry
(1994) for details. Complex multifactorial balanced experiments are rare
in medical research, and they can be analysed by regression anyway to
get identical results. Most computer programs in fact use the regression
method to calculate analyses of variance.
For another example, consider Table 17.11, which shows the results of a
study of the production of Tumour Necrosis Factor (TNF) by cells in vitro.
Two different potential stimulating factors, Mycobacterium tuberculosis
(MTB) and Fixed Activated T-cells (FAT), have been added, singly and
together. Cells from the same 11 donors have been used throughout.
Thus we have three factors, MTB, FAT, and donor. Three measurements
were made at each combination of factors; Figure 17.5(a) shows the
means of these sets of three. Every possible combination of factors is
used the same number of times in a perfect three-way factorial
arrangement. There are two missing observations. These things happen,
even in the best regulated laboratories. There are some negative values
of TNF.
Table 17.11. TNF measured under four different condit
donors (data of Dr. Jan Davies
No MTB MTB
FAT Donor TNF, 3 replicates FAT
No 1 -0.01 -0.01 -0.13 No
No 2 16.13 -9.62 -14.88 No
No 3 Missing -0.3 -0.95 No
No 4 3.63 47.5 55.2 no
No 5 -3.21 -5.64 -5.32 No
No 6 16.26 52.21 17.93 No
No 7 -12.74 -5.23 -4.06 No
No 8 -4.67 20.1 110 No
No 9 -5.4 20 10.3 No
No 10 -10.94 -5.26 -2.73 No
No 11 -4.19 -11.83 -6.29 No
Yes 1 88.16 97.58 66.27 Yes
Yes 2 196.5 114.1 134.2 Yes
Yes 3 6.02 1.19 3.38 Yes
Yes 4 935.4 1011 951.2 Yes
Yes 5 606 592.7 608.4 Yes
Yes 6 1 457 1 349 1 625 Yes
Yes 7 1 457 1 349 1 625 Yes
Yes 8 196.7 270.8 160.7 Yes
Yes 9 135.2 221.5 268 Yes
Yes 10 -14.47 79.62 304.1 Yes
Yes 11 516.3 585.9 562.6 Yes

Fig. 17.5. Tumour Necrosis Factor (TNF) measured in the presence
and absence of Fixed Activated T-cells (FAT) and Mycobacterium
tuberculosis (MTB), the natural and a transformed scale
This does not mean that the cells were sucking TNF in from their
environment, but was an artifact of the assay method and represents
measurement error.
The subject means are shown in Figure 17.5(a). This suggests several
things: there is a strong donor effect (donor 6 is always high, donor 3 is
always low, for example), MTB and FAT each increase TNF, both
together have a greater effect than either individually, the distribution of
TNF is highly skew, the variance of TNF varies greatly from group to
group, and increases with the mean. As the mean for MTB and FAT
combined is much greater than the sum of their
individual means, the researcher thought there was synergy, i.e. that
MTB and FAT worked together, the presence of one enhancing the effect
of the other. She was seeking statistical support for this conclusion (Jan
Davies, personal communication).
Table 17.12. Analysis of variance for the effects of MTB

and donor on transformed TNF
Source Degrees
of of
variation freedom
Total 43 194.040
30
Donor 10 38.890 3.889 3.72

00 00
MTB 1 58.493 58.493 55.88

20 20
FAT 1 65.244 65.244 62.33

82 82
MTB × 1 0.008 0.008 0.01

FAT 11 11
Residual 30 31.404 1.046

18 81
For statistical analysis, we would like Normal distributions with uniform

variances between the groups. A log transformation looks like a good bet,
but some observations are negative. As the log (or the square root) will
not work for negative numbers, we have to adjust the data further. The
easiest approach is to add a constant to all the observations before
transformation. I chose 20, which makes all the observations positive but
is small compared to most of the observations. I did this by trial and error.
As Figure 17.5(b) shows, the transformation has not been totally
successful, but the transformed data look much more amenable to a
Normal theory analysis than do the raw data.
The repeated measurements give us a more accurate measurement of
TNF, but do not contribute anything else. I therefore analysed the mean
transformed TNF. The analysis of variance is shown in Table 17.12.
Donor is a factor with 11 categories, hence has 10 degrees of freedom. It
is not of any importance to the science here, but is what we call a
nuisance variable, one we need to allow for but are not interested in. I
have included an interaction between MTB and FAT, because looking for
this is one of the objectives of the experiment. The main effects of MTB
and FAT are highly significant, but the interaction term is not. The
estimates of the effects with their confidence intervals are shown in Table
17.13. As the analysis was on a log scale, the antilogs (exponentials) are
also shown. The antilog gives us the ratio of the (geometric) mean in the
presence of the factor to the mean in the absence of the factor, i.e. the
amount by which TNF is multiplied by when the factor is present. Strictly
speaking, of course, it is the ratio of the geometric means of TNF plus 20,
but as 20 is small compared to most TNF measurements the ratio will be
approximately the increase in TNF.
The estimated interaction is small and not significant. The confidence
interval is wide (the sample is very small), so we cannot exclude the
possibility of an interaction, but there is certainly no evidence that one
exists. This was not what the researcher expected. This contradiction
comes about because the statistical model used is of additive effects on
the logarithmic scale, i.e. of multiplicative effects on the natural scale.
This is forced on us by the nature of the data. The
lack of interaction between the effects shows that the data are consistent
with this model, this view of what is happening. The lack of interaction
can be seen quite clearly in Figure 17.5(b), as the mean for MTB and FAT
looks very similar to the sum of the means for MTB alone and FAT alone.
Table 17.13. Estimated effects on TNF of MTB,
FAT and their interaction
Ratio
95% 95%
Effect (log effect
Confidence Confidence
scale) (natural
interval interval
scale)
With interaction term
MTB 2.333 (1.442 to 10.3 (4.2 to

3.224) 25.1)
FAT 2.463 (1.572 to 11.7 (4.8 to

3.354) 28.6)
MTB 0.054 (-1.206 to 1.1 (0.3 to 3.7)

× 1.314)
FAT
Without interaction term
MTB 2.306 (1.687 to 10.0 (5.4 to

2.925) 18.6)
FAT 2.435 (1.816 to 11.4 (6.1 to

3.054) 21.2)
Multiple regression in which qualitative and quantitative predictor

variables are both used is also known as analysis of covariance. For
ordinal data, there is a two-way analysis of variance using ranks, the
Friedman test (see Conover 1980, Altman 1991)
17.8 * Logistic regression

Logistic regression is used when the outcome variable is dichotomous, a
‘yes or no’, whether or not the subject has a particular characteristic such
as a symptom. We want a regression equation which will predict the
proportion of individuals who have the characteristic, or, equivalently,
estimate the probability that an individual will have the symptom. We
cannot use an ordinary linear regression equation, because this might
predict proportions less than zero or greater than one, which would be
meaningless. Instead we use the logit of the proportion as the outcome
variable. The logit of a proportion p is the log odds (§13.7):
The logit can take any value from minus infinity, when p = 0, to plus
infinity, when p = 1. We can fit regression models to the logit which are
very similar to the ordinary multiple regression and analysis of variance
models found for data from a Normal distribution. We assume that
relationships are linear on the logistic scale:
where x1,…,xm are the predictor variables and p is the proportion to be

predicted. The method is called logistic regression, and the calculation
is computer intensive. The effects of the predictor variables are found as
log odds ratios. We will look at the interpretation in an example.
Fig. 17.6. Body mass index (BMI) in women undergoing trial of scar
Table 17.14. Coefficients in the logistic regression for

predicting caesarian section
95%
Std.
Coef. z P Confidence
Err.
interval
BMI 0.088 0.020 4.42 <0.001 0.049 2 to

3 0 0.127 5
Induction 0.647 0.214 3.02 0.003 0.227 6 to

1 1 1.066 7
Prev. -1.796 0.298 -6.03 <0.001 -2.380 5 to

vag. del. 3 1 -1.212 0
Intercept -3.700 0.534 -6.93 <0.001 -4.747 3 to

0 3 -2.652 8
When giving birth, women who have had a previous caesarian section
usually have a trial of scar, that is, they attempt a natural labour with
vaginal delivery and only have another caesarian if this is deemed
necessary. Several factors may increase the risk of a caesarian, and in
this study the factor of interest was obesity, as measured by the body
mass index or BMI, defined as weight/height2. The distribution of BMI is
shown in Figure 17.6 (data of Andreas Papadopoulos). For caesarians
the mean BMI was 26.4 kg/m2 and for vaginal deliveries the mean was
24.9 kg/m2. Two other variables had a strong relationship with a
subsequent caesarian. Women who had had a previous vaginal delivery
(PVD) were less likely to need a caesarian, odds ratio = 0.18, 95%
confidence interval 0.10 to 0.32. Women whose labour was induced had
an increased risk of a caesarian, odds ratio = 2.11, 95% confidence
interval 1.44 to 3.08. All these relationships were highly significant. The
question to be answered was whether the relationship between BMI and
caesarian section remained when the effects of induction and previous
deliveries were allowed for.
The results of the logistic regression are shown in Table 17.14. We have
the coefficients for the equation predicting the log odds of a caesarian:
log(o) = -3.700 0 + 0.088 3 × BMI + 0.647 1 × induction - 1.796 3 × PVD
where induction and PVD are 1 if present, 0 if not. Thus for woman who
had BMI = 25 kg/m2, not been induced and had a previous vaginal
delivery the log
odds of a caesarian is estimated to be

Table 17.15. Odds ratios from the logistic
regression for predicting caesarian section
95%
Odds ratio P Confidence
interval
BMI 1.092 <0.001 1.050 to 1.136
Induction 1.910 0.003 1.256 to 2.906
Prev. vag. 0.166 <0.001 0.096 to 0.298

del.
log(o) = -3.700 0 + 0.088 3 × 25 + 0.647 1 × 0 - 1.7963 × 1 = -3.288 8

The odds is exp(-3.288 8) = 0.0373 0 and the probability is given by p =
o/(1 + o) = 0.037 30/(1 + 0.037 30) = 0.036. If labour had been induced,
the log odds would rise to
log(o) = -3.700 0 + 0.088 3 × 25 + 0.647 1 × 1 - 1.796 3 × 1 = -2.6417
giving odds exp(-2.6417) = 0.071 24 and hence probability 0.071 24/(1 +
0.071 24) = 0.067.
Because the logistic regression equation predicts the log odds, the
coefficients represent the difference between two log odds, a log odds
ratio. The antilog of the coefficients is thus an odds ratio. Some programs
will print these odds ratios directly, as in Table 17.15. We can see that
induction increases the odds of a caesarian by a factor of 1.910 and a
previous vaginal delivery reduces the odds by a factor of 0.166. These
are often called adjusted odds ratios. In this example they and their
confidence intervals are similar to the unadjusted odds ratios given
above, because the three predictor variables happen not to be closely
related to each other.
For a continuous predictor variable, such as BMI, the coefficient is the
change in log odds for an increase of one unit in the predictor variable.
The antilog of the coefficient, the odds ratio, is the factor by which the
odds must be multiplied for a unit increase in the predictor. Two units
increase in the predictor increases the odds by the square of the odds
ratio, and so on. A difference of 5 kg/m2 in BMI gives an odds ratio for a
caesarian of 1.0925 = 1.55, thus the odds of a caesarian are multiplied by
1.55. See §11.8 for a similar interpretation and fuller discussion when a
continuous outcome variable is log transformed.
When we have a case control study, we can analyse the data by using
the case or control status as the outcome variable in a logistic regression.
The coefficients are then the approximate log relative risks due to the
factors (§13.7). There is a variant called conditional logistic
regression, which can be used when the cases and controls are in
matched pairs, triples, etc.
Logistic regression is a large sample method. A rule of thumb is that
there should be at least 10 ‘yes's and 10 ‘no's, and preferably 20, for
each predictor variable (Peduzzi et al. 1996).
17.9 * Survival data using Cox regression

One problem of survival data, the censoring of individuals who have not
died at the time of analysis, has been discussed in §15.6. There is
another which is important for multifactorial analysis. We often have no
suitable mathematical model of the way survival is related to time, i.e. the
survival curve. The solution now widely adopted to this problem was
proposed by Cox (1972), and is known as Cox regression or the
proportional hazards model. In this approach, we say that for subjects
who have lived to time t, the probability of an endpoint (e.g. dying)
instantaneously at time t is h(t), which is an unknown function of time. We
call the probability of an endpoint the hazard, and h(t) is the hazard
function. We then assume that anything which affects the hazard does
so by the same ratio at all times. Thus, something which doubles the risk
of an endpoint on day one will also double the risk of an endpoint on day
two, day three and so on. Thus, if h0(t) is the hazard function for subjects
with all the predictor variables equal to zero, and h(t) is the hazard
function for a subject with some other values for the predictor variables,
h(t)/h0(t) depends only on the predictor variables, not on time t. We call
h(t)/h0(t) the hazard ratio. It is the relative risk of an endpoint occurring
at any given time.
In statistics, it is convenient to work with differences rather than ratios, so
we take the logarithm of the ratio (see §5A) and have a regression-like
equation:
where x1,…, xp are the predictor variables and b1,…, bp are the
coefficients which we estimate from the data. This is Cox's proportional
hazards model. Cox regression enables us to estimate the values of b1,
…, bp which best predict the observed survival. There is no constant term
b0, its place being taken by the baseline hazard function h0(t).
Table 15.7 shows the time to recurrence of gallstones, or the time for
which patients are known to have been gallstone-free, following
dissolution by bile acid treatment or lithotrypsy, with the number of
previous gallstones, their maximum diameter, and the time required for
their dissolution. The difference between patients with a single and with
multiple previous gallstones was tested using the logrank test (§15.6).
Cox regression enables us to look at continuous predictor variables, such
as diameter of gallstone, and to examine several predictor variables at
once. Table 17.16 shows the result of the Cox regression. We can earn-
out an approximate test of significance dividing the coefficient by its
standard error, and if the null hypothesis that the coefficient would be
zero in the population is true, this follows a Standard Normal distribution.
The chi-squared statistic tests the relationship between the time to
recurrence and the three variables together. The maximum diameter has
no significant relationship to time to recurrence, so we can try a model
without it (Table 17.17). As the change in overall chi-squared shows,
removing diameter has had very little effect.
The coefficients in Table 17.17 are the log hazard ratios. The coefficient
for
multiple gallstones is 0.963. If we antilog this, we get exp(0.963) = 2.62.

As multiple gallstones is a 0 or 1 variable, the coefficient measures the
difference between those with single and multiple stones. A patient with
multiple gallstones is 2.62 times as likely to have a recurrence at any
time than a patient with a single stone. The 95% confidence interval for
this estimate is found from the antilogs of the confidence interval in Table
17.17, 1.30 to 5.26. Note that a positive coefficient means an increased
risk of the event, in this case recurrence. The coefficient for months to
dissolution is 0.043, which has antilog = 1.04. This is a quantitative
variable, and for each month to dissolve the hazard ratio increases by a
factor of 1.04. Thus a patient whose stone took two months to dissolve
has a risk of recurrence 1.04 times that for a patient whose stone took
one month, a patient whose stone took three months has a risk 1.042
times that for a one month patient, and so on.
Table 17.16. Cox regression of time to recurrence of

gallstones on presence of multiple stones,
maximum diameter of stone and months to
dissolution
95%
Std.
Variable Coef. z P Conf.
Err.
interval
Mult. 0.838 0.401 2.09 0.038 0.046

gallstones to
1.631
Max. -0.023 0.036 -0.63 0.532 -0.094
diam. to
0.049
Months to 0.044 0.017 2.64 0.009 0.011

dissol. to
0.078
X2 = 12.57, 3 d.f., P = 0.006.
Table 17.17. Cox regression of time to recurrence

of gallstones on presence of multiple stones and
months to dissolution
95%
Std.
Variable Coef. z P Conf.
Err.
interval
Mult. 0.963 0.353 2.73 0.007 0.266

gallstones to
1.661
Months to 0.043 0.017 2.59 0.011 0.010

dissol. to
0.076
X2 = 12.16, 2 d.f., P = 0.002.
If we have only the dichotomous variable multiple gallstones in the Cox

model, we get for the overall test statistic X2 = 6.11, 1 degrees of
freedom. In §15.6 we analysed these data by comparison of two groups
using the logrank test which gave X2 = 6.62, 1 degree of freedom. The
two methods give similar, but not identical results. The logrank test is
non-parametric, making no assumption about the distribution of survival
time. The Cox method is said to semi-parametric, because although it
makes no assumption about the shape of the distribution of survival time,
it does require assumptions about the hazard ratio.
Like logistic regression (§17.8), Cox regression is a large sample
method. A rule of thumb is that there should be at least 10, and
preferably 20, events (deaths) for each predictor variable. Fuller accounts
of Cox regression are given by Altman (1991), Matthews and Farewell
(1988), Parmar and Machin (1995), and Hosmer and Lemeshow (1999).
17.10 * Stepwise regression

Stepwise regression is a technique for choosing predictor variables from
a large set. The stepwise approach can be used with multiple linear,
logistic and Cox regression and with other, less often seen, regression
techniques (§17.12) too.
There are two basic strategies: step-up and step-down, also called
forward and backward. In step-up or forward regression, we fit all
possible one-way regression equations. Having found the one which
accounts for the greatest variance, all two-way regressions including this
variable are fitted. The equation accounting for the most variation is
chosen, and all three-way regressions including these are fitted, and so
on. This continues until no significant increase, in variation accounted for
is found. In the step-down or backward method, we first fit the
regression with all the predictor variables, and then the variable is
removed which reduces the amount of variation accounted for by the
least amount, and so on. There are also more complex methods, in which
variables can both enter and leave the regression equation.
These methods must be treated with care. Different stepwise techniques
may produce different sets of predictor variables in the regression
equation. This is especially likely when the predictor variables are
correlated with one another. The technique is very useful for selecting a
small set of predictor variables for purposes of standardization and
prediction. For trying to get an understanding of the underlying system,
stepwise methods can be very misleading. When predictor variables are
highly correlated, once one has entered the equation in a step-up
analysis, the other will not enter, even though it is related to the outcome.
Thus it will not appear in the final equation.
17.11 * Meta-analysis: Data from several studies

Meta-analysis is the combination of data from several studies to produce
a single estimate. From the statistical point of view, meta-analysis is a
straightforward application of multifactorial methods. We have several
studies of the same thing, which might be clinical trials or epidemiological
studies, perhaps carried out in different countries. Each trial gives us an
estimate of an effect. We assume that these are estimates of the same
global population value. We check the assumptions of the analysis, and,
if these assumptions are satisfied, we combine the separate study
estimates to make a common estimate. This is a multifactorial analysis,
where the treatment or risk factor is one predictor variable and the study
is another, categorical, predictor variable.
The main problems of meta-analysis arise before we begin the analysis
of the data. First, we must have a clear definition of the question so that
we only include studies which address this. For example, if we want to
know whether lowering serum cholesterol reduces mortality from
coronary artery disease, we would not want to include a study where the
attempt to lower cholesterol failed. On the other hand, if we ask whether
dietary advice lowers mortality, we would include such a study. Which
studies we include may have a profound influence on the conclusions
(Thompson 1993). Second, we must have all the relevant
studies. A simple literature search is not enough. Not all studies which
have been started are published; studies which produce significant
differences are more likely to be published than those which do not (e.g.
Pocock and Hughes 1990; Easterbrook et al. 1991). Within a study,
results which are significant may be emphasized and parts of the data
which produce no differences may be ignored by the investigators as
uninteresting. Publication of unfavourable results may be discouraged by
the sponsors of research. Researchers who are not native English
speakers may feel that publication in the English language literature is
more prestigious as it will reach a wider audience, and so try there first,
only publishing in their own language if they cannot publish in English.
The English language literature may thus contain more positive results
than do other literatures. The phenomenon by which significant and
positive results are more likely to be reported, and reported more
prominently, than non-significant and negative ones is called publication
bias. Thus we must not only trawl the published literature for studies, but
use personal knowledge of ourselves and others to locate all the
unpublished studies. Only then should we carry out the meta-analysis.
When we have all the studies which meet the definition, we combine
them to get a common estimate of the effect of the treatment or risk
factor. We regard the studies as providing several observations of the
same population value. There are two stages in meta-analysis. First we
check that the studies do provide estimates of the same thing. Second,
we calculate the common estimate and its confidence interval. To do this
we may have the original data from all the studies, which we can
combine into one large data file with study as one of the variables, or we
may only have summary statistics obtained from publications.
If the outcome measure is continuous, such as mean fall in blood
pressure, we can check that subjects are from the same population by
analysis of variance, with treatment or risk factor, study, and interaction
between them in the model. Multiple regression can also be used,
remembering that study is a categorical variable and dummy variables
are required. We test the treatment times study interaction in the usual
way. If the interaction is significant this indicates that the treatment effect
is not the same in all studies, and so we cannot combine the studies. It is
the interaction which is important. It does not matter much if the mean
blood pressure varies from study to study. What matters is whether the
effect of the treatment on blood pressure varies more than we would
expect. We may want to examine the studies to see whether any
characteristic of the studies explains this variation. This might be a
feature of the subjects, the treatment or the data collection. If there is no
interaction, then the data are consistent with the treatment or risk factor
effect being constant. This is called a fixed effects model (see §10.12).
We can drop the interaction term from the model and the treatment or
risk factor effect is then the estimate we want. Its standard error and
confidence interval are found as described in §17.2. If there is an
interaction, we cannot estimate a single treatment effect. We can think of
the studies as a random sample of the possible trials and estimate the
mean treatment effect for this population. This is called the random
effects model (§10.12). The
confidence interval is usually much wider than that found using the fixed
effect model.
Table 17.18. Odds ratios and confidence intervals

in five studies of vitamin A supplementation in
infectious disease (Glasziou and Mackerras 1993)
Dose
Study Vitamin A Controls
regime
Deaths Number Deaths Number
1 200 101 12 991 130 12 209

000 IU
six-
monthly
2 200 39 7 076 41 7 006
000 IU
six-
monthly
3 8 333 37 7 764 80 7 755

IU
weekly
4 200 152 12 541 210 12 264

000 IU
four-
monthly
5 200 138 3 786 167 3 411

000 IU
once
Table 17.19. Odds ratios and confidence

intervals in five studies of vitamin A
supplementation in infectious disease
Study Odds ratio 95% Confidence interval
1 0.73 0.56 to 0.95

2 0.94 0.61 to 1.46
3 0.46 0.31 to 0.68
4 0.70 0.57 to 0.87
5 0.73 0.58 to 0.93
If the outcome measure is dichotomous, such as survived or died, the

estimate of the treatment or risk factor effect will be in the form of an
odds ratio (§13.7). We can proceed in the same way as for a continuous
outcome, using logistic regression (§17.8). Several other methods exist
for checking the homogeneity of the odds ratios across studies, such as
Woolf's test (see Armitage and Berry 1994) or that of Breslow and Day
(1980). They all give similar answers, and, since they are based on
different large-sample approximations, the larger the study samples the
more similar the results will be. Provided the odds ratios are
homogeneous across studies, we can then estimate the common odds
ratio. This can be done using the Mantel-Haenszel method (see Armitage
and Berry 1994) or by logistic regression.
For example, Glasziou and Mackerras (1993) carried out a meta-analysis
of vitamin A supplementation in infectious disease. Their data for five
community studies are shown in Table 17.18. We can obtain odds ratios
and confidence intervals as described in §13.7, shown in Table 17.19.
The common odds ratio can be found in several ways. To use logistic
regression, we regress the event of death on vitamin A treatment and
study. I shall treat the treatment as a dichotomous variable, set to 1 if
treated with vitamin A, 0 if control. Study is a categorical variable, so we
create dummy variables study1 to study4, which are set to one for
studies 1 to 4 respectively, and to zero otherwise. We test the interaction
by creating another set of variables, the products of study1 to study4 and
vitamin A. Logistic regression of death on vitamin A, study and interaction
gives a chi-squared statistic for the model of
496.99 with 9 degrees of freedom, which is highly significant. Logistic

regression without the interaction terms gives 490.33 with 5 degrees of
freedom. The difference is 496.99 - 490.33 = 6.66 with 9 - 5 = 4 degrees
of freedom, which has P = 0.15, so we can drop the interaction from the
model. The adjusted odds ratio for vitamin A is 0.70, 95% confidence
interval 0.62 to 0.79, P < 0.0001.
Fig. 17.7. Meta-analysis of five vitamin A trials (data of Glasziou and

Mackerras 1993). The vertical lines are the confidence intervals.
The odds ratios and their confidence intervals are shown in Figure 17.7.
The confidence interval is indicated by a line, the point estimate of the
odds ratio by a circle. In this picture the most important trial appears to be
study 2, with the widest confidence interval. In fact, it is the study with the
least effect on the whole estimate, because it is the study where the odds
ratio is least well estimated. In the second picture, the odds ratio is
indicated by the middle of a square. The area of the square is
proportional to the number of subjects in the study. This now makes
study 2 appear relatively unimportant, and makes the overall estimate
stand out.
There are many variants on this style of graph, which is sometimes called
a forest diagram. The graph is often shown with the studies on the
vertical axis
and the odds ratio or difference in mean on the horizontal axis (Figure
17.8). The combined estimate of the effect may be shown as a lozenge
or diamond shape and for odds ratios a logarithmic scale is often
employed, as in Figure 17.8.
Fig. 17.8. Meta-analysis of five vitamin A trials, vertical version
17.12 * Other multifactorial methods

The choice of multiple, logistic or Cox regression is determined by the
nature of the outcome variable: continuous, dichotomous, or survival
times respectively. There are other types of outcome variable and
corresponding multifactorial techniques. I shall not go into any details, but
this list may help should you come across any of them. I would
recommend you consult a statistician should you actually need to use
one of these methods. The techniques for dealing with predictor variables
described in §17.2–17.4 and §17.6 apply to all of them.
If the outcome variable is categorical with more than two categories, e.g.
several diagnostic groups, we use a procedure called multinomial
logistic regression. This estimates for a subject with given values of the
predictor variable the probability that the subject will be in each category.
If the categories are ordered, e.g. tumour stage, we can take the ordering
into account using ordered logistic regression. Both these techniques
are closely related to logistic regression (§17.8).
If the outcome is a count, such as hospital admissions in a day or deaths
related to a specific cause per week or month, we can use Poisson
regression. This is particularly useful when we have many time intervals
but the numbers of events per interval is small, so that the assumptions
of multiple regression (§17.5) do not apply.
A slightly different problem arises with multi-way contingency tables
where there is no obvious outcome variable. We can use a technique
called log linear modelling. This enables us to test the relationship
between any two of the variables in the table holding the others constant.

93. In multiple regression, R2:
(a) is the square of the multiple correlation coefficient;
(b) would be unchanged if we exchanged the outcome
(dependent) variable and one of the predictor (independent)
variables;
(c) is called the proportion of variability explained by the
regression;
(d) is the ratio of the error sum of squares to the total sum of
squares;
(e) would increase if more predictor variables were added to the
model.
View Answer
Table 17.20. Analysis of variance for the effects of age,
ethnic group (Afro-Caribbean versus White) on inter-
distance (Imafedon, personal communication)
Source Degrees
of of
variation freedom
Total 37 603.586
Age 2 124.587 62.293 6.81

group
Sex 1 1.072 1.072 0.12
Ethnic 1 134.783 134.783 14.74

group
Residual 33 301.782 9.145
94. The analysis of variance table for a study of the distance

between the pupils of the eyes is shown in Table 17.20:
(a) there were 34 observations;
(b) there is good evidence of an ethnic group difference in the
population:
(c) we can conclude that there is no difference in inter-pupil
distance between men and women;
(d) there were two age groups;
(e) the difference between ethnic groups is likely to be due to a
relationship between ethnicity and age in the sample.
View Answer
Table 17.21. Logistic regression of graft failure after 6

(Thomas et al. 1993)
Std. z 95%
Variable Coef. P
Err. =coef/se Conf.
White 1.238 0.273 4.539 < 0.695

cell 0.001
count
Graft 0.175 0.876 0.200 0.842 -1.570

type 1
Graft 0.973 1.030 0.944 0.348 -1.080

type 2
Graft 0.038 1.518 0.025 0.980 -2.986

type 3
Female -0.289 0.767 -0.377 0.708 -1.816

Age 0.022 0.035 0.633 0.528 -0.048
Smoker 0.998 0.754 1.323 0.190 -0.504
Diabetic 1.023 0.709 1.443 0.153 -0.389
Constant -13.726 3.836 -3.578 0.001 -21.369
Number of observations = 84, chi-squared = 38.05, d.f. = 8,

0.0001.
95. Table 17.21 shows the logistic regression of vein graft failure on
some potential explanatory variables. From this analysis:
(a) patients with high white cell counts were more likely to have
graft failure;
(b) the log odds of graft failure for a diabetic is between 0.389 less
and 2.435 greater than that for a non-diabetic;
(c) grafts were more likely to fail in female subjects, though this is
not significant;
(d) there were four types of graft;
(e) any relationship between white cell count and graft failure may
be due to smokers having higher white cell counts.
View Answer
Fig. 17.9. Oral and forehead temperature measurements made in a
group of pyrexic patients
96. For the data in Figure 17.9:

(a) the relationship could be investigated by linear regression;
(b) an ‘oral squared’ term could be used to test whether there is
any evidence that the relationship is not a straight line;
(c) if an ‘oral squared’ term were included there would be 2
degrees of freedom for the model;
(d) the coefficients of an ‘oral’ and an ‘oral squared’ term would be
uncorrelated;
(e) the estimation of the coefficient of a quadratic term would be
improved by subtracting the mean from the oral temperature
before squaring.
View Answer
Table 17.22. Cox regression of time to

readmission for asthmatic children following
discharge from hospital (Mitchell et al. 1994)
Std.
Variable Coef. coef/se P
err.
Boy -0.197 0.088 -2.234 0.026
Age -0.126 0.017 -7.229 <

0.001
Previous 0.395 0.034 11.695 <

admissions 0.001
(square
root)
Inpatient 0.267 0.093 2.876 0.004

i.v. therapy
Inpatient -0.728 0.295 -2.467 0.014

theophyline
Number of observations = 1 024, X2 = 167.15, 5

d.f., P < 0.0001.
97. Table 17.22 shows the results of an observational study

following up asthmatic children discharged from hospital. From this
table:
(a) the analysis could only have been done if all children had been
readmitted to hospital;
(b) the proportional hazards model would have been better than
Cox regression;
(c) Boys have a shorter average time before readmission than do
girls;
(d) the use of theophyline prevents readmission to hospital;
(e) children with several previous admissions have an increased
risk of readmission.
View Answer
Fig. 17.10. Cushion volume against number of pairs of somites for

two groups of mouse embryos (Webb and Brown, personal
communication)
Table 17.23. Number of somites and cushion volume
mouse embryos
Normal Trisomy-16
som. c.vol. som. c.vol. som. c.vol. som.
17 2.674 28 3.704 15 0.919 28
20 3.299 31 6.358 17 2.047 28
21 2.486 32 3.966 18 3.302 28
23 1.202 32 7.184 20 4.667 31
23 4.263 34 8.803 20 4.930 32
23 4.620 35 4.373 23 4.942 34
25 4.644 40 4.465 23 6.500 35
25 4.403 42 10.940 23 7.122 36
27 5.417 43 6.035 25 7.688 40

27 4.395 25 4.230 42
27 8.647
17E * Exercise: A multiple regression analysis

Trisomy-16 mice can be used as an animal model for Down's syndrome.
This analysis looks at the volume of a region of the heart, the
atrioventricular cushion, of a mouse embryo, compared between trisomic
and normal embryos. The embryos were at varying stages of
development, indicated by the number of pairs of somites (precursors of
vertebrae). Figure 17.10 and Table 17.23 show the data. The group was
coded 1=normal, 2=trisomy-16. Table 17.24 shows the results of a
regression analysis and Figure 17.11 shows residual plots.
1. Is there any evidence of a difference in volume between groups
for given stage of development?
View Answer
2. Figure 17.11 shows residual plots for the analysis of Table 17.24.
Are there any features of the data which might make the analysis
invalid?
View Answer
Table 17.24. Regression of cushion volume on number

of somites and group in mouse embryos
Degrees
of
freedom
Total 39 328.976
Due to 2 197.708 98.854 27.86

regression
Residual 37 131.268 3.548

(about
regression)
Variable Coef. Std. Err. t P 95% Conf. interval
group 2.44 0.60 4.06 < 0.001 1.29 to 3.65
somites 0.27 0.04 6.70 < 0.001 0.19 to 0.36
Fig. 17.11. Residual against number of pairs of somites and Normal

plot of residuals for the analysis of Table 17.24
3. It appears from Figure 17.10 that the relationship between volume
and number of pairs of somites may not be the same in the two
groups. Table 17.25 shows the analysis of variance for regression
analysis including an interaction term. Calculate the F-ratio to test
the evidence that the relationship is different in normal and trisomy-
16 embryos. You can find the probability from Table 10.1, using the
fact that the square root of F with 1 and n degrees of freedom is t
with n degrees of freedom.
View Answer
Table 17.25. Analysis of variance for regression with nu

pairs of somites × group interaction
Degrees
of
freedom
Total 39 328.976
Due to 3 207.139 69.046 20.40

regression
Residual 36 121.837 3.384

(about
regression)
> Table of Contents > 18 - Determination of sample size
18
Determination of sample size
18.1 * Estimation of a population mean

One of the questions most frequently asked of a medical statistician is
‘How large a sample should I take?’ In this chapter we shall see how
statistical methods for deciding sample sizes can be used in practice as
an aid in designing investigations. The methods we shall use are large
sample methods, that is, they assume that large sample methods will be
used in the analysis and so take no account of degrees of freedom.
We can use the concepts of standard error and confidence interval to
help decide how many subjects should be included in a sample. If we
want to estimate some population quantity, such as the mean, and we
know how the standard error is related to the sample size, then we can
calculate the sample size required to give a confidence interval with the
desired width. The difficulty is that the standard error may also depend
either on the quantity we wish to estimate, or on some other property of
the population, such as the standard deviation. We must estimate these
quantities from data already available, or carry out a pilot study to obtain
a rough estimate. The calculation of sample size can only be
approximate anyway, so the estimates used to do it need not be precise.
If we want to estimate the mean of a population, we can use the formula
for the standard error of a mean, s/√n, to estimate the sample size
required. For example, suppose we wish to estimate the mean FEV1 in a
population of young men. We know that in another study FEV1 had
standard deviation s = 0.67 litre (§4.8). We therefore expect the standard
error of the mean to be 0.67/√n. We can set the size of standard error we
want and choose the sample size to achieve this. We might decide that a
standard error of 0.1 litre is what we want, so that we would estimate the
mean to within 1.96 × 0.1 = 0.2 litre. Then: SE = 0.67/√n, n = 0.672/SE2 =
0.672/0.12 = 45. We can also see what the standard error and width of
the 95% confidence interval would be for different values of n:
n 10 20 50 100 200 500
standard error 0.212 0.150 0.095 0.067 0.047 0.030
95% confidence ±0.42 ±0.29 ±0.19 ±0.13 ±0.09 ±0.06

interval
So that if we had a sample size of 200, we would expect the 95%
confidence interval to be 0.09 litre on either sided of the sample mean
(1.96 standard errors) whereas with a sample of 50 the 95% confidence
interval would be 0.19 litre on
either side of the mean.
18.2 * Estimation of a population proportion

When we wish to estimate a proportion we have a further problem. The
standard error depends on the very quantity which we wish to estimate.
We must guess the proportion first. For example, suppose we wish to
estimate the prevalence of a disease, which we suspect to be about 2%,
to within 5%, i.e. to the nearest 1 per 1 000. The unknown proportion, p,
is guessed to be 0.02 and we want the 95% confidence interval to be
0.001 on either side, so the standard error must be half this, 0.000 5.
The accurate estimation of very small proportions requires very large

samples. This is a rather extreme example and we do not usually need to
estimate proportions with such accuracy. A wider confidence interval,
obtainable with a smaller sample is usually acceptable. We can also ask
‘If we can only afford a sample size of 1 000, what will be the standard
error?’
The 95% confidence limits would be, roughly, p ± 0.009. For example, if
the estimate were 0.02, the 95% confidence limits would be 0.011 to
0.029. If this accuracy were sufficient we could proceed.
These estimates of sample size are based on the assumption that the
sample is large enough to use the Normal distribution. If a very small
sample is indicated it will be inadequate and other methods must be used
which are beyond the scope of this book.
18.3 * Sample size for significance tests

We often want to demonstrate the existence of a difference or
relationship as well as wanting to estimate its magnitude, as in a clinical
trial, for example. We base these sample size calculations on
significance tests, using the power of a test (§9.9) to help choose the
sample size required to detect a difference if it exists. The power of a test
is related to the postulated difference in the population, the standard error
of the sample difference (which in turn depends on the sample size), and
the significance level, which we usually take to be α = 0.05. These
quantities are linked by an equation which enables us to determine any
one of them given the others. We can then say what sample size would
be required to detect any given difference. We then decide what
difference we need to be able
to detect. This might be a difference which would have clinical

importance of a difference which we think the treatment may produce.
Suppose we have a sample which gives an estimate d of the population
difference µd. We assume d comes from a Normal distribution with mean
µd and has standard error SE(d). Here d might be the difference between
two means two proportions, or anything else we can calculate from data.
We are interested in testing the null hypothesis that there is no difference
in the population. i.e. µd = 0. We are going to use a significance test at
the α level, and want the power, the probability of detecting a significant
difference, to be P.
I shall define uα to be the value such that the Standard Normal
distribution (mean 0 and variance 1) is less than -uα or greater than uα
with probability α. For example, u0.05 = 1.96. The probability of lying
between -uα and uα is 1 - α. Thus uα is the two sided α probability point of
the Standard Normal distribution, as shown in Table 7.2.
If the null hypothesis were true, the test statistic d/SE(d) would be from a
Standard Normal distribution. We reject the null hypothesis at the α level
if the test statistic is greater than uα or less than -uα, 1.96 for the usual
5% significance level. For significance we must have:
Let us assume that we are trying to detect a difference such that d will be
greater than 0. The first alternative is then extremely unlikely and can be
ignored. Thus we must have, for a significant difference: d/SE(d) > uα so
d > uαSE(d). The critical value which d must exceed is uαSE(d).
Now, d is a random variable, and for some samples it will be greater than
its mean, µd, for some it will be less than its mean. d is an observation
from a Normal distribution with mean µd and variance SE(d)2. We want d
to exceed the critical value with probability P, the chosen power of the
test. The value of the Standard Normal distribution which is exceeded
with probability P is -u2(1-P) (see Figure 18.1). (1 - P) is often represented
as β (beta). This is the probability of failing to obtain a significant
difference when the null hypothesis is false and the population difference
is µd. It is the probability of a Type II error (§9.4). The value which d
exceeds with probability P is the mean minus -u2(1-P) standard
deviations: µd - u2(1-P)SE(d). Hence for significance this must exceed the
critical value, uαSE(d). This gives
µd - u2(1-P)SE(d) = uαSE(d)
Putting the correct standard error formula into this will yield the required
sample size. We can rearrange it as
µ2d = (uα + u2(1-P))2SE(d)2
This is the condition which must be met if we are to have a probability P
of
detecting a significant difference at the α level. We shall use the

expression (uα2(1-P))2 a lot, so for convenience I shall denote it by f(α, P).
Table 18.1 shows the values of the factor f(α, P) for different values of α
and P. The usual value used for α is 0.05, and P is usually 0.80, 0.90, or
0.95.
Fig. 18.1. Relationship between P and u2(1-P)
Table 18.1. Values of f(α, P) = (uα + u2(1-P))2 for

different P and α
Significance level, α
Power, P
0.05 0.01
0.50 3.8 6.6
0.70 6.2 9.6
0.80 7.9 11.7
0.90 10.5 14.9
0.95 13.0 17.8
0.99 18.4 24.0
Sometimes we do not expect the new treatment to be better than the

standard treatment, but hope that it will be as good. We want to test
treatments which may be as good as the existing treatment because the
new treatment may be cheaper, have fewer side effects, be less invasive,
or under our patent. We cannot use the power method based on the
difference we want to be able to detect, because we are not looking for a
difference. What we do is specify how different the treatments might be in
the population and still be regarded as equivalent, and design our study
to detect such a difference. This can get rather complicated and
specialised, so I shall leave the details to Machin et al. (1998).
18.4 * Comparison of two means

When we are comparing the means of two samples, sample sizes n1 and
n2, from populations with means µ1 and µ2, with the variance of the
measurements being σ2, we have µd = µ1 - µ2 and
so the equation becomes:
For example, suppose we want to compare biceps skinfold in patients

with Crohn's disease and coeliac disease, following up the inconclusive
comparison of biceps skinfold in Table 10.4 with a larger study. We shall
need an estimate of the variability of biceps skinfold in the population we
are considering. We can usually get this from the medical literature, or as
here from our own data. If not we must do a pilot study, a small
preliminary investigation to collect some data and calculate the standard
deviation. For the data of Table 10.4, the within-groups standard
deviation is 2.3 mm. We must decide what difference we want to detect.
In practice this may be difficult. In my small study the mean skinfold
thickness in the Crohn's patients was 1 mm greater than in my coeliac
patients. I will design my larger study to detect a difference of 0.5 mm. I
shall take the usual significance level of 0.05. I want a fairly high power,
so that there is a high probability of detecting a difference of the chosen
size should it exist. I shall take 0.90, which gives f(α, P) = 10.5 from Table
18.1. The equation becomes:
We have one equation with two unknowns, so we must decide on the

relationship between n1 and n2. I shall try to recruit equal numbers in the
two groups:
and I need 444 subjects in each group.
It may be that we do not know exactly what size of difference we are
interested in. A useful approach is to look at the size of the difference we
could detect using different sample sizes, as in Table 18.2. This is done
by putting different values of n in the sample size equation.
Table 18.2. Difference in mean biceps skinfold

thickness (mm) detected at the 5%
significance level with power 90% for different
sample sizes, equal groups
Size of each Difference detected with

group, n probability 0.90
10 3.33
20 2.36
50 1.49
100 1.05
200 0.75
500 0.47
1000 0.33
Table 18.3. Sample size required in each group to

detect a difference between two means at the 5%
significance level with power 90%, using equally sized
samples
Difference Difference Difference

in in in
n n
standard standard standard
deviations deviations deviations
0.01 210 0.1 2100 0.6

000
0.02 52 0.2 525 0.7

500
0.03 23 0.3 233 0.8

333
0.04 13 0.4 131 0.9

125
0.05 8 0.5 84 1.0

400
If we measure the difference in terms of standard deviations, we can

make a general table. Table 18.3 gives the sample size required to detect
differences between two equally sized groups. Altman (1982) gives a
neat graphical method of calculation.
We do not need to have n1 = n2 = n. We can calculate µ1 - µ2 for different
combinations of n1 and n2. The size of difference, in terms of standard
deviations, which would be detected is given in Table 18.4. We can see
from this that what matters is the size of the smaller sample. For
example, if we have 10 in group 1 and 20 in group 2, we do not gain very
much by increasing the size of group 2: increasing group 2 from 20 to
100 produces less advantage than increasing group 1 from 10 to 20. In
this case the optimum is clearly to have samples of equal size.
Table 18.4. Difference (in standard deviations)

detectable at the 5% significance level with power
90% for different sample sizes, unequal groups
n2 n1
10 1.45 1.25 1.13 1.08 1.05 1.03
20 1.25 1.03 0.85 0.80 0.75 0.75

50 1.13 0.85 0.65 0.55 0.50 0.48
100 1.08 0.80 0.55 0.45 0.40 0.35
200 1.05 0.75 0.50 0.40 0.33 0.28
500 1.03 0.75 0.48 0.35 0.28 0.20
1000 1.03 0.73 0.48 0.35 0.25 0.18
18.5 * Comparison of two proportions

Using the same approach, we can also calculate the sample sizes for
comparing two proportions. If we have two samples with sizes n1 and n2
from Binomial populations with proportions p1 and p2 the difference is µd
= p1 - p2, the standard error of the difference between the sample
proportions (§8.6) is:
If we put these into the previous formula we have:
The size of the proportions, p1 and p2, is important, as well as their

difference. (The significance test implied here is similar to the chi-
squared test for a 2 by 2 table). When the sample sizes are equal, i.e. n1
= n2 = n, we have
There are several slight variations on this formula. Different computer

programs may therefore give slightly different sample size estimates
Suppose we wish to compare the survival rate with a new treatment with
that with an old treatment, where it is about 60%. What values of n1 and
n2 will have 90% chance of giving significant difference at the 5% level for
different values of p2? For P = 0.90 and α = 0.05, f(α, P) = 10.5. Suppose
we wish to detect an increase in the survival rate on the new treatment to
80%, so p2 = 0.80, and p1 = 0.60.
Table 18.5. Sample size in each group required

to detect different proportions p2 when p1 =
0.6 at the 5% significance level with power
90%, equal groups
p2 n
0.90 39
0.80 105
0.70 473
0.65 1964
Table 18.6. n2 for different n1 and p2 when p1 = 0.05 at th

level with power 90%
n1
p2
50 100 200 500 1000 2000
0.06 . . . . . .
0.07 . . . . . 4500
0.08 . . . . 1900 1200
0.10 . . 1500 630 472 420
0.15 5400 270 180 150 140 140
0.20 134 96 84 78 76 76
We would require 105 in each group to have a 90% chance of showing a

significant difference if the population proportions were 0.6 and 0.8.
When we do not have a clear idea of the value of p2 in which we are
interested, we can calculate the sample size required for several
proportions, as in Table 18.5. It is immediately apparent that to detect
small differences between proportions we need very large samples.
The case where samples are of equal size is usual in experimental
studies, but not in observational studies. Suppose we wish to compare
the prevalence of a certain condition in two populations. We expect that
in one population it will be 5% and that it may be more common the
second. We can rearrange the equation:
Table 18.6 shows n2 for different n1 and p2. For some values of n1 we get
a negative value of n2. This means that no value of n2 is large enough. It
is clear
that when the proportions themselves are small, the detection of small
differences requires very large samples indeed.
18.6 * Detecting a correlation

Investigations are often set up to look for a relationship between two
continuous variables. It is convenient to treat this as an estimation of or
test of a correlation coefficient. The correlation coefficient has an
awkward distribution, which tends only very slowly to the Normal, even
when both variables themselves follow a Normal distribution. We can use
Fisher's z transformation:
which follows a Normal distribution with mean
and variance 1/(n - 3) approximately, where ρ is the population

correlation coefficient and n is the sample size (§11.10). For sample size
calculations we can approximate zρ by
Thus we have
and we can estimate n, ρ or P given the other two. Table 18.7 shows the
sample size required to detect a correlation coefficient with a power of P
= 0.9 and a significance level α = 0.05.
Table 18.7. Approximate sample size required

to detect a correlation at the 5% significance
level with power 90%
ρ n ρ n ρ n
0.01 100 000 0.1 1 000 0.6 25

0.02 26 000 0.2 260 0.7 17
0.03 12 000 0.3 110 0.8 12
0.04 6 600 0.4 62 0.9 8
0.05 4 200 0.5 38
18.7 * Accuracy of the estimated sample size

In this chapter I have assumed that samples are sufficiently large for
sampling distributions to be approximately Normal and for estimates of
variance to be good estimates. With very small samples this may not be
the case. Various more accurate methods exist, but any sample size
calculation is approximate and except for very small samples, say less
than 10, the methods described above should be adequate. When the
sample is very small, we might need to replace the significance test
component of f(α, P) by the corresponding number from the t distribution.
These methods depend on assumptions about the size of difference
sought and the variability of the observations. It may be that the
population to be studied may not have exactly the same characteristics
as those from which the standard deviation or proportions were
estimated. The likely effects of changes in these can be examined by
putting different values of them in the formula. However, there is always
an element of venturing into the unknown when embarking on a study
and we can never by sure that the sample and population will be as we
expect. The determination of sample size as described above is thus only
a guide, and it is probably as well always to err on the side of a larger
sample when coming to a final decision.
The choice of power is arbitrary, in that there is not optimum choice of
power for a study. I usually recommend 90%, but 80% is often quoted.
This gives smaller estimated sample sizes, but, of course, a greater
chance of failing to detect effects.
For a fuller treatment of sample size estimation and fuller tables see
Machin et al. (1998) and Lemeshow et al. (1990).
18.8 * Trials randomized in clusters

When we randomize by cluster rather than individual (§2.11) we lose
power compared to an individually-randomized trial of the same size.
Hence to get the power we want, we must increase the sample size from
that required for an individually randomized trial. The ratio of the number
of patients required for a cluster trial to that for a simply randomized trial
is called the design effect of the study. It depends on the number of
subjects per cluster. For the purpose of sample size calculations we
usually assume this is constant.
If the outcome measurement is continuous, e.g. serum cholesterol, a
simple
method of analysis is based on the mean of the observations for all

subjects in the cluster, and compares these means between the
treatment groups (§10.13). We will denote the variance of observations
within one cluster by s2w and assume that this variance is the same for all
clusters. If there are m subjects in each cluster then the variance of a
single sample mean is s2w/m. The true cluster mean (unknown) will vary
from cluster to cluster, with variance s2c (see §10.12). The observed
variance of the cluster means will be the sum of the variance between
clusters and the variance within clusters, i.e. variance of outcome = s2c +
s2w/m. Hence the standard error for the difference between means is
given by
where n1 and n2 are the numbers of clusters in the two groups. For most
trials n1 = n2 = n. so
Hence, using the general method of §18.3, we can calculate the required
number of clusters by
When the outcome is a dichotomous, ‘yes or no’ variable, we replace s2w

by p(1 - p), where p is the probability of a ‘yes’.
For example, in a proposed study of a behavioural intervention to lower
cholesterol in general practice, practices were to be randomised into two
groups, one to offer intensive dietary intervention by specially trained
practice nurses using a behavioural approach and the other to usual
general practice care. The outcome measure would be mean cholesterol
levels in patients attending each practice one year later. Estimates of
between practice variance and within practice variance were obtained
from the MRC thrombosis prevention trial (Meade et al. 1992) and were
s2c = 0.004 6 and s2w = 1.28 respectively. The minimum difference
considered to be clinically relevant was 0.1 mmol/l. If we recruit 50
patients per practice, we would have s2 = s2w + s2w/m = 0.004 6 +
1.28/50 = 0.030 2. If we choose power P = 0.90 and and significance
level α = 0.05, from Table 18.1 f(P, α) = 10.5. The number of practices
required to detect a difference of 0.1 mmol/l is given by n = 10.5 × 0.030
2 × 2/0.12 = 63 in each group. This would give us 63 × 50 = 3 150
patients in each group. A completely randomized trial without clusters
would have s2 = 0.004 6 + 1.28 = 1.284 6 and we would need n = 10.5 ×
1.284 6 × 2/0.12 = 2 698 patients per group. Thus the design effect of
having clusters of 50 patients is 3 150/2 698 = 1.17.
The equation for the design effect is
If we calculate an intra-class correlation coefficient (ICC) for these

clusters (§11.13), we have
In this context, the ICC is called the intra-cluster correlation

coefficient. By a bit of algebra we get
DEEF = 1 + (m - 1)ICC
If there is only one observation per cluster, m = 1 and the design effect is
1.0 and the two designs are the same. Otherwise, the larger the ICC, i.e.
the more important the variation between clusters is, the bigger the
design effect and the more subjects we will need to get the same power
as a simply-randomized study. Even a small ICC will have an impact if
the cluster size is large. The X-ray guidelines study (§10.13) had ICC =
0.019. A study with the same ICC and m = 50 referrals per practice would
have design effect D = 1 + (50 - 1) × 0.019 = 1.93. Thus it would require
almost twice as many subjects as a trial where patients were randomized
to treatment individually.
The main difficulty in calculating sample size for cluster randomized
studies is obtaining an estimate of the between cluster variation or ICC.
Estimates of variation between individuals can often be obtained from the
literature but even studies that use the cluster as the unit of analysis may
not publish their results in such a way that the between practice variation
can be estimated. Donner et al. (1990), recognizing this problem,
recommended that authors publish the cluster-specific event rates
observed in their trial. This would enable other workers to use this
information to plan further studies.
In some trials, where the intervention is directed at the individual subjects
and the number of subjects per cluster is small, we may judge that the
design effect can be ignored. On the other hand, where the number of
subjects per cluster is large, an estimate of the variability between
clusters will be very important. When the number of clusters is very small,
we may have to use small sample adjustments mentioned in §18.7.

98. * The power of a two-sample t test:
(a) increases if the sample sizes are increased;
(b) depends on the difference between the population means
which we wish to detect;
(c) depends on the difference between the sample means;
(d) is the probability that the test will detect a given population
difference;
(e) cannot be zero.
View Answer
99. * The sample size required for a study to compare two

proportions:
(a) depends on the magnitude of the effect we wish to detect;
(b) depends on the significance level we wish to employ;
(c) depends on the power we wish to have;
(d) depends on the anticipated values of the proportions
themselves;
(e) should be decided by adding subjects until the difference is
significant.
View Answer
100. * The sample size required for a study to estimate a mean:

(a) depends on the width of the confidence interval which we want;
(b) depends on the variability of the quantity being studied;
(c) depends on the power we wish to have;
(d) depends on the anticipated value of the mean;
(e) depends on the anticipated value of the standard deviation.
View Answer
18E * Exercise: Estimation of sample sizes
1. What sample size would be required to estimate a 95% reference
interval using the Normal distribution method, so that the 95%
confidence interval for the reference limits were at most 20% of the
reference interval size?
View Answer
2. How big a sample would be required for an opinion pollster to

estimate voter preferences to within two percentage points?
View Answer
3. Mortality from myocardial infarction after admission to hospital is

about 15%. How many patients would be required for a clinical trial
to detect a 10% reduction in mortality, i.e. to 13.5%, if the power
required was 90%? How many would be needed if the power were
only 80%?
View Answer
4. How many patients would be required in a clinical study to

compare an enzyme concentration in patients with a particular
disease and controls, if differences of less than one standard
deviation would not be clinically important? If there was already a
sample of measurements from 100 healthy controls, how many
disease cases would be required?
View Answer
5. In a proposed trial of a health promotion programme, the

programme was to be implemented across a whole county. The plan
was to use four counties, two counties to be allocated to receive the
programme and two counties to act as controls. The programme
would be evaluated by a survey of samples of about 750 subjects
drawn from the at-risk populations in each county. A conventional
sample size calculation, which ignored the clustering, had indicated
that 1 500 subjects in each treatment group would be required to
give power 80% to detect the required difference. The applicants
were aware of the problem of cluster randomisation and the need to
take it into account in the analysis, e.g. by analysis at the level of
the cluster (county). They had an estimate of the intracluster
correlation = 0.005, based on a previous study. They argued that
this was so small that they could ignore the clustering. Were they
correct?
View Answer
> Table of Contents > 19 - Solutions to exercises
19
Solutions to exercises
Some of the multiple choice questions are quite hard. If you score +1 for
a correct answer, -1 for an incorrect answer, and 0 for a part which you
omitted, I would regard 40% as the pass level, 50% as good, 60% as
very good, and 70% as excellent. These questions are hard to set and
some may be ambiguous, so you will not score 100%.
Solution to Exercise 2M: Multiple choice

questions 1 to 6
1. FFFFF. Controls should be treated in the same place at the same time,
under the same conditions other than the treatment under test (§2.1). All
must be willing and eligible to receive either treatment (§2.4).
2. FTFTF. Random allocation is done to achieve comparable groups,
allocation being unrelated to the subjects' characteristics (§2.2). The use
of random numbers helps to prevent bias in recruitment (§2.3).
3. TFFFT. Patients do not know their treatment, but they usually do know
that they are in a trial (§2.9). Not the same as a cross-over trial (§2.6).
4. FFFFF. Vaccinated and refusing children are self-selected (§2.4). We
analyse by intention to treat (§2.5). We can compare effect of a
vaccination programme by comparing whole vaccination group,
vaccinated and refusers to the controls.
5. TFTTT. §2.6. The order is randomized.
6. FFTTT. §2.8, §2.9. The purpose of placebos is make dissimilar
treatments appear similar. Only in randomized trials can we rely on
comparability, and then only within the limits of random variation (§2.2).
Solution to Exercise 2E
1. It was hoped that women in the KYM group would be more satisfied
with their care. The knowledge that they would receive continuity of care
was an important part of the treatment, and so the lack of blindness is
essential. More difficult is that KYM women were given a choice and so
may have felt more committed to whichever scheme, KYM or standard,
they had chosen, than did the control group. We must accept this
element of patient control as part of the treatment.
2. The study should be (and was) analysed by intention to treat (§2.5). As
often happens, the refusers did worse than did the acceptors of KYM,
and worse than
the control group. When we compare all those allocated to KYM with
those allocated to control, there is very little difference (Table 19.1).
Table 19.1. Method of delivery in the KYM

study
Allocated to Allocated to
Method of KYM control
delivery
% n % n
Normal 79.7 382 74.8 354
Instrumental 12.5 60 17.8 84
Caesarian 7.7 37 7.4 35

3. Women had booked for hospital antenatal care expecting the standard
service. Those allocated to this therefore received what they had
requested. Those allocated to the KYM scheme were offered a treatment
which they could refuse if they wished, refusers getting the care for which
they had originally booked. No extra examinations were carried out for
research purposes, the only special data being the questionnaires, which
could be refused. There was therefore no need to get the women's
permission for the randomization. I thought this was a convincing
argument.

questions 7 to 13
7. FTTTT. A population can be anything (§3.3).
8. TFFFT. A census tells us who is there on that day, and only applies to
current in-patients. The hospital could be quite unusual. Some diagnoses
are less likely than others to lead to admission or to long stay (§3.2).
9. TFFTF. All members and all samples have equal chances of being
chosen (§3.4). We must stick to the sample the random process
produces. Errors can be estimated using confidence intervals and
significance tests. Choice does not depend on the subject's
characteristics at all, except for its being in the population.
10. FTTFT. Some populations are unidentifiable and some cannot be
listed easily (§3.4).
11. FFFTF. In a case-control study we start with a group with the disease,
the cases, and a group without the disease, the controls (§3.8).
12. FTFTT. We must have a cohort or case control study to get enough
cases (§3.7, §3.8).
13. TTTTF. This is a random cluster sample (§3.4). Each patient had the
same chance of their hospital being chosen and then the same chance of
being chosen within the hospital. This would not be so if we chose a fixed
number from each hospital rather than a fixed proportion, as those in
small hospitals would be more likely to be chosen than those in large
hospitals. In part (e). what about a sample with patients in every hospital?
1. Many cases of infection may be unreported, but there is not much that
could be done about that. Many organisms produce similar symptoms,
hence the
need for laboratory confirmation. There are many sources of infection,

including direct transmission, hence the exclusion of cases exposed to
other water supplies and to infected people.
2. Controls must be matched for age and sex as these may be related to
their exposure to risk factors such as handling raw meat. Inclusion of
controls who may have had the disease would have weakened any
relationships with the cause, and the same exclusion criteria were
applied as for the cases, to keep them comparable.
3. Data are obtained by recall. Cases may remember events in relation to
the disease more easily that than controls in relation to the same time.
Cases may have been thinking about possible causes of the disease and
so be more likely to recall milk attacks. The lack of positive association
with any other risk factors suggests that this is not important here.
4. I was convinced. The relationship is very strong and these scavenging
birds are known to carry the organism. There was no relationship with
any other risk factor. The only problem is that there was little evidence
that these birds had actually attacked the milk. Others have suggested
that cats may also remove the tops of milk bottles to drink the milk and
may be the real culprits (Balfour 1991).
5. Further studies: testing of attacked milk bottles for Campylobacter
(have to wait for the next year). Possibly a cohort study, asking people
about history of bird attacks and drinking attacked milk, then follow for
future Campylobacter (and other) infections. Possibly an intervention
study. Advise people to protect their milk and observe the subsequent
pattern of infection.

questions 14 to 19
14. TFFTF. §4.1. Parity is quantitative and discrete, height and blood
pressure are continuous.
15. TTFTF. §4.1. Age last birthday is discrete, exact age includes years
and fraction of a year.
16. FFTFT. §4.4, §4.6. It could have more than one mode, we cannot say.
Standard deviation is less than variance if the variance is greater than
one (§4.7,8).
17. TTTFT. §4.2,3,4. Mean and variance only tell us the location and
spread of the distribution (§4.6,7).
18. TFTFT. §4.5,6,7. Median = 2, the observations must be ordered
before the central one is found, mode = 2, range = 7 - 1 = 6, variance =
22/4 = 5.5.
19. FFFFT. §4.6,7,8. There would be more observations below the mean
than above, because the median would be less than the mean. Most
observations will be within one standard deviation of the mean whatever
the shape. The standard deviation measures how widely the blood
pressure is spread between people, not for a single person, which would
be needed to estimate accuracy. See also §15.2.
Fig. 19.1. Stem and leaf plot of blood glucose

Fig. 19.2. Box and whisker plot of blood glucose
1. The stem and leaf plot is shown in Figure 19.1:
2. Minimum = 2.2, maximum = 6.0. The median is the average of the 20th
and 21st ordered observations, since the number of observations is even.
These are both 4.0, so the median is 4.0. The first quartile is between the
10th and 11th, which are both 3.6. The third quartile is between the 30th
and 31st observations, which are 4.5 and 4.6. We have q = 0.75, i = 0.75
× 41 = 30.75, and the quartile is given by 4.5 + (4.6 - 4.5) × 0.75 = 4.575
(§4.5). The box and whisker plot is shown in Figure 19.2.
Fig. 19.3. Histogram of blood glucose
3. The frequency distribution is derived easily from the stem and leaf plot:
Interval Frequency
2.0–2.4 1
2.5–2.9 1
3.0–3.4 6
3.5–3.9 10
4.0–4.4 11
4.5–4.9 8
5.0–5.4 2
5.5–5.9 0
6.0–6.4 1
Total 40
4. The histogram is shown in Figure 19.3. The distribution is symmetrical.
5. The mean is given by
The deviations and their squares are as follows:
xi xi-[x with bar above] (xi-[x with bar above])2
4.7 0.65 0.422 5
4.2 0.15 0.022 5
3.9 -0.15 0.022 5
3.4 -0.65 0.422 5
Total 16.2 0.00 0.890 0

There are n - 1 = 4 - 1 = 3 degrees of freedom. The variance is given by
6. As before, the sum is ∑xi = 16.2, The sum of squares about the mean
is then given by ∑xi2 = 66.5 and
This is the same as found in 5 above, so, as before,
7. For the mean we have ∑xi = 162.2,

The sum of squares about the mean is given by:
There are n - 1 - 40 - 1 = 39 degrees of freedom. The variance is given

by
9. For the limits, [x with bar above]-2s = 4.055 - 2 × 0.698 = 2.659, [x with
bar above] - s = 4.055 - 0.698 = 3.357, [x with bar above] = 4.055, [x with
bar above] + s = 4.055 + 0.698 = 4.753, and [x with bar above] + 2s =
4.055 + 2 × 0.698 = 5.451. Figure 19.3 shows the mean and standard
deviation marked on the histogram. The majority of points fall within one
standard deviation of the mean and nearly all within two standard
deviations of the mean. Because the distribution is symmetrical, it
extends just beyond the [x with bar above] ± 2s points on either side.

questions 20 to 24
20. FTTTT. §5.1, §5.2. Without a control group we have no idea how
many would get better anyway (§2.1). 66.67% is 2/3. We may only have
3 patients.
21. TFFTT. §5.2. To three significant figures, it should be 1 730. We
round up because of the 9. To six decimal places it is 1 729.543 710.
22. FTTFT. This is a bar chart showing the relationship between two
variables (§5.5). See Figure 19.4. Calendar time has no true zero to
show.
23. TTFFT. §5.9, §5A. There is no logarithm of zero.
24. FFTTT. §5.5,6,7. A histogram (§4.3) and a pie chart (§5.4) each show
the distribution of a single variable.
Fig. 19.4. A dubious graph revised
Table 19.2. Calculations for a pie chart for the

Tooting Bec data
Relative
Category Frequency Angle
frequency
Schizophrenia 474 0.323 11 116
Affective 277 0.188 82 68

illness
Organic brain 405 0.276 07 99

syndrome
Subnormality 58 0.039 54 14
Alcoholism 57 0.038 85 14
Other 196 0.133 61 48
Total 1 467 1.000 00 359
1. This is the frequency distribution of a qualitative variable, so a pie chart
can be used to display it. The calculations are set out in Table 19.2.
Notice that we have lost one degree through rounding errors. We could
work to fractions of a degree, but the eye is unlikely to spot the
difference. The pie chart is shown in Figure 19.5.
2. See Figure 19.6.
3. There are several possibilities. In the original paper, Doll and Hill used
a separate bar chart for each disease, similar to Figure 19.7.
4. Line graphs can be used here, as we have simple time series (Figure
19.8). For an explanation of the difference between years, see §13E.

questions 25 to 31
25. TTFFF. §6.2. If they are mutually exclusive they cannot both happen.
There is no reason why they should be equiprobable or exhaustive, the
only events which can happen (§6.3).
26. TFTFT. For both, the probabilities are multiplied, 0.2 × 0.05 = 0.01
(§6.2).
Clearly the probability of both must be less than that for each one. The
probability of both is 0.01, so the probability of X alone is 0.20 - 0.01 =
0.19 and the probability of Y alone is 0.05 - 0.01 = 0.04. The probability of
having X or Y is the probability of X alone + probability of Y alone +
probability of X and Y together, because these are three mutually
exclusive events. Having X and having Y are not mutually exclusive as
she can have both. Having X tells us nothing about whether she has Y. If
she has X the probability of having Y is still 0.05, because X and Y are
independent.
Fig. 19.5. Pie chart showing the distribution of patients in Tooting

Bec Hospital by diagnostic group
Fig. 19.6. Bar chart showing the results of the Salk vaccine trial
27. TFTFF. §6.4. Weight is continuous. Patients respond or not with

equal probability, being selected at random from a population where the
probability of responding varies. The number of red cells might follow a
Poisson distribution (§6.7); there is no set of independent trials. The
number of hypertensives follows
a Binominal distribution, not the proportion

Fig. 19.7. Mortality in British doctors by smoking habits, after Doll
and Hill (1956)
Fig. 19.8. Line graphs for geriatric admissions in Wandsworth in the

summers of 1982 and 1983
28. TTTTF. The probability of clinical disease is 0.5 × 0.5 = 0.25. The
probability of carrier status = probability that father passes the gene and
mother does not + probability that mother passes the gene and father
does not = 0.5 × 0.5 + 0.5 ×0.5 = 0.5. Probability of not inheriting the
gene = 0.5 × 0.5 = 0.25. Probability of not having clinical disease = 1 -
0.25 = 0.75. Successive chidren are independent, so the probabilities for
the second child are unaffected by the first (§6.2)
29. FTTFT. §6.3,4. The expected number is one (§6.6). The spins are
independent (§6.2). At least one tail means one tail (PROB = 0.5) or two
tails (PROB = 0.25). These are mutually exclusive, so the probability of at
least one tail is 0.5 + 0.25 = 0.75.
Table 19.3. Probability of surviving to different
ages
Survive Survive
Probability Probability
to age to age
10 0.959 60 0.758
20 0.952 70 0.524
30 0.938 80 0.211
40 0.920 90 0.022
50 0.876 100 0.000
30. FTTFT. §6.6. E(X = 2) = µ + 2, VAR(2X) = 4σ2.

31. TTTFF. §6.6. The variance of a difference is the sum of the variances.
Variances cannot be negative. VAR(-X) = (-1)2 × VAR(X) = VAR(X).
1. Probability of survival to age 10. This illustrates the frequency
definition of probability. 959 out of 1000 survive, so the probability is
959/1000 = 0.959.
2. Survival and death are mutually exclusive, exhaustive events, so
PROB(survives) + PROB(dies) = 1. Hence PROB(dies) = 1 - 0.959 =
0.041.
3. These are the number surviving divided by 1000 (Table 19.3). The
events are not mutually exclusive, e.g. a man cannot survive to age 20 if
he does not survive to age 10. This does not form a probability
distribution.
4. The probability is found by
5. Independent events. PROB(survival 60 to 70) = 0.691,

PROB(both survive) = 0.691 × 0.691 = 0.477.
6. The proportion surviving on average is the probability of survival =
0.691. So a proportion of 0.691 of the 100 survive. We expect 0.691 ×
100 = 69.1 to survive.
7. The probability is found by
8. As in 7, we find probabilities of dying for each decade (Table 19.4).

This is a set of mutually exclusive events and they are exhaustive–there
is no other decade in which death can take place. The sum of the
probabilities is therefore 1.0. The distribution is shown in Figure 19.9.
9. We find the expected values or mean of a probability distribution by
summing each value times its probability (§6.4), to give life expectancy at
birth = 66.6
years (Table 19.5).
Table 19.4. Probability of dying in each decade

Decade Probability of Decade Probability
dying of dying
1st 0.041 6th 0.118
2nd 0.007 7th 0.234
3rd 0.014 8th 0.313
4th 0.018 9th 0.189
5th 0.044 10th 0.022
Fig. 19.9. Probability distribution of decade of death

questions 32 to 37
32. TTTFT. §7.2,3,4.
33. FFFTT. Symmetrical, µ = 0,σ = 1 (§7.3, §4.6).
34. TTFFF. §7.2. Median = mean. The Normal distribution has nothing to
do with normal physiology. 2.5% will be less than 260, 2.5% will be
greater than 340 litres/min.
Table 19.5. Calculation of expectation of life
5 × 0.041 = 0.205
15 × 0.007 = 0.105
25 × 0.014 = 0.350
35 × 0.018 = 0.630
45 × 0.044 = 1.980
55 × 0.118 = 6.490
65 × 0.234 = 15.210
75 × 0.313 = 23.475
85 × 0.189 = 16.065
95 × 0.022 = 2.090
Total 66.600
Fig. 19.10. Histogram of the blood glucose data with the
corresponding Normal distribution curve, and Normal plot
35. FTTFF. §4.6, §7.3. The sample size should not affect the mean. The
relative sizes of mean, median and standard deviation depend on the
shape of the frequency distribution.
36. TFTTF. §7.2, §7.3. Adding, subtracting or multiplying by a constant,
or adding or subtracting an independent Normal variable gives a Normal
distribution. X2 follows a very skew Chi-squared distribution with one
degree of freedom and X/Y follows a t distribution with one degree of
freedom (§7A).
37. TTTTT. A gentle slope indicates that observations are far apart, a
steep slope that there are many observations close together. Hence
gentle-steep-gentle (‘S’ shaped) indicates long tails (§7.5).
1. The box and whisker plot shows a very slight degree of skewness, the
lower whisker being shorter than the upper and the lower half of the box
smaller than the upper. From the histogram it appears that the tails are a
little longer than the Normal curve of Figure 7.10 would suggest. Figure
19.10 shows the Normal distribution with the same mean and variance
superimposed on the histogram, which also indicates this.
2. We have n = 40. For i = 1 to 40 we want to calculate (i-0.5)/n = (2i-
1)/2n. This gives us a probability. We use Table 7.1 to find the value of
the Normal distribution corresponding to this probability. For example, for
i = 1 we have
From Table 7.1 we cannot find the value of x corresponding to Φ(x) =

0.0125 directly, but we see that x = -2.3 corresponds to Φ(x) = 0.011 and
x = -2.2 to Φ(x) = 0.014. Φ(x) = 0.0125 is mid-way between these
probabilities so we can estimate the value of x as mid-way between -2.3
and -2.2, giving -2.25. This corresponds to the lowest blood glucose, 2.2.
For i = 2 we have Φ(x) = 0.0375. Referring to the table we have x = -1.8,
Φ(x) = 0.036 and x = -1.7, Φ(x) = 0.045. For Φ(x) = 0.0375 we must have
x just above -1.8, about -1.78. The
corresponding blood glucose is 2.9. We do not have to be very accurate

because we are only using this plot for a rough guide. We get a set of
probabilities as follows:
i (2i - 1)/2n = Φ(x) x Blood glucose
1 1/80 = 0.0125 -2.25 2.2
2 3/80 = 0.0375 -1.78 2.9
3 5/80 = 0.625 -1.53 3.3
4 7/80 = 0.0875 -1.36 3.3

and so on. Because of the symmetry of the Normal distribution, from i =
21 onwards the values of x are those corresponding to 40 - i + 1, but with
a positive sign. The Normal plot is shown in Figure 19.10.
3. The points do not lie on a straight line. There are pronounced bends
near each end. These bends reflect rather long tails of the distribution of
blood glucose. If the line showed a steady curve, rising less steeply as
the blood glucose value increased, this would show simple skewness
which can often be corrected by a log transformation. This would not
work here; the bend at the lower end would be made worse.
The deviation from a straight line is not very great, compared, say, to the
vitamin D measurements in Figure 7.12. As we see in Chapter 10, such
small deviations from the Normal do not usually matter.

questions 38 to 43
39. FTFTF. §8.3. The sample mean is always in the middle of the limits.
41. TTTFF. §8.1,2, §6.4) Variance is p(1-p)/n = 0.1 × 0.9/100 = 0.0009.

The number in the sample with the condition follows a Binomial
distribution, not the proportion.
42. FFTTT. It depends on the variability of FEV1 and the number in the
sample (§8.2). The sample should be random (§3.3,4).
43. FFTTF. §8.3,4. It is unlikely that we would get these data if the
population rate were 10%, but not impossible.
1. The interval will be 1.96 standard deviations less than and greater than
the mean. The lower limit is 0.810 - 1.96 × 0.057 = 0.698 mmol/litre. The
upper limit is 0.810 + 1.96 × 0.057 = 0.922 mmol/litre.
2. For the diabetics, the mean is 0.719 and the standard deviation 0.068,
so the lower limit of 0.698 will be (0.698 - 0.719)/0.068 = -0.309 standard
deviations from the mean. From Table 7.1, the probability of being below
this is 0.38, so the probability of being above is 1 - 0.38 = 0.62. Thus the
probability that an insulin-dependent diabetic would be within the
reference interval would be 0.62 or 62%. This is the proportion we
require.
4. The 95% confidence interval is the mean ± 1.96 standard errors. For
the controls, 0.810 - 1.96 × 0.00482 to 0.810 + 1.96 × 0.00482 gives us
0.801 to 0.819 mmol/litre. This is much narrower than the interval of part
1. This is because the confidence interval tells us how far the sample
mean might be from the population mean. The 95% reference interval
tells us how far an individual observation might be from the population
mean.
5. The groups are independent, so the standard error of the difference
between means is given by:
6. The difference between the means is 0.719 - 0.810 = -0.091 mmol/litre.

The 95% confidence interval is thus -0.091 - 1.96 × 0.00660 to -0.091 +
1.96 × 0.00660, giving -0.104 to -0.078. Hence the mean plasma
magnesium level for insulin dependent diabetics is between 0.078 and
0.104 mmol/litre below that of non-diabetics.
7. Although the difference is significant, this would not be a good test
because the majority of diabetics are within the 95% reference interval.

questions 44 to 49
44. FTFFF. There is evidence for a relationship (§9.6), which is not
necessarily causal. There may be other differences related to coffee
drinking, such as smoking (§3.8).
46. TTFTT. §9.2. It is quite possible for either to be higher and deviations
in either direction are important (§9.5). n = 16 because the subject giving
the same reading on both gives no information about the difference and
is excluded from the test. The order should be random, as in a cross-over
trial (§2.6).
47. FFFFT. The trial is small and the difference may be due to chance,
but there may also be a large treatment effect. We must do a bigger trial
to increase the power (§9.9). Adding cases would completely invalidate
the test. If the null hypothesis is true, the test will give a ‘significant’ result
one in 20 times. If we keep adding cases and doing many tests we have
a very high chance of getting a ‘significant’ result on one of them, even
though there is no treatment effect (§9.10).
48. TFTTF. Large sample methods depend on estimates of variance
obtained
from the data. This estimate gets closer to the population value as the
sample size increases (§9.7, §9.8). The chance of an error of the first
kind is the significance level set in advance, say 5%. The larger the
sample the more likely we are to detect a difference should one exist
(§9.9). The null hypothesis depends on the phenomena we are
investigating, not on the sample size.
49. FTFFT. We cannot conclude causation in an observational study
(§3.6,7,8), but we can conclude that there is evidence of a difference
(§9.6). 0.001 is the probability of getting so large a difference if the null
hypothesis were true (§9.3).
1. Both control groups are drawn from populations which were easy to
get to, one being hospital patients without gastro-intestinal symptoms, the
other being fracture patients and their relatives. Both are matched for age
and sex; Mayberry et al. (1978) also matched for social class and marital
status. Apart from the matching factors, we have no way of knowing
whether cases and controls are comparable, or any way of knowing
whether controls are representative of the general population. This is
usual in case control studies and is a major problem with this design.
2. There are two obvious sources of bias: interviews were not blind and
information is being recalled by the subject. The latter is particularly a
problem for data about the past. In James' study subjects were asked
what they used to eat several years in the past. For the cases this was
before a definite event, onset of Crohn's disease, for the controls it was
not, the time being time of onset of the disease in the matched case.
3. The question in James' study was ‘what did you to eat in the past?’,
that in Mayberry et al. (1978) was ‘what do you eat now?’
4. Of the 100 patients with Crohn's disease, 29 were current eaters of
cornflakes. Of 29 cases who knew of the cornflakes association, 12 were
ex-eaters of cornflakes, and among the other 71 cases 21 were ex-eaters
of cornflakes, giving a total of 33 past but not present eaters of
cornflakes. Combining these with the 29 current consumers, we get 62
cases who had at some time been regular eaters of cornflakes. If we
carry out the same calculation for the controls, we obtain 3 + 10 = 13 past
eaters and with 22 current eaters this gives 35 sometime regular
cornflakes eaters. Cases were more likely than controls to have eaten
cornflakes regularly at some time, the proportion of cases reporting
having eaten cornflakes being almost twice as great as for controls.
Compare this to James' data, where 17/68 = 25% of controls and 23/34 =
68% of cases, 2.7 times as many, had eaten cornflakes regularly. The
results are similar.
5. The relationship between Crohn's disease and reported consumption
of cornflakes had a much smaller probability for the significance test and
hence stronger evidence that a relationship existed. Also, only one case
had never eaten cornflakes (it was also the most popular cereal among
controls).
6. Of the Crohn's cases, 67.6% (i.e. 23/34) reported having eaten
cornflakes regularly compared to 25.0% of controls. Thus cases were
67.6/25.0 = 2.7 times
as likely as controls to report having eaten cornflakes. The corresponding

ratios for the other cereals are: wheat, 2.7; porridge, 1.5; rice, 1.6; bran,
6.1; muesli, 2.7. Cornflakes does not stand out when we look at the data
in this way. The small probability simply arises because it is the most
popular cereal. The P value is a property of the sample, not of the
population.
7. We can conclude that there is no evidence that eating cornflakes is
more closely related to Crohn's disease than is consumption of other
cereals. The tendency for Crohn's cases to report excessive eating of
breakfast foods before onset of the disease may be the result of greater
variation in diet than in controls, as they try different foods in response to
their symptoms. They may also be more likely to recall what they used to
eat, being more aware of the effects of diet because of their disease.

questions 50 to 56
50. FFTFT. §10.2. It is equivalent to the Normal distribution method
(§8.3).
51. FTFTF. §10.3. Whether the (population) means are equal is what we
are trying to find out. The large sample case is like the Normal test of
(§9.7), except for the common variance estimate. It is valid for any
sample size.
52. FTTFF. The assumption of Normality would not be met for a small
sample t test (§10.3) without transformation (§10.4), but for a large
sample the distribution followed by the data would not matter (§9.7). The
sign test is for paired data. We have measurements, not qualitative data.
53. FTTFF. §10.5. The more different the sample sizes are, the worse is
the approximation to the t distribution. When both samples are large, this
becomes a large sample Normal distribution test (§9.7). Grouping of data
is not a serious problem.
54. TFFTT. A P value conveys more information than a statement that the
difference is significant or not significant. A confidence interval would be
even better. What is important is how well the diagnostic test
discriminates, i.e. by how much the distributions overlap, not any
difference in mean. Semen count cannot follow a Normal distribution
because two standard deviations exceed the mean and some
observations would be negative (§7.4). Approximately equal numbers
make the t test very robust but skewness reduces the power (§10.5).
56. FTTFT. §10.9. Sums of squares and degrees of freedom add up,
mean squares do not. Three groups gives two degrees of freedom. We
can have any sizes of groups.

1. The differences for compliance are shown in Table 19.6. The stem and
leaf plot is shown in Figure 19.11.
Table 19.6. Differences and means for static

compliance
Patient Constant Decelerating Difference Mean
1 65.4 72.9 -7.5 69.15
2 73.7 94.4 -20.7 84.05
3 37.4 43.3 -5.9 40.35
4 26.3 29.0 -2.7 27.65
5 65.0 66.4 -1.4 65.70
6 35.2 36.4 -1.2 35.80
7 24.7 27.7 -3.0 26.20
8 23.0 27.5 -4.5 25.25

9 133.2 178.2 -45.0 155.70
10 38.4 39.3 -0.9 38.85
11 29.2 31.8 -2.6 30.50
12 28.3 26.9 1.4 27.60
13 46.6 45.0 1.6 45.80
14 61.5 58.2 3.3 59.85
15 25.7 25.7 0.0 25.70
16 48.7 42.3 6.4 45.50
Fig. 19.11. Stem and leaf plot for compliance

2. The plot of difference against mean is Figure 19.12. The distribution is
highly skewed and the difference closely related to the mean.
3. The sum and sum of the squared differences are ∑di = -82.7 and ∑di2
= 2648.43, hence the mean is [d with bar above] = -82.7/16 = -5.16875.
For the sum of squares about the mean
4. We have 15 degrees of freedom and from Table 7.1 the 5% point of

the t distribution is 2.13. The 95% confidence interval is -5.16875 - 2.13 ×
3.04205 to -5.16875 + 2.13 × 3.04205, giving -11.6 to +1.3.
Fig. 19.12. Difference versus mean for compliance
6. The 95% confidence interval is -0.028688 - 2.13 × 0.012503 to

-0.028688 + 2.13 × 0.012503 which gives -0.055312 to -0.002057. This
has not been rounded, because we need to transform them first. If we
transform these limits back by taking the antilogs we get 0.880 to 0.995.
This means that the compliance with a decelerating waveform is between
0.880 and 0.995 times that with a constant waveform. There is some
evidence that waveform has an effect, whereas with the untransformed
data the confidence interval for the difference included zero. Because of
the skewness of the raw data the confidence interval was too wide.
7. We can conclude that there is some evidence of a reduction in mean
compliance, which could be up to 12% (from (1 - 0.880) × 100), but could
also be negligibly small.

questions 57 to 61
57. FFTTF. Outcome and predictor variables are perfectly related but do
not lie on a straight line, so r < 1 (§11.9).
Fig. 19.13. Stem and leaf plots for log compliance
Table 19.7. Difference and mean for log transformed

compliance (to base 10)
Patient Constant Decelerating Difference Mean
1 1.816 1.863 -0.047 1.8395
2 1.867 1.975 -0.108 1.9210
3 1.573 1.636 -0.063 1.6045

4 1.420 1.462 -0.042 1.4410
5 1.813 1.822 -0.009 1.8175
6 1.547 1.561 -0.014 1.5540
7 1.393 1.442 -0.049 1.4175
8 1.362 1.439 -0.077 1.4005
9 2.125 2.251 -0.126 2.1880
10 1.584 1.594 -0.010 1.5890
11 1.465 1.502 -0.037 1.4835
12 1.452 1.430 0.022 1.4410
13 1.668 1.653 0.015 1.6605
14 1.789 1.765 0.024 1.7770
15 1.410 1.410 0.000 1.4100
16 1.688 1.626 0.062 1.6570

Fig. 19.14. Difference versus mean for log compliance
58. FTFFF. Knowledge of the predictor tells us something about the

outcome variable (§6.2). This is not a straight line relationship. For part of
the scale the outcome variable decreases as the predictor increases,
then the outcome variable increases again. The correlation coefficient will
be close to zero (§11.9). A logarithmic transformation would work if the
outcome increased more and more rapidly as the predictor increased
(§5.9).
59. FFFTT. A regression line usually has non-zero intercept and slope,
which have dimensions (§11.3). Exchanging X and Y changes the line
(§11.4).
60. FTTFF. The predictor variable has no error in the regression model
(§11.3). Transformations are only used if needed to meet the
assumptions (§11.8). There is a scatter about the line (§11.3).
61. TTFFF. §11.9,10. There is no distinction between predictor and
outcome. r should not be confused with the regression coefficient (§11.3).
1. The slope is found by
For females,
For males,
2. For the standard error, we first need the variances about the line:
then the standard error is
For females:
For males:
3. The standard error of the difference between two independent

variables is the square root of the sum of their standard errors squared
(§8.5):
The sample is reasonably large, almost attaining 50 in each group, so
this standard error should be fairly well estimated and we can use a large
sample Normal approximation. The 95% confidence interval is thus 1.96
standard errors on either side of the estimate. The observed difference is
bf - bm = 2.8710 - 3.9477 = -1.0767. The 95% confidence interval is thus
-1.0767 - 1.96 × 1.7225 = -4.5 to -1.0767 + 1.96 × 1.7225 = 2.3. If the
samples were small, we could do this using the t Distribution, but we
would need to estimate a common variance. It would be better to use
multiple regression, testing the height × sex interaction (§17.3).
4. For the test of significance the test statistic is observed difference over
standard error:
If the null hypothesis were true, this would be an observation from a

Standard Normal distribution. From Table 7.2, P > 0.5.

questions 62 to 66
62. TFTFF. §10.3, §12.2. The sign and Wilcoxon tests are for paired data
(§9.2, §12.3). Rank correlation looks for the existence of relationships
between two ordinal variables, not a comparison between two groups
(§12.4, §12.5).
63. TTFFT. §9.2, §12.2, §10.3, §12.5. The Wilcoxon test is for interval
data (§12.3).
64. FTFTT. §12.5. There is no predictor variable in correlation. Log
transformation would not affect the rank order of the observations.
65. FTFFT. If Normal assumptions are met the methods using them are
better (§12.7). Estimation of confidence intervals using rank methods is
difficult. Rank methods require the assumption that the scale is ordinal,
i.e. that the data can be ranked.
66. TFTTF. We need a paired test: t, sign or Wilcoxon (§10.2, §9.2,
§12.3).

1. The differences are shown in Table 19.6. We have 4 positive, 11
negative and 1 zero. Under the null hypothesis of no difference, the
number of positives is from the Binomial distribution with p = 0.5, n = 15.
We have n = 15 because the single zero contributes no information about
the direction of the difference. For PROB(r ≤ 4) we have
If we double this for a two-sided test we get 0.11848, again not

significant.
2. Using the Wilcoxon matched pairs test we get
Diff. -0.9 -1.2 -1.4 1.4 1.6 -2.6 -2.7 -3.0
Rank 1 2 3.5 3.5 5 6 7 8
Diff. 3.3 -4.5 -5.9 6.4 -7.5 -20.7 -45.0
Rank 9 10 11 12 13 14 15
As for the sign test, the zero is omitted. Sum of ranks for positive
differences is T = 3.5 + 5 + 9 + 12 = 29.5. From Table 12.5 the 5% point
for n = 15 is 25, which T exceeds, so the difference is not significant at
the 5% level. The three tests give similar answers.
3. Using the log transformed differences in Table 19.7, we still have 4
positives, 11 negatives and 1 zero, with a sign test probability of 0.11848.
The transformation does not alter the direction of the changes and so
does not affect the sign test.
4. For the Wilcoxon matched pairs test on the log compliance:
Diff. -0.009 -0.010 -0.014 0.015 0.022 0.024
Rank 1 2 3 4 5 6
Diff. -0.037 -0.042 -0.047 -0.049 0.062 -0.063
Rank 7 8 9 10 11 12
Diff. -0.077 -0.108 -0.126
Rank 13 14 15
Hence T = 4 + 5 + 6 + 11 - 26. This is just above the 5% point of 25 and

is different from that in the untransformed data. This is because the
transformation has altered the relative size of the differences. This test
assumes interval data. By changing to a log scale we have moved to a
scale where the differences are more comparable, because the change
does depend on the magnitude of the original value. This does not
happen with the other rank tests, the Mann–Whitney U test and rank
correlation coefficients, which involve no differencing.
5. Although there is a possibility of a reduction in compliance it does not
reach the conventional level of significance.
6. The conclusions are broadly similar, but the effect on compliance is
more strongly suggested by the t method. Provided the data can be
transformed to approximate Normality the t distribution analysis is more
powerful, and as it also gives confidence intervals more easily, I would
prefer it.

questions 67 to 73
67. TFFFF. §13.3. 80% of 4 is greater than 3, so all expected frequencies
must exceed 5. The sample size can be as small as 20, if all row and
column totals are 10.
68. FTFTF. §13.1, §13.3. (5-1) × (3-1) = 8 degrees of freedom, 80% × 15
= 12 cells must have expected frequencies > 5. It is O.K. for an observed
frequency to be zero.
69. TTFTF. §13.1, §13.9. The two tests are independent. There are (2-1)
× (2-1) = 1 degree of freedom. With such large numbers Yates' correction
does not make much difference. Without it we get χ2 = 124.5, with it we
get χ2 = 119.4 (§13.5.).
70. TTTTT. §13.4,5. The factorials of large numbers can be difficult to
calculate.
71. TTTTF. §13.7.
72. TTFTT. Chi-squared for trend and τb will both test the null hypothesis
of no trend in the table, but an ordinary chi-squared test will not (§13.8).
The odds ratio (OR) is an estimate of the relative risk for a case-control
study (§13.7).
73. TTFFF. The test compares proportions in matched samples (§13.9).
For a relationship, we use the chi-squared test (§13.1). PEFR is a
continuous variable, we use the paired t method (§10.2). For two
independent samples we use the chi-squared test (§13.1).

1. The heatwave appears to begin in week 10 and continue to include
week 17. This period was much hotter than the corresponding period of
1982.
Table 19.8. Cross-tabulation of time period by

year for geriatric admissions
Period
Year Before During After Total
heatwave heatwave heatwave
1982 190 110 82 382
1983 180 178 110 468
Total 370 288 192 850
Table 19.9. Expected frequencies for Table 19.8
Period
Year Before During After Total
heatwave heatwave heatwave
1982 166.3 129.4 86.3 382.0
1983 203.7 158.6 105.7 468.0
Total 370.0 288.0 192.0 850.0
2. There were 178 admissions during the heatwave in 1983 and 110 in
the corresponding weeks of 1982. We could test the null hypothesis that
these came from distributions with the same admission rate and we
would get a significant difference. This would not be convincing, however.
It could be due to other factors, such as the closure of another hospital
with resulting changes in catchment area.
3. The cross-tabulation is shown in Table 19.8.
4. The null hypothesis is that there is no association between year and
period, in other words that the distribution of admissions between the
periods will be the same for each year. The expected values are shown in
Table 19.9.
5. The chi-squared statistic is given by:
There are 2 rows and 3 columns, giving us (2-1) × (3-1) = 2 degrees of

freedom. Thus we have chi-squared = 11.8 with 2 degrees of freedom.
From Table 13.3 we see that this has probability of less than 0.01. The
data are not consistent with the null hypothesis. The evidence supports
the view that admissions rose by more than could be ascribed to chance
during the 1983 heatwave. We cannot be certain that this was due to the
heatwave and not some other factor which happened to operate at the
same time.
6. We could see whether the same effect occurred in other districts
between 1982 and 1983. We could also look at older records to see
whether there was a similar increase in admissions, say for the
heatwaves of 1975 and 1976.

questions 74 to 80
74. TFFTT. Table 14.2.
75. FTTTT. §14.5.
76. FFFFT. Regression, correlation and paired t methods need
continuous data (§11.3, §11.9, §10.2). Kendall's τ can be used for
ordered categories.
77. TFTFF. §14.2.
78. TFTTT. A t test could not be used because the data do not follow a
Normal distribution (10.3). The expected frequencies will be too small for
a chi-squared test (§13.3), but a trend test would be O.K. (§13.8). A
goodness of fit test could be used (§13.10).
79. FTTFT. A small-sample, paired method is needed (Table 14.4).
80. TFTFF. For a two by two table with small expected frequencies we
can use Fisher's exact test or Yates' correction (§13.4,5). McNemar's test
is inappropriate because the groups are not matched (§13.9).

1. Overall preference: we have one sample of patients so we use (Table
14.2). Of these 12 preferred A, 14 preferred B and 4 did not express a
preference. We can use a Binomial or sign test (§9.2), only considering
those who expressed a preference. Those for A are positives, those for B
are negatives. We get two-sided P = 0.85, not significant.
Preference and order: we have the relationship between two variables
(Table 14.3), preference and order, both nominal. We set up a two way
table and do a chi-squared test. For the 3 by 2 table we have two
expected frequencies less than five, so we must edit the table. There are
no obvious combinations, but we can delete those who expressed no
preference, leaving a 2 by 2 table, χ2 = 1.3, 1 degree of freedom, P >
0.05.
2. The data are paired (Table 14.2) so we use a paired t test (§10.2). The
assumption of a Normal distribution for the differences should be met as
PEFR itself follows a Normal distribution fairly well. We get t = 6.45/5.05
= 1.3, degrees of freedom = 31, which is not significant. Using t = 2.04
(Table 10.1) we get a 95% confidence interval of -3.85 to 16.75 litres/min.
3. We have two independent samples (Table 14.1). We must use the total
number of patients we randomized to treatments, in an intention to treat
analysis (§2.5). Thus we have 1721 active treatment patients including
15 deaths, and 1706 placebo patients with 35 deaths. A chi-squared test
gives us χ2 = 8.3, d.f. = 1, P < 0.01. A comparison of two proportions
gives a difference of -0.0118 with 95% confidence interval -0.0198 to
-0.0038 (§8.6) and test of significance using the Standard Normal
distribution gives a value of 2.88, P < 0.01, (§9.8).
4. We are looking at the relationship between two variables (Table 14.3).
Both variables have very non-Normal distributions. Nitrite is highly skew
and pH is bimodal. It might be possible to transform the nitrites to a
Normal distribution but the transformation would not be a simple one. The
zero prevents a simple logarithmic transformation, for example. Because
of this, regression and
correlation are not appropriate and rank correlation can be used.

Spearman's ρ = 0.58 and Kendall's τ = 0.40, both giving a probability of
0.004.
5. We have two independent samples (Table 14.1). We have two large
samples and can do the Normal comparison of two means (§8.5). The
standard error of the difference is 0.0178 s and the observed difference is
0.02 s, giving a 95% confidence interval of -0.015 to 0.055 for the excess
mean transit time in the controls. If we had all the data, for each case we
could calculate the mean MTT for the two controls matched to each case,
find the difference between case MTT and control mean MTT, and use
the one sample method of §8.3.
6. These are paired data, so we refer to Table 14.2. The unequal steps in
the visual acuity scale suggest that it is best treated as an ordinal scale,
so the sign test is appropriate. Pre minus post, there are 10 positive
differences, no negative differences and 7 zeros. Thus we refer 0 to the
Binomial distribution with p = 0.5 and n = 10. The probability is given by
7. We want to test for the relationship between two variables, which are
both presented as categorical (Table 14.3). We use a chi-squared test for
a contingency table, χ2 = 38.1, d.f. = 6, P < 0.001. One possibility is that
some other variable, such as the mother's smoking or poverty, is related
to both maternal age and asthma. Another is that there is a cohort effect.
All the age 14–19 mothers were born during the second world war, and
some common historical experience may have produced the asthma in
their children.
8. The serial measurements of thyroid hormone could be summarized
using the area under the curve (§10.7). The oxygen dependence is tricky.
The babies who died had the worst outcome, but if we took their survival
time as the time they were oxygen dependent, we would be treating them
as if they had a good outcome. We must also allow for the babies who
went home on oxygen having a long but unknown oxygen dependence.
My solution was to assign an arbitrary large number of days, larger than
any for the babies sent home without oxygen, to the babies sent home on
oxygen. I assigned an even larger number of days to
the babies who died. I then used Kendall's tau b (§12.5) to assess the
relationship with thyroid hormone AUC. Kendall's rank correlation was
chosen in preference to Spearman's because of the large number of ties
which the arbitrary assignment of large numbers produced.
9. This is a comparison of two independent samples, so we use Table
14.1. The variable is interval and the samples are small. We could either
use the two sample t method (10.3) or the Mann–Whitney U test (§12.2).
The groups have similar variances, but the distribution shows a slight
negative skewness. As the two sample t method is fairly robust to
deviations from the Normal distribution and as I wanted a confidence
interval for the difference I chose this option. I did not think that the slight
skewness was sufficient to cause any problems.
By the two sample t method we get the difference between the means,
immobile - mobile, to be 7.06, standard error = 5.74, t = 1.23, P = 0.23,
95% confidence interval = -4.54 to 18.66 hours. By the Mann-Whitney,
we get U = 178.5, z = -1.06, P = 0.29. The two methods give very similar
results and lead to similar conclusions, as we expect them to do when
both methods are valid.

questions 81 to 86
81. TFTTF. §15.2. Unless the measurement process changes the
subject, we would expect the difference in mean to be zero.
82. TFTFF. §15.4. We need the sensitivity as well as specificity. There
are other things, dependent on the population studied, which may be
important too, like the positive predictive value.
83. FTTTF. §15.4. Specificity, not sensitivity, measures how well people
without the disease are eliminated.
84. TTFFF. §15.5. The 95% reference interval should not depend on the
sample size.
85. FFFFT. §15.5. We expect 5% of ‘normal’ men to be outside these
limits. The patient may have a disease which does not produce an
abnormal haematocrit. This reference interval is for men, not women who
may have a different distribution of haematocrit. It is dangerous to
extrapolate the reference interval to a different population. In fact, for
women the reference interval is 35.8 to 45.4, putting a woman with a
haematocrit of 48 outside the reference interval. A haematocrit outside
the 95% reference interval suggests that the man may be ill, although it
does not prove it.
86. TFTTT. §15.6. As time increases, rates are based on fewer potential
survivors. Withdrawals during the first interval contribute half an interval
at risk. If survival rates change those subjects starting later in calendar
time, and so more likely to be withdrawn, will have a different survival to
those starting earlier. The first part of the curve will represent a different
population to the second. The longest survivor may still be alive and so
become a withdrawal.

1. The blood donors were used because it was easy to get the blood.
This would produce a sample deficient in older people, so it was
supplemented by people attending day centres. This would ensure that
these were reasonably active, healthy people for their age. Given the
problem of getting blood and the limited resources available, this seems
a fairly satisfactory sample for the purpose. The alternative would be to
take a random sample from the local population and try to persuade them
to give the blood. There might have been so many refusals that volunteer
bias would make the sample unrepresentative anyway. The sample is
also biased geographically, being drawn from one part of London. In the
context of the study, where we wanted to compare diabetics with
normals, this did not matter so much, as both groups came from the
same place. For a reference interval which would apply nationally, if there
were a geographical factor the interval would be biassed in other places.
To look at this we would have to repeat the study in several places,
compare the resulting reference intervals and pool as appropriate.
2. We want normal, healthy people for the sample, so we want to exclude
people with obvious pathology and especially those with disease known
to affect the quantity being measured. However, if we excluded all elderly
people currently receiving drug therapy we would find it very difficult to a
sufficiently large sample. It is indeed ‘normal’ for the elderly to be taking
analgesics and hypnotics, so these were permitted.
3. From the shape of the histogram and the Normal plot, the distribution
of plasma magnesium does indeed appear Normal.
4. The reference interval, outside which about 5% of normal values are
expected to lie, is [x with bar above] - 2s to [x with bar above] + 2s, or
0.810 - × 0.057 to 0.810 + 2 × 0.057, which is 0.696 to 0.924, or 0.70 to
0.92 mmol/litre.
5. As the sample is large and the data Normally distributed the standard
error of the limits is approximately
For the 95% confidence interval we take 1.96 standard errors on either
side of the limit, 1.96 × 0.0083439 = 0.016. The 95% confidence interval
for the lower reference limit is 0.696 - 0.016 to 0.696 + 0.016 = 0.680 to
0.712 or 0.68 to 0.71 mmol/litre. The confidence interval for the upper
limit is 0.924 - 0.016 to 0.696 + 0.016 = 0.908 to 0.940 or 0.91 to 0.94
mmol/litre. The reference interval is well estimated as far as sampling
errors are concerned.
6. Plasma magnesium did indeed increase with age. The variability did
not. This would mean that for older people the lower limit would be too
low and the upper limit too high, as the few above this would all be
elderly. We could simply estimate the reference interval separately at
different ages. We could do this using separate means but a common
estimate of variance, obtained by one-way analysis of variance (§10.9).
Or we could use the regression of magnesium on
age to get a formula which would predict the reference interval for any
age. The method chosen would depend on the nature of the relationship.

questions 87 to 92
87. FTFFF. §16.1. It is for a specific age group, not age adjusted. It
measures the number of deaths per person at risk, not the total number.
It tells us nothing about age structure.
88. FTTTT. §16.4. The life table is calculated from age specific death
rates. Expectation of life is the expected value of the distribution of age at
death if these mortality rates apply (§6E). It usually increases with age.
89. TFTTF. The SMR (§16.3) for women who had just had a baby is
lower than 100 (all women) and 105 (stillbirth women). The confidence
intervals do not overlap so there is good evidence for a difference.
Women who had had a stillbirth may be less or more likely than all
women to commit suicide, we cannot tell. We cannot conclude that giving
birth prevents suicide–it may be that optimists conceive, for example.
90. TFFFF. §16.3. Age effects have been adjusted for. It may also be that
heavy drinkers become publicans. It is difficult to infer causation from
observational data. Men at high risk of cirrhosis of the liver, i.e. heavy
drinkers, may not become window cleaners, or window cleaners who
drink may change their occupation, which requires good balance.
Window cleaners have low risk. The ‘average’ ratio is 100, not 1.0.
91. FFFTF. §16.6. A life table tells us about mortality, not population
structure. A bar chart shows the relationship between two variables, not
their frequency distribution (§5.5).
92. TFFFT. §16.1, §16.2, §16.5. Expectation of life does not depend on
age distribution (§16.4).
1. We obtain the rates for the whole period by dividing the number of
deaths in an age group by the population size. Thus for ages 10–14 we
have 44/4271 = 0.01030 cases per thousand population. This is for a 13
year period so the rate per year is 0.01030/13 = 0.00079 per 1000 per
year, or 0.79 per million per year. Table 19.10 shows the rates for each
age group. The rates are unusual because they are highest among the
adolescent group, where mortality rates for most causes are low.
Anderson et al. (1985) note that ‘… our results suggest that among
adolescent males abuse of volatile substances currently account for 2%
of deaths from all causes…’. The rates are also unusual because we
have not calculated them separately for each sex. This is partly for
simplicity and partly because the number of cases in most age groups is
small as it is.
2. The expected number of deaths by multiplying the number in the age
group in Scotland by the death rate for the period, i.e. per 13 years, for
Great Britain. We then add these to get 27.19 deaths expected
altogether. We observed 48, so the SMR is 48/27.19 = 1.77, or 177 with
Great Britain as 100.
Table 19.10. Age-specific mortality rates for volatile

substance abuse, Great Britain, and calculation of
SMR for Scotland
Great Britain
ASMRs
Scotland Scotland
Age Per Per population expected
group million thousand
(thousands) deaths
per per 13
year years
0–9 0.00 0.00000 653 0.00000
10– 0.79 0.01030 425 4.37750

14
15– 2.58 0.03358 447 15.01026

19
20– 0.87 0.01137 394 4.47978

24
25– 0.32 0.00415 342 1.41930

29
30– 0.08 0.00108 659 0.71172

39
40– 0.03 0.00033 574 0.18942

49
50– 0.09 0.00112 579 0.64848

59
60+ 0.03 0.00037 962 0.35594

Total 27.19240
3. We find the standard error of the SMR by
The 95% confidence interval is then 1.77 - 1.96 × 0.254 8 to 1.77 +

1.96 × 0.254 8, or 1.27 to 2.27. Multiplying by 100 as usual, we get
127 to 227. The observed number is quite large enough for the Normal
approximation to the Poisson distribution to be used.
4. Yes, the confidence interval is well away from zero. Other factors relate
to the data collection, which was from newspapers, coroners, death
registrations etc. Scotland has different newspapers and other news
media and a different legal system to the rest of Great Britain. It may be
that the association of deaths with VSA is more likely to be reported there
than in England and Wales.

questions 93 to 97
93. TFTFT. §17.2. It is the ratio of the regression sum of squares to the
total sum of squares.
94. FTFFF. §17.2. There were 37 + 1 = 38 observations. There is a highly
significant ethnic group effect. The non-significant sex effect does not
mean that there is no difference (§9.6). There are three age groups, so
two degrees of freedom. If the effect of ethnicity were due entirely to age,
it would have disappeared when age was included in the model.
95. TTTTF. §17.8. A four-level factor has three dummy variables (§17.6).
If the effect of white cell count were due entirely to smoking, it would
have disappeared when smoking was included in the model.
96. TTTFT. §17.4
97. FFFFT. §17.9. Boys have a lower risk of readmission than girls,
shown by the negative coefficient, and hence a longer time before being
readmitted. Theophiline is related to a lower risk of readmission but we
cannot conclude causation. Treatment may depend on the type and
severity of asthma.

1. The difference is highly significant (P < 0.001) and is estimated to be
between 1.3 and 3.7, i.e. volumes are higher in group 2, the trisomy-16
group.
2. From both the Normal plot and the plot against number of pairs of
somites there appears to be one point which may be rather separate from
the rest of the data, an outlier. Inspection of the data showed no reason
to suppose that the point was an error, so it was retained. Otherwise the
fit to the Normal distribution seems quite good. The plot against number
of pairs of somites shows that there may be a relationship between mean
and variability, but this very small and will not affect the analysis too
much. There is also a possible non-linear relationship, which should be
investigated. (The addition of a quadratic term did not improve the fit
significantly.)
3. Model difference in sum of squares = 207.139 - 197.708 = 9.431,
residual sum of squares = 3.384, F ratio = 9.431/3.384 = 2.79 with 1 and
36 degrees of freedom, corresponding to t = 1.67, P > 0.1, not significant.

questions 98 to 100
98. TTFTT. §9.9. Power is a property of the test, not the sample. It cannot
be zero, as even when there is no population difference at all the test
may be significant.
99. TTTTF. §18.5. If we keep on adding observations and testing, we are
carrying out multiple testing and so invalidate the test (§9.10).
100. TTFFT. §18.1. Power is not involved in estimation.

3. This is a comparison of two proportions (§18.5). We have p1 = 0.15
and p2 = 0.15 × 0.9 = 0.135, a reduction of 10%. With a power of 90%
and a significance level of 5%, we have
Hence we need 11400 in each group, 22800 patients altogether. With a

power of 80% and a significance level of 5%, we have
Hence we need 8577 in each group, 17154 patients altogether. Lowering

the power reduces the required sample size, but, of course, reduces the
chance of detecting a difference if there really is one.
4. This is the comparison of two means (§18.4). We estimate the sample
size for a difference of one standard deviation, µ1 - µ2 = σ. With a power
of 90% and a significance level of 5%, the number in each group is given
by
Hence we need 21 in each group. If we have unequal samples and n1 =

100, n2 is given by
and so we need 12 subjects in the disease group.

5. When the number of clusters is very small and the number of
individuals within a cluster is large, as in this study, clustering can have a
major effect. The design effect, by which the estimated sample size
should be multiplied, is DEFF = 1 + (750 - 1) × 0.005 = 4.745. Thus the
estimated sample size for any given comparison should be multiplied by
4.745. Looking at it another way, the effective sample size is the actual
sample size, 3000, divided by 4.745, about 632. Further, sample size
calculations should take into account degrees of freedom. In large
sample approximation sample size calculations, power 80% and alpha
5% are embodied in the multiplier f (α, P) = f (0.05, 0.80) = (1.96 + 0.85)2
= 7.90. For a small sample calculation using the t test, 1.96 must be
replaced by the corresponding 5% point of the t distribution with the
appropriate degrees of freedom, here 2 degrees of freedom giving t =
4.30. Hence the multiplier is (4.30 + 0.85)2 = 26.52, 3.36 times that for
the large sample.
The effect of the small number of clusters would reduce the effective
sample size even more, down to 630/3.36 = 188. Thus the 3000 men in
two groups of two clusters would give the same power to detect the same
difference as 188 men randomized individually. The applicants
resubmitted a proposal with many more clusters.
> Back of Book > References
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on milk bottles: possible mode of transmission of Campylobacter
jejuni to man. Lancet, 336, 1425–7.
Streiner, D.L. and Norman, G.R. (1996). Health Measurement Scales:

A Practical Guide to Their Development and Use, second edition,
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Stuart, A. (1955). A test for homogeneity of the marginal distributions
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‘Student’ (1931). The Lanarkshire Milk Experiment. Biometrika, 23,

398–406.
Thomas, P.R.S., Queraishy, M.S., Bowyer, R., Scott, R.A.P., Bland,

J.M., and Dormandy, J.A. (1993). Leucocyte count: a predictor of
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Thompson, S.G. (1993). Controversies in meta-analysis: the case of

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> Back of Book > Index > A
A
abridged life table 200–1
absolute difference 271–2
absolute value 239
accepting null hypothesis 140
accidents 53
acute myocardial infarction 277
addition rule 88
adjusted odds ratio 323
admissions to hospital 86 255–6 354 356 370–1
age 53 56–7 267 308–14 316 373
age, gestational 56–7
age in life table see life table
age-specific mortality rate 295–6 299–300 302 307 376–7
age-standardized mortality rate 74 296 302
age-standardized mortality ratio 297–9 303 307 376–7
agreement 272–5
AIDS 58 77–8 169–71 172 174–8 317–8
alpha spending 152
albumin 76–7
alcoholics 76–7 308–17
allocation to treatment 6–13 15 20–1 23
alterations to 11–13 21
alternate 6–7 11
alternate dates 11–12
by general practice 21 23
by ward 21
cheating in 12–13
known in advance 11
in clusters 21–2 179–81 344–6
minimization 13
non-random 11–13 21–2
physical randomization 12
random 7–11 15 17 20–1 25
systematic 11–12
using envelopes 12
using hospital number 11
alpha error 140
alternate allocation 6–7 11
alternative hypothesis 137 139–42
ambiguous questions 40–1
analgesics 15 18
analysis of covariance 321
analysis of variance 172–9 261–2 267–8 318–21
assumptions 173 175–6
balanced 318
in estimation of measurement error 271
fixed effects 177
Friedman 321
Kruskal–Wallis 217 261–2
in meta-analysis 327
multi-way 318–21
one-way 172–9 261–2
random effects 177–9
in regression 310–15 315
two-way 318
using ranks 217 261–2 321
angina pectoris 15–16 138–9 218–20
animal experiments 5 16–17 20–1 33
anticoagulant therapy 11–12 19 142
antidiuretic hormone 196–7
antilogarithm 83
appropriate confidence intervals for comparison 134
appropriate significance tests for comparison 142–3
anxiety 18 143 210
ARC 58 172 174–7
arcsine square root transformation 165
area under the curve 104–5 109–11 169–71 278 373–4
probability 104–5 109–11
serial data 169–71 373–4
ROC curve 278
arithmetic mean 59
arterial oxygen tension 183–4
arthritis 15 18 37 40
Asian women 35
assessment 19–20
ascertainment bias 38
association 230–2
asthma 21 265 267 332 372 373
atrophy of spinal chord 37
attack rate 303
attribute 47
AUC see area under the curve
average see mean
AVP 196–7
AZT (zidovudine) 77–8 169–71
> Back of Book > Index > B
B
babies 267 373–4
back-transformation 166–7 271
backwards regression 326
bar chart 73–5 354–6
bar notation 59
Bartlett's test 172
base of logarithm 82–4
baseline 79
baseline hazard 324
BASIC 107
Bayesian probability 87
Bayes' theorem 289
BCG vaccine 6–7 11 17 33 81
beta error 140 337
between groups sum of squares 174
between cluster variance 345–6
between subjects variance 178–9 204
bias 6 11–14 17–20 28 39–42 283–4 327 350 363
ascertainment 38
in allocation 11–13
ascertainment 38
in assessment 19–20
publication 327
in question wording 40–2
recall 39 350 363
in reporting 17–19
response 17–19
in sampling 28 31
volunteer 6 13–14 32
biceps skinfold 165–7 213–15 339
bimodal distribution 54–5
binary variable see dichotomous variable
Binomial distribution 89–91 94 103 106–8 110 128 130–1 132–3 180
and Normal distribution 91 106–8
mean and variance 94
probability 90–1
in sign test 138–9 247
biological variation 269
birds 45–6 255 350
birth rate 303 305
birthweight 150
blind assessment 19–20
blocks 9
blood pressure 19 28 117 191 268–9
BMI see body mass index
body mass index (BMI) 322–3
Bonferroni method 148–51
box and whisker plot 58 66 351 359
boxers 264
boxes 93–4
breast cancer 37 216–17
breast feeding 153
breathlessness 74–5
British Standards Institution 270
bronchitis 130–2 146 233–4
> Back of Book > Index > C
C
Campylobacter jejuni 44–6 255 350
C-T scanner 5–6 68
caesarian section 25 349
calculation error 70
calibration 194
cancer 23 32–9 41 69–74 216–17 241–3
breast 37 216–17
cervical cancer 23
lung 32 35–9 68–70 241–3 299
oesophagus 74 78–80
parathyroid
cancer registry 39
capillary density 159–64 174
cards 7 12 50
carry-over effect 15
case-control study 37–40 45–6 153–5 241–3 248 323 349–50 362–3
case fatality rate 303
case report 33–4
case series 33–4
cataracts 266 373
categorical data 47–8 373 see nominal data
cats 350
cause of death 70–3 75
cell of table 230
censored observations 281 308 324–5
census 27 47–8 86 294
decennial 27 294
hospital 27 47–8 86
local 27
national 27 294
years 294 299
centile 57–8 279–81
central limit theorem 107–8
cervical cancer 23
cervical smear 275
cervical cytology 22
chart
bar see bar chart
pie see pie chart
cheating in allocation 12–13
Chi-squared distribution 118–20 232–3
and sample variance 119–20 132
contingency tables 231–3 249–51
degrees of freedom 118–19 231–2 251
table 233
chi-squared test 230–6 238–40 243–51 249–51 258–9 261–2 371
372 373
contingency table 230–6 238–40 243–7 249–51 258–9 261–2 371
372 373
continuity correction 238–40 247 259 261
degrees of freedom 231–2 251
goodness of fit 248–9
logrank test 287–8
sample size 341
trend 243–5 259 261–2
validity 234–6 239–40 245
children see school
children choice of statistical method 257–267
cholesterol 55 326 345
cigarette smoking see smoking
cirrhosis 297–9 306 317
class interval 49–50
class variable 317
clinical trials 5–25 32–3 326–30
allocation to treatment 6–15 20–1 23
assessment 19–20
combining results from 326–30
cluster randomized 21–2 179–81 205 344–6 380
consent of subjects 22–4
cross-over 15–16 341
double blind 19–20
double masked see double blind
ethics 19 22–4
grouped sequential 152
informed consent 22–4
intention to treat 14–15 23 348 372
meta-analysis 326–30
placebo effect 17–19
randomized 7–11
sample size 336–42 344–6 347
selection of subjects 16–17
sequential 151–2
volunteer bias 13–14
Clinstat computer program 3 9 30 93 248 298
cluster randomization 21–2 179–81 205 344–6 380
cluster sampling 31 344–6
Cochran, W.G. 230
coefficient of variation 271
coefficients in regression 189 191–2 310–12 314 317 322–3 325
Cox 325
and interaction 314
logistic 322–3
multiple 310–12 314 317
simple linear 189 191–2
coeliac disease 34 165–7 213–15 339
cohort study 36–7 350
cohort, hypothetical in life table 299
coins 7 28 87–92
colds 69 241–3
colon transit time 267
combinations 97–8
combining data from different studies 326–30
common cold see colds
common estimate 326–30
common odds ratio 328–30
common proportion 145–7
common variance 162–4 173
comparison
multiple see multiple comparisons
of means 12–19 143–5 162–4 170–6 338–41 347 361 379–80
of methods of measurement 269–73
of proportions 130–2 145–7 233–4 245–7 259 341–3 347 372 379
of regression lines 208 9 367–8
of two groups 128–32 143–7 162–4 211–17 233–4 254 255–7 338–
43 344–6 347 361 372 379–80
of variances 172 260
within one group 159–62 217–20 245–7 257 260–1 341
compliance 183–4 228–9 363–7 369–70
computer 2 8–9 30 107 166 174 201 238 288–90 298 308 310
318
diagnosis 288–90
random number generation 8–9 107
program for confidence interval of proportion 132
programs for sampling 30
statistical analysis 2 174 201 298 308 310 318
conception 142
conditional logistic regression 323
conditional odds ratio 248
conditional probability 96–7
conditional test 250
confidence interval 126–34
appropriate for comparison 134
centile 133 280–1
correlation coefficient 200–1
difference between two means 128–9 136 162–4 361
difference between two proportions 130–1 243
difference between two regression coefficients 208–9 368
hazard ratio 288 325
mean 126–7 136 159–60 335–6 361
median 133
number needed to treat 290–1
odds ratio 241–3 248
percentile 133 280–1
predicted value in regression 194–5
proportion 128 132–3 336
quantile 133 280–1
ratio of two proportions 131–2
reference interval 280–1 290 375 378
regression coefficient 191–2
regression estimate 192–4
and sample size 335–6
or significance test 142 145 227
SMR 298–9 307 376–7
sensitivity 276
sensitivity 276
survival probability 283
transformed data 166–7
using rank order 216 220
confidence limits 126–34
confounding 34–5
consent of research subjects 22–4
conservative methods 15
constraint 118–19 250–1
contingency table 230 330
continuity correction 225–6 238–40 247

chi-squared test 238–40
Kendall's rank correlation coefficient 226
Mann-Whitney U test 225
McNemars test 247
continuous variable 47–50 75 87–8 93 103–6 276–8 323
in diagnostic test 276–8
contrast sensitivity 266 373
control group
case control study 37–9 350 362–3
clinical trial 5–7
controlled trial see clinical trial
cornflakes 153–5 362–3
coronary artery disease 34 149 326
coronary thrombosis 11–12 36
correlation 197–205 220 260–2 309–11
assumptions 200–1
between repeated measurements 341
coefficient 197–204
confidence interval 200–1
Fisher's z transformation 201 339–40 343
intra-class 179 204–5 272 346
intra-cluster 346 347
linear relationship 199
matrix 202 309–10
multiple 311
negative 198
positive 197
product moment 198
r 198–200
r2 199–200
rank see rank correlation
and regression 199–200 311
repeated observations 202–4
sample size 343–4
significance test 200–1
table of 200
table of sample size 344
zero 198
cough 34–5 41 128–32 144–7 233–4 240–1 254
counselling 41–2
counties 347
covariance analysis 321
Cox regression 324–5
crime 97
Crohn's disease 153–5 165–7 213–15 339 362–3
cross-classification 230 370–1
cross-over trial 15–16 137 341
cross-sectional study 34–5
cross-tabulation 230 370–1
crude death rate 294–5
crude mortality rate 294–5 302
cumulative frequency 48–51 56
cumulative survival probability 282–3 299
cushion volume 333–4 378
cut-off point 277–8 281
> Back of Book > Index > D
D
death 27 70–3 96 101–2 281
death certificate 27 294
death rate see mortality rate
decennial census 27 294
decimal dice 8
decimal places 70 268
decimal point 70
decimal system 69–70
decision tree 289–90
Declaration of Helsinki 22
degrees of freedom 61 67 118–20 153–4 159 169 171–2 191 231–2
251 288 309 311 319 331
analysis of variance 173–5
Chi-squared distribution 118–20
chi-squared test 231–2 251
F distribution 120
F test 171 173–5
goodness of fit test 248–9
logrank test 288
regression 191 310 313
sample size calculations 335
t distribution 120 157–8
t method 157–8 160–4
variance estimate 61 67 94–5 119 352–3
delivery 25 230–1 322–3 349
demography 299
denominator 68–9
dependent variable 187
depressive symptoms 18
Derbyshire 128
design effect 344–6 380
detection, below limit of 281
deviation from assumptions 161–2 164 167–8 175–6 196–7
deviations from mean 61 352
deviations from regression line 187–8
dexamethasone 290–1
diabetes 135–6 360–1
diagnosis 47–8 86 275–9 288–90 317
diagnostic test 136 275–9 361
diagrams 72–82 85–6
bar see bar chart
pie see pie chart
scatter see scatter diagram
diarrhoea 172 318
diastolic blood pressure see blood pressure
dice 7–8 87–9 122
dichotomous variable 258–62 308 317 321–3 325 328
difference against mean plot 161–2 184 271–5 364–5 367
differences 129–30 138–9 159–62 184 217–20 271–5 341 364–5
369–70
differences between two groups 128–31 136 143–7 162–7 211–17
258–9 338–43 344–6 347 362–3
digit preference 269
direct standardization 296
discharge from hospital 48
discrete data 47 49
discriminant analysis 289
distribution
Binomial see Binomial distribution
Chi-squared see Chi-squared distribution
cumulative frequency see cumulative frequency distribution
F see F distribution
frequency see frequency distribution
Normal see Normal distribution
Poisson see Poisson distribution
probability see probability distribution
Rectangular see Rectangular distribution
t see t distribution
Uniform see Uniform distribution
distribution-free methods 210
diurnal variation 249
DNA 97
doctors 36 68 86 297–9 356
Doppler ultrasound 150
dot plot 77
double blind 19–20
double dummy 18
double masked see double blind
double placebo see double dummy
drug 69
dummy treatment see placebo
dummy variables 317 328
Duncan's multiple range test 176
> Back of Book > Index > E
E
e, mathematical constant 83–4 95
ecological fallacy 42–3
ecological studies 42–3
eczema 97
election 28 32 41
electoral roll 30 32
embryos 333–4 378
enumeration district 27
envelopes 12
enzyme concentration 347 379–80
epidemiological studies 32 34–40 42–3 45–6 326
equality, line of 273–4
error 70 140 187 192 269–72 337
alpha 140
beta 140 337
calculation 70
first kind 140
measurement 269–72
second kind 140 337
term in regression model 187 192
type I 140
type II 140 337
estimate 61 122–36 326–30
estimation 122–36 335–6
ethical approval 32
ethics 4 19 22–4 32
evidence-based practice 1
expectation 92–4
of a distribution 92–3
of Binomial distribution 94
of Chi-squared distribution 118
of life 102 300–2 305 357–8
of sum of squares 60–4 98–9 119
expected frequency 230–31 26 250
expected number of deaths 297–9
expected value see expectation, expected frequency
experimental unit 21–2 180
experiments 5–25
animal 5 16–17 20–1 33
clinical see clinical trials
design of 5–25
factorial 10–11
laboratory 5 16–17 20–1
expert system 288–90
ex-prisoners 128
> Back of Book > Index > F
F
F distribution 118 120 334
F test 171 173–5 311 313–15 317–18 320 334 378
face-lifts 23
factor 317–18
factorial 90 97
factorial experiment 10–11
false negative 277–9
false positive 277–9
family of distributions 90 96
Farr, William 1
FAT see fixed activated T-cells
fat absorption 78 169–71
fatality rate 303
feet, ulcerated 159–64 174
fertility 142 302–3
fertility rate 303
FEVl 49–54 57–60 62–3 125–7 133 185–6 188–95 197–9 201 279–
80 310–11 335–6
fever tree 26
Fisher 1
Fisher's exact test 236–40 251–2 259 262
Fisher's z transformation 201 343
five figure summary 58
five year survival rate 283
fixed activated T-cells (FAT) 318–21
fixed effects 177–9 328
follow-up, lost to or withdrawn from 282
foot ulcers 159–64 174
forced expiratory volume see FEV1
forest diagram 330
forward regression 326
fourths 57
frequency 48–56 68–9 230–1 250
cumulative 48–51
density 52–4 104–5
distribution 48–56 66–7 103–5 351–2 354
expected 230–1 250
per unit 52–4
polygon 54
and probability 87 103–5
proportion 68
relative 48–50 53–4 104–5
tally system 50 54
in tables 71 230–1
> Back of Book > Index > G
G
G.P. 41
Gabriel's test 177
gallstones 284–8 324–5
Galton 186
gastric pH 265–6 372–3
Gaussian distribution see Normal distribution
gee whiz graph 79–80
geometric mean 113 167 320
geriatric admissions 86 255–6 354 356 370–1
gestational age 196–7
glucose 35 66–7 121–2 351–3 359–60
glue sniffing see volatile substance abuse
goodness of fit test 248–9
Gossett see Student
gradient 185–6
graft failure 331
graphs 72–82 85–6
group comparison see comparisons
grouped sequential trials 152
grouping of data 167
guidelines 179–81
> Back of Book > Index > H
H
harmonic mean 113
hay fever 97
hazard ratio 288 324–25
health 40–1
health centre 220–1
health promotion 347
healthy population 279 292–3
heart transplants 264
heatwave 86 255–6 356 370–1
height 75–6 87–8 93–4 112 159 185–6 188–95 197–9 201 208–9
308–17 367–9
Helsinki, Declaration of 22
heteroscedasticity 175
heterogeneity test 249 328–9
Hill, Bradford 1
histogram 50–7 67 72 75 103–4 267 303–4 352 354 356 359
historical controls 6
HIV 58 128 172 174–7
holes 93–4
homogeneity of odds ratios 328–9
homogeneity of variance see uniform variance
homoscedasticity 175
hospital admissions 86 255–6 356 370–1
hospital census 27 47–8 85
hospital controls 38–9
house-dust mite 265 372
housing tenure 230–1 317
Huff 79 81
human immunodeficiency virus see HIV
hypercholesterolaemia 55
hypertension 43 91 265 372
hypocalcaemia 34
hypothesis, alternative see alternative hypothesis
hypothesis, null see null hypothesis
> Back of Book > Index > I
I
ICC see intra-class correlation
ICD see International Classification of Disease
ileostomy 265 372
incidence 303
independent events 88 357
independent groups 128–32 143–7 162–4 172–7 211–17
independent random variables 93–4
independent trials 90
independent variable in regression 187
India 17 33
indirect standardization 296–9
induction of labour 322–3
infant mortality rate 303
infinity (∞) 291
inflammatory arthritis 40
informed consent 22–3
instrumental delivery 25 349
intention to treat 14–15 348–9 372
interaction 310 313–14 320–1 327–9 334 378
intercept 185–6
International Classification of Disease 70–72
inter-pupil distance 331
interquartile range 60
interval, class 49
interval estimate 126
interval scale 210 217 258–62 373
intra-class correlation coefficient 179 204–5 272 380
intra-cluster correlation coefficient 272 380
> Back of Book > Index > J
J
jittering in scatter diagrams 77
> Back of Book > Index > K
K
Kaplan-Meier survival curve 283
Kaposi's sarcoma 69 220–1
Kendall's rank correlation coefficient 222–6 245 261–2 373 374
continuity correction 226
in contingency tables 245
τ 222
table 225
tau 222
ties 23–4
compared to Spearman's 224–5
Kendall's test for two groups 217
Kent 245–7
Know Your Midwife trial 25 348–9
knowledge based system 289–90
Korotkov sounds 268–9
Kruskal-Wallis test 217 261–2
> Back of Book > Index > L
L
labour 322–3 348–9
laboratory experiment 5 16 20–1
lactulose 172 175–7
Lanarkshire milk experiment 12
laparoscopy 142
large sample 126 128–32 143–7 168–9 258–60 335–6
least squares 187–90 205–6 310
left censored data 281
Levene test 172
life expectancy 102 300–2 305 357–8
life table 101–2 282–3 296 299–302
limits of agreement 274–5
line graph 77–80 354 356
line of equality 273–4
linear constraint 118–19 243–5 250–1
linear regression see regression, multiple regression
linear relationship 185–209 243–5
linear trend in contingency table 243–5
Literary Digest 31
lithotrypsy 284
log see logarithm, logarithmic
log hazard 324–5
log-linear model 330
log odds 240 252–3 321–3
log odds ratio 241–2 252–3 323
logarithm 82–4 131
base of 82–4
logarithm of proportion 131
logarithm of ratio 131
logarithmic scale 81–2
logarithmic transformation 113–14 116 164–7 175–6 184
and coefficient of variation 271
and confidence interval 167
geometric mean 113 167
to equal variance 164–7 175–6 196–7 271
to Normal distribution 113–14 116 164–7 175–6 184 360 364–5
372
standard deviation 113–14
variance of 131 248
logistic regression 289 321–3 326 328–9 330
conditional 323
multinomial 330
ordinal 330
logit transformation 235 248–9 321–3
Lognormal distribution 83 113
logrank test 284 287–9 325
longitudinal study 36–7
loss to follow-up 282
Louis, Pierre-Charles-Alexandre 1
lung cancer 32 35–9 68–70 96 242–3 299
lung function see FEV1, PEFR, mean transit time, vital capacity
lymphatic drainage 40
> Back of Book > Index > M
M
magnesium 135–6 292–3 360–1 375–6
malaria 26
mannitol 58 172 174–7 317–18
Mann–Whitney U test 164 211–17 225–7 258–9 259 278 373–4
and two-sample t method 211 215–17
continuity correction 225–6
Normal approximation 215 225–6
and ROC curve 278
table 212
tables of 217
ties 213 215
Mantel's method for survival data 288
Mantel-Haenszel
method for combining, 2 by 2 tables 328
method for trend 245
marginal totals 230–1
matched samples 159–62 217–20 245–7 260 341 363–7 369–70
matching 39 45–6
maternal age 267 373
maternal mortality rate 303
maternity care 25
mathematics 2
matrix 309
maximum 58 65 169 345
maximum voluntary contraction 308–16
McNemar's test 245–7 260
mean transit time 265 368
mean 59–60 67
arithmetic 59
comparison of two 128–9 143–5 162–4 338–41 361 378–9
confidence interval for 126–7 132 335 361
deviations from 60
geometric 113 167
harmonic 113
of population 126–7 335–6
of probability distribution 92–4 105–6
of a sample 56–8 65–6 352–3
sample size 335–6 338–41
sampling distribution of 122–5
standard error of 126–7 136 156 335 361
sum of squares about 60–65
measurement 268–9
measurement error 269–72
measurement methods 272–5
median 56–9 133 216–7 220 351
confidence interval for 133 220
Medical Research Council 9
mercury 34
meta-analysis 326–30
methods of measurement 269–73
mice 21 33 333–4 378
midwives 25 342–3
mild hypertension 265 368
milk 12–13 45–6 255 349–50
mini Wright peak flow meter see peak flow meter
minimization 13
minimum 58 66 351
misleading graphs 78–81
missing denominator 69
missing zero 79–80
mites 265 372
MLn 3
MLWin 3
mode 55
modulus 239
Monte carlo methods 238
mortality 15 36 70–6 86 294–6 302–3 347 356 357–8 376–7
mortality rate 36 294–6 302–3
age-specific 295–6 299–300 302 307 376–7
age-standardized 296 302
crude 294–5 302
infant 303 305
neonatal 303
perinatal 303
mosquitos 26
MTB see mycobaterium tuberculosis
MTT see mean transit time
multifactorial methods 308–34
multi-level modelling 3
multinomial logistic regression 330
multiple comparisons 175–7
multiple regression 308–18 333–4
analysis of variance for 310–15
and analysis of variance 318
assumptions 310 315–16
backward 326
class variable 317–18
coefficients 310–12 314 378
computer programs 308 310 318
correlated predictor variables 312
degrees of freedom 310 312
dichotomous predictor 317
dummy variables 317–18
F test 311 313 317
factor 317–18
forward 326
interaction 310 313–14 333–4 378
least squares 310
linear 310 314
in meta-analysis 327
non-linear 310 314–15 378
Normal assumption 315–16
outcome variable 308
polynomial 314–15
predictor variable 308 312–13 316–18
quadratic term 315 316 378
qualitative predictors 316–18
R2 311
reference class 317
residual variance 310
residuals 315–16 333–4 378
significance tests 310–13
standard errors 311–12
stepwise 326
sum of squares 310 313–14 378
t tests 310–12 317
transformations 316
uniform variance 316
variance ratio 311
variation explained 311
multiple significance tests 148–52 169
multiplicative rule 88 90 92–4 96
multi-way analysis of variance 318–21
multi-way contingency tables 330
muscle strength 308–16
mutually exclusive events 88 90 357
mycobaterium tuberculosis (MTB) 318–21
myocardial infarction 277 347 379
> Back of Book > Index > N
N
Napier 83
natural history 26 33
natural logarithm 83
natural scale 81–2
nausea and vomiting 290–1
Nazi death camps 22
negative predictive value 278–9
neonatal mortality rate 300
New York 6–7 10
Newman-Keuls test 176
Nightingale, Florence 1
nitrite 265 372–3
NNH see number needed to harm
NNT see number needed to treat
nodes in breast cancer 216–17
nominal scale 210 258–62
non-parametric methods 210 226–7
non-significant 140–1 142–3 149
none detectable 281
Normal curve 106–9
normal delivery 25 349
Normal distribution 91 101–20
and Binomial 91 106–8
in confidence intervals 126–7 258–60 262 373
in correlation 200–1
derived distributions 118–20
independence of sample mean and variance 119–20
as limit 106–8
and normal range 279–81 293
of observations 112–18 156 210 258–62 359–60
and reference interval 279–81 293 375 378
in regression 187 192 194 315–16
in significance tests 143–7 258–60 262 368
standard error of sample standard deviation 132
in t method 156–8
tables 109–10
Normal plot 114–19 121–2 161 163 165–7 170–3 175–6 180–1 267
359–60
Normal probability paper 114
normal range see reference interval
null hypothesis 137 139–42
number needed to harm 290
number needed to treat 290–1
Nuremburg trials 22
nuisance variable 320
> Back of Book > Index > O
O
observational studies 5 26–46
observed and expected frequencies 230–1
occupation 96
odds 240 321–3
odds ratio 240–2 248 252–3 259 323 328–9
oesophogeal cancer 74 77–80
Office of National Statistics 294
on treatment analysis 15
one-sided percentage point 110
one-sided test 141–2 237
one-tailed test 141–2 237
opinion poll 29 32 41 347 378–9
ordered nominal scale 258–62
ordinal logistic regression 330
ordinal scale 210 220 258–62 373
outcome variable 187 190 308 321
outliers 58 196 378
overview 326–30
oxygen dependence 267 373–4
> Back of Book > Index > P
P
pa(O2) 183–4
pain 15–16 18
pain relief score 18
paired data 129–30 138–9 159–62 167–8 217–20 245–7 260 341
363–7 369–70 372
in large sample 129–30
McNemar's test see McNemar's test
sample size 341
sign test see sign test
t method see t methods
Wilcoxon see test Wilcoxon test
parameter 90
parametric methods 210 226–7
parathyroid cancer 282–4
parity 49 52–3 248–9
passive smoking 34–5
PCO see polycystic ovary disease
peak expiratory flow rate see PEFR
peak flow meter 269–75
peak value 169
Pearson's correlation coefficient see correlation coefficient
PEFR 54 128–9 144–5 147–8 208–9 265 269–75 363–4 368
percentage 68 71
percentage point 109–10 347 378
percentile 57 279–81
perinatal mortality rate 303
permutation 97–8
pH 265 372–3
phlegm 145 147–8
phosphomycin 69
physical mixing 12
pictogram 80–1
pie chart 72–3 80 354–5
pie diagram see pie chart
pilot study 335 339 341
Pitman's test 260
placebo 17–20 22
point estimate 125
Poisson distribution 95–6 108 165 248–50 252 298–9
Poisson heterogeneity test 249
Poisson regression 330
poliomyelitis 13–14 19 68 86 355
polycystic ovary disease 35
polygon see frequency polygon
polynomial regression 314–15
population 27–34 36 39 87 335–6
census 27 294
estimate 294
mean 126–7 335–6
national 27 294
projection 302
pyramid 303–5
restricted 33
standard deviation 124–5
statistical usage 28
variance 124–5
positive predictive value 278
power 147–8 337–46
p–p plot 117–18
precision 268–9
predictor variable 187 190 308 312–13 316–18 321 323 324
pregnancy 25 49 348–9
premature babies 267
presenting data 68–86
presenting tables 71–2
prevalence 35 90 278–9 303
probability 87–122
addition rule 88
conditional 96–7
density function 104–6
distribution 88–9 92–4 103–6 357–8
of dying 101–2 299–300 357–8
multiplication rule 88 96
paper 114
in significance tests 137 9
of survival 101–2 357–8
that null hypothesis is true 140
product moment correlation coefficient see correlation coefficient
pronethalol 15–16 138–9 217–20
proportion 68–9 71 128 130–3 165 321–3
arcsine square root transformation 165
confidence interval for 128 132–3 336
denominator 69
difference between two 130–1 145–7 233–4 245–7 341–3 347
as outcome variable 321–3
ratio of two 131–2 147
sample size 336 341–3 347
standard error 128 336
in tables 71
of variability explained 191 200
proportional frequency 48
proportional hazards model 324–5
prosecutor's fallacy 97
prospective study 36–7
protocol 268
pseudo-random 8
publication bias 327
pulmonary tuberculosis see tuberculosis
pulse rate 178–9 190–1 204
P value 1 139–41
P value spending 152
pyramid, population 303–5
> Back of Book > Index > Q
Q
q–q plot see quantile–quantile plot
quadratic term 315 316 378
qualitative data 47 258–62 316–18
quantile 56–8 116–18 133 279–81
confidence interval 133 280–1
quantile-quantile plot 116–18
quantitative data 47 49
quartile 57–8 66 351
quasi-random sampling 31
questionnaires 36 40–2
quota sampling 28–29
> Back of Book > Index > R
R
r, correlation coefficient 198–9
r2 199–20 311
rS, Spearman rank correlation 220
R, multiple correlation coefficient 311
R2 311
radiological appearance 20
RAGE 23
random allocation 7–11 15 17 20–3 25
by general practice 21 23
by ward 21
in clusters 21–2 344–6
random blood glucose 66–7
random effects 177–9 328
random numbers 8 10 29–30
random sampling 9 29–32 38 90
random variable 87–118
addition of a constant 93
difference between two 94
expected value of 92–4
mean of 92–4
multiplied by a constant 92
sum of two 92–3
variance of 92–4
randomization see random allocation
randomized consent 23
randomizing devices 7–8 87 90
range 59–60 279
interquartile 59–60
normal see reference interval
reference see reference interval
rank 211 213–14 218 221 223
rank correlation 220–6 261–2 373 374
choice of 226 261–2
Kendall's 222–6 261–2 373 374
Spearman's 220–2 226 261–2 374
rank order 211 213–14 221
rank sum test 210–20
one sample see Wilcoxon
two sample see Mann Whitney
rate 68–9 71
age specific mortality 295–6 299–300 302 307
age standardized mortality 296 302
attack 303
birth 303 305
case fatality 303
crude mortality 294–5 302
denominator 69
fertility 303
five year survival 283
incidence 303
infant mortality 303 305
maternal mortality 303
mortality 294–6 302–3
multiplier 68 295
neonatal mortality 303
perinatal mortality 303
prevalence 303
response 31–2
stillbirth 303
survival 283
ratio
odds see odds ratio
of proportions 131–2 147
scale 257–8
standardized mortality see standardized mortality ratio
rats 20
raw data 167
recall bias 39 350 363
receiver operating characteristic curve see ROC curve
reciprocal transformation 165–7
Rectangular distribution 107–8
reference class 317
reference interval 33 136 279–81 293 361 375 378
confidence interval 280–1 293 375 378
by direct estimation 280–1
sample size 347 378
using Normal distribution 279–80 293 361 375 378
using transformation 280
refusing treatment 13–15 25
register of deaths 27
regression 185–9 199–200 205–7 208–9 261–2 308–18 312–30
333–4
analysis of variance for 310–15
assumptions 187 191–2 194–5 196–7
backward 326
coefficient 189 191–2
comparing two lines 208–9 367–8
confidence interval 192
in contingency table 234–5
and correlation 199–200
Cox 324–5
dependent variable 187
deviations from 187
deviations from assumptions 196–7
equation 189
error term 187 192
estimate 192–3
explanatory variable 187
forward 326
gradient 185–6
independent variable 187
intercept 185–6
least squares 187–90 205–6
line 187
linear 189
logistic 321–3 326 328–9
multinomial logistic 330
multiple see multiple regression
ordinal logistic 330
outcome variable 187 190
outliers 196
perpendicular distance from line 187–8
Poisson 330
polynomial see polynomial regression
prediction 192–4
predictor variable 187 190
proportional hazards 324–5
residual sum of squares 191
residual variance 191
residuals 194–6
significance test 192
simple linear 189
slope 185–6
standard error 191–4
stepwise 326
sum of products 189
sum of squares about 191–2 310
sum of squares due to 191–2
towards the mean 186–7 191
variability explained 191 200
variance about line 191–2 205–6
X on Y 190–1
rejecting null hypothesis 140–1
relationship between variables 33 73–8 185–209 220–6 230–45 257
261–2 308–34
relative frequency 48–50 53 103–5
relative risk 132 241–3 248 323
reliability 272
repeatability 33 269–72
repeated observations 169–71 202–3
repeated significance tests 151–2 169
replicates 177
representative sample 28–32 34
residual mean square 174 270
residual standard deviation 191–2 270
residual sum of squares 174 310 312
residual variance 173 310
residuals 165–6 175–6 267 315–16 333–4
about regression line 194–6 315–16
plots of 162–4 173–4 194–6 315–16 333–4 378
within groups 165–6 175–6
respiratory disease 32 34–5
respiratory symptoms 32 34–5 41 125–9 142–7 233–4 240–1 243–7
254
response bias 17–19
response rate 31–2
response variable see outcome variable
retrospective study 39
rheumatoid arthritis 37
Richter scale 114
risk 131–2
risk factor 39 326–7 350
RND (X) 107
robustness to deviations from assumptions 167–9
ROC curve 277–8
> Back of Book > Index > S
S
s2, symbol for variance 61
saline 13–14
Salk vaccine 13–14 17 19 68 355
salt 43
sample 87
large 127–31 168–9 258–60 262 335–6
mean see mean
size see size of sample
small 130–1 132–3 156–69 227 258–60 262 344
variance see variance
sampling 27–34
in clinical studies 32–4 293 375
cluster 31
distribution 122–5 127
in epidemiological studies 32 34–9
experiment 63–4 122–5
frame 29
multi-stage 30
quasi-random 31
quota 29
random 29–31
simple random 29–30
stratified 31
systematic 31
scanner 5–6
scatter diagram 75–7 185–6
scattergram see scatter diagram
schoolchildren 12–13 17 22 31 34–5 41 43 128–32 143–7 233–4
240–1 243–7 254
schools 22 31 34
screening 15 22 81 216–7 265 275–9
selection of subjects 16–17 32–3 37–9
in case control studies 37–9
in clinical trials 16–17
self 31–2
self selection 31–2
semen analysis 183
semi-parametric 325
sensitivity 276–8
sequential analysis 151–2
sequential trials 151–2
serial measurements 169–71
sex 71–2
sign test 138–9 161 210 217 219–20 228 246–7 260 369–70 372
373
signed-rank test see Wilcoxon
significance and importance 142–3
significance and publication 327
significance level 140–1 147
significance tests 137–55
multiple 148–52 169
and sample size 336–8
in subsets 149–50
inferior to confidence intervals 142 145
significant difference 140
significant digits see significant figures
significant figures 69–72 268–9
size of sample 32 147–8 335–47
accuracy of estimation 344
in cluster randomization 344–6
correlation coefficient 343–4
and estimation 335–0
paired samples 6–341
reference interval 347 378
and significance tests 147–8 336–8
single mean 335–6
single proportion 336 378–9
two means 338–41 379–80
two proportions 341–3 379
skew distribution 56 59 67 112–14 116–17 165 167–8 360
skinfold thickness 165–7 213–15 335
slope 185–6
small samples 156–67 227 258–60
smoking 22 26 31–2 34–9 41 67 74–5 241–3 356
SMR 297–9 303 307 376–7
Snow, John 1
sodium 116–17
somites 333–4 378
South East London Screening Study 15
Spearman's rank correlation coefficient 220–2 226 261–2 373
table 219
ties 219
specificity 276–8
spinal chord atrophy 37
square root transformation 165–7 175–7
squares, sum of see sum of squares
standard age specific mortality rates 297–8
standard deviation 60 62–4 67 92–4 119–21
degrees of freedom for 63–4 67 119
of diiferences 159–62
of population 123–4
of probability distribution 92–4 105
of sample 62–4 67 119 353
of sampling distribution 123–4
and transformation 113–14
and standard error 126
standard error of 132
within subjects 269–70
standard error 122–5
and confidence intervals 126–7
centite 280
correlation coefficient 201 343
difference between two means 128–9 136 338–41 361 379–80
difference between two proportions 130–1 145–7 341–3 379
difference between two regression coefficients 208 367–8
different in significance test and confidence interval 147
log hazard ratio 325
log odds ratio 241–2 252–3
logistic regression coefficient 322
mean 123–5 136 335–6 361
percentile 280
predicted value in regression 192–4
proportion 128 336 378–9
quantile 280
ratio of two proportions 121–2
reference interval 280 370–1 378
regression coefficient 191–2 311–12 317
regression estimate 192–3
SMR 298–9 377
standard deviation 132
survival rate 283–4 341
Standard Normal deviate 114–17 225–6

Standard Normal distribution 108–11 143 156–8 337–8
standard population 296
standardized mortality rate 74 296
standardized mortality ratio 296–9 303 307
standardized Normal probability plot 117–18
Stata 118
StatExact 238
statistic 47 139 302–3 337
test 139 337
vital 302–3
Statistics 1
statistical significance see significance test
stem and leaf plot 54 57 66 184 351 364–6
step function 51 283
step-down 326
step-up 326
stepwise regression 326
stillbirth rate 303
stratification 31
strength 308–16
strength of evidence 137 140 362
streptomycin 9–10 17 19–20 81 235–6 290
stroke 5–6 23 249
Stuart test 248 260
Student 12–13 156 158–9
Student's t distribution see t distribution
Studentized range 176
subsets 149–50
success 90
suicide 306
sum of products about mean 189 198–200
sum of squares 60–1 63–5 98–9 119 173–4 310 313–14
about mean 60–1 63–5 119 352–3
about regression 191–2 310 313–14
due to regression 191–2 310 313–14
expected value of 63–4 98–9 119
summary statistics 169 180–1 327
summation 59
survey 28–9 42 90
survival 10 101–2 281–8 324–5
analysis 281–8 324–5
curve 283–4 286 324
probability 101–2 282–4 287
rate 283
time 162 281–8
symmetrical distribution 54 56 59
synergy 320
syphilis 22
systolic blood pressure see blood pressure
> Back of Book > Index > T
T
t distribution 120 156–9
degrees of freedom 120 153–4 157–8
and Normal distribution 120 156–8
shape of 157
table 158
t methods 114 156–69
assumptions 161–8 184 365–7
confidence intervals 159–63 164 167
deviations from assumptions 161–2 164 167–8
difference between means in matched sample 159–61 184 260 363–
7 370 372
difference between means in two samples 162–7 258–9 262
one sample 159–62 184 260 363–7 370 372
paired 159–62 167–8 184 217 220 260 363–7 370 372
regression coefficient 191–2 310–12 317
single mean 159–62 176
two sample 162–7 217 258–9 262 317 373–4
unpaired same as two sample
table of probability distribution
Chi-squared 233
correlation coefficient 200
Kendall's τ 225
Mann–Whitney U 212
Normal 109–10
Spearman's ρ 222
t 158
Wilcoxon matched pairs 219
table of sample size for correlation coefficient 344
tables of random numbers 8–9 29–30
tables, presentation of 71–2
tables, two way 230–48
tails of distributions 56 359–60
tally system 49–50 54
Tanzania 69 220–4
TB see tuberculosis
telephone survey 42
temperature 10 70 86 210 255–6 332
test, diagnostic 136 275–9 361
test, significance see significance test
test statistic 136 337
three dimensional effect in graphs 80
thrombosis 11–12 36 345
thyroid hormone 267 373–4
ties in rank tests 213 215 218–19 222–4
ties in sign test 138
time 324–5
time series 77–8 169–71 354 356
time to peak 169
time, survival see survival time
TNF see tumour necrosis factor
total sum of squares 174
transformations 112–14 163–7 320
arcsine square root 165
and confidence intervals 167
Fisher's z 201 343
logarithmic 112–14 116 163–7 170–1 175–6 184 320 364–7 369–
70
logit 240 252–3
to Normal distribution 112–14 116 164–7 175–6 184
reciprocal 113 165–7
and significant figures 269
square root 165–7 175–7
to uniform variance 163–7 168 175–6 196–7 271
treated group 5–7
treatment 5–7 326–7
treatment guidelines 179–81
trend in contingency tables 243–5
chi-squared test 243–5
Kendall's τb 245
Mantel–Haenzsel 245
trial, clinical see clinical trial
trial of scar 322–3
triglyceride 55–6 58–59 63 112–13 280–2
trisomy-16 333–4 378
true difference 147
true negative 278
true positive 278
tuberculosis 6–7 9–10 17 81–2 290
Tukey 54 58
Tukey's Honestly Significant Difference 176
tumour growth 20
tumour necrosis factor (TNF) 318–21
Tuskegee Study 22
twins 204
two-sample t test see t methods
two-sample trial 16
two-sided percentage point 110
two-sided test 141–2
two-tailed test 141–2
type I error 140
type II error 140 337
> Back of Book > Index > U
U
ulcerated feet 159–64 174
ultrasonography 134
unemployment 42
Uniform distribution 107–8 249
uniform variance 159 162–4 167–8 175–6 187 191 196–7 316 319–
20
unimodal distribution 55
unit of analysis 21–2 179–81
urinary infection 69
urinary nitrite 265 372–3
> Back of Book > Index > V
V
vaccine 6–7 11 13–14 17 19
validity of chi-squared test 234–6 239–40 245
variability 59–64 269
variability explained by regression 191 200
variable 47
categorical 47
continuous 47 49
dependent 187
dichotomous 259–62
discrete 47 49
explanatory 187
independent 187
nominal 210 259–62
nuisance 320
ordinal 210 259–62
outcome 187 190 308 321
predictor 187 190 308 312–13 316–18 321 323 324
qualitative 47 316–18
quantitative 47
random see random variable
variance 59–64 67
about regression line 191–2 205–6
analysis of see analysis of variance
between clusters 345–6
between subjects 178–9 204
common 162–4 170 173
comparison in paired data 260
comparison of several 172
comparison of two 171 260
degrees of freedom for 61 63–4 352–3
estimate 59–64 124–5
of logarithm 131 252
population 123–4
of probability distribution 91–4 105
of random variable 91–4
ratio 120 311
residual 192 205–6 310
sample 59–64 67 94 98–9 119 352–3
uniform 162 163–7 168 174–6 187 196–7 316
within clusters 345–6
within subjects 178–9 204 269–72
variation, coefficient of 271
visual acuity 266 373
vital capacity 75–6
vital statistics 302–3
vitamin A 328–9
vitamin D 115–16
volatile substance abuse 42 307 376–7
volunteer bias 6 13–14 32
volunteers 5–6 13–14 16–17
VSA see volatile substance abuse
> Back of Book > Index > W
W
Wandsworth Health District 86 255–6 356
website 3 4
weight gain 20–1
wheeze 267
whooping cough 265 373
Wilcoxon test 217–20 260 369–70 373
matched pairs 217–20 260 369–70 373
one sample 217–20 260 369–70 373
signed rank 217–20 260 369–70 373
table 219
ties 218–19
two sample 217 see Mann-Whitney
withdrawn from follow-up 282
within cluster variance 345–6
within group residuals see residuals
within groups sum of squares 173
within groups variance 173
within subjects variance 178–9 204
within subjects variation 178–9 269–72
Wooif's test 328
Wright peak flow meter see peak flow meter
> Back of Book > Index > X
X
[x with bar above], symbol for mean 59
X-ray 19–20 81 179–81
> Back of Book > Index > Y
Y
Yates' correction 238–40 247 259 261
> Back of Book > Index > Z
Z
z test 143–7 258–9 262 234
z transformation 201 343
zero, missing 78–80
zidovudine see AZT
% symbol 71
! (symbol for factorial) 90 97
∞ (symbol for infinity) 291
| (symbol for given) 96
| (symbol for absolute value) 239
α (symbol for alpha) 140
β (symbol for beta) 140
χ (symbol for chi) 118–19
µ (symbol for mu) 92–3
φ (symbol for phi) 108
Φ (symbol for Phi) 109
ρ (symbol for rho) 220–2
Σ (symbol for summation) 57
σ (symbol for sigma) 92–3
τ (symbol for tau) 222–5

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To the memory of Ernest and Phyllis Bland, my parents

Preface to the Third Edition

In preparing this third edition of An Introduction to Medical Statistics, I

Sections marked * contain material

1.1 Statistics and medicine

these circumstances we make mistakes. No piece of research can be

1.2 Statistics and mathematics

1.3 Statistics and computing

software to carry them out. The theory of multilevel modelling (Goldstein

1.4 The scope of this book

2.1 Comparing treatments

Table 2.1. Analysis of the difference in survival

Pairs with 1978 9 (31%) 34 (38%)

Pairs with same 18 (62%) 38 (43%)

Pairs with 1978 2 (7%) 17 (19%)

Second, we could obtain concurrent groups by comparing our own

The difference in degree of cooperation between the parents of the two

Table 2.2. Results of studies of BCG vaccine in New Yo

BCG 445 3 0.67% 3.6 43%

Control 545 18 3.30% 1.7 24%

1933–44 Alternative selection carried out centrally

BCG 566 8 1.41% 2.8 40%

Control 528 8 1.52% 2.4 34%

Different methods of allocation to treatment can produce different results.

2.2 Random allocation

an aid to gambling. For large randomizations we use a different, non-

Table 2.4. Allocation of 20 subjects to two

Subject Digit Group

The system described above gave us unequal numbers in the two

Table 2.5. Condition of patients on admission

General condition Good 8 8

Max. evening temperature 98- 4 4

Sedimentation rate 0-10 0 0

Table 2.6. Survival at six months in the MRC

101°F and Alive 20 11

no drug at all. We can carry out experiments to compare several factors

2.3 * Methods of allocation without random

allocated) and even managed to do marginally better than those who

Table 2.7. Outcome of a clinical trial using

Even dates Odd dates

Survived 125 39 10 125

Total 164 50 10 206

secondarily, both measurements.

2.4 Volunteer bias

population was exposed to it at some time, usually in childhood. In the

Table 2.8. Result of the field trial of Salk

Control 201229 115 57

Not 338778 121 36

Control 1st 725173 330 46

2.5 Intention to treat

of health checkups (South-east London Screening Study Group 1977).

2.6 Cross-over designs

in Table 2.9. The advantage in favour of pronethalol is shown by 11 of the

Table 2.9. Results of a trial of pronethalol for

Number of attacks Difference

2 323 348 –25

Cross-over designs can be useful for laboratory experiments on animals

2.7 Selection of subjects for clinical trials