Statistical Methods in Quality Management 21
15. The distribution center manager at internet distributor CyberAuto Warehouse wants to
find a confidence interval for the average time required for an associate to fill an order
for shipment. A sample of 16 orders is taken and the mean time was found to be 8.5
minutes, with a standard deviation of 2.8 minutes. Compute 95 percent and 99 percent
confidence intervals. Which one is larger? Explain why.
Answer
15. a) 95 percent Confidence interval =
𝑥̅ ± 𝑡 𝜎
√𝑛=8.5 ± 2.131 (2.8
√16)=8.5 ± 1.4917= 7.008 to 9.992
16. A new product is being tested by Zed Electronics to determine if it will continue to
operate in a stable fashion under a variety of conditions. A sample of 400 items were
tested, and 60 failed the test. Determine a 90 percent confidence interval for the
population proportion.
Statistical Methods in Quality Management 22
Answer
16. For this calculation, to find the z value, the Excel function NORM.S.INV can be used:
17. Tessler Electric utility requires service operators to answer telephone calls from
customers in an average time of 0.1 minute or less. A sample of 25 actual operator
times was drawn, and the results are given in the following table. In addition, operators
are expected to determine customer needs and either respond to them or refer the
customer to the proper department within 0.5 minute. Another sample of 25 times was
taken for this job component and is also given in the table. If these variables can be
considered to be independent, are the average times taken to perform each component
statistically different from the standards?
Component
Mean Time
Standard Deviation
Answer
0.1023
0.0183
Service
0.5290
0.0902
Answer
Specification for answer time for the Tessler utility is :
Statistical Methods in Quality Management 23
18. A quality manager at Newvis Pharmaceutical Company is monitoring a process that
fills vials with a liquid medication designed to prevent glaucoma in the eyes of the user.
The company wants to ensure that each vial contains at least 60 ml. (2.03 fluid oz.) of
the product. A sample of 25 vials is tested, and a mean of 63 ml. and a sample standard
deviation of 10 ml. are found. The quality manager wishes to test the null hypothesis
that vials contain less than or equal to 60 ml. using an α = 0.05 significance level
(rejecting this hypothesis provides evidence that the vials contain the required amount.)
Conduct the test and explain your results.
Answer
The hypothesis to be tested is:
Statistical Methods in Quality Management 24
19. The quality manager at Newvis Pharmaceutical Company is certifying a new process
that must produce 90 percent (or better) good product before certification can be
completed. A sample of 49 containers from the process line are tested, and 87percent
are found to be good. Formulate the appropriate hypotheses and test them using an =
0.05 significance level. Explain your results.
Answer
The hypothesis which requires testing is:
H0: Proportion of good product ≥ 90
Statistical Methods in Quality Management 25
20. Rabbitfoot Community Bank makes a large number of home equity loans each year.
The vice president of loan administration wishes to determine if their time for
paperwork processing is lower than the average time of their top competitor. A sample
of 30 loans taken at Rabbitfoot Bank yielded a mean of 38.10 minutes and a standard
deviation of 2.58 minutes. (see the data in the Ch06Data Excel workbook). Data
obtained from competitor of 36 applications indicates that the average time for
processing an application is 39.48 minutes, with a standard deviation of 2.48 minutes.
a) Verify the calculation of the mean, standard deviation, and variance for each set of
data using the Descriptive Statistics tool.
b) Test the null hypothesis that his bank’s processing time is greater than or equal to the
competitor’s average, versus the alternative hypothesis that the bank’s time is less than
the competitor at the 5 percent significance level. Use the z-Test: Two Sample
Assuming Equal Variances from the Data Analysis menu in Excel.
Answer
To conduct this hypothesis test for comparing the processing time, select t–test: Two-
Sample Assuming Equal Variances from the Data Analysis menu.
Rabbitfoot
Industry Data
Mean
38.0967
39.4750
Variance
6.6307
6.1534
Statistical Methods in Quality Management 26
21. Softswift, a software developer, is trying to determine if any of three potential
subcontractors has better programmers in order to outsource a development project.
The three subcontractors agreed to test 5 programmers, using a standardized test
provided by Softswift, as provided in the data in the Ch06Data Excel workbook. Use
the single factor ANOVA Excel tool to determine if there is a significant difference
between the scores of programmers at the three contractors at the 5 percent level.
Answer
The null hypothesis is: H0: µ1 = µ2 = µ3 versus
The alternative hypothesis: H1: Not all µ1 = µ2 = µ3
Statistical Methods in Quality Management 27
22. At Rockglass, Inc. a kiln is used to bake ceramic pottery. The production manager
wishes to determine the relationship between temperature and brittleness, so he takes
measurements of the brittleness of test items versus the temperature of the oven in
degrees C. Use the Excel Regression tool to determine the regression equation and the
R2 value. Explain the output. If the oven is heated to 975 deg. C, what would you
predict that the brittleness measure will be?
Answer
22. See spreadsheet P06-22REGRESS.xlsx for details. From the table below, we can see
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.96085
R Square
0.92324
Adjusted R Square
0.91734
Standard Error
0.52237
Observations
15.00000
ANOVA
df
SS
MS
F
Significance
F
Regression
1
42.66604
42.66604
156.3608
1.27E-08
Statistical Methods in Quality Management 28
Coefficients
Standard
Error
t Stat
P-value
Lower
95%
Upper
95%
Lower
95%
Upper
95%
23. The process engineer at Sival Electronics was trying to determine whether three suppliers
would be equally capable of supplying the mounting boards for the new figold plated”
components that she was
testing. The table found in the worksheet Prob. 6-23 in the Excel workbook C06Data
shows the coded defect levels for the suppliers, according to the finishes that were tested.
Lower defect levels are preferable to higher levels. Using ANOVA, analyze these results.
What conclusion can be reached?
Answer
23. The process engineer at Sival Electronics can develop a one-way ANOVA spreadsheet
(see spreadsheet P06-23Anova.xlsx for details) that shows:
Supplier 1
Supplier 2
Supplier 3
Finish 1
11.9
6.8
13.5
Finish 2
10.3
5.9
10.9
8.0
9.0
0 200 400 600 800 1000 1200 1400
Temp-deg C
Temp-deg C Line Fit Plot
Statistical Methods in Quality Management 29
SUMMARY
Groups
Count
Sum
Average
Variance
Supplier 1
5
54.6
10.92
4.762
24. A quality analyst at Paintfast Manufacturing Co. wants to determine if a new paint
formulation, used to paint parts for a customer’s assembly operation will dry fast enough
to meet the customer’s needs. The customer would prefer to obtain a high level of
fidryability” at low temperatures, even if it requires a higher level of drying agent. He
hypothesizes that a high level of drying agent will result in high dryability, high
temperature – alone – will result in a moderately high level of dryability, and low
temperature or a low level of drying agent will result in a low level of dryability. He
hopes that the main and interaction effects with the temperature, which is expensive
(because an oven would need to be used), will be minimal. The data found in the
worksheet Prob. 6-24 in the Excel workbook C06Data were gathered in testing all
combinations. What recommendation would you make?
Answer
24. As can be seen, from the spreadsheet P06-24-2X2FactExp.xlsx results, shown below,
Factor 1 (the drying agent) has the greatest impact on dryability. This shows that the best
Supplier 2
5
35.6
0.697
Supplier 3
5
64.1
12.82
1.812
Between Groups
42.117
3.885
Within Groups
2.424
Total
Statistical Methods in Quality Management 30
Treatment
Factor 1
Factor 2
Response
A
Low
Low
78
25. The process engineer at Sival Electronics is also trying to determine whether a newer,
more costly design involving a gold alloy in a computer chip is more effective than the
present, less expensive silicon design. She wants to obtain an effective output voltage at
both high and low temperatures, when tested with high and low signal strength. She
hypothesizes that high signal strength will result in higher voltage output, low
temperature will result in higher output, and the gold alloy will result in higher output
than the silicon material. She hopes that the main and interaction effects with the
expensive gold will be minimal. The data found in the worksheet Prob. 6-25 in the Excel
workbook C06Data were gathered in testing of all combinations. What recommendation
would you make?
Answer
25. Using the following input data:
Signal
Material
Temperature
Output Voltage
Low
Silicon
Low
7
Silicon
Low
Low
Low
8
Low
Low
Silicon
Silicon
Low
The process engineer at Sival Electronics can calculate the main effects as follows:
Signal
B
High
Low
91
C
Low
High
85
D
High
High
87
Main Effects
Factor 1
Factor 2
Interaction
Factor 1 x Factor 2
Statistical Methods in Quality Management 31
Material
Gold (18 + 12 + 8 + 11)/ 4 = 12.25
SUGGESTIONS FOR PROJECTS, ETC.
1. This experiment is designed to give the student fihands on” experience in elementary
experimental design.
2. This experiment in aerodynamics will give the student some fihands on” experience in
dealing with design problems in which multiple variables can have a significant impact
on the operability of the mechanism.
ANSWERS TO CASE QUESTIONS
Case – Sizzlegrill Burrito House
1. Portions of the spreadsheet Ch06 Sizzlegrill Case Soln.xlsx are shown below. A frequency
Statistical Methods in Quality Management 32
around a 4. Tastiness of the food and overall satisfaction averaged 3.9 for all respondents.
Customer survey responses
Avg
Std. dev.
Menu was easy to read
4.56
1.06
Food was tasty
3.90
1.08
Value for price paid
4.01
1.19
2. The average responses to the first seven questions by customers, are moderately
correlated with their satisfaction scores. The R2 = 0.700, which does not indicate an
extremely close correlation [correlation coefficient = √0.700=0.837] between the
average score and the overall satisfaction score, can be visualized on the scatter chart,
below.
0.0
6.0
0 1 2 3 4 5 6
Overall satisfaction
Overall satisfaction Line Fit Plot
Statistical Methods in Quality Management 33
3. The likelihood of the customer dining again at Sizzlegrill can be predicted by using the
satisfaction score and regression analysis by customer. The likelihood of customer’s
dining again is moderately correlated to the satisfaction score. The R2 = 0.779, which
does not indicate an extremely close correlation between the average score and the
overall satisfaction score, as seen on the scatter chart, below.
4. The descriptive statistics for burrito weights show that the mean 𝑥̅ =1.509 and standard
deviation, s = 0.108. The frequency distribution and histogram show that the sample is
somewhat normal in shape. The range and standard deviation show that the food
servings are somewhat variable. The range is 0.55, or ½ pound difference between the
lowest and highest values. This could be due to the nature of the burrito product, where
the customer specifies ingredients, which add more or less weight to the burrito.
Descriptive Statistics – Sizzlegrill
Bin
Frequency
1.25
0
Mean
1.509
1.30
3
Standard Error
0.009
1.35
9
Median
1.500
1.40
16
Mode
1.570
1.45
17
Standard Deviation
0.108
1.50
34
Sample Variance
0.012
1.55
22
Kurtosis
1.60
23
Skewness
0.332
1.65
11
Range
0.550
1.70
7
Minimum
1.290
1.75
6
Maximum
1.840
1.80
1
1.85
1
Count
More
0
Confidence Level (95.0 percent)
0.017
Likely to dine with us again? Line Fit Plot
Statistical Methods in Quality Management 34
5. Recommendations for improvement to Juan Niceley include:
a. Work to ensure that food is served hot.
b. Develop a panel to do taste-testing of various existing and new products.
Case – Berton Card Company
The main effects, interactions and interaction charts (see Ch06-2x3BertonCardCo.xlsx
for details) for the 8 experiments are seen below.
Main Effects
Line Speed
0.87
0.51
Interactions
0.07
0.34
Interaction Charts
20
25
30
35
40
Histogram
Statistical Methods in Quality Management 35
6.5
Low 1000 High 1150
Factor 1
Factor 1 x Factor 2
7
9.5
Low 1000 High 1150
Factor 1
Factor 1 x Factor 3
7.6
7.8
Low 1000 High 1150
Factor 2
Factor 2 x Factor 3
Factor 3 Low
Statistical Methods in Quality Management 36
Conclusions based on Data Analysis
The line speeds [Factor 1] and Front (Face) roller pressure [Factor 2] appear to be the
significant variables affecting roughness. There is negligible interaction between those
Recommendations
Management should be advised that as line speed and Front roller pressure are increased,
surface roughness increases. In regard to the interaction effect between speed and Back
roller pressure, as speed is increased, while holding Front roller pressure constant,
management should be advised to have operators to keep the Back roller pressure low, in
order to have a smaller effect on roughness. In regard to the interaction effect between the
Case – The Battery Experiment
1. Using the data in Table 6.6 and Excel template 2×3 Battery Experiment.xlsx in the
Instructor Reserve folder we find the main effects, interactions, and interaction plots for
the three factors as follows.
To test the hypothesis that spending more money on high-quality batteries, using
expensive gold-plated connectors, and storing batteries at low temperatures will improve
battery life performance in a race, an electrical test circuit was constructed to measure
Battery cost
Statistical Methods in Quality Management 37
High = (612 + 490 + 493 + 489)/4 = 521 minutes
Low = (72 + 93 + 75 + 94)/4 = 83.5 minutes
Main effect = High – Low = 437.5 minutes
Connector type
Gold-plated = (94 + 75 + 490 + 493)/4 = 288 minutes
The plots for each of the three factors and their interactions are shown here.
Factor 1
Factor 1 x Factor 2
600
Low cost High Cost
Factor 1
Factor 1 x Factor 3
Statistical Methods in Quality Management 38
2. Using the data in Table 6.7 and the Excel Data Analysis tool an ANOVA analysis was
conducted to determine whether a significant difference exists between battery types. An
examination of the SUMMARY and ANOVA parts of the table, below, shows that the
mean 𝑥̅ =521 and the variance, s2 = 3683.333, which are considerably larger than the
0
50
400
Low cost High Cost
Factor 2
Factor 2 x Factor 3
Statistical Methods in Quality Management 39
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
A
4
2084
521
3683.333
Instructor Reserve Cases
Case: The HMO Pharmacy Crisis
It appears that, because of the crisis, top management at Dover is reacting in a typical
fired Bead” manner. The importance of statistical concepts in quality management cannot
be overemphasized. Indeed, statistics is essential in implementing a continuous
improvement philosophy.
Statistical thinking is a philosophy of learning and action based on the principles that:
1. All work occurs in a system of interconnected processes.
B
4
100
212.6667
ANOVA
Between Groups
491892.7
245946.3
183.0412
Within Groups
1343.667
Total
503985.7
Statistical Methods in Quality Management 40
The complex interactions of these variations in drugs, equipment, computer systems,
professional, clerical, and technical staff, and the environment are not easily understood.
Variation due to any of these individual sources could be random; individual sources may
not be identified or explainable. However their combined effect in the pharmaceutical
system is probably stable and might be predicted statistically. These common causes of
variation that are present as a natural part of the process need to be understood before
special causes can be separated and eliminated.
To address the problem, Dover should consider using the following steps:
• Form a cross-functional group consisting of pharmacists, assistant pharmacists,
physicians, nurses, health care insurance experts, and administrative support people