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Chapter 10 - Quality Control
CHAPTER 10
QUALITY CONTROL
Teaching Notes
As a result of increased global competition, a rapidly growing number of companies of all sizes are
paying much more attention to issues involving quality and productivity. Many statistical techniques are
available to assist organizations in improving the quality of their products and services. It is important for
companies to use these techniques in the context of an overall quality system (Total Quality Management)
which requires quality awareness, careful planning and commitment to quality at all levels of the
organization. Many companies are not only utilizing these statistical techniques themselves, but are also
requiring their suppliers to meet certain standards of quality based on various statistical measures. This
chapter covers the statistical applications of quality control. Control charts are given the primary
emphasis, but other quality control topics such as process capability and inspection are also discussed.
When covering the material in this chapter, we need to stress that through the use of control charts, the
nonrandom (special) causes of variation must be controlled before random (common) causes of variation
and process capability can be analyzed.
Answers to Discussion and Review Questions
1.
The elements in the control process are:
a. Define
b. Measure
c. Compare to standard
d. Evaluate
e. Take corrective action if needed
f.
Evaluate corrective action to insure it is working
2.
Control charts are based on the premise that a process which is stable will reflect randomness:
statistics of samples taken from the process (means, number of defects, etc.) will conform to a
sampling distribution with known characteristics, so that statistical significance tests can be
performed on sample statistics; and successive samples will not reveal any patterns which will
enable prediction of future values other than specification of range of variability.
3.
Control charts are used to judge whether the sample data reflects a change in the parameters (e.g.,
mean) of the process. This involves a yes/no decision and not an estimation of process
parameters.
4.
Order of observation of process output is necessary if patterns (e.g., trends, cycles) in the output
are to be detected.
5.
a.
x chart—A control chart used to monitor process variables by focusing on the central
tendency of a process.
b. Range control charts are used to monitor process variables, focusing on the dispersion of a
process.
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Chapter 10 - Quality Control
c. p-chart—is a control chart for attributes, used to monitor the proportion of defectives in a
process.
d. c-chart—is a control chart for attributes, used to monitor the number of defects per unit.
6.
A run is a sequence of observations with a given characteristic. Run tests are helpful in detecting
patterns in time series (e.g., control chart) data.
7.
All points can be within control limits but with certain patterns developing in the data which
would suggest the output is not random, and hence, not in control for long.
8.
It is usually desirable to use both an up/down and a median run test on a given set of data because
the tests are sensitive to different things. For example, one test can be more sensitive to trend and
the other to bias.
9.
No, there is always the possibility of a Beta or Type II error which is the probability of calling
something random when in fact it is non-random or concluding that non-randomness is not
present, when it actually is.
10.
Specifications are limits on the range of variation of output which are set by design (e.g.,
engineering, customers). Control limits are statistical bounds on a sampling distribution. They
indicate the extent to which summary values such as sample means or sample ranges will tend to
vary solely on a chance basis. Process variability refers to the inherent variability of a
processthe extent to which the output of a process will tend to vary due to chance. Control
limits are a function of process variability as well as sample size and confidence level. Both are
essentially independent of tolerances.
11.
The problem is that even when the machine is functioning as well as it can, unacceptable output
will result. Among the possible options that should be considered are:
a. Use 100 percent inspection to weed out the defectives. If destructive testing is required, this
may not be feasible.
b. Attempt to convince customers (offer a lower price?) to widen tolerances or engineering
(communicate the cost of 100 percent inspection if relevant). The problem is that engineering
may resent this suggestiondepending on how it is handled. Moreover, it may be that the
tolerance is necessary for proper functioning of the final product or service.
c. Attempt to substitute a different machine (e.g., a newer one) which has the capability to
handle the job.
d. Hope for a miracle.
12.
a. This “problem” often goes undetected since there are no complaints from customers about
output not within specs. However, it is quite possible to realize decreased costs or more
profits by taking certain actions.
b. A “marketing approach” to this problem might be to see if the customer is willing to pay
more for output that meets narrower tolerances. If not, perhaps the job could be shifted to
another, less capable machine, freeing up this equipment for more demanding work. Still a
third option would be to cut back on inspection since virtually 100 percent of the output will
be acceptable, and even a slight out of control situation will not warrant corrective action.
13.
a. An optimum level of inspection is one where the cost and effort of inspection equals the
benefits derived from inspection, or the point (number of units inspected) at which the
marginal cost of inspection equals the marginal benefit from inspection.
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Chapter 10 - Quality Control
b. Cost of product or service, volume, costs of inspection, cost of letting undetected defects slip
through, degree of human involvement, stability of process, and the number and size of lots.
c. The main issues in the decision of whether to inspect on site or in a central location are the
situation (size and mobility), inspection time, costs of process interruption, need for quick
decision, importance to avoid extraneous factors affecting samples or tests, need for
specialized equipment, and the need for a more favorable testing environment.
d. Raw materials and purchased parts, finished products, before a costly operation, before an
irreversible process, and before a covering process.
14.
Two basic assumptions that must be satisfied in order to use a process capability index are:
a. The process is stable (Non-random causes of variation have been identified and corrected);
b. The process distribution is normal.
15.
It is very important. The company’s (and manager’s) reputation is at stake, and there may be cost,
liability, legal and safety issues. Although the risks may differ substantially for different products
and services, ethical standards should be maintained “across the board.”
16.
a. Type I error
b. Type II error
c. Eating a dirty cookie is a Type II error
Not eating a clean cookie is a Type I error
d. Type II error
Taking Stock
1.
a. In deciding whether to use 2-sigma or 3-sigma limits, the quality control people should be
involved as well as the accounting /record keeping personnel because it will be critical to
determine the cost of unnecessarily stopping the process vs. cost of not correcting a special
cause of variation. In addition, we may want to involve the customers’ quality control
personnel since they will be ultimately affected by this decision.
b. The quality control department should make this decision. However, input from the
production control department may be very useful to estimate the cost of sampling.
c. Increasing the capability of the process is a significant and potentially costly adventure.
Therefore, the upper management in consultation with the quality and production
departments should determine what type of systematic improvements need to be made to
improve the capability of the process.
2.
In setting the quality standards, customers should definitely be involved since they will ultimately
be using the product. In consultation with the upper management, the quality and production
departments should work as a team in establishing the quality standards.
3.
The technology had a profound impact on quality. Improvement in measurement systems
drastically improved the measurement of quality. The computer technology has enabled many
companies to perform on-line, real-time statistical process control, which enabled companies to
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Chapter 10 - Quality Control
4.
respond to quality problems faster. Due to technological improvements in computerized design, the
products are designed better, thus have significantly fewer quality problems. The artificial
intelligence systems forewarn potential problems before they occur.
Critical Thinking
1. If the analysis of the output of a process suggests that there is an unusual occurrence, but the
result of the investigation cannot pinpoint or determine the assignable causes, the limits may be
set too tight. Loosening the limits will allow us to determine the possible assignable causes of
variation. If the record keeping is poor it may be very difficult to identify the assignable causes of
variation. Improving the record keeping may assist us in identifying non-random or assignable
causes of variation.
2. A single standard would be easier to work with, and everyone would know what the standard
was. Multiple standards might be used if the cost of errors differed significantly across products
or services. For instance, not meeting the specifications for a product such as paper clips
wouldn’t have the same degree of importance as meeting the specifications for heart-monitoring
equipment. Also, differing standards might reflect progress in continuous improvement; as
processes are improved, the standard would be increase to reflect that, and perhaps be used as a
competitive strength in B2B dealings.
3. Student answers will vary
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Chapter 10 - Quality Control
Memo Writing Exercises
1.
A p-chart is used to monitor the proportion of defective units generated by a process,
while an x chart is used to monitor the central tendency of a process (i.e. change in the
mean or the nominal value of a process). A p-chart classifies the observations into one
of two mutually exclusive categories (good vs. bad pass vs. fail, etc.). An x chart
usually requires taking measurements in data collection to monitor the average of a
process. Examples of characteristics requiring an x chart include measurement of a
diameter of a tire, length of a bolt, tensile strength of a rubber product, and weight of a
cereal box. In using a p-chart data collection is usually easier because instead of taking
actual measurements, we would simply record whether the item is conforming or not
conforming. In addition, the p-chart requires a considerably larger sample size than the
x chart. On the other hand, if workers do the charting on the line, the training required
for p-chart is simpler than the training required for x and R chart.
x chart is usually preferred over p-chart for characteristics that require taking actual
measurements because the lower cost of sampling and higher information content will
outweigh the extra cost of measurement. However, when the characteristic in question
is a dichotomous classification (defective vs. nondefective, on vs. off) the x chart is not
applicable and p-chart should be used.
2.
In order to monitor the capability of the process to control the number of defective units,
we must first make sure that the special (assignable) causes of variation are eliminated
with the use of control charts. However, control charts will not pinpoint the cause of
defective units because the natural process variability may exceed the specification
limits (tolerances). In other words, even if a process is in control using control charts it
may still be producing defective units. Therefore, after the process is deemed to be in
control using control charts, we need to determine whether the process is capable by
making sure that the natural variation of the process is within the specification limits,
and the percentage of expected number of defective units is acceptable.
Additional Experiential Learning Exercises
Sampling Demonstrations. Bowls of colored beads (e.g., 1,000 beads, 40% white, 30% green,
20% red, 5% black, 3% yellow, etc.) are available with paddles that have indentations :that
facilitate quickly obtaining samples of various sizes. Use to demonstrate sampling variability –
take repeated small samples (or have a student do it) to demonstrate that different percentages of
a color appear in different samples. Then increase the sample size, focusing on a specific color,
and have students recognize that there is less variability (i.e., sampling variability distribution)
as the sample size increases. Afterward, explain why in process sampling, small samples are
used even though large samples are more “accurate.” One is the cost, time, and disruption
caused by taking large samples. Even more important is the ability to capture process changes
that might occur between small samples (e.g. 6 samples of n = 10 taken over time could reveal a
trend, whereas 1 sample of 60 could not).
Process Control Demonstrations. Obtain 30 clear plastic “zipper” bags and an ample supply
of colored beads, marbles, or jelly beans.
a.
Place 10-20 beads/beans in each of 10 bags, focusing, say on red beads/beans. Number
the bags 1 to 10. Put 0, 1, or 2 red beads/beans in each bag, along with other colors of
beads/beans. Pass the bags out to the class and ask them what kind of control chart
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Chapter 10 - Quality Control
b.
c.
would be appropriate (p-chart or c-chart). Then say, let’s suppose the red beads/beans
are defects, so we shall construct a c-chart. Have then report (in order), the number of
defects they have counted. Next, have the class calculate 2s control limits, and then see
that all are within the control limits.
In a second set of bags numbered 11-21, arrange it so that bag #14 has too many red
beads/beans, making it “out of control.” Pass out those bags and have the students report
the numbers of red they find. When the student with bag #14 reports a large number of
reds, tell the class that the process would be halted while efforts were made to find and
correct the problem. Then resume “sampling” with the process now back in control.
Arrange the third set so that a trend begins to appear at about the fifth bag, but with all
results still within the limits. Point out that the process doesn’t appear to be random, so
it would be halted at the ninth or tenth sample to find and correct the problem.
Solutions
1.
specs: 24 oz. to 25 oz.
.0062
.0062
=
 = 24.5 oz. [assume  = x ]
 = .2 oz.
-2.5
0
24
24.5
a. [refers to population]
 .5
z
 2.5  2(.0062 )  .0124
.2

.2
b.   2
 24 .5  2
 24 .5  .10 or 24 .40 to 24 .60
n
16
2.
 = 1.0 liter
 = .01 liter
n = 25
a.
Control limits :   z

1.0043
n
[z = 2.17 for 97%]
.01
UCL is 1.0  2.17
 1.0043
25
LCL is 1.0  2.17
1.006
b.
(liters) 1.002
Mean
+2.5
z-scale
25
16
out
UCL
*


1.000
.998
.01
 .9957
25


LCL
.9957
*
.994
out
3. a. n = 20
A2 = 0.18
D3 = 0.41
D4 = 1.59
=
=

X = 3.10 Mean Chart: X ± A2 R = 3.1 ± 0.18(0.45)

= 3.1 ± .081
R = 0.45
Hence, UCL is 3.181
and LCL is 3.019. All means are within these limits.

Range Chart: UCL is D4R = 1.59(0.45) = .7155

LCL is D3R = 0.41(0.45) = .1845
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Chapter 10 - Quality Control
4.
5.
In control since all points are within these limits.
Range
=

Mean Chart: X ± A2R = 79.96 ± 0.58(1.87)
2.6
Sample
Mean
1
79.48
2
80.14
2.3
3
80.14
1.2
4
79.60
1.7
5
80.02
2.0
6
80.38
1.4
= 79.96 ± 1.08
UCL = 81.04, LCL = 78.88

Range Chart: UCL = D4R = 2.11(1.87) = 3.95

LCL = D3R = 0(1.87) = 0
[Both charts suggest the process is in control: Neither has any
points outside the limits.]
n = 200
a.
1
2
3
4
.020 .010 .025 .045
b. (2.0 + 1.0 + 2.5 + 4.5)/4 = 2.5%
c. mean = .025
Std. dev. 
p (1  p )
.025 (.975 )

 .011
n
200
d. z = 2.17
.025 ± 2.17(0.011) = .025 ± .0239 = .0011 to .0489.
e. .025 + z(.011) = .047
Solving, z = 2, leaving .0228 in each tail. Hence, alpha = 2(.0228)
= .0456.
f.
Yes.
g. mean = .02
Std. dev. 
.02 (.98)
 .0099 [round to .01]
200
h. .02 ± 2(.01) = 0 to .04
The last sample is beyond the upper limit.
6.
n = 200
p
25
 .0096
13(200 )
Control Limits = p  2
p (1  p )
n
 .0096  2
.0096 (.9904 )
200
 .0096  .0138
Thus, UCL is .0234 and LCL becomes 0.
Since n = 200, the fraction represented by each data point is half the
amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc.
10-7
Chapter 10 - Quality Control
Sample 10 is too large.
7.
c
110
 7.857
14
Control limits: c  3 c  7.857  8.409
UCL is 16.266, LCL becomes 0.
All values are within the limits.
10-8
Chapter 10 - Quality Control
8.
c
21
 1.5
14
Control limits: c  3 c  1.5  3.67
UCL is 5.17, LCL becomes 0.
All values are within the limits.
9.
p
total number of defectives
87

 .054
total number of observatio ns 16 (100 )
p(1  p)
.054 (.946 )
 .054  1.96
n
100
 .054  .044 . Hence, UCL = 0.10
LCL = 0.01
Note that observations must be converted to fraction defective, or control limits must be
converted to number of defectives. In the latter case, the upper control limit would be 10
defectives and the lower control limit would be 1 defective. Even though all points are within
these limits, the process appears to be out of control because 75% of the values are above 4%.
Control limits are p  z
10.
There are several slightly different ways to solve this problem. The most straightforward seems to
be the following:
1) Observe that the upper control limit is six standard deviations above the lower control limit.
2) Compute the value of the upper control limit at the start:
15  6
.02
 15.12 cm.
1
3) Determine how many pieces can be produced before the upper control limit just touches the
upper tolerance, given that the upper limit increases by .004 cm. per piece:
15.2cm. – 15.12cm.
= 20 pieces.
.004 cm./piece
11.
Out of the 30 observations, only one value exceeds the tolerances, or 3.3%. [This case is
essentially the one portrayed in the text in Figure 10–9A.] Thus, it seems that the tolerances are
being met: approximately 97 percent of the output will be acceptable.
12.
a.  = .146
n = 14
x
Control limits are x  3

n
 x  150 .15  3.85
39
39
 .385  3
0.146
 3.85  .117
14
So UCL is 3.97, LCL is 3.73. Sample 29 is outside the UCL, so the process is
not in control. Because sample 29 lies outside the control limits, the control
chart limits should be recalculated without sample 29.
10-9
Chapter 10 - Quality Control
b. [median is 3.85]
Sample
A/B
13.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
A
A
B
B
B
B
A
A
B
B
A
A
A
B
B
A
B
A
B
A
Test
Median
Up/down
obs.
18
29
Mean
U/D
Sample
A/B
Mean
U/D
3.86
3.90
3.83
3.81
3.84
3.83
3.87
3.88
3.84
3.80
3.88
3.86
3.88
3.81
3.83
3.86
3.82
3.86
3.84
3.87

U
D
D
U
D
U
U
D
D
U
D
U
D
U
U
D
U
D
U
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
B
B
A
A
A
A
B
A
A
A
A
B
B
B
A
B
B
B
[B]
3.84
3.82
3.89
3.86
3.88
3.90
3.81
3.86
3.98
3.96
3.88
3.76
3.83
3.77
3.86
3.80
3.84
3.79
3.85
D
D
U
D
U
U
D
U
U
D
D
D
U
D
U
D
U
D
U
exp.
20.5
25.7

3.08
2.57
Z
–.81
1.28
Conclusion
random
random
a.
A A B B A B A B B A B A A B A A A B B B A B A B A B


























D D D U D U D D U D U U D U U D D D U U D U D U D
b.
A A A A B A A B B B B B A B B B A A A A B B B B B B














 


 






U D U D U D D U D U D U D U D U D U D D U D U D D
10-10
Chapter 10 - Quality Control
14.

z
14
2.50
1.6
random
17
17
2.07
0.0
random
8
22
14
17
2.50
2.07
Summary:
obs.
exp.
a. median
18
up/down
b. median
up/down
Conclusion
–2.40 nonrandom
2.41 nonrandom
a. Because neither z-score exceeds +2.00, the process output is probably random.
b. N = 20
Test
Median
Up/Down
r
expected
Std. dev.
z-score
Conclusion
14
11
2.18
+1.38
random
8
13
1.80
-2.78
nonrandom
Because the Up/Down test had a z-score outside of z = +2.00, the analyst can conclude the sequence is
probably nonrandom, so the process should be investigated for a possible assignable cause of variation.
c. [Data from Chapter 10, Problem 8]
Median is 1.5 A = Above, B = Below, U = Up, D = Down.
Sample:
1
2
3
4
5
6
7
8
9
10 11 12 13 14
Median:
A
A
B
B
B
A
A
B
A
B
A
B
A
B
Data:
Up/down:
2
–
3
U
1
D
0
D
1
U
3
U
2
D
0
D
2
U
1
D
3
U
1
D
2
U
0
D
d. [Data from #7] Median is 7.5.
Day:
Median:
1
B
2
A
3
A
4
A
5
A
6
B
7
B
8
A
9
A
10
B
11
B
12
B
13
B
14
A
Data:
4
10
14
8
9
6
5
12
13
7
6
4
2
10
Up/down
–
U
U
D
U
D
D
U
U
D
D
D
D
U
10-11
Chapter 10 - Quality Control
For part c and d:
E(r)med =
N
14
 1   1  8 runs
2
2
E(r)u/d =
2 N  1 2(14 )  1

 9 runs
3
3
med 
N 1
14  1

 1.803 runs
4
4
16 N  29
224  29

 1.472 runs
90
90
For part c:
u / d 
10  8
 1.109
1.803
For part d:
Zup/down 
Zmed =
10  9
 .679
1.472
68
79
 1.109
Zup/down 
 1.36
1.803
1.472
Since the absolute values of all Z statistics calculated above are less than 2, all patterns appear to
be random.
Zmed =
Summary:
Test
a. median
up/down
b. median
up/down
Exp.
8

1.80
z
1.11
Conclusion
random
10
9
1.47
.68
random
6
8
1.80
–1.11
random
7
9
1.47
–1.36
random
obs.
10
10-12
Chapter 10 - Quality Control
15.
Day
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Amount
B 27.69
B 28.13
A 33.02
B 30.31
A 31.59
A 33.64
A 34.73
A 35.09
A 33.39
A 32.51
B 27.98
A 31.25
A 33.98
B 25.56
B 24.46
B 29.65
A 31.08
A 33.03
B 29.10
B 25.19

U
U
D
U
U
U
U
D
D
D
U
U
D
D
U
U
U
D
D
Day
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Amount
B 28.60
B 20.02
B 26.67
A 36.40
A 32.07
A 44.10
A 41.44
B 29.62
B 30.12
B 26.39
A 40.54
A 36.31
B 27.14
B 30.38
A 31.96
A 32.03
A 34.40
B 25.67
A 35.80
A 32.23
U
D
U
U
D
U
D
D
U
D
U
D
D
U
U
U
U
D
U
D
Summary:
Test
median
obs. exp.
22 31

3.84
z
–2.34
Conclusion
non-random
up/down
35
3.22
–1.45
random
39.67
Day
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Amount
B 26.76
B 30.51
B 29.35
B 24.09
B 22.45
B 25.16
B 26.11
B 29.84
A 31.75
B 29.14
A 37.78
A 34.16
A 38.28
B 29.49
B 30.81
B 30.60
A 34.46
A 35.10
A 31.76
A 34.90
D
U
D
D
D
U
U
U
U
D
U
D
U
D
U
D
U
U
D
U
Since one of the tests suggests non-randomness, the conclusion must be that the process is not in
control. In other words, the variation in daily expenses is not random. Further investigation would
be necessary in order to determine what sort of pattern is present.
10-13
Chapter 10 - Quality Control
16.
3.5
cm
UCL
n=1
 = 0.05 cm
3.44
Mean
LCL
3 sigma
3 sigma
3.0 cm
(i) The upper control limit is 6 standard deviations above the lower control limits.
(ii) When UCL = 3.5 cm, the LCL = 3.5 – 6
0.05
 3.5  0.30  3.2 cm
1
(iii) Determine how many pieces can be produced before the LCL just crosses the lower
tolerance of 3 cm.
3.20 – 3.00
0.20
200
=
=
= 200 pieces
0.001
0.001
1
17.
It is necessary to see if the process variability is within 9.96 and 10.35. Two observations have
values above the specified limits, i.e., 10% of the 20 observations fall outside the limits. Perhaps
the process mean should be set a bit lower.
18.
1 Step 10% scrap, 2nd 6%, and 3rd 6%.
a. Let x be the number of units started initially at Step 1. With a scrap rate of 10% in Step 1 the
input to Step 2 is 0.9x. The input to Step 3 is (1 – 0.06) (1 – 0.10)x. With a scrap rate of 6% at
Step 3 the number of good units after Step 3 = (1 – 0.06) (1 – 0.06) (1 – 0.10)x =
(0.94)2(0.90)x.
The required output is 450 units
(0.94)2(0.90)x = 450 units
x = 565.87  566 units
b. (1 – 0.03)2(1 – 0.05)x = 450 units
(0.97)2(0.95)x = 450 units
x = 503.44  504 units
Savings of 566 – 504 = 62 units
c. From (a)
The scrap = 566 – 450 = 116 units
@ $10 per unit, The Total Cost = $1,160.00
10-14
Chapter 10 - Quality Control
19.
Sample #
1
3
Median = 2.5 A/B A
Up/Down
–
N = 20
2
2
B
D
3 4 5
4 5 1
A A
U U
Observed Expected
Median
13
11
Up/down
14
13
20.
a.
1
4.3
2
4.5
3
4.5
6
2
B
D
7
4
B
U
8 9 10 11 12 13 14 15
1 2 1 3 4 2 4 2
A B B B A A B A
U D U D U U D U
Standard
Deviation
2.1794
1.7981
16
1
B
D
17 18 19 20
3 1 3 4
B A B A A
D U D U U
Z
Conclude
0.9177 Random
0.5561 Random
4
4.7
b. =
x = (4.3 + 4.5 + 4.5 + 4.7)/4 = 4.5
std. dev. (of data set) = .192
c. mean = 4.5, std. dev. = .192 / 5  .086
d. 4.5 ± 3(.086) = 4.5 ± .258 = 4.242 to 4.758
The risk is 2(.0013) = .0026
e. 4.5 + z(.086) = 4.86
Solving, z = 4.19, so the risk is close to zero
f.
None

g. R = (.3 + .4 + .2 + .4)/4 = .325
n=5
Means: A2 = 0.58

=
x ± A2R = 4.5 ± 0.58(.325) = 4.3115 to 4.6885.
The last mean is above the upper limit.
Ranges: D3 = 0
0 to 2.11(.325) = 0 to .6875
All ranges are within the limits.
h. Two different measures of dispersion are being used, the standard deviation and the range.
i.
21.
4.4  3
0.18
5
 4.4  .241  4.16 to 4.64. The last value is above the upper limit.
Solution
a.
Cp 
specificat ion width
.02
.02


 1.11
process width
6(.003 ) .018
b. In order to be capable, the process capability ratio must be at least 1.33. In this instance, the
index is 1.11, so the process is not capable.
10-15
Chapter 10 - Quality Control
22.
23.
Process Standard Deviation (in.) Job Specification (in.) Cp
Capable ?
001
0.02
0.05
0.833
No
002
0.04
0.07
0.583
No
003
0.10
0.18
0.600
No
004
0.05
0.15
1.000
No
005
0.01
0.04
1.333
Yes
Process
A
Cost per unit ($) Standard Deviation (mm.) Cp
20
0.059
1.355
B
12
0.060
1.333
C
11
0.063
1.27
D
10
0.061
1.311
You can narrow the choice to processes A and B because they are the only ones with a capability
ratio of at least 1.33. You would need to know if the slight additional capability of process A is
worth an extra cost of $8 per unit.
24.
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
X = Process mean,  = Process standard deviation
For process H:
X  LSL 15  14 .1

 .93
3
(3)(.32 )
USL  X 16  15

 1.04
3
(3)(.32 )
C pk  min .938 , 1.04  .93
.93  1.0, not capable
For process K:
X  LSL 33  30

 1.0
3
(3)(1)
USL  X 36 .5  33

 1.17
3
(3)(1)
C pk  min{1.0, 1.17}  1.0
Assuming the minimum acceptable C pk is 1.33, since 1.0 < 1.33, the process is not capable.
10-16
Chapter 10 - Quality Control
For process T:
X  LSL 18 .5  16 .5

 1.67
3
(3)(0.4)
USL  X 20 .1  18 .5

 1.33
3
(3)(0.4)
C pk  min{ 1.67 , 1.33}  1.33
Since 1.33 = 1.33, the process is capable.
25.
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
X = Process mean,  = Process standard deviation.
USL = 90 minutes,
X 1 = 74 minutes,
LSL = 50 minutes,
 1 = 4.0 minutes
X 2 = 72 minutes,
 2 = 5.1 minutes
For the first repair firm:
X  LSL 74  50

 2.0
3
(3)(4.0)
USL  X 90  74

 1.333
3
(3)(4.0)
C pk  min{ 2.0, 1.33}  1.333
Since 1.333 = 1.333, the firm 1 is capable.
For the second repair firm:
X  LSL 72  50

 1.44
3
(3)(5.1)
USL  X 90  72

 1.18
3
(3)(5.1)
C pk  min{1.44,1.88}  1.18
Assuming the minimum acceptable C pk is 1.33, since 1.18 < 1.33, the firm 2 is not capable.
10-17
Chapter 10 - Quality Control
26.
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
X = Process mean,  = Process standard deviation.
USL = 30 minutes,
X Armand = 38 minutes,
X Jerry = 37 minutes,
X Melissa = 37.5 minutes,
LSL = 45 minutes,
 Armand = 3 minutes
 Jerry = 2.5 minutes
 Melissa = 2.5 minutes
For Armand:
X  LSL 38  30

 .89
3
(3)(3)
USL  X 45  38

 .78
3
(3)(3)
C pk  min{. 89, .78}  .78
Since .78 < 1.33, Armand is not capable.
For Jerry:
X  LSL 37  30

 .93
3
(3)(2.5)
USL  X 45  37

 1.07
3
(3)(2.5)
C pk  min{. 93, 1.07}  .93
Since .93 < 1.33, Jerry is not capable.
For Melissa, since USL  X  X  LSL  7.5 , the process is centered, therefore we will use Cp
to measure process capability.
USL  LSL 45  30

 1.39
6
(6)(1.8)
Since 1.39 > 1.33, Melissa is capable.
Cp 
10-18
Chapter 10 - Quality Control
27. a. Cp = spec width = 20 = 1.33. Solving,  = 2.506.
6
6
b. Box variance = 3.84; box  = 1.96. Average box weight = 6(1.01) = 6.06 ounces. Students can
deduce that 1 ounce = 28.33 grams, making the box weight in grams: 6.06(28.33) = 171.70 gr.
Cpk =
Upper spec  171.70
3
=
180  171.70
3(1.96)
= 1.41 (capable).
c. The lowest setting implies a mean that is less than average:
Cpk =
Mean  Lower spec
3(1.96)
= 1.33.
Solving, mean = 167.82. Mean/6 = 27.97 gr. = .987 ounces.
28.
Note that all points are within the control limits, so the process is apparently in control.
Run tests:
Test
Median
Up/down
Observed
15
19
Expected
12
14.33
Std. dev.
2.29
1.89
Z
1.31
2.46
Conclusion
random
non-random
Because one of the run tests indicated that the output is not random, the process is probably not random,
and should be investigated to determine the cause.
10-19
Chapter 10 - Quality Control
29.
Step 1: a. A c chart is appropriate.
b. The mean of the data is c = 18/12 = 1.50. Control limits are
c + z c = 1.50 + 2(1.225) = 1.50 + 2.45. UCL = 3.95 and LCL = −.95 → 0
c. All observations are within the control limits (the zeros are considered to be within the
limits).
Step 2: Conduct run tests:
Test
r
expected
Std. dev.
z-score
Conclusion
Median
6
7.00
1.66
−0.60
random
Up/Down
6
7.67
1.35
−1.24
random
Step 3: Plot the data:
There is obvious cycling in the data. The process output is not random.
10-20
Chapter 10 - Quality Control
Case: Toys Inc.
A consultant must consider the long-term implications of decisions suggested by management.
1.
Cutting cost in design and product development may not be beneficial to the company in the long
run.
2.
The trade-in and repair program, while appeasing customers in the short run, may be too costly
and will not be correcting the root cause of the problem.
3.
Since the company thrives on its reputation of high quality products, it needs to continue to
design products of high quality that fulfils the needs of the market place. Manufacturing needs to
place greater emphasis on preventive quality management/control rather than inspecting already
completed parts. The company may want to consider investing more in R&D.
4.
If implemented well, this strategy will enable the company to become more competitive in the
long run.
Case: Tiger Tools
1.
For the first data set R = .873. From Table 10–2, for n = 20, A2 = .18. Using the hint, the
estimated standard deviation is .234:
A2 R  3
AR

n 
. Rearranging terms, we have 2
3
n
Solving, we obtain  
(.18)(.873 )
3
The process capability is
 20   .234
1.44
 1.03 . Because this is less than 1.33, the process is not
(6)(.234 )
capable.
2.
The process seems to be cycling, as indicated by the control chart for the smaller sample size.
Taking large samples probably resulted in combining the results of several different process
means, and therefore did not reveal the changes that were occurring. By taking smaller samples
more frequently, the pattern was easier to discern.
10-21
Chapter 10 - Quality Control
Control chart for n = 20:
UCL = 5.16














10
12




LCL = 4.86
2
4
6
8
sample number
14
16
18
Control chart for n = 5:
UCL = 5.24
 




 



















LCL = 4.76
2
3.
4
6
8
10
12
Sample number
14
16
18
20
22
24
26
If the cycling can be removed. The true process standard deviation is probably much smaller than
the apparent process standard deviation. For the second data set, R = .411. From Table 10–2,
A2=.58. Performing the same calculations as in #1, we obtain an estimated standard deviation of
.178.

A2 R
(.58)(.411)
n
5  .178
3
3
The process capability is
1.44
 1.35 . Because this is more than 1.33, the process is capable.
6(.178 )
10-22
Chapter 10 - Quality Control
4.
Small samples tend to be less reliable than large samples (the standard deviation of the sampling
distribution of means decreases as the sample size increases). Also, a manager must weigh the
cost of inspecting each item and cost of taking a sample. If the cost to obtain a sample is high, but
the cost to inspect an item is low, larger samples might be the better choice.
10-23