Minitab14 Manual

MINITAB 14 Supplement for: Biostatistics for the Health Sciences
R. Cliord Blair and Richard A. Taylor December 30, 2007
2 c R. Cliord Blair and Richard A. Taylor 2005 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of R. Cliord Blair or Richard A. Taylor.
i Preface This manual is not an introduction to MININITAB and contains only a portion of the material that would be contained therein.1 Rather, this manual teaches the reader how to use MINITAB Release 14 for Windows to analyze data related to the problems addressed in Biostatistics For The Health Sciences by R. Cliord Blair and Richard A. Taylor. To this end, MINITAB is used to analyze various of the data sets and problems contained in that text. The reader will be required, especially in the early part of the manual, to input selected data to the system. Additional data sets accompany this manual so that the reader will not be burdened with data entry once this skill is acquired. r It is assumed that the reader is familiar with the Microsoft Windows operating system so that little instruction in this regard will be provided. It is further assumed that the reader is using MINITAB Release 14 rather than some earlier version. Chapter 1 familiarizes the reader with some basic MINITAB concepts. Key among these is the entry and saving of data which is prerequisite to all that follows. Chapters 2 through 10 address the corresponding chapters in the text. That is, Chapter 2 in the manual addresses problems in Chapter 2 in the text and so on. Section headings in these latter chapters reect the sections/pages in the text addressed in that section of the manual. Thus, while studying the text readers can quickly locate the comparable area in the manual by referring to the manuals table of contents. In addition, a manual appendix relates specic examples, data tables etc. in the text to specic pages in the manual. These provisions allow easy integration of MINITAB analyses as study of the text progresses. Finally, we would caution that data analysis, whether by hand calculations or software is no substitute for a clear understanding of the underlying statistical concept. Such analysis can, however, be an invaluable aid to mastery of such concepts and is indispensable for research purposes.
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1 The denitive introduction to the MINITAB analysis system is found in MINITAB Handbook by Barbara Ryan, Brian Joiner, and Jonathon Cryer published by Thomson.
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Contents
1 Preliminaries 1.1 Introduction . . . . . . . . 1.2 Getting Started . . . . . . 1.3 Inputting Data . . . . . . 1.4 Performing Analyses . . . 1.5 Editing and Printing Your 1.6 Using the Calculator . . . 1.7 Graphs . . . . . . . . . . . Exercises . . . . . . . . . . . . 2 Descriptive Methods 2.1 Introduction . . . . 2.4 Distributions . . . 2.5 Graphs . . . . . . . 2.6 Numerical Methods Exercises . . . . . . . . 1 1 1 2 4 6 6 8 10 11 11 11 15 24 30 31 31 31 31 33 35 35 35 36 42 44
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3 Probability 3.1 Introduction . . . . . 3.3 Contingency Tables . 3.4 The Normal Curve . Exercises . . . . . . . . .
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4 Introduction to Inference and One Sample Methods 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Sampling Distributions . . . . . . . . . . . . . . . . . . 4.3 Hypothesis Testing . . . . . . . . . . . . . . . . . . . . 4.4 Condence Intervals . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5 Paired Samples Methods 45 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 5.2 Methods Related to Mean Dierence . . . . . . . . . . . . . . . . 45 5.3 Methods Related to Proportions . . . . . . . . . . . . . . . . . . 46 iii
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CONTENTS 5.4 Methods Related to Paired Samples Risk Ratios . . . . . . . . . 5.5 Methods Related to Paired Samples Odds Ratios . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 49 52 53 53 53 55 56 56 57 59 59 59 60 61 63 65 65 65 66 67 69 69 69 70 73
6 Two Independent Samples Methods 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Methods Related to Dierences Between Means . . . . 6.3 Methods Related to Proportions . . . . . . . . . . . . 6.4 Methods Related to Independent Samples Risk Ratios 6.5 Independent Samples Odds Ratios . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Multi-Sample Methods 7.1 Introduction . . . . . . . . . . . . . . . . . . . 7.2 The One-Way Analysis of Variance (ANOVA) 7.3 The 2 By k Chi-Square Test . . . . . . . . . . 7.4 Multiple Comparison Procedures . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . 8 The Assessment of Relationships 8.1 Introduction . . . . . . . . . . . . . . . . . 8.2 The Pearson Product-Moment Correlation 8.3 The Chi-Square Test for Independence . . Exercises . . . . . . . . . . . . . . . . . . . . . 9 Linear Regression 9.1 Introduction . . . . . . . . . 9.2 Simple Linear Regression . 9.3 Multiple Linear Regression Exercises . . . . . . . . . . . . .
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10 Methods Based on the Permutation Principle 75 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 10.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 A Table Relating Text Items to Tasks 79
Chapter 1
Preliminaries
1.1 Introduction
The hand calculations you perform in conjunction with your studies of the statistical concepts in Biostatistics For The Health Sciences by R. Cliord Blair and Richard A. Taylor1 are designed to enhance your understanding of the concepts under study. Modern statistical practice only rarely resorts to such methods for the analysis of data collected for research purposes. This is because (1) data sets collected in applied research contexts are often too large for hand calculations and (2) such calculations are prone to errors. Instead, researchers resort to tried and true statistical software that can be relied upon to provide quick, accurate results. The MINITAB system is one of the most popular of these. In this manual, you will learn to use MINITAB to analyze data associated with the problems outlined in the text. You will learn some basic MINITAB concepts in this chapter. Chapters 2 through 10 will address problems in the like numbered chapters in the text.
1.2
Getting Started
Before starting MINITAB, you should make yourself a directory in which to store your work. Ive called my directory bhs dir. You should then copy the les that accompany this manual into your directory. To start MINITAB you can either double click the MINITAB shortcut icon on the desktop or select the sequence StartProgramsMINITAB 14. The notation StartProgramsMINITAB 14 means that you are to rst select the Start menu, then Programs, and nally MINITAB 14. We will use this notation throughout this manual.
1 Henceforth
referred to as the text.
CHAPTER 1. PRELIMINARIES
Figure 1.1: Screen appearance when MINITAB is started.
Unless some of the default settings for MINITAB have been altered, the windows shown in Figure 1.1 will appear. These are but two of several windows utilized by MINITAB. You can move between these two windows by simply clicking on the desired window. Some of the commands available to you will depend on which window you are currently addressing. The upper panel in Figure 1.1 is termed the Session Window and we will have more to say about it later. Before doing so, lets examine the lower panel which is called the Data Window.
1.3
Inputting Data
Logically, before you can analyze data, you must input it to the analysis system. This is done by either typing the data into a worksheet in the Data Window or loading a preexisting worksheet into the Data Window. Worksheets contain the data and may be viewed in the Data Window or stored in a directory. As you can see, the worksheet is similar to a spreadsheet and has many, though not all, of the characteristics of a spreadsheet. The rows of the worksheet are numbered 1,2, 3, etc. while the columns are designated C1, C2, C3 etc. The intersection of a row and column is termed a cell. We will use the notation RiCj to designate the cell dened by the intersection of the ith row and jth column. Thus, R2C3
1.3. INPUTTING DATA
Figure 1.2: Screen appearance when data from Table 2.6 are entered.
is the cell dened by the intersection of the second row and third column. Lets use the data in Table 2.6 on page 34 of the text to demonstrate data entry. To begin, click cell R1C1. Notice that a dark frame appears around this cell. The frame indicates the cell in which data will be entered when you type. If you click on some other cell, the frame will appear around that cell. Now type the number 3,2 which is the rst value in Table 2.6, in cell R1C1. Because variables are entered columnwise, we will type the second entry, also a 3, in R2C1. To do this, you can click this cell and type the entry. You can also use the four keyboard direction keys to move from cell to cell. Alternatively, we can use an automatic feature that allows the frame to advance to the next cell for data entry each time we press Enter. To use this feature, notice the small arrow at the left of the C1 designation. This arrow, called the data-direction arrow, will move the frame one cell downward when it points downward or one cell to the right when it points to the right. To change the direction in which the arrow points, simply click it. Now enter the remaining values from the rst column of Table 2.6. It is a good idea to name the variables in your worksheet. We will name our variable x because that was the only designation used in table 2.6 but we would usually want to use a more informative name such as blood pressure or age. To enter the variable name, click on the empty cell just below the C1 designation. Type x in this cell. Your Data Window should now appear as shown in Figure 1.2. It is wise to save ones work at frequent intervals so lets save our data at this point. To do this, we choose FileSave Current Worksheet.3 Lets name this worksheet table 2.6 data. Be sure you are saving into the directory you created earlier for this purpose (bhs dir in my case). This worksheet is now available to us at any time we choose to load it into the Data Window. To see that this is true, close the worksheet by selecting FileClose Worksheet. The table 2.6 data worksheet is replaced by a new blank worksheet. We can load table 2.6 data back into the Data Window by selecting File
2 We 3 Do
will use italics to indicate things you are to type or choose in windows. not use the save button for this purpose.
Open Worksheet..., and choosing table 2.6 data.MTW from the le list. Note that MINITAB has appended the sux MTW to the le. Now click Open and table 2.6 data.MTW reappears in the Data Window.
1.4
Performing Analyses
Now that we have entered our data, we can use MINITAB to perform analyses on these data. We will provide a simple example in this section but will give numerous examples in the chapters that follow. MINITAB uses two methods for performing analyses and other tasks. The rst is menu driven while the second makes use of Session commands. We will cover only the use of menus for analyses in this manual. To learn the Session command method, consult the MINITAB Handbook (see footnote 1 on page i). Before beginning the analysis, be sure your data are displayed. If they are not displayed,4 select FileOpen Worksheet..., navigate to the directory in which you saved table 2.6 data and open it. Lets begin by calculating a few descriptive statistics on the variable weve labeled x. To do this, select StatBasic StatisticsDisplay Descriptive Statistics.... A window opens. You can determine which statistics will be displayed by selecting Statistics.... If you select this option, a window opens listing all the statistics that may be displayed with this command. Those that are checked will be displayed. Close this window by selecting OK or Cancel. Though you only have one variable (x) in this data set, you will often have more. For this reason, you must tell MINITAB which variable(s) you wish to enter into the analysis. You can de this by double-clicking on x in the left hand window or typing x into the Variables: window. You can also enter the column number (C1 in this case). It is important to remember that names containing a space must be enclosed in single quotes. MINITAB will put these quotes around the variable name for you if you use the double-click method for entering the variable name into the Variables: window. Your screen appears as in Figure 1.3. Now select OK to perform the analysis. Because we didnt change any of the statistics in the Statistics... window, results are displayed in the Session Window as shown in Figure 1.4. Displayed there are (a) the variable name (x), (b) the number of variable observations (11), (c) the number of missing variable observations (0), (d) the mean of x (4.000), (e) the standard error of the mean (0.234), (f) the standard deviation of x (0.775), (g) the minimum value of x (3.000), (h) the rst quartile (3.000), (i) the median (4.000), (j) the third quartile (5.000) and, (k) the maximum value of x (5.000). You will learn about each of these in your text. If other statistics were checked in the Statistics... window you will have these statistics displayed.
4 Your data will not be displayed if youve ended the MINITAB session or closed the worksheet.
1.4. PERFORMING ANALYSES
Figure 1.3: Screen appearance for display of basic statistics.
Figure 1.4: Display of basic statistics from x variable in text Table 2.6.
Figure 1.5: Edited output of basic statistics from x variable in text Table 2.6.
1.5
Editing and Printing Your Output
Once you have produced output in the Session Window, you will likely want to edit and/or print it. For example, you may wish to add your name, date and a class assignment number. You may then want to print out the result to turn in to your instructor. To begin, be sure your editor is set to allow editing of output. To do this, click anywhere in the Session Window. (You can easily move from the Data Window to the Session Window and visa versa by simply clicking in the desired window.) Then select Editor. If, in the list of options, Output Editable has a check mark by it, the output is already amenable for editing and you need do nothing more. If not, click by this option to place a check mark there. You can now carry out routine editing tasks such as typing, highlighting, cutting, copying, pasting etc. You can also cut and copy portions of the output to r Microsoft Word or other word processors if you so desire. If you wish to print portions of the output, simply highlight the part you want to print and use the usual printing commands. Figure 1.5 shows our edited output.
1.6
Using the Calculator
From time to time you may wish to use the variables in your data set to create new variables. This can be done through use of the calculator provided by MINITAB. To demonstrate its use, lets use it to create the second column in Table 2.6 which, as you can see, is the square of the rst column. To begin, be sure that table 2.6 data.MTW is the current worksheet. Then select CalcCalculator.... The Calculator window opens. In the Store result in variable: window type x square. This will be the name of the new variable which will be placed in the next available column which is C2 in this case. In the Expression: window type x**2. The symbol ** means raise to the power of. Had we typed x**3 the new variable would be the cube of x. Thus, what we have typed in this case is equivalent to x square=x**2. Figure 1.6 shows the calculator screen used for the calculation. Now click OK
1.6. USING THE CALCULATOR
Figure 1.6: Calculator screen used to create the square of x.
to perform the calculation. Notice that the worksheet now contains the variable x square in C2. We can save our new data set by selecting FileSave Current Worksheet. As a second example, lets generate the values in the column headed x x in Table 2.6. These values are termed deviation scores and will be explained in text Chapter 2. To calculate these deviation scores, we must subtract the average (i.e. mean) of x from each value of x. We previously found that the average of x was 4.000 so we could write an expression to subtract this value from each of the x scores. But lets do this in a slightly more sophisticated way. We will rst calculate the mean of x by selecting CalcColumn Statistics.... A window opens. Choose the button next to mean from the list of statistics. In the Input variable: window type x. This tells MINITAB that we are calculating the statistic on the x variable. In the Store result in: window type xmean. This tells MINITAB to name the mean value xmean. Your screen should appear as shown in Figure 1.7. Now click OK. Notice that the mean of x is displayed in the Session window. Lets save our work by selecting FileSave Current Worksheet. We can now use xmean in the calculator to compute deviation scores. Select CalcCalculator.... Notice that in addition to C1 and C2, the calculator window now shows K1 with the associated name of xmean. MINITAB refers to columns of data as C1, C2, etc. but refers to a single value (a constant) such as the mean of x as K1, K2, etc. Lets name the new variable deviation by typing this in the Store result in variable: window. Now type x-xmean in the Expression window and click OK. C3 now contains the deviation scores
Figure 1.7: Calculating the mean of x.
as shown in Table 2.6. Save the worksheet.
1.7
Graphs
MINITAB can be used to create a wide variety of graph types ranging from the simple to the complex. You will employ some of these capabilities in Chapter 2. For the moment, we will provide only a simple example of one graph type by constructing a histogram of the x variable in Table 2.6. To begin, be sure that table 2.6 data.MTW is the current worksheet. Now select GraphHistogram.... Four windows appear. Click on the rst one labeled simple to be sure it is darkened then select OK. A window labeled Histogram-Simple opens. Either double-click on x or type it into the Graph variables: window and select OK. A histogram of x appears as shown in Figure 1.8. Close the window without saving changes.
1.7. GRAPHS
Figure 1.8: Histogram of x.
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Figure 1.9: Reconstruction of Table 2.6.
Exercises
1.1 Use the Calculator... to construct the remaining variables in Table 2.6. The result should appear as in Figure 1.9. (Hint: double-clicking on a function in the Functions: window places it in the Expression: window. 1.2 Use Display Descriptive Statistics... to nd the means of all the variables in Table 2.6. Dont generate any other statistics. Use Display Descriptive Statistics... only once. (Hint: You can place multiple variables in the Variables window by double-clicking on each. 1.3 Label the output from Exercise 1.2 with your name and the current date. Print out the result. 1.4 Enter the data from text Table 7.1 into a new worksheet. Save the worksheet with the name table 7.1 data. 1.5 Find the means of the three diet groups in Table 7.1. Label them mean one, mean two and mean three. Save them in the table 7.1 data worksheet. (Dont forget that names containing spaces must be enclosed in single quotes.)
Chapter 2
Descriptive Methods
2.1 Introduction
In this chapter you will learn to use MINITAB to construct the graphs and calculate most of the statistics presented in text Chapter 2. You will also learn that in a few instances MINITAB and the text do not agree on the calculated values of certain descriptive statistics because dierent denitions of these statistics are used by the software and text. You will also be aorded an opportunity to practice data entry and data transformations.
2.4
Distributions (page 15)
Task 2.1
Use the data in Table 2.1 (table 2.1 data.MTW) to construct frequency, relative frequency, cumulative frequency, and cumulative relative frequency distributions as shown in Table 2.2.
Solution
Load the worksheet containing the Table 2.1 data by selecting FileOpen Worksheet... and opening le table 2.1 data.MTW. Note rst that the Pain Level variable is expressed in text rather than numeric form. Because the cumulative frequency and cumulative relative frequency distributions require that the Pain Level variable be ordered, we must inform MINITAB as to the low to high ordering of this variable before we can form these distributions. To do this, right-click on the Pain Level variable in the worksheet. A series of choices is displayed. Select ColumnValue Order. A window opens. Select the option button for User specied Order:. New Order is automatically highlighted in the Choose an order: window. The text values appear in a window labeled Dene an order (one value per line):. Place these values in 11
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CHAPTER 2. DESCRIPTIVE METHODS
Figure 2.1: Screen appearance for ordering pain level variable.
Figure 2.2: Frequency, cumulative frequency, relative frequency and cumulative relative frequency of perceived pain scores.
order from lowest to highest by editing the window entries. The screen should appear as shown in Figure 2.1. Now click Add Order and then OK. Now MINITAB understands, at least in terms of order of magnitude, the meaning of none, mild, moderate and severe and will use this order when creating graphs and tables involving this variable. Save the worksheet. To create the desired distributions select StatTablesTally Individual Variables.... Now double-click on the Pain Level variable in the Tally Individual Variables window to place this variable in the Variables: window. Now check all four boxes in the Display list and click the OK button. The output is displayed in Figure 2.2. Several points regarding this output should be noted. First, what your text refers to as Frequency, Relative Frequency, Cumulative Frequency and Cumulative Relative Frequency are labeled Count, Cumulative Count, Percent and Cumulative Percent by MINITAB. This is not unusual as many terms and styles of presentation are commonly used by both textbook authors and software developers. This may present some diculties, but they tend to diminish as you
2.4. DISTRIBUTIONS
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Figure 2.3: Simple frequency distribution of BP scores as shown in Table 2.3.
gain more experience with dierent software packages. Second, with the exception that MINITAB uses percents where the text uses proportions and the number of decimal places reported, the distributions provided by MINITAB are the same as those in text Table 2.2. Third, MINITAB orders the distributions in descending order while the text uses an ascending order. The style used by MINITAB is the more common. The text authors prefer the ascending ordering but either is correct and provides the same information.
Task 2.2
Use the data in worksheet table 2.3 data.MTW to construct the simple frequency distribution depicted in the text Table 2.3.
Solution
Because the blood pressure data in worksheet table 2.3 data.MTW are numeric as opposed to text, we need not go through the process of dening the variable order as was necessary with Task 2.1. Thus, we can construct the simple frequency distribution without any preliminaries. After loading table 2.3 data.MTW we select StatTablesTally Individual Variables..., then double-click on BP to place it in the Variables: window. Place a check by Counts in the Display window and remove any checks by any other options in this list. Now select OK to produce the frequency distribution. The resultant frequency distribution is shown in Figure 2.3. Note that I have edited the output into three columns to save page space.
Task 2.3
Use the data in worksheet table 2.3 data.MTW to construct the grouped distributions depicted in the text Table 2.4.
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Figure 2.4: Partial conversion table used to relate BP values to group intervals.
Solution
Unfortunately, MINITAB does not provide internal methods for constructing grouped distributions. So, in order to complete this task, we must rst code the numeric values in table 2.3 data.MTW to form the discrete intervals shown in text Table 2.4. Because these categories are text based (even though they contain numbers) we will follow the same procedure used for Task 2.1 to order the text categories. We will then be able to construct the desired distributions as with previous tasks. To begin the coding, we construct a conversion table that relates all BP values to specic category intervals. We will use C2 and C3 in our worksheet for this purpose. For convenience, lets label C2 as x and C3 as y. We now list all possible values of BP in x and the categories to which they relate in y. The (partial) result is shown in Figure 2.4. We are now ready to use the conversion table to create the Interval variable. To accomplish this, select DataCodeUse Conversion Table.... The Code-Use Conversion Table window opens. Enter BP in the Input column: window. This tells MINITAB that it is the BP values that are to be converted into intervals. Now type Interval in the Output column: window. This names the new variable Interval. Now type x in the Column of Original Values: window and y in the Column of New Values: window. This tells MINITAB the values of BP that correspond to the specied intervals. Now click OK. A new variable, named Interval, consisting of the specied intervals is created and placed in column C4 of the worksheet. Notice that each BP value is related to its corresponding interval.
2.5. GRAPHS
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Figure 2.5: Screen appearance for ordering Interval variable.
Weve now reached the point where we began Task 2.1. That is, we must order the categories and construct the desired distributions. To do this, rightclick on the Interval variable in the worksheet. A series of choices is displayed. Select ColumnValue Order. A window opens. Select the option button for User specied Order:. New Order is automatically highlighted in the Choose an order: window. The text values appear in a window labeled Dene an order (one value per line):. Place these values in order from lowest to highest by editing the window entries as shown in Figure 2.5. Now click Add Order and OK. MINITAB now understands that 85-89 is a lesser category than 90-94 which is less than 95-99 etc. Save the worksheet. We are, now ready to construct the desired distributions. To create the distributions, select StatTablesTally Individual Variables.... Now double-click on the Interval variable in the Tally Individual Variables window to place this variable in the Variables: window. Check all four boxes in the Display list and click the OK button. The output is displayed in Figure 2.6. Save the worksheet, we will use it again later.
2.5
Graphs (page 19)
Task 2.4
Use the data in Table 2.1 (table 2.1 data.MTW) to construct the relative frequency bar graph depicted in text Figure 2.1.
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Figure 2.6: Grouped frequency distributions for BP variable in 12 categories.
Figure 2.7: Bar graph of pain level scores.
Solution
Open table 2.1 data.MTW as the active worksheet. Select GraphBar Chart.... If the bar chart in the window labeled Simple is not selected, click on that window to select it and click OK. The Bar Chart-Counts of unique values, Simple window opens. Double-click on Pain Level to place it in the Categorical variables: window. Because the task specied that a relative frequency bar graph is to be constructed, we must inform MINITAB that the bars are to be shown in terms of proportions (or percents in this case). To do this, select Bar Chart Options... and place a check in the Show Y as percent box and click OK. You are now back at the Bar Chart-Counts of unique values, Simple window. Click OK. The graph appears as shown in Figure 2.7. Print this graph by selecting FilePrint Graph. MINITAB allows graphs to be edited if so desired. As one example, suppose we wish the title of the graph to be Bar Chart of Pain Level rather than
2.5. GRAPHS
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Figure 2.8: Informing MINITAB that a relative frequency histogram is to be constructed.
simply Chart of Pain Level. Just double-click on Chart of Pain Level at the top of the graph. The Edit Title window opens. Be sure the Font tab is selected. Now type in the desired title in the Text: window and click OK. The graph appears with the new title. You should experiment a bit to see what other alterations you can make to the graph. Notice that the pain levels are ordered. How do you think MINITAB knew the order of this variable? (Hint: go back and look at Task 2.1.)
Task 2.5
Use the data in Table 2.3 (table 2.3 data.MTW) to construct the relative frequency histogram depicted in text Figure 2.2.
Solution
Be sure table 2.3 data.MTW is the active worksheet. Then select Graph Histogram.... The Histograms window opens. Select the Simple histogram by clicking on the depiction in the Simple window. Click on OK. In the Histogram-Simple window, double-click on BP to move this variable into the Graph variables: window. If we wished to construct a (default) frequency histogram we would click OK at this point and continue, but because we wish to form a relative frequency histogram we must inform MINITAB that this is the case. To do so, select Scale... in the Histogram-Simple window. Then choose the Y-Scale Type tab and then click the Percent option. Your screen should appear as in Figure 2.8. Now click OK then click OK again. A relative frequency1 histogram appears but it is not grouped as in text Figure 2.2.
1 Once again, MINITAB expresses the relative frequency as a percent rather than as a proportion as is done in the text. This will generally be the case so we will not continue to
18
Figure 2.9: Specifying intervals for a relative frequency histogram.
To form the appropriate grouping, double-click anywhere along the horizontal (i.e. x) axis of the histogram. The Edit Scale window opens. Choose the Scale tab and darken the Position of ticks: button. Remove any entries in the associated window and enter 84.5:144.5/5. Click the Binning tab. Choose Cutpoint in the Interval Type list. This tells MINITAB to label the interval rather than the midpoint for each bar. Now choose Midpoint/Cutpoint positions. In the accompanying window you could type 84.5 89.5 94.5 etc. to label the graph but MINITAB provides a shortcut that will be easier in this case. Type 84.5:144.5/5 in the window. This provides MINITAB with the range of values for the x axis (84.5 to 144.5) as well as the interval size (5). The screen appears as in Figure 2.9. Click OK and the histogram depicted in Figure 2.10 appears.
Task 2.6
Use the data in Table 2.3 (table 2.3 data.MTW) to construct the relative frequency polygon depicted in text Figure 2.4.
Solution
This task can be approached in essentially the same manner as was used with Task 2.5 with the major dierence being that we will ask MINITAB to provide connected dots rather than bars in the depiction. To this end, select GraphHistogram.... Choose the Simple histogram and click OK. In the Histogram-Simple window double-click on the BP variable to place it in the Graph-variables: window if it isnt there already. Now select Scale... and in the Histogram-Scale window choose the Y-Scale Type tab. Under this tab darken the Percent button so that MINITAB will plot percents rather than frequencies. Click OK. You are now back in the Histogram-Simple window.
remind you of this fact.
2.5. GRAPHS
19
Figure 2.10: MINITAB rendering of the histogram depicted in text Figure 2.2.
From this window select Data View.... Under the Data Display tab place a check by Symbols and uncheck any other checks. Now select the Smoother tab. Choose the Lowess button in the Smoother list and set the Degree of smoothing to 0 (zero). This causes MINITAB to connect the dots with line segments. Click OK then click OK again. A relative frequency polygon appears. But it does not reect the twelve groupings. Double-click anywhere along the x axis. The Edit Scale window opens. Under the Scale tab, choose the Position of ticks: button. In the accompanying window type 84.5:144.5/5. This tells MINITAB to place the rst tick mark at 84.5, the last at 144.5 and every ve units in between. Now choose the Binning tab. Be sure that Midpoint is selected under Interval Type so that the polygon dots will be placed at the midpoints of the intervals. Now choose the Midpoint/Cutpoint positions: button. We now enter the midpoints of our desired intervals as 82:147/5 and click OK. The desired polygon appears. Lets edit the graph just a bit. Double-click on the Histogram of BP title at the top of the graph. In the Text: box type Relative Frequency Polygon of Blood Pressures in 12 Categories and click OK. The graph appears with the new title. Now double-click on the BP designation at the bottom of the graph. Type Systolic Blood Pressure in the Text: box and click OK. The relative frequency polygon appears as in Figure 2.11.
Task 2.7
Use the data in Table 2.3 (table 2.3 data.MTW) to construct the cumulative relative frequency polygon depicted in text Figure 2.6.
20
Figure 2.11: MINITAB rendering of the polygon depicted in text Figure 2.4.
Solution
In formulating a solution for Task 2.3, we used a conversion table to assign subjects to specied intervals. We will use this technique again here to group subjects into the 12 intervals depicted in Figure 2.6. More specically, we will assign the value 84.5 to all subjects with blood pressures of 85, 86, 87, 88 or 89, a value of 89.5 to all subjects with pressures of 90, 91, 92, 93 or 94 and so on with the ve highest scale values being assigned 139.5. You will see the underlying rationale shortly. The conversion table employed for the solution of Task 2.3 consisted of two variables which we named x and y. The rst of these listed all possible blood pressure values while the second indicated the interval to which each of these should be assigned. We will use the x variable again here but we will replace the y variable with a new set of intervals. Be sure that table 2.3 data.MTW is the current worksheet. If you saved it at the completion of Task 2.3, it should appear as in Figure 2.4 except that the variable named Interval has been added in C4-T. Lets clean up this worksheet by deleting the y and Interval variables. To do this, click and hold down the mouse button on C3-T and drag across to C4-T. This will select (darken) these two columns. Now select EditDelete Cells. The C3 and C4 columns are now empty. To construct the new conversion table begin by labeling the C3 column y. Now enter the interval values in y as shown in Figure 2.12. This table indicates that the blood pressures listed in x will be converted to the values specied in y. Now lets perform the conversion. Select DataCodeUse Conversion Table.... The Code-Use Conversion Table window opens. Double-click on BP to place it in the Input column: window. This tells MINITAB that it is the blood pressure values that
2.5. GRAPHS
21
Figure 2.12: Specifying intervals for a cumulative relative frequency polygon.
are being coded. Now type Interval in the Output column: window. This tells MINITAB to name the new variable Interval. Place an x in the Column of Original Values: window and a y in the Column of New Values: window. This tells MINITAB which x values correspond to which y values. Click OK and the Interval variable appears in C4. Our strategy now involves calculating the cumulative relative frequency for the Interval variable. To this end, select StatTablesTally Individual Variables.... In the Tally Individual Variables window double-click on the Interval variable to place it in the Variables window. In the Display list put a check by Cumulative percents and uncheck any others that have a check by them. Click OK. The resultant distribution appears in the Session Window and is shown in Figure 2.13. We must now paste this result into the worksheet. To accomplish this, click and drag the cursor in the Session Window so as to select (darken) all of the output depicted in Figure 2.13. Now select EditCopy to copy the selected portion of the Session Window to the clipboard. Now click and hold down the mouse button on C5 in the worksheet and drag the cursor across to C6 so that both the C5 and C6 columns are selected (darkened). Select EditPaste Cells. Be sure that the Use spaces as delimiters button is selected and click OK. The cumulative relative frequency distribution now appears in columns C5 and C6 as shown in Figure 2.14.2 Lets now round o the CumPct values to the nearest integer. To do this, select CalcCalculator.... Double-click CumPct to place it in the Store result in variable: window. Place the cursor in the Expression: window and doubleclick on round in the Functions: window. (You will have to scroll down to nd this function.) This enters ROUND(number, num digits) in the Functions:
2 You
can paste this distribution in C1 and C2 of a new worksheet if you prefer.
22
Figure 2.13: Cumulative relative frequency distribution for blood pressure measures in 12 intervals.
Figure 2.14: Worksheet after pasting the cumulative relative frequency distribution into columns C5 and C6.
2.5. GRAPHS
23
Figure 2.15: MINITAB rendering of the cumulative relative frequency polygon depicted in text Figure 2.6.
window. Edit this expression so it appears as ROUND(CumPct,0). This tells MINITAB to round o the CumPct variable to the nearest whole number. Click OK and the CumPct values are rounded to the nearest whole number. We are now ready (at last) to create the graph. Choose GraphScatter Plot.... In the Scatterplots window choose the With Connect Line gure and click OK. Put CumPct in the Y variables window and Interval 1 in the X variables window and click OK. The cumulative distribution is presented on the screen. In order to have the x axis scale conform with the scale in Figure 2.6, doubleclick on the x axis of the graph. In the Edit Scale wondow choose the Position of ticks: button. In the accompanying window type 84.5:144.5/5 which tells MINITAB to label the x axis from 84.5 to 144.5 in ve unit increments. Click OK. The graph appears with the x axis properly scaled. Finally, lets bring the y scale down so that zero will be closer to the x axis as in Figure 2.6. Double-click on the y axis. In the Edit Scale window be sure that the Scale tab is selected. Under Scale Range uncheck Minimum: and type .1 in the accompanying window. This puts the bottom of the y scale near zero. Now click OK and the graph depicted in Figure 2.15 appears. Feel free to edit the graph titles to make them more meaningful.
Task 2.8
Construct a stem-and-leaf-display for the blood pressure data in text Table 2.3.
Solution
Be sure that table 2.3 data.MTW is the current worksheet. Select GraphStemand-leaf.... Double-click on BP to place it in the Graph variables... window and click OK. A stem-and-leaf-display appears in the Session Window as shown
24
Figure 2.16: MINITAB rendering of a stem-and-leaf display for the BP variable in text Table 2.3.
Figure 2.17: Edited MINITAB rendering of a stem-and-leaf display for the BP variable in text Table 2.3.
in Figure 2.16. Note that MINITAB divides leaves in half so that digits 0 through 4 are placed on one line and 5 through 9 are placed on a second line. Also, the numbers on the far left are what MINITAB refers to as the depth of the stem. This column gives the cumulative frequency from the top of the gure down and from the bottom of the gure up. The number in parentheses gives the leaf count for the row containing the median. Unlike other graphs produced by MINITAB, stem-and-leaf-displays are not provided in special graph windows but are printed in the Session Window. This means that they can be edited with ordinary cut and paste methods. Figure 2.17 shows the display after weve edited it to more closely conform with text Figure 2.7.
2.6
Numerical Methods (page 23)
Task 2.9
In text Example 2.4 on page 27 you are asked to nd the median of the numbers 3, 2, 0, 2, 1, 1, 2, and 2. Two dierent methods are used that produce two dierent answers, namely, 2.00 and 1.75. It is argued that the better answer is provided by the method that produces the median value of 1.75. What answer is provided by MINITAB?
2.6. NUMERICAL METHODS

Solution
25
We begin by entering the numbers in C1 of a new worksheet. Name the variable x. Select StatBasic StatisticsDisplay Descriptive Statistics... Double-click on x to place it in the Variables... window. Now click on Statistics.... Place a check by Median if there isnt one there already. Uncheck all other statistics and click OK then click OK again. The answer provided in the Session Window is 2.00.
Task 2.10
Find the mean, median, and standard deviation of the data in the table associated with Example 2.7 on page 30. Is the median reported by MINITAB the same as that reported in the text? Why do you think this is the case?
Solution
We will have to enter this data into a worksheet since none is provided. This process can be somewhat simplied when data are given in the form of a simple frequency distribution as in this case. Open a new worksheet, if necessary, by selecting FileNew.... Be sure MINITAB Worksheet is highlighted and click OK. From the table on text page 30 we notice that there are three 2.4s in the table. To enter these select CalcMake Patterned DataArbitrary Set of Numbers.... In the Store patterned data in: window type C1 to store the 2.4s in C1. Type 2.4 in the Arbitrary set of numbers: window and 3 in the List each value <blank> times window. There should be a 1 in the List the whole sequence <blank> times window. Thus, we have informed MINITAB to place three 2.4 values in C1. Now click OK. Three values of 2.4 appear in C1. The next value in the table is 2.3 and there are 40 of those so we will place 40 2.3s in C2. To enter these select CalcMake Patterned DataArbitrary Set of Numbers.... In the Store patterned data in: window type C2. Type 2.3 in the Arbitrary set of numbers: window and 40 in the List each value <blank> times window. There should still be a 1 in the List the whole sequence <blank> times window. Now click OK. C2 now contains 40 2.3s. We continue this process by placing 36 2.2s in C3, 21 2.1s in C4, 18 2.0s in C5, 8 1.9s in C6, 14 1.8s in C7 and 5 1.7s in C8. The rst 18 rows of your worksheet should appear as in Figure 2.18. In order to perform analyses, we must form these eight columns into a single column variable. To do this, select DataStackColumns.... Now doubleclick C1 through C8 in the Stack Columns window in order to enter these columns in the Stack the following columns: window. Now choose the Column of current worksheet button and enter C1 in the accompanying window. This set of instructions tells MINITAB to stack the values in C1 through C8 in C1. Now click OK. The values in C1 through C8 are now entered into C1. Lets clean up the worksheet by deleting C2 through C8 which we no longer need. To accomplish this, click on the C2 heading in column C2
26
Figure 2.18: Data from example 2.7.
and drag across to C8. This selects (darkens) columns C2 through C8. Now select EditDelete Cells and these columns disappear. The rst 18 rows of your worksheet should now appear as in Figure 2.19. To complete the task of nding the mean, median, and standard deviation of these data, select StatBasic StatisticsDisplay Descriptive Statistics.... Double-click on C1 to place it in the variables: window. Click on the Statistics... bar to open the list of statistics to be calculated. Be sure that only Mean, Standard deviation and Median have check marks. Click OK then click OK again. The output is shown in Figure 2.20. As may be seen, the median reported here is 2.2 which contrasts with the value of 2.17 reported in the text as the answer for Example 2.7. Thus, it appears that MINITAB calculates the median as the middle score (or average of the two middle scores) rather than by the more sophisticated method given by Equation 2.5.
Task 2.11
Find the variance of the x variable in Table 2.6 (worksheet Table 2.6 data.MTW). Is the result provided by MINITAB the same as that given in the text for example 2.15 on page 36?
Solution
After loading worksheet Table 2.6 data.MTW, select StatBasic StatisticsDisplay Descriptive Statistics.... Double-click on x in the Display Descriptive Statistics window in order to place it in the Variables: window. Now click on the Statistics... bar to open the Descriptive Statistics-Statistics window. Put a check by Variance and uncheck all others. Click OK then click OK again. The result, .60, which is the answer given in the text, is displayed in the Session Window.
27
Figure 2.19: Data from example 2.7 in a single column.
Figure 2.20: Descriptive statistics for data from example 2.7.
28
Task 2.12
Find the twenty-fth, sixtieth, and seventy-fth percentiles of the blood pressure data in worksheet table 2.3 data.MTW. Do the answers provided by MINITAB match those given in the solution to text Example 2.17 on page 38?
Solution
MINITAB does not provide a general method for nding percentiles.3 We can nd the twenty-fth, and seventy-fth percentiles, however, because MINITAB gives the 1st and 3rd quartiles which, by denition, are the twenty-fth and seventy-fth percentiles. To obtain these, select StatBasic StatisticsDisplay Descriptive Statistics.... Double-click on BP in the Display Descriptive Statistics window in order to place it in the Variables: window. Now click on the Statistics... bar to open the Descriptive Statistics-Statistics window. Put a check by First quartile and Third quartile. Uncheck all other statistics. Click OK then click OK again. The result is displayed in the Session Window. The rst and third quartiles (twenty-fth and seventy-fth percentiles) are given as 115 and 135 respectively. The values given in the solution to Example 2.17 in the text are 114.67 and 134.63. The discrepancy is likely because, as with the median (ftieth percentile), MINITAB employs a simpler method for nding percentiles than the one given in the text.
Task 2.13
Use MINITAB to perform the analyses outlined in Example 2.21 on page 43.
Solution
Open a new worksheet. Type the numbers 1, 3, 3, 9 in the rst column. Name this variable x. Select CalcStandardize.... Enter x in the Input columns(s): window. Type C2 in the Store results in: window. Be sure that the Subtract mean and divide by std. dev. button is selected and click OK. The z scores appear in C2. Notice that these z scores agree with those given in the text except that MINITAB gives the z score for nine as 1.44338 while the text gives 1.445. The discrepancy occurs because the text uses 3.46 as the standard deviation while MINITAB carries this calculation out to more decimal places. Now use StatBasic StatisticsDisplay Descriptive Statistics... to nd the mean and standard deviation of the z scores in C2. As expected, MINITAB gives these values as zero and one respectively.
Task 2.14
Use MINITAB to conduct the analyses alluded to in Example 2.23 on page 44.
3 Percentiles can be found by use of macros but that is a topic beyond the scope of this manual.

Solution
29
Open the table 2.8 data.MWS worksheet. The data for distributions A, B and C are in columns with the like letter. Select StatBasic StatisticsDisplay Descriptive Statistics.... Double-click A, B and C to place them in the Variables: window. Click Statistics... and place a check by skewness and uncheck all other statistics. Click OK and click OK again. Skew for the three distributions is displayed in the Session Window. MINITAB reports skew for the three distributions as 0.00, 0.77 and 0.77 while the solution for Example 2.23 gives skew of zero, .648, and .648. Because there are dierent denitions of skew, it appears that MINITAB uses a dierent form than does the text. The denition provided by the text is probably more common.
Task 2.15
Use MINITAB to perform the analyses outlined in Example 2.24 on page 46.
Solution
Open the table 2.9 data worksheet. Follow the same steps outlined in the solution to Task 2.14 with Kurtosis being checked rather than Skewness in the Descriptive Statistics-Statistics window. The values obtained for distributions A and B respectively are 5.50 and 1.65 which diers markedly from the values of 5.04 and 1.26 given in the text. This again occurs because MINITAB used a dierent denition of kurtosis than that used in the text. The denition given in the text appears to be the more common.
30
Exercises
2.1 Use the data in worksheet table 2.3 data.MTW to construct the grouped distributions depicted in the text Table 2.5. 2.2 Construct the histogram depicted in text Figure 2.3. 2.3 Construct a relative frequency polygon similar to that depicted in Figure 2.4 but with eight instead of 12 categories. 2.4 Construct a cumulative relative frequency polygon similar to that depicted in Figure 2.6 but with eight instead of 12 categories. 2.5 Use MINITAB to nd the mean and median of the following data. Score 9 8 7 6 frequency 134 442 391 777
2.6 Calculate z scores for the BP variable in table 2.3 data.MTW. Find the mean, standard deviation and variance of the resultant z scores. 2.7 Use MINITAB to nd the skew of the BP data in table 2.3 data.MTW.
Chapter 3
Probability
3.1 Introduction
In this chapter you will learn to use MINITAB to form contingency tables as well as to nd areas under the normal curve.
3.3
Contingency Tables (page 52)
Task 3.1
Use worksheet table 3.1 data.MWS to construct text Table 3.1 located on page 54.
Solution
Notice that C1 of worksheet table 3.1 data.MWS characterizes each subject as being a smoker (S) or a non-smoker (SN) while C2 characterizes each subject as having some disease (D) or not having the disease (DN). In order to form these data into a contingency table, select StatTablesCross Tabulation and Chi-Square.... Use double-clicks to enter Smoking Status in the For rows: window and Disease Status in the For columns: window. Check Counts in the Display list and click OK. The contingency table, with cell entries as depicted in text Figure 3.1, appears in the Session Window.
3.4
The Normal Curve (page 62)
Task 3.2
Use MINITAB to nd the area of the normal curve indicated in Example 3.9 on page 64. 31
32
CHAPTER 3. PROBABILITY
Figure 3.1: MINITAB rendering of Table 3.1.
Figure 3.2: Screen appearance for nding the area below 220 of a normal curve with mean 250 and standard deviation 25.
Solution
MINITAB can nd the area below any point on a normal distribution. To nd the indicated area for this task, select CalcProbability DistributionsNormal.... The Normal Distribution window opens. Be sure the Cumulative probability button is selected. This tells MINITAB that the area below the specied value is to be calculated. Now enter the mean of 250 and standard deviation of 25 in the appropriate windows. Be sure the Input constant: button is selected and type 220 in the accompanying window. Your screen should appear as in Figure 3.2. Click OK and the area, 0.115070, appears in the Session Window. This agrees with the answer of .1151 given in the text.
Exercises
33
Exercises
3.1 Construct a worksheet to represent the data in the accompanying contingency table. Then use the worksheet to construct the contingency table. A A B 1 3 4 B 3 3 6 4 6
3.2 Use MINITAB in lieu of the normal curve table (text Appendix A) to nd the areas of the normal curve indicated in Examples 3.10, 3.11 and 3.12.
34
CHAPTER 3. PROBABILITY
Chapter 4
Introduction to Inference and One Sample Methods

4.1 Introduction
In this chapter you will use MINITAB to nd binomial probabilities as well as probabilities associated with the normal curve. You will also learn to perform various hypothesis tests, including equivalence tests, and form certain condence intervals. In addition, you will perform power calculations and compute required sample sizes to attain specied levels of power.
4.2
Sampling Distributions (page 75)
Task 4.1
Generate the binomial probabilities in Table 4.1 on page 83.

Solution
Enter the values 0 through 5 in C1 of a new worksheet. Then select CalcProbability DistributionsBinomial.... In the Binomial Distribution window select the Probability button. Type 5 in the Number of trials: window and .10 in the Probability of success: window. Select the Input column: button and type C1 in the accompanying window. This tells MINITAB the number of successes for which probabilities are to be calculated. If you wish to have the probabilities placed in C2, type C2 in the accompanying Optional storage: window or, if you prefer, leave this window blank and the probabilities will be printed in the Session Window. Now click OK and the probabilities, as shown in Figure 4.1, appear in the Session Window.1
1 Or
in C2 if you chose that option.
35
36
CHAPTER 4. INFERENCE AND ONE SAMPLE METHODS
Figure 4.1: MINITAB rendering of binomial probabilities in Table 4.1.
Task 4.2
The Z value calculated in connection with Example 4.7 on page 86 was 5.33. Use MINITAB to nd the associated probability.
Solution
Select CalcProbability DistributionsNormal.... In the Normal Distribution window be sure the Cumulative probability button is selected. This tells MINITAB to give us the proportion of the curve that lies below the indicated Z value. Now enter 0.0 and 1.0 for the mean and standard deviation respectively. This is because we are dealing with a distribution of Z scores (i.e. the standard normal curve) which has mean zero and standard deviation one. Select the Input constant: button and enter 5.33 in the accompanying window and click OK. The result appears in the Session Window as 0.0000000. While the result cannot be exactly zero, it is zero to seven decimal places.
4.3
Hypothesis Testing (page 86)
Task 4.3
Use MINITAB to perform the one mean Z test outlined in Example 4.9 on page 92.
Solution
Select StatBasic Statistics1-Sample Z.... In the 1-Sample Z (Test and Condence Interval) window select the Summarized data button. This tells MINITAB that you will supply the sample size and mean rather than having MINITAB calculate it from a data sample. In the Sample size: window type 150 and in the Mean: window type 128.2. In the Standard deviation: window type 40 then type 120 in the Test mean: window. We must now inform MINITAB as to the nature of the alternative hypothesis. To do this, select Options... and in the Alternative: window select greater than. Now click OK and click OK again. MINITAB displays obtained Z of 2.51 with
4.3. HYPOTHESIS TESTING
37
a one-tailed p-value of .006. Both numbers agree with those reported in the solution to Example 4.9. Normally, we would not bother performing the critical Z versus obtained Z version of the test but in the spirit of learning more about MINITAB we will next nd critical Z. To do this select CalcProbability DistributionsNormal.... Select the Inverse cumulative probability button in the Normal Distribution window. This tells MINITAB that you want a Z value dened by having a specied proportion of the normal curve lying below it. Because we are dealing with a one-tailed test with alternative of the form HA : > 120 and = .01, the desired Z value cuts o .01 in the upper tail of the curve which means that .99 of the curve lies below it. Choose the Input constant: button and enter .99 in the accompanying window. Be sure the Mean: and Standard deviation: are specied as 0.0 and 1.0 respectively and click OK. MINITAB gives critical Z (labeled X in the output) as 2.32635 which agrees with that given in the solution to Example 4.9.
Task 4.4
Use MINITAB to perform the one mean Z test outlined in Example 4.12 on page 97.
Solution
Select StatBasic Statistics1-Sample Z.... In the 1-Sample Z (Test and Condence Interval) window select the Summarized data button. This tells MINITAB that you will supply the sample size and mean rather than having MINITAB calculate it from a data sample. In the Sample size: window type 40 and in the Mean: window type 87.1. In the Standard deviation: window type 21 then type 80 in the Test mean: window. We must now inform MINITAB as to the nature of the alternative hypothesis. To do this, select Options... and in the Alternative: window select not equal which species a two-tailed test. Now click OK and click OK again. MINITAB displays obtained Z of 2.14 with a two-tailed p-value of 0.032. Both numbers agree with those reported in the solution to Example 4.12. To nd critical Z, select CalcProbability DistributionsNormal.... Select the Inverse cumulative probability button in the Normal Distribution window. This tells MINITAB that you want a Z value dened by having a specied proportion of the normal curve lying below it. Because we are dealing with a two-tailed test with = .10, one of the desired Z values cuts o .05 in the upper tail of the curve which means that .95 of the curve lies below it. Choose the Input constant: button and enter .95 in the accompanying window. Be sure the Mean: and Standard deviation: are specied as 0.0 and 1.0 respectively and click OK. MINITAB gives critical Z as 1.64485. Because the test is two-tailed we know the critical values are 1.64485 which, with the exception of the number of decimal places employed, agrees with those given in the solution to Example 4.12.2
2 Note
that 1.64485 rounds to 1.64 when rounded to two decimal places. However, the value
38
Task 4.5
Use MINITAB to perform the one mean t test outlined in Example 4.14 on page 104.
Solution
Enter the data from Example 4.14 into C1 of a new worksheet. Label this variable x. Now select StatBasic Statistics1-Sample t.... In the 1Sample t (Test and Condence Interval) window, be sure the Samples in columns: button is chosen. Type C1 in the accompanying window. Now type 8 in the Test mean: window and then click on Options.... Specify Alternative: as less than. Now click OK and click OK again. MINITAB reports obtained t as 3.13 which agrees with the calculation in the text. The associated p-value is given as 0.010. Normally, we would not bother performing the critical t versus obtained t version of the test but in the spirit of learning more about MINITAB we will next nd critical t. To do this select CalcProbability Distributionst.... Select the Inverse cumulative probability button in the t Distribution window. This tells MINITAB that you want a t value dened by having a specied proportion of the t curve lying below it. Because we are dealing with a one-tailed test with alternative of the form HA : < 8 and = .01, the desired t value cuts o .01 in the lower tail of the curve. Type 6 in the Degrees of freedom: window. Choose the Input constant: button and enter .01 in the accompanying window and click OK. MINITAB gives critical t as 3.14267 which agrees with that given in the solution to Example 4.14. Notice that this is a rare instance in which the decision appears to depend on whether you perform the p-value versus alpha version of the test or the obtained t versus critical t version of the test. Because the reported p-value is .01, you would reject on the basis of this criterion. However, MINITAB rounded this p-value to .01. The actual value is closer to 0.0102012 which would result in failure to reject the null hypothesis.
Task 4.6
Use MINITAB to perform the one sample test for a proportion outlined under the heading one-tailed test on page 108.
Solution
Select StatBasic Statistics1 Proportion.... Choose the Summarized data button and enter 10 for Number of trials: and 6 for Number of events:. Now click on Options... and enter .38 in the Test proportion: window. Select greater than in the Alternative: window and click OK twice. MINITAB reports an exact p-value of 0.135 which agrees with that reported in the text.
1.65 is used by convention because it is slightly more conservative.

Task 4.7
39
Use MINITAB to perform the one sample test for a proportion outlined in Example 4.19 on page 113.
Solution
Select StatBasic Statistics1 Proportion.... Choose the Summarized data button and enter 8 for Number of trials: and 2 for Number of events:. Now click on Options... and enter .35 in the Test proportion: window. Select not equal in the Alternative: window and click OK twice. MINITAB reports an exact p-value of 0.721 which does not agree with the value of .85562 reported in the text. This is because MINITAB uses a much more complicated method for nding the p-value for two-tailed tests based on the binomial distribution.
Task 4.8
Use MINITAB to carry out the two-tailed equivalence test outlined in Example 4.25 on page 120.
Solution
Begin by entering the sample data into C1 of a new worksheet. We can now use MINITAB to conduct two one-tailed t tests as is done in the text. However, there is a simpler way to establish two-tailed equivalence which we will also demonstrate.3 We begin by conducting the tests as was done in the text. To conduct Test One select StatBasic Statistics1 Sample t.... Be sure the Samples in columns: button is selected and type C1 in the accompanying window. Type 2 in the Test mean: window. Now click on the Options... button and choose less than in the Alternative: window. Click OK and click OK again. MINITAB reports obtained t in the Session Window as 3.64 which agrees with the result given in the text. The reported p-value is 0.003 which results in rejection of the null hypothesis. Test Two is carried out in the same manner with -2 replacing 2 in the Test mean: window and greater than replacing less than in the Alternative: window. Obtained t is reported as 2.18 which agrees with that given in the text. The p-value is given as 0.028 which results in rejection of the null hypothesis. Because both tests are signicant, the null hypothesis of non-equivalence is rejected in favor of the alternative of equivalence. A much simpler way of conducting a two-tailed equivalence test4 is by forming a 100 (1 2) percent condence interval and noting the relationship of L and U to EIL and EIU . If both L and U are between EIL and EIU , the
3 The authors of your text presented two-tailed equivalence testing in the manner they did because they believed that, while more tedious, this method led to a better understanding of the equivalence concept. 4 You may wish to delay learning this form of testing until youve studied Sections 4.4 and 4.5.
40
Figure 4.2: Condence interval used to perform two-tailed equivalence test.
two-tailed equivalence null hypothesis is rejected otherwise, it is not rejected. You should be able to determine why this is true after you study Section 4.4 and especially Section 4.5. Let us demonstrate this technique on the current example. To perform the shortcut two-tailed equivalence test, select StatBasic Statistics1 Sample t.... Be sure the Samples in columns: button is selected and type C1 in the accompanying window. Click the Options... button and choose not equal in the Alternative: window. Now type 90 in the Condence level: window. This gure is arrived at by noting that 100 (1 2) = 100 (1 (2) (.05)) = 90. Click OK and click OK again. The output is shown in Figure 4.2. Weve marked the lower (L) and upper (U) limits of the condence interval. Because L of 1.759686 is greater than or equal to EIL of 2 and U of 0.759686 is less than or equal to EIU of 2, the equivalence null hypothesis is rejected. Had either of these conditions not been met, the equivalence null hypothesis would not have been rejected.
Task 4.9
Use MINITAB to carry out the exact two-tailed equivalence test outlined in Example 4.26 on page 122.
Solution
We will rst use MINITAB to test the equivalence null hypothesis by conducting Test One and Test Two as is done in the text. We will then use a shortcut method for conducting this test based on an exact condence interval.5 To conduct Test One Select StatBasic Statistics1 Proportion.... Choose the Summarized data button and enter 8 for Number of trials: and 4 for Number of events:. Now click on Options... and enter .70 in the Test proportion: window. Select less than in the Alternative: window and click OK twice. MINITAB reports an exact p-value of 0.194 which agrees with the value reported in the text. Conduct Test Two in the same manner as was done for Test One with the
5 You
will learn how to construct such intervals when you reach the discussion on page 149.
41
exceptions that .30 is substituted for .70 in the Test proportion: window and greater than is substituted for less than in the Alternative: window. The reported p-value is again 0.194. Because both p-values are less than = .20, the equivalence null hypothesis is rejected. For the shortcut method, Select StatBasic Statistics1 Proportion.... Choose the Summarized data button and enter 8 for Number of trials: and 4 for Number of events:. Now click on Options... and enter 100 (1 2) = 100 (1 (2) (20)) = 60 in the Condence level: window. Now enter not equal in the Alternative: window and click OK twice. The reported condence interval is (0.303226, 0.696774). Because this interval is completely contained in the equivalence interval of .3 to .7, the two-tailed equivalence null hypothesis is rejected.
Task 4.10
Compute power for the situation described in Example 4.32.

Solution
Select StatPower and Sample Size1-Sample Z.... In the Power and Sample Size for 1-Sample Z window, type 25 in the Sample sizes: window and -2 in the Dierences: window. This is the dierence 0 . Be sure the Power values: window is blank. Type 20 in the Standard deviation: window then click the Options... button. Be sure the Not equal button is selected in the Alternative hypothesis list. Be sure the Signicance level: is set to .05 and click OK twice. MINITAB gives the power calculation as 0.0790975 which is slightly dierent from the value of .0721 calculated in the text. The discrepancy of .0790975.0721 = .0069975 can be explained by reference to Panel B of Figure 4.25 on page 134 of the text. In this gure power is depicted as the darkly shaded portion of the alternative curve that lies in the lower tail critical region of the null curve. MINITAB adds to this gure the portion of the alternative curve that lies in the right hand critical region of the null curve the reasoning being that this too would bring about rejection of the false null hypothesis albeit for the wrong reason. That is, while < 0 , rejection in the right hand tail would lead one to believe that > 0.6 We will leave calculation of the portion of the alternative curve that lies in the right hand critical region of the null curve as an exercise.
Task 4.11
Use MINITAB to nd the power alluded to in Example 4.34 on page 135.

Solution
We will begin by demonstrating a method that seems a natural way to proceed

6 Some authors refer to this as a Type III error while others simply treat it as part of the power calculation as does MINITAB.
42
but provides poor results and then show how better results can be achieved. To begin, select StatPower and Sample Size1 Proportion.... The Power and Sample Size for 1 Proportion window appears. Type 8 in the Sample sizes: window, .5 in the Alternative values of p: window and .35 in the Hypothesized p: window. Be sure the Power values: window is blank. Now click Options... and choose the Greater than button. Type .05 in the Signicance level: window and click OK twice. Power is shown in the Session Window to be 0.235589 which is dramatically dierent from the value of .14454 calculated in the text. The problem arises because MINITAB uses an approximate method based on the normal curve for its calculation while the calculation in the text is exact. MINITAB will give us a better approximation if we replace the .05 entry in the Signicance level: window with .02533 which is the exact level of signicance calculated in the text. When we do this, MINITAB gives the power estimate as 0.154862 which is much closer to the value of .14454 calculated in the text. The lesson to be learned here is that if we are using an exact test for a proportion, we should calculate and use the exact level of signicance rather than simply inputting the intended level.
Task 4.12
Use MINITAB to calculate the sample size described in Example 4.36 on page 136.
Solution
Select StatPower and Sample Size1-Sample Z.... The Power and Sample Size for 1-Sample Z window appears. Be sure the Sample sizes: window is blank. Type 2 in the Dierences: window. This is the dierence 0 . Now type .8 in the Power values: window and 4 in the Standard deviation: window. Click the Options... button and be sure the Not equal button is selected. Type .01 in the Signicance level: window and click OK twice. MINITAB prints the sample size of 47 in the Session Window which is the solution calculated in t he text.
4.4
Condence Intervals (page 137)
Task 4.13
Use MINITAB to construct the 95 percent condence interval outlined in Example 4.37 on page 143.
Solution
Select StatBasic Statistics1-Sample Z.... Select the Summarized data button in the 1-Sample Z (Test and Condence Interval) window. Type 60 in the Sample size: window, 90.1 in the Mean: window and 16 in the Standard deviation: window. Now click the Options... button. Be
4.4. CONFIDENCE INTERVALS
43
sure that the Condence level: is set to 95.0 and the Alternative: is set to not equal. Click OK twice. The interval, L = 86.0515 and U = 94.1485 are provided in the Session Window. Both of these values agree with the results provided in the text.
Task 4.14
Use MINITAB to construct the condence interval described in Example 4.41 on page 147.
Solution
Enter the data from Example 4.41 into C1 of a new worksheet. Select StatBasic Statistics1-Sample t.... Select the Samples in columns: button and type C1 in the accompanying window. Now click the Options... button and type 99.0 in the Condence level: window. Be sure not equal is selected in the Alternative: window and click OK twice. MINITAB gives the condence interval as (94.654, 120.146) which agrees with the result given in the text.
Task 4.15
Use MINITAB to form the exact and approximate condence intervals constructed in Example 4.42 on page 150.
Solution
Select StatBasic Statistics1 Proportion.... Select the Summarized data button. Enter 10 in the Number of trials: window and 4 in the number of event: window. Click the Options... button. Be sure that 95.0 is entered in the Condence level: window and not equal is selected for the Alternative: window and click OK twice. MINITAB displays the condence interval as (0.121552, 0.737622) which, given rounding, agrees with the exact computations carried out in the text. To obtain the approximate result, repeat the above steps but check the Use test and interval based on normal distribution box under the Options... button. MINITAB gives the interval (0.096364, 0.703636) which, again, agrees with the approximate interval given in the text. It is noteworthy that MINITAB provides the following admonition in association with this result. * NOTE * The normal approximation may be inaccurate for small samples.
44
Exercises
4.1 Use MINITAB to reproduce the binomial probabilities in Table 4.5 on text page 109 on page 143. 4.2 In Example 4.8 on page 86, the solution to the problem involves nding the area of the normal curve that lies above a Z value of 1.36. The solution gives this area as .0895. Use MINITAB and, if necessary a calculator, to nd this area. 4.3 Use MINITAB to perform the one mean Z test outlined in Example 4.11 on page 95 as well as to nd critical Z for the test. 4.4 Use MINITAB to perform the one mean Z test outlined in Example 4.13 on page 98 as well as to nd critical Z values for the test. 4.5 Use MINITAB to perform the one mean t test outlined in Example 4.16 on page 106 as well as to nd critical t for the test. 4.6 Use MINITAB to perform the test outlined in Example 4.18 on page 111. 4.7 Use MINITAB to perform the test outlined in Example 4.20 on page 114. 4.8 Use MINITAB to perform the following test of hypothesis using both the method outlined in the text (i.e. Test One and Test Two) and the shortcut (i.e. condence interval) method. H0E : .45 or .55 HAE : .45 < < .55 where p = .47 and n = 500. Use = .05. 4.9 In reference to Example 4.32, calculate the portion of the alternative curve that lies in the right hand critical region of the null curve thereby reconciling the dierence of approximately .0069975 between the power calculation provided by MINITAB and that calculated in the text. 4.10 Use MINITAB to obtain a good estimate of the exact power calculated in Example 4.35. 4.11 Use MINITAB to form the condence interval described in Example 4.38 on page 144. 4.12 Use MINITAB to form the condence intervals described in Example 4.39 on page 144. 4.13 Use MINITAB to construct the condence interval outlined in Exercise 4.22 on page 157. 4.14 Use MINITAB to construct the two-sided and one-sided condence intervals alluded to in Example 4.43 on page 151.
Chapter 5
Paired Samples Methods

5.1 Introduction
In this chapter you will use MINITAB to perform tests of hypotheses and form condence intervals for various paired samples methods. You will conduct paired samples t tests, McNemars test, as well as tests on risk and odds ratios. You will also form condence intervals for each of these statistics and perform equivalence tests.
5.2
Methods Related to Mean Dierence (page 160)
Task 5.1
Use MINITAB to perform the paired samples t test described in Example 5.1 on page 162.
Solution
Load the worksheet containing the Table 5.1 data by selecting FileOpen Worksheet... and opening le table 5.1 data.MTW. Now select StatBasic StatisticsPaired t.... In the Paired t (Test and Condence Interval) window be sure the Samples in columns button is selected. In the First sample: window type C2 then type C1 in the Second sample: window. The sample designations are done in this fashion because MINITAB subtracts the second sample from the rst sample when performing computations. Now click the Options... button and be sure that Test mean: is set to 0.0 and Alternative: is set to not equal. Click OK twice. MINITAB reports obtained t as T-Value = 2.93 which is the value computed in the text. The p-value is given as 0.011 which is less than = .05 so that the null hypothesis is rejected. 45
46
Task 5.2
CHAPTER 5. PAIRED SAMPLES METHODS
Use MINITAB to perform the equivalence test described in Example 5.4 on page 168.
Solution
Load the worksheet containing the Table 5.5 data by selecting FileOpen Worksheet... and opening le table 5.5 data.MTW. The one-tailed equivalence test may be carried out by conducting a one-tailed paired samples t test as follows. Select StatBasic StatisticsPaired t.... In the Paired t (Test and Condence Interval) window be sure the Samples in columns button is selected. Type C2 in the First sample: window and C1 in the Second sample: window. Now click the Options... button. Type 6.0 in the Test mean: window and select less than in the Alternative: window and click OK twice. MINITAB gives obtained t as 2.61 which agrees with the value calculated in the text. The p-value is given as 0.009 which results in rejection of the null hypothesis.
Task 5.3
Use the paired data in Table 5.1 to construct a one-sided 95 percent condence interval for the lower bound of d .
Solution
Load the worksheet containing the Table 5.1 data by selecting FileOpen Worksheet... and opening le table 5.1 data.MTW. Now select StatBasic StatisticsPaired t.... In the Paired t (Test and Condence Interval) window be sure the Samples in columns button is selected. In the First sample: window type C2 then type C1 in the Second sample: window. Now click the Options... button and be sure that Condence level: is set to 95.0 and Alternative: is set to greater than.1 Click OK twice. MINITAB reports the lower bound estimate as 2.55343 which is the value reported in the solution to Example 5.6.
5.3
Methods Related to Proportions (page 174)
Task 5.4
Use both approximate and exact methods to perform a two-tailed McNemars test on the data in Table 5.7 on page 175 at the = .05 level.
1 Read the material under One-Tailed Hypothesis Tests And One-Sided Condence Intervals on page 154 if you are not sure why the greater than option was chosen.
5.3. METHODS RELATED TO PROPORTIONS
47
Figure 5.1: Cross tabulation of disease status for vaccine treated twins.
Solution
Unfortunately, MINITAB does not provide a direct method for performing this test from the paired data. However, recalling that McNemars test is simply a test for a proportion, we can use this facility of MINITAB to perform the test. We begin by nding the number of noninformative pairs, the number of pairs in which the vaccine one treated twin had disease but the vaccine two treated twin did not and the number of pairs in which the vaccine one treated twin had no disease while the vaccine two treated twin did develop the disease. To begin, load the worksheet containing the Table 5.7 data by selecting FileOpen Worksheet... and opening le table 5.7 data.MTW. Columns C1 and C2 contain the paired designations as to whether the twin contracted the disease (D) or did not contract the disease (D bar) for the two vaccine groups. Now select StatTablesCross Tabulation and Chi-Square.... Doubleclick on Vaccine Two to place this variable in the For rows: window (or alternatively, type C2 in the window) then double-click Vaccine One to place it in the For columns: window (or alternatively, type C1 in the window). Put a check in the Counts window and be sure the other items under Display are unchecked. Click OK. The output is shown in Figure 5.1. Notice that for two pairs of twins both had the disease while for six pairs neither had the disease. Thus, there are eight noninformative pairs. Of the remaining eight pairs, there were six for whom the Vaccine One twin had disease while the Vaccine Two twin was disease free. We now have sucient information to conduct the test. We are now ready to perform McNemars test. To this end, select StatBasic Statistics1 Proportion.... Choose the Summarized data: button then type 8 in the Number of trials: window and 6 in the Number of events: window. Now click the Options... button and be sure that .5 is specied in the Test proportion: window and not equal is specied in the Alternative: window. In order to perform the approximate test we place a check in the Use test and interval based on normal distribution window and click OK twice. MINITAB gives obtained Z as 1.41 which is the value computed in the
48
text on page 176. The exact test is performed by following the above steps with the exception that the Use test and interval based on normal distribution window is not checked. In this case MINITAB gives the two-tailed p-value as .289 which is the value calculated in conjunction with Example 5.11 on page 179 of the text.
Task 5.5
Use the Outcome variable in Table 5.14 to perform the exact test outlined in Example 5.13 on page 183.
Solution
Enter the Outcome variable, minus the noninformative observations, from Table 5.14 into C1 of a new worksheet. Select StatBasic Statistics1 Proportion.... Select the Samples in columns: button and type C1 in the accompanying window. Click the Options... button and enter .4 in the Test proportion: window. Select greater than in the Alternative: window and click OK twice. MINITAB gives the p-value for the test as 0.514 which is the value calculated in the solution to Example 5.13.
Task 5.6
Construct the condence intervals described in Example 5.15 on page 186.

Solution
Select StatBasic Statistics1 Proportion.... Select the Summarized data button and enter 10 in the Number of trials: window and 4 in the Number of events: window. Click on Options... and be sure that the Condence level: window contains 95.0. Enter not equal in the Alternative: window and check the Use test and interval based on normal distribution: box to obtain the approximate interval. Now click OK twice. MINITAB displays the interval as (0.096364, 0.703636) which agrees with the calculations obtained in the solution to Example 5.15. Repeat the above steps with the Use test and interval based on normal distribution: box unchecked. MINITAB gives the exact interval as (0.121552, 0.737622 which, except for rounding, again agrees with the calculations provided in the solution to 5.15.
5.4
Methods Related to Paired Samples Risk Ratios (page 190)
Task 5.7
Carry out the tasks outlined in Example 5.17 on page 191.
5.5. METHODS RELATED TO PAIRED SAMPLES ODDS RATIOS

Solution
49
MINITAB does not provide a direct means for calculating a paired samples risk ratio. However, it can still be useful in such an endeavor in that it can be used to form the data into a two by two table thereby facilitating the calculation. This can be particularly useful for large data sets. For the problem at hand, generate the table shown in Figure 5.1 of this manual by following the steps given in Task 5.4. You can then use the information given in this table to calculate the risk ratio via Equation 5.7 on page 191. The test of signicance conducted via Equation 5.4 was carried out in the solution to Task 5.4. To obtain the result provided by Equation 5.5, simply square the result given by Equation 5.4.
Task 5.8
Use MINITAB to perform the equivalence test as outlined in Example 5.20 on page 195.
Solution
Unfortunately, MINITAB does not provide a straightforward means for performing this test.
Task 5.9
Use MINITAB to form the condence interval described in Example 5.22 on page 197.
Solution
Unfortunately, MINITAB does not provide a straightforward means for constructing this interval.
5.5
Methods Related to Paired Samples Odds Ratios (page 199)
Task 5.10
Use MINITAB to test the hypothesis H0 : OR = 1 as described in Example 5.24 on page 201.
Solution
Using the fact that a test of H0 : OR = 1 is equivalent to a test of H0 : = .5, we conduct the test as follows. Select StatBasic Statistics1 Proportion.... Choose the Summarized data button and enter 30 in the Number of trials: window and 25 in the Number of events: window. The number of trials is specied as 30 because after eliminating the noninformative pairs there are
50
25 + 5 = 30 remaining pairs that do provide information. Of these, in twentyve pairs, the member with cancer had been exposed while the pair member without cancer was not exposed. We dene this as the number of events. Now click Options... and be sure that the Test proportion: is set to 0.5. Put a check in the Use test and interval based on normal distribution box and click OK twice. MINITAB performs the test as a Z test while a chi-square statistic is reported in the text. By squaring obtained Z of 3.65 we obtain, except for rounding, the chi-square statistic of 13.33 reported in the text.
Task 5.11
Perform the exact two-tailed equivalence test described in Example 5.27 on page 206.
Solution
Unfortunately, MINITAB does not provide a straightforward means of performing this test. However, by using the relationship between the paired samples odds ratio and proportions along with the so called shortcut method of performing equivalence tests introduced on page 39, the task can be performed without too much diculty. We begin by constructing a 90 (not 95) percent condence interval for the estimation of . To this end, select StatBasic Statistics1 Proportion.... Choose the Summarized data button and enter 15 in the Number of trials: window and 8 in the Number of events: window. Now click Options... and be sure that the Condence level: is set to .90 and Alternative: is not equal. Because we are conducting an exact test, be sure the Use test and interval based on normal distribution box is not checked. Click OK twice. The condence interval (0.299986, 0.756273) is displayed in the Session Window. Because this condence interval is not contained in the equivalence interval EIL = .454 and EIU = .545, the equivalence null hypothesis is not rejected. This is the same conclusion reached in the Solution to Example 5.27.
Task 5.12
Construct the exact condence interval described in Example 5.30 on page 211.
Solution
Unfortunately, MINITAB does not provide a straightforward means of constructing this interval. However, by using the relationship between the paired samples odds ratio and proportions, we can construct the CI for the associated proportion and then convert this interval for the estimation of into an interval for estimating the paired samples OR. To this end, select StatBasic Statistics1 Proportion.... Choose the Summarized data button and enter 24 in the Number of trials: window and 13 in the Number of events: window. Now click Options... and be sure that the Condence level: is set to .95 and Alternative: is not equal. Because
5.5. METHODS RELATED TO PAIRED SAMPLES ODDS RATIOS
51
we are conducting an exact test, be sure the Use test and interval based on normal distribution box is not checked. Click OK twice. The condence interval (0.328208, 0.744470) is displayed in the Session Window. This is the same interval reported as part of the solution to Example 5.30. In order to convert this interval into an interval for the estimation of OR, you must apply Equation 5.17 (page 208) as is shown in the solution to Example 5.30.
52
Exercises
5.1 Use MINITAB to perform the test of hypothesis outlined in Example 5.2 on page 163. 5.2 Use MINITAB to perform the equivalence test outlined in Example 5.5 on page 169. 5.3 Use the paired data in Table 5.2 to form a 95 percent condence interval for the estimation of 1 2 where 1 represents the mean for Treatment One and 2 represents the mean for Treatment Two. 5.4 Use MINITAB to perform the test outlined in Example 5.12 on page 180. 5.5 Use the Outcome variable in Table 5.16 to perform the tests alluded to in Example 5.14 on page 185. 5.6 Use MINITAB to construct the condence intervals discussed in Example 5.16 on page 188. 5.7 Conduct the test outlined in Example 5.18. 5.8 Conduct the test of signicance alluded to in Example 5.19 on page 192. 5.9 Use MINITAB to perform the exact two-tailed test of signicance outlined in Example 5.25 on page 202. 5.10 Use MINITAB to construct the approximate condence interval discussed in Example 5.28 on page 208.
Chapter 6
Two Independent Samples Methods

6.1 Introduction
In this chapter you will use MINITAB to perform tests of hypotheses and form condence intervals for various two (independent) sample methods. You will conduct independent samples t tests and tests for dierences between proportions. You will also form condence intervals for each of these statistics and perform equivalence tests.
6.2
Methods Related to Dierences Between Means (page 215)
Task 6.1
Conduct the independent samples t test described in Example 6.1 on page 220.
Solution
Open the table 6.1 data.MTW worksheet. Select StatBasic Statistics2Sample t.... The 2-Sample t (Test and Condence Interval) window opens. Select the Samples in dierent columns button. Enter C1 in the First: window and C2 in the Second: window. Put a check in the Assume equal variances box. Click the Options... button. Be sure that the Test dierence: window is set to 0.0 and the Alternative: window is set to not equal. Click OK twice. The result shown in Figure 6.1 appears in the Session Window. Notice that obtained t is 0.81 which is the value calculated in the text. The associated p-value is 0.426 which results in failure to reject the null hypothesis 53
54
CHAPTER 6. TWO INDEPENDENT SAMPLES METHODS
Figure 6.1: MINITAB output for an independent samples t test.
at the .05 level.

Task 6.2
Conduct the equivalence test described in Example 6.3 on page 225.

Solution
Open the table 6.1 data.MTW worksheet. Select StatBasic Statistics2Sample t.... The 2-Sample t (Test and Condence Interval) window opens. Select the Samples in dierent columns button. Enter C1 in the First: window and C2 in the Second: window. Put a check in the Assume equal variances box. Click the Options... button. Set the Test dierence: window to 5.0 and the Alternative: window to greater than. Click OK twice. The result appears in the Session Window. Notice that obtained t of 0.38 is the same value calculated in the text. The associated p-value is 0.353 so that the equivalence null hypothesis is not rejected.
Task 6.3
Form the condence interval described in Example 6.5 on page 228.

Solution
Open the table 6.1 data.MTW worksheet. Select StatBasic Statistics2Sample t.... The 2-Sample t (Test and Condence Interval) window opens. Select the Samples in dierent columns button. Enter C1 in the First: window and C2 in the Second: window. Put a check in the Assume equal variances box. Click the Options... button. Be sure the Condence level: is set to 95.0 and the Alternative: is set to not equal. Click OK twice. MINITAB prints the following in the Session Window. 95% CI for dierence: (-12.02002, 5.22002) This is the interval calculated in the text.
6.3. METHODS RELATED TO PROPORTIONS
55
6.3
Methods Related to Proportions (page 230)
Task 6.4
Perform the two sample test for a dierence between proportions addressed in Example 6.7 on page 233.
Solution
Select StatBasic Statistics2 Proportions.... In the 2 Proportions (Test and Condence Interval) window, choose the Summarized data button. In the First: window enter 314 for Trials: and 23 for Events:. In the Second: window enter 316 for Trials: and 39 for Events:. Now click the Options... button. Be sure that Test dierence: is set to 0.0, Alternative: is not equal and the Use pooled estimate of p for test box is not checked. Now click OK twice. MINITAB reports obtained Z as 2.12, which is the result reported in the text, with an associated p-value of 0.034. This value would be appropriate for a two-tailed test. For the one-tailed test addressed here we calculate .034/2 = .017 which causes the null hypothesis to be rejected.
Task 6.5
Perform the two-tailed equivalence test described in Example 6.9 on page 235.
Solution
For Test One, select StatBasic Statistics2 Proportions.... In the 2 Proportions (Test and Condence Interval) window, choose the Summarized data button. In the First: window enter 100 for Trials: and 7 for Events:. In the Second: window enter 100 for Trials: and 6 for Events:. Now click the Options... button. Set Test dierence: to 0.04, Alternative: to less than and be sure the Use pooled estimate of p for test box is not checked. Now click OK twice. MINITAB reports obtained Z as .86, which is the result reported in the text, with an associated p-value of 0.195. For Test Two, follow the same steps as for Test One with the exception that Test dierence: is set to -.04 and Alternative: to greater than. MINITAB gives obtained Z as 1.43, which is the value calculated in the text, with an associated p-value of 0.076. Thus, the equivalence null hypothesis is not rejected.
Task 6.6
Construct the condence interval described in Example 6.11 on page 236.

Solution
Select StatBasic Statistics2 Proportions.... In the 2 Proportions (Test and Condence Interval) window, choose the Summarized data button. In the First: window enter 299 for Trials: and 106 for Events:.
56
In the Second: window enter 313 for Trials: and 66 for Events:. Now click the Options... button. Be sure that Condence level: is set to 95.0, Alternative: is not equal and the Use pooled estimate of p for test box is not checked. Now click OK twice. MINITAB reports the condence interval as (0.0730676, 0.214237). Notice that these values dier slightly from those calculated in the text. This is because the methods used by MINITAB and the text for constructing such intervals dier slightly. The method used in the text is slightly more conservative and, in the opinion of the authors, is preferable.
6.4
Methods Related to Independent Samples Risk Ratios (page 238)
Unfortunately, MINITAB does not implement the methods discussed in the text for dealing with independent samples risk ratios.
6.5
Methods Related to Independent Samples Odds Ratios (page 247)
Unfortunately, MINITAB does not implement the methods discussed in the text for dealing with independent samples odds ratios.
Exercises
57
Exercises
6.1 The MINITAB output associated with Task 6.1 contains the statement Both use Pooled StDev = 11.5245. How does the value 11.5245 relate to the calculations carried out in the text in conjunction with Example 6.1? 6.2 Use MINITAB to conduct the test of hypothesis described in Example 6.2. 6.3 Perform the equivalence test described in Example 6.4 on page 226. 6.4 Find the condence interval discussed in Example 6.6 on page 229. 6.5 Perform the test described in Example 6.8 on page 233. 6.6 Perform the equivalence test outlined in Example 6.10 on page 235. 6.7 Construct the one-sided condence interval described in Example 6.12 on page 237.
58
Chapter 7
Multi-Sample Methods
7.1 Introduction
In this chapter you will use MINITAB to perform a One-Way ANOVA, a 2 by k chi-square test and Tukeys HSD test.
7.2
The One-Way Analysis of Variance (ANOVA) F Test (page 264)
Task 7.1
Use the data in Table 7.1 to compute M Sw , M Sb , obtained F and to produce an ANOVA table.
Solution
Load the table 7.1b data.MTW worksheet. Notice that C2 contains the weights of the 15 subjects while C1 contains an indicator (1, 2, or 3) as to which group the weight represents. Select StatANOVAOne-Way.... The One-Way Analysis of Variance window opens. Type C2 in the Response: window and C1 in the Factor: window. Click OK. The relevant portion of the MINITAB output is shown in Figure 7.1. In the ANOVA table provided by MINITAB, between groups elements are labeled Factor while within group elements are labeled Error. From the table we see that M Sw and M Sb are given as 625 and 935 respectively. Except for rounding, these are the values calculated in the text. Obtained F is given as 1.50 which is the value given in the text. The associated p-value is .263. 59
60
CHAPTER 7. MULTI-SAMPLE METHODS
Figure 7.1: MINITAB rendering of ANOVA table for data in text Table 7.1.
Figure 7.2: Nutritional data arranged for chi-square analysis.
7.3
The 2 By k Chi-Square Test (page 276)
Task 7.2
Perform the chi-square test outlined in Example 7.6 on page 280.

Solution
Open a new worksheet. Label C1 Instruction and C2 No Instruction. Now enter the number of Low Birth Weight babies for the Instruction group in R1C1 and the number of No Low Birth Weight babies in R2C1. Enter the corresponding values for the No Instruction group in R1C2 and R2C2. Your worksheet should appear as in Figure 7.2. To perform the analysis, select StatTablesChi-Square Test (Table in Worksheet).... Enter C1 C2 in the Columns containing the table: window and click OK. The output is shown in Figure 7.3. Notice that in addition to obtained chi-square with its associated p-value, the output also shows the observed and expected frequencies for each cell along with the contribution made by that cell to the overall chi-square value.1
1 That
is, the values summed in the calculation of obtained chi-square.
7.4. MULTIPLE COMPARISON PROCEDURES
61
Figure 7.3: Chi-square analysis of nutritional data.
7.4
Multiple Comparison Procedures (page 282)
Task 7.3
Perform the Tukeys HSD tests outlined in Example 7.9 on page 289.
Solution
Load the table 7.1b data.MTW worksheet. Notice that C2 contains the weights of the 15 subjects while C1 contains an indicator (1, 2, or 3) as to which group the weight represents. Select StatANOVAOne-Way.... The One-Way Analysis of Variance window opens. Type C2 in the Response: window and C1 in the Factor: window. Click the Comparisons... button. Put a check in the Tukeys, family error rate: box and enter .05 in the accompanying window. Click OK twice. In addition to the ANOVA table, the output shown in Figure 7.4 appears in the Session Window. Notice that comparisons are made in terms of condence intervals rather than hypothesis tests as is done in the text. Thus, the comparison 2 1 2 is represented by the interval L = 34.35, U = 49.95. Because zero is in the interval, the null hypothesis is not rejected. This is true of the remaining comparisons as well so that the conclusions provided by MINITAB are the same as those reached in the text.
2 This
comparison is represented as 1 2 in the text.
62
Figure 7.4: Results of Tukeys HSD tests conducted on the dieting study data in Table 7.1.
Exercises
63
Exercises
7.1 Carry out the ANOVA analysis outlined in Example 7.3 on page 269. 7.2 Carry out the ANOVA analysis outlined in Example 7.4 on page 271. 7.3 Perform the chi-square analysis outlined in Example 7.5 on page 279. 7.4 Perform the multiple comparison tests outlined in Example 7.10 on page 290.
64
Chapter 8
The Assessment of Relationships

8.1 Introduction
In this chapter you will use MINITAB to calculate the Pearson product-moment correlation coecient and to test the hypothesis H0 : = 0. Additionally, you will conduct the chi-square test for independence.
8.2
The Pearson Product-Moment Correlation Coecient (page 295)
Task 8.1
Calculate the Pearson product-moment correlation coecient for the data in Table 8.1 on page 297. Perform a two-tailed test of the hypothesis H0 : = 0 at the .05 level.
Solution
Load the table 8.1 data.MTW worksheet. Select StatBasic StatisticsCorrelation.... Enter x and y in the Variables: window and be sure the Display p-values box is checked. Click OK. MINITAB displays the correlation as 0.964 which is the value calculated in the text. The associated p-value of 0.000 results in reject of the null hypothesis H0 : = 0. 65
66
CHAPTER 8. THE ASSESSMENT OF RELATIONSHIPS
Figure 8.1: Chi-square analysis of data described in Example 8.7.
8.3
The Chi-Square Test For Independence (page 312)
Task 8.2
Perform the chi-square test described in Example 8.7 on page 314.

Solution
Enter the observed frequencies into a new worksheet as shown in manual Figure 8.1. Then select StatTablesChi-Square Test (Table in Worksheet).... Enter C1 C2 C3 in the Columns containing the table: window and click OK. The observed and expected frequencies for each cell of the table are printed along with the contribution made by each cell to the overall chisquare statistic. Obtained chi-square of 206.811 is reported with 4 degrees of freedom and an associated p-value 0.000.
Exercises
67
Exercises
8.1 Calculate the correlation coecient discussed in Example 8.2 on page 298.
68
CHAPTER 8. THE ASSESSMENT OF RELATIONSHIPS
Chapter 9
Linear Regression
9.1 Introduction
In this chapter, you will use MINITAB to construct simple and multiple regression models, generate R2 values and conduct tests of signicance.
9.2
Simple Linear Regression (page 320)
Task 9.1
Construct the simple linear regression model alluded to in Example 9.1 on page 321.
Solution
Open the table 8.2 data worksheet. Select StatRegressionRegression.... In the Regression window, type C2 in the Response: window and C1 in the predictors: window. Click Options... and be sure there is a check in the Fit intercept box. Click OK twice. MINITAB gives the regression equation as y = 0.874 + 0.706 x which is the result given in the text.
Task 9.2
Find SSreg , SSres , and R2 for the data in Table 9.1 on page 323.
Solution
Open the table 9.1 data worksheet. Select StatRegressionRegression.... In the Regression window, type C1 in the Response: window and C2 in the predictors: window. Click Options... and be sure there is a check in the Fit intercept box. Click OK twice. MINITAB gives the Regression sum of squares (SSreg ) as 107.44 and the Residual Error sum of squares (SSres ) as 69
70
CHAPTER 9. LINEAR REGRESSION
8.29 which agrees with the values calculated in the text. R2 is given as R-Sq = 92.8% which is the value calculated in the solution to Example 9.2 on page 324.
Task 9.3
Test the null hypothesis R2 = 0 for a model used to predict y from x for the data in Table 9.1 on page 323.
Solution
Using the output from the solution to Task 9.2, we see that obtained F is reported as 168.49 with an associated p-value of 0.000. The value of 168.49 reported by MINITAB diers slightly from the value of 167.56 calculated in the text. This small dierence appears to be the result of dierences in rounding.
9.3
Multiple Linear Regression (page 329)
Task 9.4
Use the data in Table 9.2 on page 330 to construct a two predictor model with x1 and x2 being used to predict y.
Solution
Open the table 9.2 data worksheet. Select StatRegressionRegression.... In the Regression window, type C1 in the Response: window and C2 C3 in the predictors: window. Click Options... and be sure there is a check in the Fit intercept box. Click OK twice. MINITAB gives the regression model as y = 2.69 + 0.404 x1 + 1.00 x2 which agrees with the results calculated in the text.
Task 9.5
Test the null hypothesis H0 : 1 = 2 = 0 for the model you constructed in Task 9.4.
Solution
From the output for the solution to Task 9.4, we note that obtained F for the regression model is 4.73 which is the value calculated in the text. The associated p-value is 0.031 which results in rejection of the null hypothesis for a test conducted at = .05. This is the same conclusion reached in the text.
Task 9.6
Use the data in Table 9.2 on page 330 to determine whether adding x2 to a model that contains x1 adds signicantly to R2 .
9.3. MULTIPLE LINEAR REGRESSION
71
Figure 9.1: Sequential General Linear Model analysis of Table 9.2 data.
Solution
This problem can be approached in a number of dierent ways. For example, we could use MINITAB to nd R2 and R2 then use a calculator to apply y.1 y.12 Equation 9.24. We will use a more direct, if somewhat less obvious, approach. Select StatANOVAGeneral Linear Model.... The General Linear Model screen opens. Type y in the Responses: window and x1 x2 in the Model: window. To this point we have set up a model in which y is the dependent variable with x1 and x2 being predictors. Now click on Covariates... and enter x1 x2 in the Covariates: window. Among other things, this last action cause MINITAB to treat x1 and x2 as continuous variables rather than transforming them as might be done in other forms of analysis. Click OK. Now click Options.... In the Sum of Squares: column, darken the Sequential (Type I) button. Click OK twice. The relevant portion of the output is shown in manual Figure 9.1. This output will require some explanation. Of interest for the problem at hand are the headings labeled Seq SS (Sequential Sum of Squares), F (obtained F), and P (p-value). The Seq SS values represent the increment in SSreg that occurs as each variable is added to the model.1 Thus, with only x1 in the model, SSreg was 150.83. Adding x2 to the model increased this value by 90.78. It will be instructive to convert these values into expressions of R2. This can be done by dividing each value by SSy which is labeled in the printout as Total. For x1 we have R2 = 150.83 = .2750 548.40
which is equal to R2 . Of more interest is the R2 value obtained from x2 which y.1 is R2 = 90.78 = .1655 548.40
which is equal to R2 R2 . The values of F reported in the output are the y.12 y.1 values of obtained F for a test of signicance on each of the incremental sums of squares. It follows that obtained F of 3.55 is obtained F for a test conducted on
1 The order in which variables are added to the model is determined by the order in which they are listed in the Model: window.
72
R2 R2 . Indeed, this is the value calculated on page 337 of the text. The y.12 y.1 associated p-value of 0.084 means that we would fail to reject the null hypothesis R2 R2 = 0 at the .05 level. This is the conclusion reached on page 337 of y.12 y.1 the text. Therefore, we are unable to demonstrate that adding x2 to a model that contains x1 adds to R2 . To further elucidate, suppose we were working with four predictors, x1, x2 , x3 and x4 and had entered them in that order in the Model: box. The obtained F and associated p-values for each predictor would represent tests of the following null hypotheses. Source x1 x2 x3 x4 Hypothesis R2 = 0 y.1 R2 R2 = 0 y.12 y.1
2 R2 y.123 Ry.12 = 0 2 R2 y.1234 Ry.123 = 0
Exercises
73
Exercises
9.1 Find the regression equation for predicting y from x for the data in Table 9.1 on page 323. 9.2 Use MINITAB to construct the model described in Exercise 9.7 on page 341 of the text. 9.3 Test the null hypothesis H0 : R2 = 0 for the model you constructed in Exercise 9.2. 9.4 Use the data in Table 9.2 on page 330 to determine whether adding x1 to a model that contains x2 adds signicantly to R2 .
74
Chapter 10
Methods Based on the Permutation Principle

10.1 Introduction
In this chapter you will use MINITAB to perform Wilcoxons signed-ranks test, Wilcoxons rank-sum (Mann-Whitney) test, the Kruskal-Wallis test and Fishers exact test.
10.3
Task 10.1
Applications (page 348)
Perform the Wilcoxons signed-ranks test described in Example 10.15 on page 369.
Solution
Open the table 10.7 data.MTW worksheet. Notice that this le contains dierence scores rather than the pre- and post-treatment values. Select Stat Nonparametrics1-Sample Wilcoxon.... Type C1 or d in the Variables: window. Darken the Test median: button and type 0.0 in the accompanying window if necessary. Be sure the Alternative: window is set to not equal. Click OK. MINITAB reports the test p-value as 0.262 which results in failure to reject the null hypothesis at = .05. This is the same conclusion reached in the text. It should be noted, however, that p-values reported by MINITAB for Wilcoxons signed-ranks are approximations rather than exact values. 75
76
CHAPTER 10. PERMUTATION BASED METHODS
Figure 10.1: Data arranged for Kruskal-Wallis test analysis.
Task 10.2
Perform the Wilcoxons rank-sum test described in Example 10.20 on page 382.
Solution
Enter the data accompanying Example 10.20 into a new worksheet. Select StatNonparametricsMann-Whitney....1 The Mann-Whitney page opens. Type C1 in the First Sample: window and C2 in the Second Sample: window. Set Alternative: to less than and click OK. MINITAB gives the one-tailed p-value as 0.3313. As noted in the solution to Example 10.20, the exact p-value is .35 so that the approximation provided by MINITAB is quite good even for this case in which sample size is small.
Task 10.3
Perform a Kruskal-Wallis test on the raw data in the table on page 393.
Solution
Enter the group number, i.e. 1, 2, or 3, of each of the 6 subjects in C1 and their outcome measure in C2. The entries should appear as in Figure 10.1. Now select StatNonparametricsKruskal-Wallis.... The Kruskal-Wallis page opens. Enter C2 in the Response: window and C1 in the Factor: window. Click OK. MINITAB gives the p-value for the test as 0.156. As is shown in the solution to this problem (page 395), the exact p-value is .20 so that the MINITAB approximation is not too good. But this is to be expected with such small sample sizes.
1 Recall
that the Mann-Whitney test is just a dierent form of the Wilcoxon rank-sum test.
10.3. APPLICATIONS
Task 10.4
77
Perform a two-tailed Fishers exact test on the data in Table 10.23 on page 405.
Solution
Open the table 10.23 data.MTW worksheet. Notice that C1 contains row numbers and C2 the column numbers of the cells that makeup Table 10.23. Taken together, these columns uniquely identify each cell of the table. Column C3 provides the frequency in each of the four cells. Select StatTablesCross Tabulation and Chi-Square.... The Cross Tabulation and Chi-Square page opens. Type C1 in the For rows: window, C2 in the For columns: window and C3 in the Frequencies are in: window. Now click Other Stats... and place a check in the Fishers exact test for 2x2 tables box. Click OK twice. MINITAB gives the p-value for the test as 0.0889111 which is the value calculated for a two-tailed test in the solution to Example 10.30 on page 405. Notice that MINITAB did not provide the option of performing a one-tailed test.
78
CHAPTER 10. PERMUTATION BASED METHODS
Exercises
10.1 Conduct the test described in Example 10.16 on page 371. 10.2 Perform the test alluded to in Example 10.21 on page 383. 10.3 Conduct the test described in Example 10.22 on page 385. 10.4 Conduct the Kruskal-Wallis test described in Example 10.27 on page 395. 10.5 Conduct the test described in Example 10.31 on page 407.
Appendix A
Table Relating Text Items to Tasks

This table relates various portions of the text to specic tasks and the task page number. Text Table 2.2 Table 2.3* Table 2.4 Figure 2.1 Table 2.3* Figure 2.4 Figure 2.6 Example 2.4 Example 2.7 Table 2.6 Example 2.17 Example 2.21 Example 2.23 Example 2.24 Table 3.1 Example 3.9 Table 4.1 Example 4.7 Example 4.9 Example 4.12 Task 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.9 2.10 2.11 2.12 2.13 2.14 2.15 3.1 3.2 4.1 4.2 4.3 4.4 Page 11 13 13 15 17 18 19 24 25 26 28 28 28 29 31 31 35 36 36 37
79
80 Text
APPENDIX A. TABLE RELATING TEXT ITEMS TO TASKS Task 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 6.1 6.2 6.3 6.4 6.5 6.6 7.1 7.2 7.3 8.1 8.2 9.1 9.2 9.3 9.4 9.5 Page 38 38 39 39 40 41 41 42 42 43 43 45 46 46 46 48 48 48 49 49 49 50 50 53 54 54 55 55 55 59 60 61 65 66 69 69 70 70 70
Example 4.14 page 108 Example 4.19 Example 4.25 Example 4.26 Example 4.32 Example 4.34 Example 4.36 Example 4.37 Example 4.41 Example 4.42 Example 5.1 Example 5.4 Table 5.1 Table 5.7 Example 5.13 Example 5.15 Example 5.17 Example 5.20 Example 5.22 Example 5.24 Example 5.27 Example 5.30 Example 6.1 Example 6.3 Example .6.5 Example 6.7 Example 6.9 Example 6.11 Table 7.1 Example 7.6 Example 7.9 Table 8.1 Example 8.7 Example 9.1 Table 9.1* Table 9.1* Table 9.2* Table 9.2*
81 Text Table 9.2* Example 10.15 Example 10.20 page 393 Table 10.23 Task 9.6 10.1 10.2 10.3 10.4 Page 70 75 76 76 77
* Appears more than once in list.

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MINITAB 14 Supplement for: Biostatistics for the Health Sciences

R. Cliord Blair and Richard A. Taylor December 30, 2007

referred to as the text.

Figure 1.1: Screen appearance when MINITAB is started.

1.3. INPUTTING DATA

1.4. PERFORMING ANALYSES

Figure 1.3: Screen appearance for display of basic statistics.

Editing and Printing Your Output

Using the Calculator

1.6. USING THE CALCULATOR

Figure 1.6: Calculator screen used to create the square of x.

Figure 1.7: Calculating the mean of x.

as shown in Table 2.6. Save the worksheet.

Figure 1.8: Histogram of x.

Figure 1.9: Reconstruction of Table 2.6.

Distributions (page 15)

CHAPTER 2. DESCRIPTIVE METHODS

Figure 2.1: Screen appearance for ordering pain level variable.

Figure 2.3: Simple frequency distribution of BP scores as shown in Table 2.3.

CHAPTER 2. DESCRIPTIVE METHODS

Figure 2.5: Screen appearance for ordering Interval variable.

Graphs (page 19)

CHAPTER 2. DESCRIPTIVE METHODS

Figure 2.6: Grouped frequency distributions for BP variable in 12 categories.

Figure 2.7: Bar graph of pain level scores.

Figure 2.8: Informing MINITAB that a relative frequency histogram is to be constructed.

CHAPTER 2. DESCRIPTIVE METHODS

Figure 2.9: Specifying intervals for a relative frequency histogram.

CHAPTER 2. DESCRIPTIVE METHODS

Figure 2.12: Specifying intervals for a cumulative relative frequency polygon.

can paste this distribution in C1 and C2 of a new worksheet if you prefer.

CHAPTER 2. DESCRIPTIVE METHODS

CHAPTER 2. DESCRIPTIVE METHODS

Numerical Methods (page 23)

2.6. NUMERICAL METHODS

CHAPTER 2. DESCRIPTIVE METHODS

Figure 2.18: Data from example 2.7.

2.6. NUMERICAL METHODS

Figure 2.19: Data from example 2.7 in a single column.

Figure 2.20: Descriptive statistics for data from example 2.7.

CHAPTER 2. DESCRIPTIVE METHODS

2.6. NUMERICAL METHODS

CHAPTER 2. DESCRIPTIVE METHODS

Contingency Tables (page 52)

The Normal Curve (page 62)

Figure 3.1: MINITAB rendering of Table 3.1.

Introduction to Inference and One Sample Methods

Sampling Distributions (page 75)

Generate the binomial probabilities in Table 4.1 on page 83.

in C2 if you chose that option.

CHAPTER 4. INFERENCE AND ONE SAMPLE METHODS

Figure 4.1: MINITAB rendering of binomial probabilities in Table 4.1.

Hypothesis Testing (page 86)

4.3. HYPOTHESIS TESTING

CHAPTER 4. INFERENCE AND ONE SAMPLE METHODS

4.3. HYPOTHESIS TESTING

CHAPTER 4. INFERENCE AND ONE SAMPLE METHODS

Figure 4.2: Condence interval used to perform two-tailed equivalence test.

4.3. HYPOTHESIS TESTING