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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 analysis is also important.
Through the use of control charts, the nonrandom (special) causes of variation will be controlled, and
random (common) causes of variation will be analyzed through process capability.
Answers to Discussion & 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 ensure it is working
2.
Control charts are based on the premise that a process which is stable will reflect only
randomness. Statistics of samples taken from the process (means, number of defects, etc.) will
conform to a sampling distribution with known characteristics. From this, control limits are
determined. Observing a sample statistic outside the control limits or a trend in sample statistic
indicates the process has significantly changed, hence there must be an assignable cause.
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.
3.
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 by focusing on the central tendency of a
process.
b. Range control charts are used to monitor process, focusing on the dispersion of a process.
c. p-chart - is a control chart used to monitor the proportion of defectives in a process.
d. c-chart - is a control chart used to monitor the number of defects per unit.
6.
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
Instructor’s Manual, Chapter 10
175
limits are a function of process variability as well as sample size and confidence level. Both are
essentially independent of tolerances.
7.
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 customer (offer a lower price?) to widen tolerances or engineering
(communicate the cost of 100 percent inspection if relevant). The problem is that
customers/engineers 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.
8.
9.
This “problem” often goes undetected since there are no complaints from customers about output
being within specs. However, it is quite possible to realize decreased costs or more profits by
taking certain actions. 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.
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.
b. Cost of product or service, volume, cost 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 & mobility), inspection time, cost 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 favourable testing environment.
d. Raw materials & purchased parts, finished products, before a costly operation, before an
irreversible process, and before a covering process.
10.
a.
Type I error
b.
Type II error
Memo Writing Exercise
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 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 number of observations and higher information content will outweigh
the extra cost of measurement. However, when the characteristic in question is a dichotomous
176
Operations Management, 2/ce
classification (defective vs. nondefective, on vs. off) the x chart is not applicable and p chart should be
used.
Solutions
1.
spec’s: 24 kg. to 25 kg.
.0062
.0062
=
 = 24.5 kg. [assume µ = x ]
 = .2 kg.
-2.5
a.
24
+.5
z=
= 2.5  2(.0062) = .0124 or 1.24%
.2
.2

b.  ± 2
= 24.5 ± 2
= 24.5 ± .1 or 24.4 to 24.6
n
16
2.
From
Appendix B,
Table B
0
+2.5
z-scale
24.5
25
16
 = 2.0 litres
 = .01 litre
n=5
a. Control limits:  ± 2

n
b.
UCL
2.010
[z = 2.0 for 95.5%]
.01
UCL is 2.0 + 2
= 2.009 litres
5
LCL is 2.0 - 2
2.015
.01
= 1.991 litres
5
(litres)2.005
Mean 2.000
1.995
*
*
*
*
1.990
*
*
LCL
1.895
Yes, the process is in control.
n = 10
=
=

A2 = 0.31 X
= 3.10 a.
Mean Chart: X ± A2 R = 3.1 ± 0.31(0.45)

D3 = 0.22 R
= 3.1 ± .14
= 0.45
D4 = 1.78
Hence, UCL is 3.24
and LCL is 2.96

Range Chart: UCL is D4R = 1.78(0.45) = .801

LCL is D3R = 0.22(0.45) = .099
b.
In control since all points are within these limits.
4.
Sample
Mean
Range
=

Mean Chart: X ± A2R = 79.96 ± 0.58(1.90)
1
79.48
2.6
= 79.96 ± 1.1
2
80.14
2.3
3.
3
80.14
1.4
4
79.60
1.7
5
80.02
2.0
6
80.38
79.96
1.4
1.9
Instructor’s Manual, Chapter 10
UCL = 81.06, LCL = 78.86

Range Chart: UCL = D4R = 2.11(1.90) = 4.009

LCL = D3R = 0(1.90) = 0
[Both charts suggest that the process is in control: Neither has any
points outside the limits.]
177
Solutions (continued)
Embedded Excel Worksheet
5.
obs #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Octane
rating
89.20
86.50
88.40
91.80
90.30
87.50
92.60
87.00
89.80
92.20
85.40
91.60
87.70
85.00
91.50
90.30
85.60
90.90
82.10
85.80
Moving
Range
obs #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
Octane
rating
16.20
13.80
17.00
15.80
13.50
14.70
14.00
14.80
13.20
16.80
14.90
13.00
12.50
16.70
15.90
14.60
16.50
18.40
15.20
14.60
17.20
16.10
14.40
17.00
13.80
15.50
Moving
Range
2.70
1.90
3.40
1.50
2.80
5.10
5.60
2.80
2.40
6.80
6.20
3.90
2.70
6.50
1.20
4.70
5.30
8.80
3.70
avg=
std dev=
88.56
2.91
UCLx=
LCLx=
97.29
79.83
All observations fall within above limits
avg moving range=
UCLmr=
LCLmr=
4.11
13.42
0
All moving ranges fall within above limits
Therefore, process is in control.
6.
178
2.40
3.20
1.20
2.30
1.20
0.70
0.80
1.60
3.60
1.90
1.90
0.50
4.20
0.80
1.30
1.90
1.90
3.20
0.60
2.60
1.10
1.70
2.60
3.20
1.70
avg=
std dev=
15.23
1.50
UCLx=
LCLx=
19.73
10.74
All observations fall within above limits
avg moving range=
UCLmr=
LCLmr=
1.92
6.29
0
All moving ranges fall within above limits
Therefore, process is in control.
Operations Management, 2/ce
Solutions (continued)
7.
8.
n = 200
a.
1
2
3
4
4
12
5
9
= .020
= .010
= .025
= .045
200
200
200
200
b. (0.02 + 0.01 + 0.025 + 0.045)/4 = 0.025
c. mean = .025
p(1  p)
.025(.975)
Std. dev. 

 .011
n
200
d. α = 0.03
confidence = .97
z = 2.17
.025 ± 2.17(0.011) = .025 ± .024 = .001 to .049.
e. .025 + z(.011) = .047
Solving, z = 2, leaving .0228 in each tail. Hence, alpha
= 2(.0228)
= .0456.
f. Yes, all sample proportions in part (a) fall within these control limits.
g. mean = .02
.02(.98)
Std. dev. 
 .0099
200
h. .02 ± 2(.0099) = 0.0002 to .0399.
No, the last sample is beyond the upper limit.
p(1  p)
n = 200
Control Limits = p  2
n
p
25
 .0096
13(200 )
 .0096  2
.0096 (.9904 )
200
 .0096  .0138
Thus, UCL is .0234 and LCL is 0 (because it can't be negative).
Since n = 200, the fraction represented by each data point is half the
amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc.
Sample 10 is too large. Omitting that value and recomputing
18
= .0075 yields
limits with 
p=
12(200)
UCL = .0075 + .0122 = .0197 and LCL = 0.
9.

c=
110
= 7.857
14
Control Limits: c ± 3 c = 7.857 ± 8.409
UCL is 16.266, LCL, if negative, should be changed to 0.
All numbers of daily compaints are within the limits.
10.

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.
Instructor’s Manual, Chapter 10
179
Solutions (continued)

p=
11.
total number of defectives
87
= .054
=
Total number of observations
16(100)


.054(.946)
p (1 - p ) = .054 ± 1.96
Control limits are 
p±z
n
100
= .054 ± .044. Hence, UCL = .098
LCL = .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 9.8
defectives and the lower control limit would be 1 defective. Even though all points are within
these limits, the process error rate can be shown statistically to be different from 4% error rate
because 5.4% is too far from 4% given 1600 observations.
12.
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:
.01
15  6
 15.06 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. - 15.06cm.
= 35 pieces.
.004 cm./piece
13.
a. 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 10-11A.] Thus, it seems that the tolerances
are being met: approximately 97 percent of the output will be acceptable.

b. R = 1.90 from problem 4.
A2 = .58
n
5
A2R 
(.58)(1.90)  .82
σ≈
3
3
max spec  min spec 81  78
CP 

 .61 < 1 => not capable
6
6(.82)
No, part (a) and (b) give different results. It is best to use CP Index because it is based on
population, not just a sample of 30.
14.
a.  = .146
n=6
x
 x  150.15  3.85.
39
.146
 .385  2
 3.85  .119
Control limits are x  2
n
6

39
So UCL is 3.97, LCL is 3.73. Sample 29 is outside the UCL, so the process is
not in control.
180
Operations Management, 2/ce
Solutions (continued)
3.5
Cm
15.
UCL
n=1
 = 0.01 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 0.01 = 3.5 - 0.06 = 3.44 Cm.
1
(iii) Determine how many pieces can be produced before the LCL just crosses the lower
tolerance of 3 Cm.
3.44 - 3.00
0.44
440
=
=
= 440 pieces
0.001
0.001
1
16.
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.
n
4
A 2 R from (10-4) 
(.73)(.52)  .253
σ≈
3
3
max spec  min spec 10.35  9.65
CP 

 .46 < 1 => process is not capable
6
6(.253)
One should reduce the process variability, e.g., using experimental design or use more accurate
machines.
17. a.
1
4.3
2
4.5
3
4.5
4
4.7
b. =
x = (4.3 + 4.5 + 4.5 + 4.7)/4 = 4.5
std. dev. (of data set) = .192 using Excel
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: A = 0.58
x ± A R = 4.5 ± 0.58(.325) = 4.3115 to 4.6885.
2
Ranges: D3 = 0
Instructor’s Manual, Chapter 10
2
The first mean is below the lowest limit, and the last mean is above the
upper limit.
0 to 2.11(.325) = 0 to .68575
All ranges are within the limits.
181
Solutions (continued)
h. Two different measures of dispersion are being used, the standard deviation and the range, and
the process distribution is not quite Normal.
i. 4.4 ± 3 0.18 = 4.4 ± .241 = 4.16 to 4.64. The last sample mean is above the upper limit.
5
Embedded Excel Worksheet
4.5
4.2
4.2
4.3
4.3
mean=
std dev=
4.6
4.5
4.4
4.7
4.3
4.5
4.6
4.4
4.4
4.6
4.7
4.6
4.8
4.5
4.9
4.5
0.191943
Bin
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
More
Frequency
0
2
3
3
4
4
2
1
1
0
18. Process mean = 0.03 cm, σ = 0.003 cm, tolerance = (0.02, 0.04) cm.
a. C = specification width
.04 - .02 = .02
.02
=
=
= 1.11
p
process width
6(.003)
.018
b. In order to be capable, the process capability ratio must be at least 1.00. In this instance, the index
is 1.11, so the process is capable. (If 1.33 were used, the process would not be capable.)
Machine Standard Deviation (cm) Job Specification (cm)
Cp
Capable ?
0.833
0.583
0.600
1.000
1.333
No
No
No
Yes
Yes
19.
001
002
003
004
005
20.
182
0.02
0.04
0.10
0.05
0.01
0.05
0.07
0.18
0.15
0.04
Machine Cost per unit ($) Standard Deviation (mm.)
A
20
0.079
B
12
0.080
C
11
0.084
D
10
0.081
Cp
1.013
1.000
0.952
0.988
Operations Management, 2/ce
Solutions (continued)
You can narrow the choice to machines A and B because they are the only ones with a capability
ratio of at least 1.00. You would need to know if the slight additional capability of machine A is
worth an extra cost of $8 per unit.
21.
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
X = Process mean,  = Process standard deviation
For process H:
X  LSL 15  14.1

 .9375
3
(3)(.32)
USL  X 16  15

 1.04
3
(3)(.32)
Cpk  min .9375, 1.04  .9375
.9375  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
Since1.17  1.0, the processis capable.
For process T:
X  LSL 18.5  16.5

 1.33
3
(3)(0.5)
USL  X 20.1  18.5

 1.06
3
(3)(0.5)
C pk  min 1.33, 1.06  1.06
Since1.06  1.0, the processis capable.
Instructor’s Manual, Chapter 10
183
22.
No, the maximum value of CPK occurs when the process mean is centred in the specification
range, and then CPK will equal CP because
max spec  process mean process mean  min spec 1  max spec  min spec 
CPK =

 
 = CP.
3
3
2
3

Therefore CPK ≤ CP.
23.
Let USL = Upper Specification Limit
X = Process mean,  = Process standard deviation.
USL = 45 minutes,
X Armand  38 min .,  Armand  3 min .
X Jerry  37 min .,  Jerry  2.5 min .
X Melissa  37.5 min .,  Melissa  2.5 min .
For Armand:
USL  X 45  38

 .78
3
(3)(3)
Since .78  1.0, Armand is not capable.
C pk 
For Jerry:
USL  X 45  37

 1.07
3
(3)(2.5)
Since 1.07  1.0, Jerry is capable.
C pk 
For Melissa:
USL  X 45  37.5

 1.0
3
(3)(2.5)
Since 1.0  1.0, Melissa is capable.
C pk 
Jerry is most capable.
184
Operations Management, 2/ce
Solutions (continued)
Embedded Excel Worksheet
24.
a.
Stroh Brewery
Sample
1
0.6
0.55
0.5
2
0.4
0.45
0.35
3
0.5
0.3
0.4
Process mean = 0.49
Process stan. dev.=
0.10
5
0.65
0.45
0.45
6
0.6
0.5
0.55
7
0.7
0.55
0.55
8
Freq.
0.3
0.4
0.5
0.6
0.7
0.8
More
1
5
7
6
2
0
0
Tolerance =(0,1.45) cc
Cpk =
3.18 >1
Frequency
Bin
0.3
0.4
0.5
0.6
0.7
0.8
4
0.4
0.4
0.5
6
4
2
0
0.3
0.4
0.5
0.6
0.7
0.8
More
Bin
Capable
b.
Sample
1
0.6
0.55
0.5
Sample mean: 0.55
Sample Range: 0.10
2
0.4
0.45
0.35
0.40
0.10
Sample Mean control chart:
UCL =
0.63
LCL =
0.35
3
0.5
0.3
0.4
0.40
0.20
4
0.4
0.4
0.5
0.43
0.10
5
0.65
0.45
0.45
0.52
0.20
6
0.6
0.5
0.55
0.55
0.10
7
0.7
0.55
0.55
0.60
0.15
avg.
0.49
0.14
= grand mean
= average range
Mean chart
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
1
Sample Range control chart:
UCL =
0.35
LCL =
0.00
2
3
4
5
6
7
5
6
7
Range chart
0.50
0.40
0.30
0.20
0.10
0.00
1
Instructor’s Manual, Chapter 10
2
3
4
185
Solutions (continued)
Embedded Excel Worksheet
c.
Sample
1
0.6
0.55
0.5
Sample mean: 0.55
Sample Range: 0.10
2
0.4
0.45
0.35
0.40
0.10
3
0.5
0.3
0.4
0.40
0.20
Time period
4
0.4
0.4
0.5
0.43
0.10
5
0.65
0.45
0.45
0.52
0.20
6
0.6
0.5
0.55
0.55
0.10
7
0.7
0.55
0.55
0.60
0.15
avg.
0.49 = grand mean
0.14 = average range
Mean chart
1.60
1.40
1.20
Out of control
1.00
0.80
0.60
0.40
0.20
0.00
1
2
3
4
5
6
7
8
Range chart
0.50
0.45
0.40
0.35
0.30
0.25
0.20
In control
0.15
0.10
0.05
0.00
1
2
3
4
5
6
7
8
Case: Toys Inc.
A consultant must consider the long-term implications of decisions suggested.
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. 100% inspection may be
too expensive. Manufacturing needs to place greater emphasis on preventive quality
management/control (e.g., use of control charts) rather than inspecting already completed parts.
The company may want to consider investing more in R&D.
Case: Tiger Tools
Answers to Questions:

1. For the first data set R = .873. From Table 10-2 , for n = 20, A2 = .18. Using Equation 10.4, the
estimated standard deviation is:
186
Operations Management, 2/ce
n
20
A2R 
(.18)(.873)  .234
3
3

1.20
6(.234)
The process capability is
= .854. Because this is less than 1.00, the process is not capable

2. For the second data set, R = .401. From Table 10-2 , A2=.58. Performing the same calculations as in
#1, we obtain an estimated standard deviation of:
A R
.58(.401)
 2
n
5  .173
3
3
1.2
The process capability is
= 1.153. Because this is more than 1.00, the process is capable.
6(.173)
The process seems to be cycling, as indicated by the plot of the smaller sample size below. Taking
large samples probably resulted in combining the results of several different process means, and
therefore resulted in a large estimate for σ.
For n = 5
 




 












2
4
6
8
10
12
Sample number
14
16
18
20







22
24
26
Even though process is capable, it is not in control (because of the wave-like pattern of the average
lengths of prybars). The cause for this should be investigated and fixed.
Instructor’s Manual, Chapter 10
187
Returned
bottles
Case: Canadian Springs
Visual
Inspection
Answers to Questions:
1.
Cleansing &
Sterilization
Plant
aquifer
Tanker
trucks
Spring
water
Carbon
filters & . . .
membranes
Conveyor
belt
Holding
tanks
Reverse
Osmosis
Filling
Premium
water
Capping
Holding
tanks
2. Quality for Canadian Springs means that the product, water, is free of any impurities or most bacteria.
Also, the water's taste, smell, and colour should be acceptable.
Canadian Springs draws the water from closed aquifers that have almost pure water. By using
sanitized tanker trucks and equipment in the plants, contamination of water is kept to a minimum. In
addition, water goes through filteration processes. Also, the returned bottles are cleaned and
sterilized. Finally, regular hourly quality tests are performed on the water in the holding tanks, and
bottled water is kept for up to 30 days and tested to see how the water keeps over time.
Case: In the Chips at Jays
Answers to Questions:
1.
big
Semi-trailers
of potatoes
Holding
bins
Sort
Conveyor
belt
Washing
Skinning
Laser check
(opti-sort
scanner) for
dark spots,
holes
Broken
chips fall
through
small
Inspection
for rotten
potatoes
Chippers
Inspection
for burnt
chips
Cooking in
corn oil,
circulated,
salted,
flavoured
Holding
bins
Bagging
188
Scales
Operations Management, 2/ce
2. Chips should taste good (i.e., from good-tasting, not rotten, potatoes), be pure (without skin), be
whole (not broken), not burnt, and consistently flavoured. First, good quality, North Dakota potatoes
are purchased, then washed and skinned, and inspected to remove the rotten ones. While frying, the
oil and flavours are circulated to provide consistency. After cooking, burnt and broken ones are
separated too.
Instructor’s Manual, Chapter 10
189