Chapter 10 – Quality Control
26. Given:
Let USL = Upper Specification Limit, LSL = Lower Specification Limit,
X
= Process mean, = Process standard deviation.
USL = 30 minutes, LSL = 45 minutes,
X
Armand = 38 minutes, Armand = 3.0 minutes
X
For Jerry:
93.0}07.1 ,93.0min{
07.1
)5.2)(3(
3745
3
93.0
)5.2)(3(
3037
3
pk
C
XUSL
LSLX
measure process capability.
39.1
)8.1)(6(
3045
6
LSLUSL
Cp
Because 1.39 ≥ 1.33, Melissa is capable.
b. No, the value of Cpk can never exceed Cp. When the process is centered exactly, Cpk will
equal Cp. When the process is not centered, Cpk always will be less than Cp.
Education.
27. Given:
a. Specifications for the 12-ounce chocolate bar are 330 grams to 350 grams. Average fill is 340
grams. What is the largest standard deviation (in grams) possible for the machine where it
1.33(6σ) = 20
(1.33)6σ = 20
7.98σ = 20
σ = 20/7.98
σ = 2.506 (round to three decimals)
b. The machine that fills the bar models for the 1-ounce bars has a standard deviation of .80
grams per bar. The filling machine is set to deliver an average of 1.01 ounces per bar,
Box variance = 6 * 0.64 = 3.84
Box standard deviation =
Average box weight = 6 *1.01 = 6.06 ounces * 28.33 = 171.68 grams for the six-bar box.
41.1}41.1,99.1min{
41.1
)96.1)(3(
68.171180
3
99.1
)96.1)(3(
16068.171
3
pk
C
XUSL
LSLX
Because 1.41 ≥ 1.33, this process is capable.
Chapter 10 – Quality Control
10–41
Education.
c. The lowest setting in ounces for the filling machine in terms of the six-bar box that will
provide capability:
= 167.82 grams per six-bar box.
167.82 grams per box /6 bars per box= 27.97 grams per bar.
27.97 grams per bar / 28.33 grams per ounce = 0.987 ounces per bar.
28. Given:
All points are within the control limits. Next, check for nonrandomness.
N = 22. Count the number of A/B and U/D on the chart in the text.
Median Test:
Observed number of runs = 15
Expected number of runs (N = total number of observations = 22):
(round to one decimal)
Standard deviation:
10–42
Education.
Up/Down Test:
Observed number of runs = 19
Expected number of runs (N = total number of observations = 22):
(round to one decimal)
Standard deviation:
Overall Conclusion: Nonrandom variations are present in the data.
29. Given:
Observation
1
2
3
4
5
6
7
8
9
10
11
12
No. of errors
1
0
3
2
0
1
3
2
1
0
2
3
Step 1: A c-chart is appropriate.
The mean of the data is c = 18/12 = 1.50.
Control limits:
674.350.150.1350.13 cc
Chapter 10 – Quality Control
All points are within the control limits.
Step 2: Conduct run tests:
Median Test:
Observed number of runs = 6
(round to two decimals)
Up/Down Test:
Observed number of runs = 6
Expected number of runs (N = total number of observations = 12):
(round to one decimal)
Standard deviation:
So far, the process appears to be random. However, there is obvious cycling in the data in the
control chart based on repeating patterns of runs in the Median Test. Therefore, the
Chapter 10 – Quality Control
10–44
Education.
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 correct the root cause of the problem.
3. Because the company thrives on its reputation of high quality products, it needs to continue to
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
= 0.873. From Table 10–2, for n = 20, A2 = 0.18. Using the hint, the
estimated standard deviation is .234:
.3
2n
RA
Rearranging terms, we have
n
RA
3
2
)873.0)(18.0(
Chapter 10 – Quality Control
10–45
2. First set of data (n = 20 per sample):
From Table 10.3 with n = 20: A2 = 0.18.
Mean Chart:
Chapter 10 – Quality Control
Second set of data (n = 5 per sample):
From Table 10.3 with n = 5: A2 = 0.58.
Mean Chart:
Upper Control Limit (UCL) =
Lower Control Limit (LCL) =
3. If the problem with cycling could be removed, the true process standard deviation probably
would be much smaller than the apparent process standard deviation. For the second set of
45.229
44.996
44.763
44.700
44.800
44.900
45.000
45.100
45.200
45.300
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 27
Mean Chart: Second Set
UCL
LCL
Mean