UNIT 11 STATISTICAL QUALITY CONTROL

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Structure Page No.

1 1.1 Introduction
Objectives
1 1.2 Concept of Quality
Nature of Quality Control
11.3 Statistical Process Control
.. Concept of Variation
Control Charts
Control Charts For Variables
Process Capability Analysis
Ir s Control Charts For Attributes
11.4 ~ c c e ~ t a n Sampling
ce
Sampling Plan Concepts
Single Sampling Plans
11.5 Summary

11.1 INTRODUCTION

A widely accepted definition of the quality of a product is its fitness for use for its
intendedpurpose. For exatnple, a ball pen should write well throughout its life.
Besides, the cap should not be loose, neither it should leak nor break easily, etc., are
some of the other features. Again, for a cricket ball, some of the quality characteristics
are like its weight, size, shining, quality of stitches, etc. And, for a water tap washer,
these are its thickness, inner diametel; outer diameter, etc.
The quality of a product is assessed by the totality of its features. Have you ever
thought of when and where products are made? We only curse the products (and the
people who made them) whenever these products gwe us trouble or displeasure. Rarely
we think how these products are made or what precautionary measures have been taken
t1 during the production process to ensure that the products are of good quality. Quality Control (QC, in
short) consists of
Who is responsible for the quality of a product? Obviously, it is the manufacturer of the procedures and
C
. product. In this unit, we shall discuss some simple statistical tools, which are methodologies which
extensively used in the production process to ensure the quality of products. The most ensure that the quality
effective use of Statistical Quality Control (SQC, in short) generally requires characteristicsof a product
cooperation among those responsible for the three different types of functions: conform to its
specification, production, and inspection. specifications.
In Sec. I I .2, you will get introduced to the concept of quality and learn about methods
used in the process of controlling and systematically improving quality of a product. In
Sec. 1I .3, we shall discuss the primary tools of statistical process control that are useful
in monitering the quality aspects of a product. In Sec. 11.4, you will learn about
acceptance sampling plans - a technique used in ensuring that the produced products
conform to the specified quality standards.
Objectives
After reading this unit, you should be able to
explain the term quality and the phrase statistical quality control;
describe the concept of variation, chance and assignable cuases; 77
 Applied Statistical construct and interpret control charts for variables and attributes;
Techniques
estimate process capability from control chart3 data;
describe acceptance sampling plans;
interpret and use oc curves in determining acceptable quality level, producer's risk,
consumer's risk, and lot tolerance percentage defective;
describe single sampling plan and construct some simple single sampling plans with
help of a binomial nomograph.

- - - -

11.2 CONCEPT OF QUALITY

Everyday, from the time we get up and till we go back to bed, each one of us is busy
with a number of activities. And, we depend on number of objects to carry out these
activities. For example, we need a tooth brush, a tooth paste, wash basin or a water tap,
the soaps and detergents, the stoves that we use for cooking, the vehicles that we use for
going to our offices, electrical bulbs, phones, medicines and what not?
Quality of any product depends upon certain characteristics related to the product and
Have you ever noticed that the materials that go into it. For example, in the case of a ball pen, if the diameter of the
there is a tiny ball in the nib ball is undersized then the pen is bound to leak. Similarly, if the refill length is
of every ball pen? oversized, then the press button may not work properly. So, ball diameter and ref ill A

length are two quality characteristics for a simple product like ball pen.
In fact, a ball pen has many more quality characteristics. But, unlike diameter and
length, not all quality characteristics are measurable. Recall, non-measurable
characteristics are called attributes and the measurable ones are called the variables.

11.2.1 Nature of Quality Control

Let us continue our discussion with the example of a ball pen. As you may agree,
certain brands of ball pens stop writing from the second or third day onwards; some
write well only on certain types of papers; some write well on almost any kind of paper
and write till the last drop of ink remains in the refi!l. When a pen writes well, we say it
h
-
is of good quality.
How we decide the quality of a product? At the time when a product is designed,
certain specifications or levels of tolerances are established on all important quality
characteristics of the product. For example, in case of a ball pen, the specifications on
refill length may be that it should lie between 9.80 crns and 10.20 crns. So, if we can
ensure that all the quality characteristics of a product are maintained within their
specified limits, then automatically the quality of that product will be good.
Cookingin hotels is an But, we know that a manufacturing process is an interaction among people,
example of a manufacturing equipments, materials, methods and environment, wherein output could be either
process. another product or a component thrat goes into an end product. Thus, the performance
of a process is indicated by the quality of its output and can be assessed by examining
I
the quality characteristics of the output. So, a process is operated in such a way that the
quality characteristics of output product is maintained at desired levels. This is called
controlling a process.
In 1924, Walter A. Shewhart of Bell Telephone Laboratories introduced statistical
control charts as a tool for controlling quality of industry products. From then onwards,
people slowly started recognising the use of statistical techniques in quality control.
Today, it is known world over that statistical techniques are not only indispensible for
quality control but also play a very crucial role in all other facets of quality related
Statistical Quality Control techniques can be broadly divided into two categories: (i) Statistical Quality Control
Statistical Process Control (SPC, in short) techniques; and (ii) Acceptance Sampling.
S P C techniques are widely used in almost any manufacturing process and are very SPC techniqries have wide
usefil in solving real situation problems, achievingprocess stability, and making applications in
contin~ioiisinzprovements in product quality. The most important among these are non-manufacturing
control charts. We shall discuss control charts in the next section. processes as well.

In many situations, however, one or Inore components of a product are bought from
outside agents and the manufacturer does not have a direct control over the quality of
the components. Then, in such cases, acceptance sampling techniques are useful in
ensuring that the bought out components conform to specified quality levels.
In the next section, we would discuss SPC techniques and acceptance sampling will be
discussed in Sec. 1 1.4.

11.3 STATISTICAL PROCESS CONTROL

Statistical Process Control is a methodology used for understanding and monitoring a
process by collecting the data on quality characteristics periodically from the process,
analysing them and taking necessary actions based on the analysis results.
From now onwards, we will focus our discussion by referring to M/s BP & Cor7lLany, a
ball pen manufacturing company.
One of the sections in this company produces refills for the ball pens. The Refill length problem
specifications on the refill length, one of the quality characteristics of refill, are
10 & 0.2 cms. While producing the refills how does one ensure that their lengths
conform to these specifications?

11.3.1 Concept of Variation

As you may agree, however well the process is maintained, certain amount of variation
in the lengths of the refills is unavoidable. But, if this process is operated under stable
conditions i.e., machine settings are same, quality of the materials used is same,
operators are equally experienced, etc., then the quality characteristics such as refill
length normally exhibit a specific patterns of variation. Pattern of variation means
statistical distribution.
That is, if a process is operating under stable conditions then the amount of variation in
the quality characteristic is usually small and is a result of several small causes. These
small causes are known as chance causes and are usually unevitable.
The resulting variation is called the chance cause variation (or the inherent variation) Chance Causes
of the process. In practice, we find that most processes are often disturbed inadvertantly
or otherwise.
On the other hand, a change in the machine settings, sudde,n drop in the quality of raw
material or induction of a new operator due to an absence of the regular operator, etc.,
are some of the causes that might disturb a stable process. The reasons for variation
outside this stable process may be discovered and corrected. These causes are known as
assignable causes.
When the assignable causes are prevailing in a process, the process becomes unstable Assignable Causes
and this is reflected in the behaviour of the quality characteristic. That is, there will be
frequent changes in the distribution of the quality characteristic. As a result, there will
be more variation in the data. This variation is called the variation diie to assignable
causes.
It is important to remember that when a process is operating only _under chance causes, 79
 Applied Statistical we say that the process is statistically stable or that the process is under statistical
Techniques
control.
The power of Shewhart control charts lies in its ability to separate out assignable causes
of quality variation. So, a control chart is used to monitor the stability of a process and
alert us as and when an assignable cause creeps into +beprocess. Also, control charts
are very useful in detecting gradual improvement or deterioration in a process.

11.3.2 Control Charts

The concept of control charts is one of the most powerful techniques for on-line
On-line process control process control. Besides monitoring, control charts are useful also in the evaluation of
means monitoring a process capability of a process and in making continuous improvements in the process.
by periodically examining
sample outputs of the Recall the ref ill length problem mentioned above. We can use control chart technique to
1 process and taking effectively solve this problem. But, firstly, let us see how a control-chartis constructed.
corrective actions as and Qpieally, a control chart is a two-dimensional graph in which x-axis represents the
when necessary.
sample numbers and y-aris represents a quality characteristic. It has a soiid center
line (CL) and two dotted lines called upper control limit (UCL) and lower control limit
(LCL) (see Fig. 1).
process out of control
I /
'
---------- UCL
.*0
Y
"l
.c
3
f
-8
-0
.* - CL
0
&

+-------------------- LCL
I , , , , , , r , l l r ~ l l l ~ .

1 5 10 15 20 sample number
Fig. 1
Thus, the construction of control charts involves collecting samples periodically from
the process, computing the quality characteristic for each sample and plotting it
against the sample numbel: The consecutive points are joined by line segments.
As long as the plotted points are within the upper and lower control limits and do not
exhibit any specific patterns, we have no evidence that the process is not under
statistical control. When a point falls outside the control limits (below LCL or above
UCL), it is a cause and indicates the presence of an assignable cause with a high
probability.
However, conrrol chart cannot tell us what went wrong with a process when something
has gone wrong. It will only indicate that possibly something has gone wrong with the
process. In fact, it is the responsibility of supervisor or QC manager to find out what
has gone wrong.
Three Sigma Limits In most situations, a quality characteristic follows a normal distribution or can be
approximated by a normal distribution. Also, we know that the probability of a
normally distributed random variable taking values below p - 30 or above p 30, +
whgre p is the mean and a is the stalidard deviation, is very low (equal to 0.0027).
Therefore, if an observation falls outside 30 limits, it is logical to suspect that possibly
something might have gone wrong. For this reason, the control limits on a control chart
are set up using 3a limits. Consequently, when a point falls outside the control limits
on a control chart, it is more likely that it is due to the presence of an assignable cause,
 rather than a mere chance cause. Statistical Quality Control

Depending on the nature of quality characteristics, control charts are divided into two
i categories: (i) control charts for variables; and (ii) control charts for attributes.
Before we proceed further with our discussion, try the following exercise.
I
i1 E l ) For what type of quality characteristic do you think is a control charts for variables
I
suitable? Cite some quality characteristics that need control charts for attributes.
I
t
I
Control charts for variables are adopted in situations where the quality characteristic is
of measurable type. In the next part of the section, we shall discuss with you X - R
charts - a type of control charts for variables.
P
11.3.3 Control Charts For Variables
r
Once again we shall refer to the refill length problem to explain these charts. Consider
the data on refill length as given in Table 1. Observe that the samples are taken at
30-minute intervals. Here, each sample corresponds to lengths of five refills produced
at the time of collecting the sample. Each sample is called a subgroup.
Table 1 : Refill length data
Subgroup
S.No. Date Time 1 2 3 4 5 AverageRange
1 23.12.97 08:OO 10.11 10.08 10.14 10.10 10.15 10.116 0.07
2 08:30 10.08 10.08 10.12 10.13 10.06 10.094 0.07
3 09:OO 10.07 10.22 10.01 10.11 10.07 10.096 0.21
4 09:30 10.21 10.1010.0910.1310.02 10.110 0.19
5 1O:OO 10.12 9.98 9.91 10.05 10.17 10.046 0.26
6 1030 10.17 10.14 10.08 10.06 10.23 10.136 0.17
7 11:OO 10.10 10.11 10.21 10.05 10.22 10.138 0.17
8 11:30 10.10 10.06 10.23 10.14 9.97 10.100 0.26
9 12:OO 10.10 9.96 10.13 10.14 10.04 10.074 0.18
10 12:30 10.05 10.19 10.13 10.10 10.08 10.110 0.14
I1 24.12.97 08:OO 10.08 10.05 10.05 10.08 10.16 10.084 0.1 1
12 08:30 9.91 10.21 10.00 10.02 10.29 10.086 0.38
13 09:OO 10.11 9.98 9.97 10.04 10.08 '10.036 0.14
14 09:30 10.08 10.21 10.13 10.16 10.04 10.124 0.17
15 1O:OO 9.99 10 14 9.96 10.09 10.07 10.050 0.18
16 10:30 10.1710.1810.04 9.9910.11 10.098 0.19
17 11:OO. 10.06 9.92 10.10 10.06 10.02 10.032 0.18
18 11:30 10.16 10.12 10.16 10.02 10.19 10.130 0.17
19 12:OO 10.14 10.04 10.14 10.02 10.07 10.082 0.12
20 12:30 10.0810.07 9.9710.0910.12 10.066 0.15

Thus, in above table, we have data for 20 subgroups. The variation in the five
observations of any subgroup can be attributed only to chance causes because the f ive
observations correspond to five refills that were produced almost at the same time and
it is very unlikely that they were affected by any assignable causes in such a short span
of time.
Next, let us talk about the frequency of sampling. Supposing the samples are taken once
in every five minutes instead of every 30 minutes, then we are not going to find big
differences or changes in consecutive subgroups. So, too frequent sampling is an
unnecessary labour.
While keeping these points in mind, one has to decide the selection of subgroups and
their frequency in such a way that the variation in observations within a subgroup is
only due to chance causes and the variation among subgroups is likely to be affected by
assignable causes. The subgroups selected in this way are called the rational
subgroups.
Applied Statistical
Techniques

Here, LSL and USL stands

Recall that the refill lenght specifications are 10 f 0.2 cms. So, we will be too happy if
the refills are produced with length exactly equal to 10 cms. But we know that this is
practically impossible. So, we try to ensure that atleast the average refill length is as
close to 10 crns as possible and the variation In the refill length as low as possible.
7
for the lower and upper
specijicatiorz limits, LSL . LSL USL
respectively. I I I
I
I
I
I
I
I
I I

9.80 10.00 10.20 9.80 10.00 10.20

Fig. 2
The four curves (a - d) in Fig. 2 describe four different distributions for refill lengths.
Observe that, for the curve (a), the mean length is high even though the variation is low.
So, if our process produces refills like this, then some percentage of the refills do not
conform to the length specifications. However, if we can adjust the process so that the
mean length is equal to 10 crns, then our process will be good. Then, in this situation,
the refill length distribution will look like the one in (b).
On the other hand, if the distribution is given by (c) then even after adjusting the mean
to the target there will be non-conformance with regard to length (see curve (d)). So, in
this case, we must improve the process to reduce the variation.
Controlling Mean and The moral of the story is that, when the quality characteristic is a variable, it is
Variability essential to control both mean and variation. For this reason, two separate control
charts - one for controlling the mean and the other for controling variability, are
maintained for a measurable quality characteristic.
In quality control terminology, when a process is stable, the mean of a quality
characteristic under study is referred to as process mean and its variability is referred
to as process variability.
Here, we denote the process mean by p and process standard deviation by u.
Now, let xl, x2, . . . , xg be five independent observations of a subgroup on the refill
x - R Charts length. If the subgroup is rational, we can expect that x l , x2, . . . ,x5 are random
variables having same mean and standard deviation.
Recall that range is the Let X and R denote the average and range, respectively, of these five observations.
difference between the Then, TI is called the subgroup average and R the subgroup range. ~

largest and smallest values

of the observations. Two separate charts are maintained in an X - R chart: (i) TI - chart; and (ii) R - chart. In
the X - chart, used for controlling the process mean, we plot the sample averages
against the sample numbers; and, in the R -chart, used for controlling process
variability, we plot the sample range against the sample number.
Recall, the control limits are worked out based on the So-limits concept. Also, from
what your read in Unit 4, we know ifxl, x2, . . . , X, are independent random variables
with mean p and standard deviation o, then jT has mean p and its standard deviation is
u
equal to - (by central limit theorem).
-\G
As such, since we plot the sample averages on the TI- chart, we should construct the
center line and control limits using the mean and standard deviation of the averages and
not of the individual observations. Thus, the CL, LCL and UCL for the X - chart are
given by Statistical Quality Control
a a
CL=p, LCL=p-3- and U C L = p + 3 -
fi fil
where n is the sample size (For example, n = 5 for the refill length problem). And, the In practice, we do not know
estimates of p and a are given by fi = Z and 3 = R/dz, where Tr is the average of all the values of p and a. So,
subgroup averages, R is the average of subgroup ranges, and d2 is a constant depending we estimate these values
on n. Substituting these estimates in Eqn.(l), we get from the data and replace p
and u by their estimates in
Eqn.(l).

3
where A2 = -
d26'
Similarly, the control limits for an R - chart are also worked out based on the 3 a limits
concept and these are given by
CL = R, LCL = d3R and UCL = d4R, (3)
where d3 and d4 are constants depending n. In our subsequent discussion, we shall
make use of the values of d2, d3, d4 and A2 as given in Table 2.
Table 2 : Control chart constanls
Sample size, n I dz I
A2 I ds I dr
2 11.128 ( 1.88 ( 0 13.27

For example, when n = 5, d2 = 2.326, .A2 = 0.58, d3 = 0 and d4 = 2.11. Let us use
these values to solve the following problem.
Problem 1. Estimate p and a for refill length data and work out the control limits for JI
and R - charts.
Solution. Using data given in Table 1, we get
jT=
+
- 10.116 + 10.094 + . . . 10.066 201.808
- ------- = 10.09, and
20 20
- 0.07 + 0.07 + . . . + 0.15 3.51
R= - - 0.175.
20 20
Then, the estimates for p and a are given by
0.175
fi = 10.09 and & = -= 0.0752
2.326
Therefore, control limits for E - chart are given by
CL = = 10.09; LCL = - A ~ = R 10.09 - 0.58 x 0.175 = 9.99; and
UCL = jT + A ~ = R 10.09 + 0.58 x 0.175 = 10.19.
Similarly, the control limits for R - chart are given by ' At this stage, we may call
CL = = 0.175; LCL = d J R = 0.0 x 0.175 = 0.0; and these control limits as trial
control limits because we
UCL = d4R = 2.11 x 0.175 = 0.369. do not know whether the
X
data we have analysed
Now, you try the following exercise. correspond to a stable
process or not.
E2) A cricket ball ~ l ~ d n u f ~ c l u rcompany
ing wants to maintain control charts for the
weight of the balls. Twenty-five szxples, csch of size 4, were co!;ecied. The ziiin
of sample averages and the sum of sample ranges were found to be 7575 grams
and 154 grams, respectively. Estimate the process mean and standard deviation,
and compute the control limits for X and R - charts.

Using data from Table 1,'X and R - charts for Problem 1 are as shown in Fig. 3.
Applied Statistical
Techniques tI - - - - - - - - - - -X -chart- - - - - - - -
-
UCL = 10.19

t-------------------- LCL = 9.99

1 5 10 15 20 sample number

1 R - chart
- - - - - - - - UCL = 0.37

0
M
C
L?
-
QJ CL = 0.175
f
V1

t-------------------- LCL = 0.00

1 5 10 15 20 sample number
Fig. 3
Note that the 12th point on the R - chart indicates an out of control situation with regard
to variability. The 12th subgroup might not be similar to the other subgroups, indicating
the process might not be stable.
Though the actual cause could not be ascertained, there must have been one or more
assignable causes which have caused this high variation. In this situation, it is better to
For this reason, we called recalculate the control limits omitting the susceptible subgroup number 12.
the earlier limits, trial
You try to do that in the following exercise.
control limits.

E3) Discard the data of 12th subgroup in Table 1 and estimate p and a. Compute the
The process of recalculating control limits for X and.R - charts with the remaining 19 subgroups. Do the data
the control limits after indicate statistical control without 12th subgroup?
eliminating the outlying
subgroups is called
While homogenizing, if we have to discard more than 25% of the subgroups, then it is
homogenization.
better to discard the entire data and collect fresh data. But, while collecting data for
control chart analysis, it is important to ensure that process conditions remain the same
throughout the data collection period.
Now, from the solution of E3, we know that if we redraw the control charts after
eliminating the 12th subgroup then both K and R- charts exhibit a state of statistical
control. But then, does it mean that the process meets the requirements?
Problem 2. Do the refills produced by this process conform to their length
specifications?
Solution. Let X stands for the length and we assume that it is distributed normally with
tne mean 'and standard deviation as 1U.09 and 0.075, respectively. Then, we find that

where Z is the standard normal distribution. This means about 7% of the refills
produced by the process are oversize. It is clear from the K - chart that the process mean
is set at 10.09 crns (see Fig. 3). Statistical Quality Control
X

After observing this, the QC manager suspected that there was a setting problem in the
refill cutting machine. But, before carrying out any investigation, he decided to analyse
the variation aspect too.

11.3.4 Process Capability Analysis

Supposing that QC manager has corrected the process so that the mean is as desired, do
you think that the refills will conform to their length specifications? The answer is NO.

9.80 10Ioo 10120 1

Fig. 4
In Fig. 4, curve (a) represents the process before adjusting for the mean and (b)
represents the process after adjusting the mean. Note that there is no change in the
variability. Thus, we will continue to have rejections even after adjusting for mean
unless the process variability itself is reduced.
f
At this stage, we have to answer two question: (1) What is the existing variability? (2)
How much should we reduce it by? To answer these questions, we need the following
definitions.
Definition. Total tolerance of a measurable quality characteristic, denoted by T, is
given by the difference T = USL - LSL, where USL and LSL are the upper and
lower specification limits, respectively.
For example, as LSL = 9.80 crns and USL = 10.20 crns for refill length problem, so,
the total tolerance T = 0.4cms in this case.
Definition. When a process is under statistical control, its process capability is given The concept of 30-limits is
by 60, where a is the process standard deviation. used to define the process
capability.
The first question that we raised above can be answered by specifying an estimate of
the process capability. Here, in our situation,
estimate of the process capability = 6E/d2.
For refill length problem, the specification limits are 10 f 0.2 crns. So, to avoid
rejections, we must necessarily have a process for which the 3a limits lie within the
specification limits after setting the mean at the target (see Fig. 5). -\

9.80 10 - 30 10.00 10 +3a 10.20

Fig. 5
In other words, the process capability must be less than the total tolerance. Therefore,
the answer to the second question we raised above is that the process capability should
not exceed the total tolerance.
Applied Statistical Definitian. The process capability ratio of a stable process is the ratio of total tolerance
Techniques
to the process capability and is denoted by C,. That is,
total tolerance - USL - LSL
C -
- -
process capability 60
Thus, by what we have said above, if Cp < 1, then the process is bound to produce
rejections even when the mean is set on target. And, if C, 2 1,the rejection percentage
will be almost zero, provided the mean is at the target.
An estimate of C, can be obtained by substituting the estimate for a in the above
fdrrr,ula. For example, an estimate of Cp for the refill length problem under study is
given by

Since the estimate of C, is less than 1, we may infer that the ref ill length process is not
capable. More generally, even if we get the estimate of C, slightly more than 1, we
Precisely for this reason, a may still consider the process incapable. This is because our estimate for Cp may be an
process is considered to be underestimate due to sm.i;iing fluctuations.
capable, only if the estimate
of C, is atleast 1.33. Try the followi.ig exzr~isenow.

E4) Give an example of

a) a process whose Cp = 1.5 but has high rejections on USL side; and
b) a process whose Cp = 1.5 but has high rejections on LSL side.
(Hint: Specify the process parameter p and a.)

With above analysis in hand, the QC manager carried out an investigation of the
process and found out that a unit of the machine used for setting the refill length was
not properly calibrated. He also found that certain parts in the machine were worn out
which were causing vibrations in the length cutting machine.
Subsequently, he got the unit recalibrated and replaced the worn out parts and collected
five samples from the process. The new data is given in Table 3.
Table 3 : Refill leng:h data after correctwe action.
-
S.No. Date Time I 1 2 3 4 5 (
Average Range
21 6.1.98 08:OO 1 9.88 10.02 9.94 9.86 10.04 1 9.925 0.18

The new points are plotted on the charts shown in Fig. 1 a d new c W now look like
as in Fig. 6.
Note the clear distinction between the behaviours of the first 20 points and the last 5
points on both X and R - charts. The difference is the effect of a change in the process
after the first 20 pointa And the change is the result of the corrective actions taken by
the manager.
Since some changes were made in the process, we must reconstruct the control charts
with new data. To complete the task, manager has collected 15 more samples. The
sample averages and ranges are summarised in Table 4.
'hble 4 : Summary of 15 more samples on refill length data
S.No. Average Range SlNo. Average Range S.No. Average Range
26 10.035 0.14 31 10.053 0.08 36 10.048 0.18
27 9.995 0.14 32 9.973 0.08 37 10.030 0.14
28 10.020 0.08 33 9.983 0.20. 38 10,048 0.22
29 9.970 0.13 34 10.020 0.16 39 10.073 0.15
30 9.970 0.10 35 9.990 0.11 40 9.985 0.23
Total 49.99 0.59 Total 50.019 0.63 Total 50.184 0.92
 Statistical Quality Control

I - - - - - - - - - - - - - - - - - - - - y- - y- -LCL = 9.99
1 ' ' ' 1 " ' ' 1 " ' ' ~ ' ' " ~ ' " ' ~ 1
1 5 10 15 20 25 sample number

I R-chart
UCL = 0.37

CL = 0.175

......................... -LCL = 0.00

I " ' 1 " " 1 " ~ ' I ' ' " I ' " ' I 1
1 5 10 15 20 25 sample number
Fig. 6
You can help and advise the manager with the above data, provided you do the
following exercise.

E5) Combining the information provided in Table 3 and Table 4, do the following.
(a) Estimate the new process mean and standard deviation;
(b) Construct the control charts;
(c) State whether the process is under statistical control;
(d) Estimate the process capability and process capability ratio; and
(e) State your advice to the manager.

Next, let us talk about some control charts used for non-measurable quality
characteristics.

11.3.5 Control Charts for Attributes

, When products are inspected, they are classified into good and defective products. A
defective product is one that has one or more defects. The performance of a pr&ess is A defect is a nonconformir?,
often assessed by the proportion of defective items produced by the prqcess or by with respect to any of the
I counting the number of defects per unit of the product. quality characteristics of the
product.
Control charts used in these situations are known as attribute control charts. Here, we
discuss the following four such type of charts.
i) p and n p charts for the control of defective products;
ii) c charts for the control of number of defects per unit.
In case of BP & Co., 100 refills are selected at random from the process each day and p-charts
are inspected for all quality characteristics. Based on the inspection results, each refill
is classified as good or defective.
So, each day's sample of 100 refills is taken as a subgroup. The results of 14 days
sample collection are given in Table 5. Here, X denotes the number of defective refills
out of 100 inspected each day.
?gWo 5 :Refill inspection data
Applied Statistical When the prc;cess is stable, it is reasonable to assume that (i) the probability of any
Techniques
ref ill being defective is same for all ref ills, and (ii) the event that any of the ref ills
being good or defective does not injluence the quality of other ref ills.
Let
1, if the ith refill inspected is defective
(i = 1 , 2 , . . . , n ) ,

where n is the sample size. For example, n = 100 in our situation. Then, the total
+ + +
number of defective refills in the sample is given by X = XI Xz ... Xn
UriJer the assumptions (i) and (ii) above, X has a binomial distribution with parameters
n and p, where p is the common probability of any refill being defective. Also, we know
that the mean and standard deviation (SD) of a binomial distribution with parameters n
and p are given by
Mean(X) = np and SD(X) = Jnp(l-p).
In order to control the proportion of defective items produced by a process, we use a
If y denotes the proportion p-chart. In a p-chart, we plot the y values against the corresponding sample numbers.
of defective items in a Unlike St- R charts, a t t r k t e control c h a ~ t shave only one chart to be plotted.To arrive at
sample of n items, then the control limits, we need the mean and standard deviation of y. These are given by
X

Recall, p is the proportion of defective items produced by the process. In quality control
terminology, p is referred to as the process average. An estimate of p is given by
p = total number of defective items in all the samples
- inspected
total number of items
Thus, if wk have m samples, then

5 di

where di is the number of

As before, the control limits in an attribute control chart are based on the 3a limits
concept. When all the subgroups are of same size (= n, say), then the control limits for
a p-chart are given by

CL=p,LCL=p-3

Since we do not know the value of p, the control limits will be obtained by replacing p
by its estimate p given by Eqn.(6). Thus, the control limits for a p-chart are given by

CL = P,LCL = fr- 3

Using data from Table 5, the control limits for the refill length problem are given by
9 + 5 + ...+2 - -
84
cL=p = = 0.06,
14 x 100 14 x 100

LCL = 0 . 0 6 - 3

UCL = 0.06 +3
When the lower control limit happens to be negative, the control line is set at zero. So,
LCL = 0 in the p-chart of refill length problem (see Fig. 7).
 UCL = 0.13 Statistical Quality Control

E LCL = 0.00

1 5 10 15 sample number
Fig. 7
We have to be cautious in interpreting a p-chart. The points which fall above the UCL
I are called the high spots and the points that fall below the LCL are called the low spots.
A high spot may be doe to deterioration of the process or it could be due to a change in
the inspection standard (more stringent inspection may result in larger number of
defective items). Similarly, a low spot may indicate an improvement in the process or a
deterioration in the inspection standards.
Try the following exercise.

E6) Let the refill inspection data collected be as given in the following table. Recall, X denotes the
Table 6 : Refill inspection data number of defective ref ills
S . N o 1 2 3 4 5 6 7 8 9 1 0 out of 100 inspected each
Date 6 7 8 9 10 12 13 14 15 16 Total day.
X 1 3 2 2 1 0 5 1 1 3 19

1
5 10 15
! Fig. 8
(a) Plot the first five samples in Fig. 7 itself (use a pencil).
(b) Do you notice any change in the process? If so, what is thc difference?
(c) Can you say, from the chart you plotted, that the process is out of control?
(d) Construct a new p-chart using only the data given in Table 6 and comment on
the process (use the blank chart given in Fig. 8 above).
(e) Was there any improvement in the process? Estimate the rejection
percentages for the two periods.

Now, suppose we are not inspecting the same number of refills each day. Then, in thisi Varying Sample Sizes
case, the subgroup sample sizes are varying. In such cases, the control limits will vary
depending upon the subgroup sample size. Let us say, we have m subgroups with ni
sample items for the ith subgroup, then the control limits for the subgroups are given by

and UCLi = p + 3
jlpYR .
Applied Statistical The CL is drawn at P and there is no change in the formula for p (Eqn.(6) takes care of
Techniques
the unequal sample sizes).
You try the following exercise now.

E7) Construct the control limits of p-chart for the following data.
Wble 7 : Data far pchart with unequal sample sizes

Draw a rough sketch of the p-chart for this data.

-
np charts Sometimes it is nevessary (or convenient) to look at the number of defective items rather
than the proportion of defective items. In such situations, we use np - charts instead of
p - charts. The only difference between p - charts and np - churts is that in the latter case
y-axis represents the number of defective items in a subgroup.
Try to convince yourself that the control limits on an np-chart should be
CL = np, LCL = np - 3 J m ' , and UCL = np +3 Jm,
where p is same as defined for p-charts above.
Try the following exercise now.

E8) If the subgroup sample sizes are not the same, will the control limits vary in an
np-chart? In particular, what can you say about the center line, CL?

Finally, let us discuss the control charts, which are used tofind the number of defects
-
c chart per unit. These charts are useful in situations when the performance of a process is
assessed by number of defects per unit. Note that a unit may be a single product or a
fixed number of products.
PCBs are used in many For example, a printed circuit board (PCB) having several hundreds of circuits built in
electronic products such as it, may be treated as a unit. On the other hand, a bunch of ball pens pached in a carton
TV, Computer, etc. may also be treated as a unit.
The quality characteristic plotted on a c-chart is the total number of defects per unit,
denoted by c. Usually, Poisson distribution is a good approximation for c. Therefore, a
c-chart is constructed based on the assumption that c follows Poisson distribution. We
know that if m is the mean of a Poisson distribution, then its standard deviation is equal
to Jm.
Hence, the control limitsfor a c-chart are given by
CL=m,LCL=m-$6, and ~ C L = m + 3 6 .
As in the case of p-churts, the control limits are obtained by replacing m by its estimate
iii in the above formulae. If k is the total number of units inspected, then the average
number of defects per unit, denoted by i%i, is given by
- total number of defects in k units
m=
k
The following problem will explain what a c-chart is and how it is applied.
Problem 3. The assembly section of MIS BP & Co. has five groups of operators. Each
group consists of 3 operators. The job of the groups is to assemble various components
into ball pens and pack them in cartons. The groups are also responsible for identifying
and setting aside the defective components while assembling. Each carton consists of
200 pens. A sample of one carton from each group is selected at random and all the
pens in the carton are inspected. The total number of defects per carton (c) is recorded
for each group. The performance of each group is monitored by maintaining a c-chart Statistical Quality Cmtrol
fix each group separately. For one of the group, find (i) the average number of defects
per &it, (ii) the cmtml litnitsfor c-chart, and (in) plot tke contrbl chart.
%htfon. We plat the c-chart for group A of BP & Co. using the data as given in
Table 8.
Table 8 : Assembly defects data for group A.

Here, the average number of defects per carton is given by

m = 3 + 3 + ...+5 + 3 - 72
- - = 4.8.
15 15
And, the contml limits for c-churt are given by
C L = 14.8, LCL = 4.8 - 3 m 3 = -1.77 and UCL = 4.8 + 3J4?i = 11.37
Since we are plotting the number of defects per carton (c)on the chart, the control
limits are drawn using LCL = 0 and UCL = 11.37. The corresponding control chart is
as shown in Fig. 9.
UCL = 11.37

CL = 4.80

LCL = 0.00
I
i " " I " " I

1 5 10 15
sample number
Fig. 9

Try the following exercise to have a comparison between groups A and E.

E9) The table below gives the defects for group E for the same period.
Table 9 ;Assembly defects data for group E

(a) Estimate the average number of defects per carton.

(b) Is the assembly process of group E under statistical control?
(c) Which of the groups A and E, do you think is better?
(d) Explain how displaying the c-charts in front of these groups will help in
improving the process.

In this section, we have discussed use of control charts in building quality into a product
while it is being produced. Next, we discuss with you the concept of acceptance
sampling - a techniques used to ensure that the produced products conform to specified
quality standards. Here, it is important to remember that process controls are used to
control and systematically improve quality, but acceptance sampling is not.

11.4 ACCEPTANCE SAMPLING

- -

You know that, in most situations, we don't buy products directly from their
Applied Statistical manufacturer. The products from the manufacturer are first supplied to dealers, the
Techniques dealers supply them to retail shops and we buy them from retail shops.
Let us consider the case of Mr. Anil who is one of the dealers for MIS BP & Company.
Mr. Anil buys ball pens from the company in large quantities. He receives the pens in
lots where each lot contains 1000 pens. Mr. Anil will be too happy if all the lots that he
receives have no defective pens at all. But, we know that this never happens in practice.
Inspection is the process of Can we then think of a procedure that will ensure that no lot has a defective pen? If so,
comparing actual at what cost? Is it worth adopting such a procedure? You might think that 100%
measurable characterstics inspection before packing is a procedure that ensures defect-free lots. In fact, 100%
with pre-determined inspection is not always 100% efficient.
standard characteristics.
And, 100% inspection In some cases, there may be slips in inspection due to fatigue or measuring equipment
means inspecting all items errors. And, in some other situations, it may not be feasible to cany out 100%
in the lot. inspection. For example, if the product is bullets and inspection involves firing the
bullets, then 100% inspection means, we will be left with nothing. So, what is the
alternative?
We use sampling techniques in most such situations. In fact, samplipg is extensively
used in our day to day life. For example, we use sampling while purchasing vegetables
and groceries, for selection of our family doctor, and so on.
A sampling procedure used to accept or reject a lot of items is known as an
acceptance sampling plan (ASP, in short). Indeed, we use acceptance sampling plan as
an audit tool to ensure that the product cf a process conforms to requirements.

11.4.1 Sampling Plan Concepts

For a smooth discussion of the topic, we need an understanding of terms defined in the
following definition.
It is assumed that there are Definition. A lot is a collection of units of product picked for the purpose of sampling.
no inspection errors i s . , an Based on the results of inspection of a random sample from the lot a decision is made to
item*On found accept or reject all the units in the lot. The act of accepting or rejecting the entire lot is
to be defective if and only if called sentencing the lot.
it is actually so.
The number of units in a lot is called the lot size. The number of units inspected to
sentence a lot is called the sample size. The proportion of defective items in a lot is
called the lot quality. Throughout, we shall use (i) N for the lot size, (ii) n for its
sample size, and (iii) p for the lor quality.
There are three approaches to lot sentencing: (1)no inspection; (2) 100% inspection;
and (3) acceptance sampling. Here, we have to discuss the acceptance sampling and, to
understand this in a better way, let us start with a simple example of an acceptance
sampling plan for Mr. Anil, which we denote by ASP1.
The maximum allowable ASPI. From each lot of IOOOpens, take 100 at random and inspect them. Accept the lot
number of defective units in ifthe inspected sample contains at most one defective pen; otherwise reject it.
a sample is called the
acceDtmce number.which Thus, for ASPI, N = 1000 and n = 100. SO,if a lot consists of 20 defective pens, then
we denote by e. the lot quality p = 0.02. Also, observe that acceptance number c = 1in this case.
Now identify these quantities in the following exexcise.

E 10) Specify N, n and p for

(a) a lot of 400 bolts having 36 defective bolts;
(b) a box of 50 cricket balls having two defective balls;
(c) a situation when a sample of 40 bolts is drawn from a lot of 400 bolts and the
lot is rejected if the sample contains two or more defectives.
For sentencing, suppose a lot is subjected to ASP1 by Mr. Anil. What do you think is Statistical Quality Control
the probability that the lot will be accepted? Of course, this is one, if all the pens in the
lot are good; and it is zero, if all the pens in the lot are defective. Thus, the probability
of accepting a lot, denoted by Pa, depends on the number of defective items in the lot.
Suppose a lot has 20 defective items under ASPI. What is Pa for such a lot? Since ASP1
allows at most one defective pen in a sample of 100 pens, this probability is given by
Pa = P[X = 01 + P[X = I.],
where X is the number of defective pens in the sample.
We know that, in general, X has hypergeometric distribution and, so, we can compute Hypergeometric
the above probabilities using this fact. However, we know that when lot size N is large distribution with
compared to sample size n, these probabilities can be closely approximated by parameters N, G and n is the
distrlbut~onof the number
assuming that X follows binomial distribution with parameters n and p.
of good objects in a simple
random sample of size n

P[X = k] = "Ckpk(l - p)n7k,

where "Ck stands for the number of ways of choosing k items out of n items. And, the
probability of acceptance (Pa)
is given by

Pa = P[X 5 k] = " cpd(l

~ - p)n-d
Thus, for n = 100 and p = 0.02, we get Pa = P[X < I ] = 0.1326 -I-0.2706 = 0.4032.

even though their quality is same as those accepted.

Since Pa depends on p, from now onwards, we shall write it more explicitly as Pa(p).
Then, by above calculations, Pa(0.02) = 0.4032, for ASP1.
Try the following exercise.

E l 1) For ASP1, compute (a) P,(0.03); and (b) Pa(0.05).

.
The three curves shown in Fig. 10 (for c = 0,1, and 2) are developed by evaluating
Eqn.(ll) for various values of p, 0 5 p _< 1. So, each point on a curve is represented
by (P, Pa(p)).
Each curve is an oc curve (operating characreristic curve) for some acceptance
sampling plan. The preference of an acceptance sampling plan is completely described
by its oc curve.
In view of Eqn.(ll), a typical oc curve is a pictorial representation of the relationship
between the lot quality (p) and the probability of acceptance (P,), for a given sampling
plan. The greater the slope of an oc curve, the greater is the discriminatory power.
However, as c decreases, the oc curve gets shifted to left (without much a change in its
slope).
In general, the exact shape of a specific oc curve depends on the values of parameters
 Applied Statistical AQL, LTPD, a in this section.
Techniques

Fig. 10. oc curves for different values of c.

Here, we ask: What will happen if we change sampling plan ASP I? Let us consider
another acceptance sampling plan with c = 3, which we denote by ASI"2.
ASP2. From each lot oj 1000pens, inspect 100 at random and accept the lot if the
inspected sample has at most three defective pens.
Once again, let u, compute Pa for a lot which has exactly 20 defective items (i.e.,
I p = 0.02). Under ASP2, it is given by I
-

d=O
On comparing, we find that a lot of the same quality (as p = 0.02, in each case) has a
chance of 0.40 (Founded off to two decimal points) of getting accepted, if it is subjected
to ASPI, and has a chance of 0.86, if it is subjected to ASP2. So, we conclude that the
probability of accepting a lot depends on both (a) the lot quality; and (b) the
acceptance sampling plan.
Just as for ASPl, we can have oc curve for ASP2. In Fig.
- 11, the oc curves for three
different accept&ce plans are shown for comparison.
P, 1.0

0.6
113200,c - 4

,
0.0 I I I I
I I I
P
0.01 0.02 0.03 0.04 0.05 0.0
Fig. 11. oc curves for drfferent sample mes.
In Table 10, we have listed Pa values under ASPl and ASP2 for some selected values of
p. Examine these values to compare the two sampling plans.
Table 10 : Lot acceptance probabilities, Pa
1 D I P, under ASP1 1 P. under ASP2 1

I 1 1
0.010 0.7358 0.9816
0.020 0.4033 .0.8590
- 0.030 0.1946 0.6472
0.0525 0.3196
0.076 0.0034 0.0490
0.0003 , 0.0078
We shall continue discussing oc curve a little later. Here, we take a break to talk about
certain important parameters, which are used while adopting an acceptance sampling
an.
 Both, MIS BP & Co. and Mr. Anil, understand that supply of completely defect-free Statistical Quality Control
lots is not possible. So, they have to come to a compromise. This is how a bargain starts
between the two. Finally, they reach to an agreement : MI:Anil will accept majority of
lots which have at most 1% defective pens (i.e., p = 0.01) and the company will take
back all those lots which contain more than 1% defective pens.
A level of quality, which is mutually agreed upon by both the buyer and the seller is Acceptable Quality Level
called acceptable quality level (AQL). Thus, in view of above agreement between
Mr. Anil and MIS BP & Co., AQL = 0.01(= p). This means ;hat Mr. Anil should
accept lots with p 5 0.01 in majority of the cases.
Now, let us examine the consequence of Mr. Anil's decision, when AQL = 0.01 and
ASPl is adopted by him.
From Table 10, we find that Pa(O.O1) = 0.7358. So, under ASP1, Mr. Anil will reject
26.42% (=100(1-0.7358)) of the lots whose quality is 0.01. This is obviously a risk to
the producer because it was agreed upon by both that lots of quality 0.01 will be
accepted in majority of the cases whereas Mr. Anil is rejecting 26.42% of them. Of
course, a lot with p < 0.01 has a smaller chance of getting rejected than 26.42% (see
Table 10). In other words, the producer's risk for any p < AQL is less than 26.42%.
So, here is an important observation to note; Among all values of p between 0 and Producer's Risk
AQL, producer's risk is maximum when p = AQL. Producer's risk is defined as the
probability of rejecting a lot whose p = AQL and is denoted by a. Usually it is
expressed as a percentage. So, producer's risk in above situation is given by
a = lOO(1 - P,(AQL))%.
What will happen if Mr. Anil adopts ASP2 instead of ASPl? Try the following exercise
to find the answer.

E12) Assuming ASP2 is adopted, find out Pa(O.O1). Also locate (roughly) the points
AQL(= 0.01) on x-axis, producer's risk a on y-axis and (AQL, Pa(AQL)) on the
two oc curves plotted under ASPl and ASP2.

With AQL = 0.01 in above exercise, you must have got the producer's risk as 1.84%
under ASP2. Thus, under ASP1, the producer's risk is more and, so, the company
would say to Anil: 26.42% is too much of a risk for us, we would bear at most 5% risk.
What is the solution? We can change the sampling plan.
. On the other hand, under ASP2, the producer's risk is only 1.84% (much less than the
. desired 5%). So, the company will be complacent. But then, what will happen to
Mr. Anil as a consunkr?
From Eqn.(l 1), we find that Pa(0.05) = 0.2578, under ASP2. This means that, if ASP2
is adopted, then lots whose quality is as bad as having 5% defective pens (this is much
worse than the agreed upon quality AQL = 0.01) will get accepted 25.78% of the time.
Obviously, this is a risk to the consumer.
Why should Mr. Anil accept such bad lots 25.78% of the time? Accepting lots whose Consumer's Risk
quality is worse than AQL, the consumer is at a loss. Thus, ASP2 is not gobd for
Mr. Anil even though it is good for the producer.
If we want a sampling plan that is best for both the consumer and the producer, then all
those lots with p 5 AQL should be accepted with probability 1 and all those lots with
p > QAL should be rejected with probability 1 (or accepted with probability 0).
The oc curve of such a sampling plan will look like the one in Fig. 12 and is called an
ideal oc curve. It is clear that such an acceptance sampling plan would call for almost
 Applied Statistical 100% irispection and the inspection costs will be high.
Techniques
P,lO.

0.8-

0.6-

0.4-

0 2-

0.0 I
I I I I I
P
0.01 0.02 0.03 0.04 0.05 0.06 0.07

Fig. 12. Ideal oc curve.

Lot Tolerance Proportion As a compromise, Mr. Anil is ready to tolerate lot qualities worse than AQL upto
- Defective certah iimit, but not beyond. Let us assume Mr. Anil is willing to tderate lots with
p 2 0.05 but not not more than 10% of the time. The tolerance limit on the lot quality
p = 0.05 that he ).?: chosen to tolelate, is called the lot tolerance proportion defective
and is denoted by LTTD. 'Thus, LTPD = 0.05 in case of Mr. Anil.
In the above paragraph, we have mentioned that Mr. Anil is willing to accept lots with
quality p = LTPD ~ n l y10% of the time. Here, 10% is what we call the consumer's risk.
The probability of accepting a lot with p = LTPD is called the consumer's risk and is
denoted by 0.In other words, consumer's risk is equal to Pa(LTPD). Like the
producer's risk, the consumer's risk is also usually expressed as a percentage.
Above we have seen that while ASP1 is good for the consumer, ASP2 is good for the
producer. But neither of the t y o plans will satisfy both of them. So, we should look out
for a sampling plan that should be acceptaMe to both consumer and producer.
It is customer=to use the AQL and LTPD points for this purpose and the corresponding
points on the oc curve, a and p, respectively, give producer's and consumer's risk (see
Fig. 13). Such an acceptance sampling plan protects the interest of both producer and
consumer.

AQL = 0.013, a = 0.95

LTPD=0.016, P = 0.10

AQL LTPD 0.07

Fig. 13
Note that Pa for any lot with p < AQL is at least 1 - a ( or probability of rejecting such
If you want to have cake a lot is at most a ) and Pa for any lot with p > LTPD is at most P.
and eat it too, then you must
have too many cakes. Obtaining acceptance sampling plans that protect both producer and consumer usually
calls for large sample sizes. So, in practice, one has to compromise somewhere.
A sampling plan never guarentees the acceptance of 100% perfect material. In fact,
some defective material will also get accepted in the process. But, by proper
96 specification of 0risk, it is possible to minimise the amount of defective material.
Thus, a sampling plans is used to have a reasonably good idea about how much Statistical Quality Control
unacceptable material will be involved.

11.4.2 Single Sampling Plans

We use some statistical and probability tools to develop sampling plans that meet the
desired a & ,f3 risks and maintain the desired AQL and LTPD quality levels. Of course,
our main objective is to determine Pa of lots with varying quality.
There are many types of acceptance sampling plans. As before, we will confine our
discussion to sigle sampling plans and learn how to design them. Other types of
sampling plans will be discussed very briefly at the end of the section.
Already, we discussed some examples like ASP1 and ASP2. &re generally, a single Single Sampling Plans
s a m p l i n ~ l p nis completely described by specifying (i) the lot size, N; (ii) sample
size, n; and (iii) acceptance number, c .
In a single sampling plan, given by the three values (N, n, c), we inspect n items at
random from a lot of N items and accept the lot if the number of defective items in the
sample is less than or equal to the acceptance number c; otherwise we reject the lot.
And, throughout, we use simple random sampling without replacement for sampling
items.
A simple graphical procedure, called a binomial nomograph, is used to constfuct single
sampling plans for a specified AQL, LTPD, a and ,f3 (see Fig. 14).

Fig. 14. (Source: D.C.Montgomely, Introduction to Statistical Quality Control (4/e),2001.)

As an illustartion, we try to find a single sampling plan when AQL = 0.02, LTPD
= 0.08, a = 0.05 and ,f3 = 0.10.
Firstly, draw a straight 1ine.joining 0.02 on the p-scale and 0.95, (= 1 - a ) on the
Pa-scale. Draw another straight line joining 0.08 on the p-scale and 0.1(= P) on the
Pa-scale. Now, read the values of n and c corresponding to the point of intersection of
the two lines drawn. You can see t h s we will get n = 90 and c = 3. Observe that we
have not talked about lot size anywhere in the process.
Infact, the lot size is implicitly assumed to be at least 10 times the sample size. Thus, if
Applied Statistical Mr. Anil wants to use the above derived plan, his lot size should be at least 900. And, as
Techniques
N = 1000 for refill length problem, the above plan can be used.
Try the following exercise.

E13) From the nomograph shown in Fig. 14, derive single sampling plans when
(i)AQL= 0.01, LTPD= 0.10, a = 0.05, 0 = 0.10;and
(ii) AQL = 0.03, LTPD = 0.08, a = 0.05, = 0.10.

Percentage Sampling In the past, it was a common practice in industry to inspect certain percentage of items
in the lots. In other words, single sampling plans of the type (100, 10, O), (500,50,0)
and (1000,100,0) were often being used. The sample size in all these plans is 10% of
the lot size.
This may make us believe that all these plans are equally good. No, they are not! On
examining the oc curves for the 3 plans mentioned in this paragraph with c = 0 (see
Fig. 15), it is sufficiently clearly that the three plans are dkastically different.
Pa '.Oh

0.01 0.02 0.03 0.h 0.05 0.06 0.07 0.08

Fig. 15
The main disadvantage in this approach is that different sample sizes offer different
levels of protection. As such, it is illogical for the level of protection the consumer
enjoys for a critical part to vary as the size of the lot varies.
You will find the following exercise more convincing.

E14) Assuming AQL = 0.05, read, roughly, the producer's risks for the 3 plans in
Fig. 15.

There are many other sampling plans such as double sampling plans, multiple sampling
plans, sequential sampling plans, chain sampling plans, and so on. A number of
published sampling plans are developed from various view points. Of course, many
computer softwares are available today in the market'with which you can desing and
evaluate various sampling plans and schemes at your finger tips.
With this we have come to the end of the block. Let us summarise what we have learnt
in this unit.

1 . 5 SUMMARY

In this unit, we have discussed with you the following aspects of SQC.
1. The concept of quality and its characteristics.
2. How quality of a product is achieved by ensuring that quality characteristics
conform to their specificptions. Z

3. Use of control charts as a primary tool for an on-line process control.

4. The construction and application of K-R charts for controlling variable
characteristics.
5. The meaning of process capability and its evaluation through TI- R charts. Statistical Quality Control
6. The construction and application of p, n p , and c charts for controlling attribute
quality characteristics.
7 . The concept of acceptance sampling plans and use of the oc curve, AQL,
producer's risk, consumer's risk, and LTPD, while arriving at a plan that protect
the interest of both consumer and producer's equally. Also, we have seen how oc
curves have been used in comparing sampling plans.
8. The construction of a single sampling plan using binomial nomograph.

11.6 SOLUTPONIANSWERS

E l ) Measurable characteristics such as diameter of a ball, length of refill, weight of

cricket ball, thickness of a washer etc., are suitable for control charts for variables.
Characteristics such as defective ball pens, defects such as scratches on a cricket
ball, neps and faded portions on a piece of cloth, etc., are suitable for control
charts for attributes.
E2) Estimate of process mean is given by
- sum of sample averages - 7575
X = - -= 303 grams.
number of samples 25 a
Estimate of process standard deviation is given by
-
R sum of sample rangeslnumber of samples - 154125
-- - = 2.992 grams.
d2 2.059 2.059
Then, the control limits for TI - chart are given by
CL=303, LCL = F - A ~ = R 303 - 0.73 x 6.16 = 298.503,
+
UCL = 303 0.73 x 6.16 = 307.497.
~ n dthe
, control limits for R-chart are given by
CL = R = 6.16, LCL = d3R = 0, UCL = d4R = 2.28 x 6.16 = 14.045.
E3) After dropping 12th subgroup,
- 201.808 - 10.086 3.51 - 0.38
x= = 10.091, R = = 0.1647.
19 19
Therefore the revised control limits, for TI - chart, are
LCL = 9.995, CL = 10.091, UCL = 10.186; and
LCL = 0 CL = 0.1647, UCL = 0.3476, for R - chart.
From Table 1, we find that none of the points (excluding 12th subgroup), either on
Tt-chart or R - chart, fall outside control limits. So, the remaining data indicate
statistical control.
E4) We may take refill length problem. Here, total tolerance (= T) = 0.4. Then,

+
Thus, (a) rejctions will occur on-USL side when p 30 > USL. For instance, if
p = 10.111 (i.e., p +20 = USL), there will be rejections on USL side. And, (b)
rejctions will occur on LSL side when p + 30 < LSL. For instance, if p = 9.889
(i.e., p - 20 = LSL), there will be rejections on LSL side.
200.177 - 2.65
E5) For the combined data, F = -= 10.009 and R = -
20 20 = 0.1325. Thus, '

(a) estimates of mean and standard deviation for the new process are given by
-
- R 0.1325
j.2 = TI = 10.009 and 6 = - = -= 0.0569.
d2 2.326
And, (b) the control limits, for TI - chart, are given by
LCL = 9.932, CL = 10.009, UCL = 10.086; and
LCL = 0, CL = 0.1325, UCL = 0.2795, for R - chart.
Applied Statistical Again. (c) the first subgroup average is below LCL. on K - chart. Possibly there
Techniques
. was an assignable cause. But for this, the data indicate statistical control. And, (d)
since R-chart indicates control, we might use all the 20 subgroups to estimate the
process variability. An estimate of the process capability is given by

and, an estimate of process capability ratio is given by

Finally, (e) since cp< 1.~33,further reduction in process variability is essential.

So, the advice is that the manager should explore the possibilities of reducing
variation further.
E6) (b) There is clear indication that the process average has shifted downwards. (c)
We cannot comment on the process control unless we redraw the control limits.
19
(d) For the data in Table 6, CL = jj = = 0.019,
10 x 100"

Since, LCL < 0, LCL is plotted at 0.0. (e) Data indicate imkovement in the
process. The rejection percentage for the first period is equal to 6% (= 100 x old
jj estimate) and for the latter paiod it is equal to 1.9% (=I00 x new P estimate).
E7) The estimate of process average is given by P = & = 0.013. In this case, the
LCL turns out be zero for all the five subgroups. T i e UCL for subgroups with
subgroup size 100 is equal to 0.013 + 3 -4 = 0.048. The iable below
summarises the UCLs and sample averages for the data. :'

samplesize 100 121 8 1 , . 100 121

sample average 0.020 0.016 0.000 0.010 0.01 6
UCL 0.048 0.045 0.052 0.048 0.045
E8) YES. The control limits will vary with varying sample sizes. Even the center line
will vary because CL = np.
E9) (a) Estimated number of defects per carton = = 10. (b) The control limits are
: CL = 10, LCL = 10 - 3 m = 0.513 and UCL = 19.487. Since what we plot
on the y-axis is the total number of defects per carton, we may take LCL = 1and
UCL = 19. Clearly, assembly process of Group E is under statistical control. (c)
Since the average number of defects per carton of Group A is only 4.8, ~ r o u p A is
better. (d) If the chart is displayed in front of the operators, they will have a
continuous feed back on their performance. So, whenever the quality deteriorates,
they can correct themselves. Psychologically, it will have good impact on the
operators.
E10) (a) N = 400, p = 0.09; (b) N = 50, p = 0.04; (c) N = 400, n = 40, p = 0.005.
El 1) (a) Pa(0.03) = 0.1946; (b) Pa(0.05) = 0.0371.
E12) Under ASP2, P,(0.01) = 0.9816 (see Table 10). The actual points are
(0.01,0.7358), for ASPI, and (0.01,0.9816), for ASP2. So, these should be the
points that you read from the graph approximately.
E13) The single sampling plans from the nomograph are (i) (N, 4 0 , l ) and
(ii) (N, 200,lO). Here, the lot size N should be at least l0,times the the
corresponding sample size.
E14) The producer's risks are: 0.2141, for (1000,100,O); 0.2578, for (500,50,0); and
0.1498, for (100,10,0).
v2 = Dee- v , = Dtyees of freedom for numerator
of freedom
for
dcnom~nator I 2 3 4 5 6 7 8 9 10 12 I5 20 24 30 40 bU 120 0.

1 161 200 216 225 230 234 237 239 241 242 224 246 248 249 250 251 252 253 254
2 1830 19.00 19.20 19.20 19.30 19.30 19.40 19.40 19.40 19.40 19.40 19.40 19.40 19.50 19.50 19.50 19.5Q. 19.50 19.50
3 10.10 955 9.n 9.12 9.01 8.94 8.89 8.85 8.81 8.79 8.74 8 70 8.66 8 . 8 62 8.59 8.57 8.55 8.53
4 7 71 6.94 6.59 6.39 6.26 6.16 6.09 6.04 6.00 5.96 5.91 5.86 5.80 5.77 5.75 5.72 5.69 5 66 5.63
5 6.61 5.79 4 5.19 5 . 0 5 4.95 4.88 4.82 4.77 4.74 468 4.62 4.56 4.53 5 0 4.46 4 4 0 4.37

6 5.99 5.14 4.76. 4.53 4.39 4.28 4 . 4 . 4 . 4.06 4.00 3.M 3.87 3.84 3.81 3.77 3.74 3.70 367
7 5.59 4.74 4.35 4.12 397 3.87 3.79 3.73 3.68 3.64 337 3.1 3.44 341 3.38 3.34 330 3.27 3.23
8 5.32 4.46 4.07 3.84 . 3.69 3.58 3.50 3 3.39 3.35 3.28 3.22 3.15 3.12 3 08 3.04 3.01 2.97 2.93
9 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.4 307 3.01 294 2.90 2.86 283 2.79 2.75 2.7r
10 4.96 4 . 3.71 3.48 3.33 3.22 3.14 3.7 3.02 2.98 2.91 2.85 Z.7l 2.74 270 2.66 2.62 2.58 2.54

II 4.84 3.98 359 3.36 3.20 3.09 3 01 2-95 2.90 2.85 2 79 2.72 265 2.61 2 57 2.53 2.49 2.45 240
12 4.75 3.89 349 3.26 -3.11 3.00 2.91 2.85 2.80 2.75 2.69 2.62 2.54 2.5 2.47 2.38 2.38 2.30 2.30
13 4.67 3.81 4 3.18 3.03 2.92 2.83 5.77 2.71 267 2.60 2.53 2.46 2.42 2.38 2.34 2.30 2.25 221
I4 460 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 2.60 2.53 2.46 2.39 2.35 231 2.27 2.22 2.18 213
15 4 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 254 2.48 2.40 2.33 2.29 2.25 2.20 216 2.11 2.07

16 4-49 3.63 3.24 3.01 2.85 2.74 2.66 2.59 2.54 249 2.42 2.35 2.28 2.24 2.19 2.15 2.11 2.06 2.01
17 3.45 359 3.20 2.96 2.81 2.70 2.61 255 2.49 2.45 2.38 231 2.23 2.19 2.15 210 2.06 2.01 1%
18 4-41 335 3.16 2.93 2.77 2.66 258 251 2.46 2.4 2.34 2.27 2.19 2.15 211 2.06 202 1.97 193
19 4.38 332 3.13 2.90 2.74 2.63 2.54 2.48 2.42 2.38 2.31 2.23 216 2.11 2.07 2.03 1.98 1.93 1.88
20 4.35 3.49 3.10 2.87 2.71 2.66 2.51 2.45 2.39 2.35 2.28 2.20 2.12 208 2.04 199 1.95 1.90 184

21 4.32 3.47 3.07 2.84 2.68 2.57 2.49 242 2.37 2.32 225 2.18 2.10 2.05 2.01 I.% 1.92 1.87 181
22 4.30 3 . 3.05 2.82 2.66 2.55 2.46 2 40 2.34 2.30 2 23 2.15 2.07 2.03 1.98 1.94 1.89 I . 1.78
23 4.28 3.42 3.03 2.80 2.64 253 2 U 2.37 2.32 227 . 220 213 2.05 201 1.96 191 1.86 1.81 1.76
24 4.26 3.40 3.01 2.78 2.62 2.51 2.42 236 2.30 2.25 218 2.11 2.03 1.98 1.94 1.89 1.84 179 173
25 4.24 3.39 2.99 2.76 2.60 2.49 2.40 2.34 2.28 2.24 2 16 2.09 2.01 196 1.92 1.87 1.82 1.77 1.71

30 4.17 3.32 2.92 2.69. 253 2.42 2.33 2.27 2.21 2.16 2.09 2.01 1.93 1.89 1.84 179 7 1.68 162
40 4 08 3.23 2.84 2.61 2 45 2.34 2.25 2.18 2.12 2.08 2.00 1.92 1.84 1.79 1.74 1.69 6 I58 1.51
60 4.00 3.15 2.76 253 2.37 2.25 2.17 2 10 2.04 . 1.99 192 1.84 1.75 1.70 165 1.59 I33 4 1.39
120 3.92 3.07 2.68 2.45 2.29 2.18 2.09 2.02 I.% 1.91 1.83 1.75 1.66 1.61 1.55 1.50 1.43 1.35 1.25
w 3.84 3.00 2.60 2.37 2.21 2.10 2.01 1.9). 1.88 8 I . 1.67 I . I52 146 1.39 1.32 1.22 1.00
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 TABLE-2
t-distribution