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from Solutions Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Chapter 3
Experiments with a Single Factor: The Analysis of Variance
Solutions
3.1. An experimenter has conducted a single-factor experiment with four levels of the factor, and each
factor level has been replicated six times. The computed value of the F-statistic is F 0 = 3.26. Find bounds
on the P-value.
Table P-value = 0.025, 0.050
Computer P-value = 0.043
3.2. An experimenter has conducted a single-factor experiment with six levels of the factor, and each
factor level has been replicated three times. The computed value of the F-statistic is F 0 = 5.81. Find
bounds on the P-value.
Table P-value < 0.010
Computer P-value = 0.006
3.3. A computer ANOVA output is shown below. Fill in the blanks. You may give bounds on the Pvalue.
One-way ANOVA
Source
DF
SS
MS
F
P
Factor
3
36.15
?
?
?
Error
?
?
?
Total
19
196.04
Completed table is:
One-way ANOVA
Source
DF
SS
MS
F
P
Factor
3
36.15
12.05
1.21
0.3395
Error
16
159.89
9.99
Total
19
196.04
3.4. A computer ANOVA output is shown below. Fill in the blanks. You may give bounds on the Pvalue.
One-way ANOVA
Source
DF
SS
MS
F
P
Factor
?
?
246.93
?
?
Error
25
186.53
?
Total
29
1174.24
3-1
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Completed table is:
One-way ANOVA
Source
DF
SS
MS
F
P
Factor
4
987.71
246.93
33.09
< 0.0001
Error
25
186.53
7.46
Total
29
1174.24
3.5. An article appeared in The Wall Street Journal on Tuesday, April 27, 2010, with the title “Eating
Chocolate Is Linked to Depression.” The article reported on a study funded by the National Heart, Lung
and Blood Institute (part of the National Institutes of Health) and conducted by the faculty at the University
of California, San Diego, and the University of California, Davis. The research was also published in the
Archives of Internal Medicine (2010, pp. 699-703). The study examined 931 adults who were not taking
antidepressants and did not have known cardiovascular disease or diabetes. The group was about 70% men
and the average age of the group was reported to be about 58. The participants were asked about chocolate
consumption and then screened for depression using a questionnaire. People who scored less than 16 on
the questionnaire are not considered depressed, while those with scores above 16 and less than or equal to
22 are considered possibly depressed, while those with scores above 22 are considered likely to be
depressed. The survey found that people who were not depressed ate an average of 8.4 servings of
chocolate per month, while those individuals who scored above 22 were likely to be depressed ate the most
chocolate, an average of 11.8 servings per month. No differentiation was made between dark and milk
chocolate. Other foods were also examined, but no patterned emerged between other foods and depression.
Is this study really a designed experiment? Does it establish a cause-and-effect link between chocolate
consumption and depression? How would the study have to be conducted to establish such a link?
This is not a designed experiment, and it does not establish a cause-and-effect link between chocolate
consumption and depression. An experiment could be run by giving a group of people a defined amount of
chocolate servings per month for several months, while not giving another group any chocolate. Ideally it
would be good to have the participants not eat any chocolate for a period of time before the experiment,
and measure depression for each participant before and after the experiment.
3.6. An article in Bioelectromagnetics (“Electromagnetic Effects on Forearm Disuse Osteopenia: A
Randomized, Double-Blind, Sham-Controlled Study,” Vol. 32, 2011, pp. 273 – 282) describes a
randomized, double-blind, sham-controlled, feasibility and dosing study to determine if a common pulsing
electromagnetic field (PEMF) treatment could moderate the substantial osteopenia that occurs after forearm
disuse. Subjects were randomized into four groups after a distal radius fracture, or carpal surgery requiring
immobilization in a cast. Active of identical sham PEMF transducers were worn on a distal forearm for 1,
2, or 4h/day for 8 weeks starting after cast removal (“baseline”) when bone density continues to decline.
Bone mineral density (BMD) and bone geometry were measured in the distal forearm by dual energy X-ray
absorptiometry (DXA) and peripheral quantitative computed tomography (pQCT). The data below are the
percent losses in BMD measurements on the radius after 16weeks for patients wearing the active or sham
PEMF transducers for 1, 2, or 4h/day (data were constructed to match the means and standard deviations
read from a graph in the paper).
Sham
PEMF
1h/day
PEMF
2h/day
PEMF
4h/day
4.51
5.32
4.73
7.03
7.95
6.00
5.81
4.65
4.97
5.12
5.69
6.65
3-2
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.00
7.08
3.86
5.49
7.97
5.48
4.06
6.98
2.23
6.52
6.56
4.85
3.95
4.09
8.34
7.26
5.64
6.28
3.01
5.92
9.35
7.77
6.71
5.58
6.52
5.68
6.51
7.91
4.96
8.47
1.70
4.90
6.10
4.58
5.89
4.54
7.19
4.11
6.55
8.18
4.03
5.72
5.34
5.42
2.72
5.91
5.88
6.03
9.19
6.89
7.50
7.04
5.17
6.99
3.28
5.17
5.70
4.98
5.38
7.60
5.85
9.94
7.30
7.90
6.45
6.38
5.46
7.91
(a) Is there evidence to support a claim that PEMF usage affects BMD loss? If so, analyze the data to
determine which specific treatments produce the differences. The ANOVA from the Minitab
output shows that there is no difference between the treatments; P=0.281.
Minitab Output
One-way ANOVA: Sham, PEMF 1h/day, PEMF 2h/day, PEMF 4h/day
Source
Factor
Error
Total
DF
3
76
79
S = 1.606
Level
Sham
PEMF 1h/day
PEMF 2h/day
PEMF 4h/day
SS
10.04
196.03
206.07
MS
3.35
2.58
F
1.30
R-Sq = 4.87%
N
20
20
20
20
Mean
5.673
6.165
5.478
6.351
P
0.281
R-Sq(adj) = 1.12%
StDev
2.002
1.444
1.645
1.232
Individual 95% CIs For Mean Based on
Pooled StDev
-+---------+---------+---------+-------(-----------*----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
-+---------+---------+---------+-------4.80
5.40
6.00
6.60
(b) Analyze the residuals from this experiment and comment on the underlying assumptions and
model adequacy. The residuals show the model is good.
3-3
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residual Plots for Sham, PEMF 1h/day, PEMF 2h/day, PEMF 4h/day
Normal Probability Plot
Versus Fits
99.9
4
90
Residual
Percent
99
50
10
1
0.1
2
0
-2
-4
-5.0
-2.5
0.0
Residual
2.5
5.0
5.6
5.8
6.0
Fitted Value
6.2
6.4
Histogram
Frequency
16
12
8
4
0
-3.2
-1.6
0.0
Residual
1.6
3.2
3.7. The tensile strength of Portland cement is being studied. Four different mixing techniques can be
used economically. A completely randomized experiment was conducted and the following data were
collected.
Mixing
Technique
1
2
3
4
3129
3200
2800
2600
Tensile Strength (lb/in2)
3000
2865
3300
2975
2900
2985
2700
2600
2890
3150
3050
2765
(a) Test the hypothesis that mixing techniques affect the strength of the cement. Use α = 0.05.
Design Expert Output
Response:
Tensile Strengthin lb/in^2
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
4.897E+005
3
1.632E+005
A
4.897E+005
3
1.632E+005
Residual
1.539E+005
12
12825.69
Lack of Fit
0.000
0
Pure Error
1.539E+005
12
12825.69
Cor Total
6.436E+005
15
F
Value
12.73
12.73
Prob > F
0.0005
0.0005
The Model F-value of 12.73 implies the model is significant. There is only
a 0.05% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
2971.00
56.63
2-2
3156.25
56.63
3-3
2933.75
56.63
4-4
2666.25
56.63
Treatment
1 vs 2
1 vs 3
1 vs 4
Mean
Difference
-185.25
37.25
304.75
DF
1
1
1
Standard
Error
80.08
80.08
80.08
t for H0
Coeff=0
-2.31
0.47
3.81
3-4
Prob > |t|
0.0392
0.6501
0.0025
significant
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
2 vs 3
2 vs 4
3 vs 4
222.50
490.00
267.50
1
1
1
80.08
80.08
80.08
2.78
6.12
3.34
0.0167
< 0.0001
0.0059
The F-value is 12.73 with a corresponding P-value of .0005. Mixing technique has an effect.
(b) Construct a graphical display as described in Section 3.5.3 to compare the mean tensile strengths for
the four mixing techniques. What are your conclusions?
S yi . =
MS E
12825.7
=
= 56.625
4
n
Scaled t Distribution
(4)
(3)
2700
2800
2900
(1)
3000
(2)
3100
Tensile Strength
Based on examination of the plot, we would conclude that µ1 and µ3 are the same; that µ 4 differs from µ1
and µ3 , that µ 2 differs from µ1 and µ3 , and that µ 2 and µ 4 are different.
(c) Use the Fisher LSD method with α=0.05 to make comparisons between pairs of means.
LSD = t α
2
,N − a
LSD = t 0.025 ,16 − 4
2 MS E
n
2( 12825.7 )
4
LSD = 2.179 6412.85 = 174.495
Treatment 2 vs. Treatment 4 = 3156.250 - 2666.250 = 490.000 > 174.495
Treatment 2 vs. Treatment 3 = 3156.250 - 2933.750 = 222.500 > 174.495
Treatment 2 vs. Treatment 1 = 3156.250 - 2971.000 = 185.250 > 174.495
Treatment 1 vs. Treatment 4 = 2971.000 - 2666.250 = 304.750 > 174.495
Treatment 1 vs. Treatment 3 = 2971.000 - 2933.750 = 37.250 < 174.495
Treatment 3 vs. Treatment 4 = 2933.750 - 2666.250 = 267.500 > 174.495
The Fisher LSD method is also presented in the Design-Expert computer output above. The results agree
with the graphical method for this experiment.
(d) Construct a normal probability plot of the residuals. What conclusion would you draw about the
validity of the normality assumption?
3-5
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
There is nothing unusual about the normal probability plot of residuals.
Normal plot of residuals
99
Normal % probability
95
90
80
70
50
30
20
10
5
1
-181.25
-96.4375
-11.625
73.1875
158
Residual
(e) Plot the residuals versus the predicted tensile strength. Comment on the plot.
There is nothing unusual about this plot.
Residuals vs. Predicted
158
Res iduals
73.1875
-11.625
2
-96.4375
-181.25
2666.25
2788.75
2911.25
3033.75
3156.25
Predicted
(f) Prepare a scatter plot of the results to aid the interpretation of the results of this experiment.
Design-Expert automatically generates the scatter plot. The plot below also shows the sample average for
each treatment and the 95 percent confidence interval on the treatment mean.
3-6
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
One Factor Plot
3300
Tens ile Strength
3119.75
2939.51
2759.26
2
2579.01
3
2
1
4
Technique
3.8. (a) Rework part (c) of Problem 3.7 using Tukey’s test with α = 0.05. Do you get the same conclusions
from Tukey’s test that you did from the graphical procedure and/or the Fisher LSD method?
Minitab Output
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0117
Critical value = 4.20
Intervals for (column level mean) - (row level mean)
1
2
3
2
-423
53
3
-201
275
-15
460
4
67
543
252
728
30
505
No, the conclusions are not the same. The mean of Treatment 4 is different than the means of Treatments
1, 2, and 3. However, the mean of Treatment 2 is not different from the means of Treatments 1 and 3
according to Tukey’s method, they were found to be different using the graphical method and the Fisher
LSD method.
(b) Explain the difference between the Tukey and Fisher procedures.
Both Tukey and Fisher utilize a single critical value; however, Tukey’s is based on the studentized range
statistic while Fisher’s is based on t distribution.
3.9. Reconsider the experiment in Problem 3.7. Find a 95 percent confidence interval on the mean
tensile strength of the portland cement produced by each of the four mixing techniques. Also find a 95
percent confidence interval on the difference in means for techniques 1 and 3. Does this aid in interpreting
the results of the experiment?
3-7
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
yi . − tα
2
,N − a
MS E
MS E
≤ µi ≤ yi . + tα ,N − a
n
n
2
Treatment 1: 2971 ± 2.179
12825.69
4
2971 ± 123.387
2847.613 ≤ µ1 ≤ 3094.387
Treatment 2: 3156.25±123.387
3032.863 ≤ µ 2 ≤ 3279.637
Treatment 3: 2933.75±123.387
2810.363 ≤ µ3 ≤ 3057.137
Treatment 4: 2666.25±123.387
2542.863 ≤ µ 4 ≤ 2789.637
Treatment 1 - Treatment 3: yi . − y j . − tα
2
,N − a
2 MS E
2 MS E
≤ µi − µ j ≤ yi . − y j . + tα ,N − a
n
n
2
2(
12825.7 )
4
−137.245 ≤ µ1 − µ3 ≤ 211.745
2971.00 − 2933.75 ± 2.179
Because the confidence interval for the difference between means 1 and 3 spans zero, we agree with the
statement in Problem 3.5 (b); there is not a statistical difference between these two means.
3.10. A product developer is investigating the tensile strength of a new synthetic fiber that will be used to
make cloth for men’s shirts. Strength is usually affected by the percentage of cotton used in the blend of
materials for the fiber. The engineer conducts a completely randomized experiment with five levels of
cotton content and replicated the experiment five times. The data are shown in the following table.
Cotton
Weight
Percentage
15
20
25
30
35
7
12
14
19
7
Observations
15
12
19
22
11
7
17
19
25
10
11
18
18
19
15
9
18
18
23
11
(a) Is there evidence to support the claim that cotton content affects the mean tensile strength? Use α =
0.05.
Minitab Output
One-way ANOVA: Tensile Strength versus Cotton Percentage
Analysis of Variance for Tensile
Source
DF
SS
MS
Cotton P
4
475.76
118.94
Error
20
161.20
8.06
Total
24
636.96
F
14.76
P
0.000
Yes, the F-value is 14.76 with a corresponding P-value of 0.000. The percentage of cotton in the fiber
appears to have an affect on the tensile strength.
3-8
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(b) Use the Fisher LSD method to make comparisons between the pairs of means. What conclusions can
you draw?
Minitab Output
Fisher's pairwise comparisons
Family error rate = 0.264
Individual error rate = 0.0500
Critical value = 2.086
Intervals for (column level mean) - (row level mean)
15
20
25
20
-9.346
-1.854
25
-11.546
-4.054
-5.946
1.546
30
-15.546
-8.054
-9.946
-2.454
-7.746
-0.254
35
-4.746
2.746
0.854
8.346
3.054
10.546
30
7.054
14.546
In the Minitab output the pairs of treatments that do not contain zero in the pair of numbers indicates that
there is a difference in the pairs of the treatments. 15% cotton is different than 20%, 25% and 30%. 20%
cotton is different than 30% and 35% cotton. 25% cotton is different than 30% and 35% cotton. 30%
cotton is different than 35%.
(c) Analyze the residuals from this experiment and comment on model adequacy.
The residual plots below show nothing unusual.
Normal Probability Plot of the Residuals
(response is Tensile Strength)
99
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-5.0
-2.5
0.0
Residual
3-9
2.5
5.0
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals Versus the Fitted Values
(response is Tensile Strength)
5.0
Residual
2.5
0.0
-2.5
-5.0
10
12
14
16
Fitted Value
18
20
22
3.11. Reconsider the experiment described in Problem 3.10. Suppose that 30 percent cotton content is a
control. Use Dunnett’s test with α = 0.05 to compare all of the other means with the control.
For this problem: a = 5, a-1 = 4, f=20, n=5 and α = 0.05
d 0.05 (4, 20)
2 MS E
2(8.06)
= 2.65
= 4.76
n
5
1 vs. 4 : y1. − y4. = 9.8 − 21.6 = −11.8*
2 vs. 4 : y2. − y4. = 15.4 − 21.6 = −6.2 *
3 vs. 4 : y3. − y4. = 17.6 − 21.6 = −4.0
5 vs. 4 : y5. − y4. = 10.8 − 21.6 = −10.8*
The control treatment, treatment 4, differs from treatments 1, 2 and 5.
3.12. A pharmaceutical manufacturer wants to investigate the bioactivity of a new drug. A completely
randomized single-factor experiment was conducted with three dosage levels, and the following results
were obtained.
Dosage
20g
30g
40g
24
37
42
Observations
28
37
44
31
47
52
30
35
38
(a) Is there evidence to indicate that dosage level affects bioactivity? Use α = 0.05.
3-10
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Minitab Output
One-way ANOVA: Activity versus Dosage
Analysis of Variance for Activity
Source
DF
SS
MS
Dosage
2
450.7
225.3
Error
9
288.3
32.0
Total
11
738.9
F
7.04
P
0.014
There appears to be a different in the dosages.
(b) If it is appropriate to do so, make comparisons between the pairs of means. What conclusions can you
draw?
Because there appears to be a difference in the dosages, the comparison of means is appropriate.
Minitab Output
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0209
Critical value = 3.95
Intervals for (column level mean) - (row level mean)
20g
30g
30g
-18.177
4.177
40g
-26.177
-3.823
-19.177
3.177
The Tukey comparison shows a difference in the means between the 20g and the 40g dosages.
(c) Analyze the residuals from this experiment and comment on the model adequacy.
There is nothing too unusual about the residual plots shown below.
Normal Probability Plot of the Residuals
(response is Activity)
99
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-8
-6
-4
-2
0
Residual
3-11
2
4
6
8
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals Versus the Fitted Values
(response is Activity)
8
6
Residual
4
2
0
-2
-4
-6
-8
30
32
34
36
38
Fitted Value
40
42
44
46
3.13. A rental car company wants to investigate whether the type of car rented affects the length of the
rental period. An experiment is run for one week at a particular location, and 10 rental contracts are
selected at random for each car type. The results are shown in the following table.
Type of Car
Sub-compact
Compact
Midsize
Full Size
3
1
4
3
5
3
1
5
3
4
3
7
Observations
7
6
5
7
5
6
5
7
1
5
10
3
3
3
2
4
2
2
4
7
1
1
2
2
6
7
7
7
(a) Is there evidence to support a claim that the type of car rented affects the length of the rental contract?
Use α = 0.05. If so, which types of cars are responsible for the difference?
Minitab Output
One-way ANOVA: Days versus Car Type
Analysis of Variance for Days
Source
DF
SS
MS
Car Type
3
16.68
5.56
Error
36
180.30
5.01
Total
39
196.98
F
1.11
P
0.358
There is no difference.
(b) Analyze the residuals from this experiment and comment on the model adequacy.
3-12
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Normal Probability Plot of the Residuals
(response is Days)
99
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-4
-3
-2
-1
0
1
Residual
2
3
4
5
Residuals Versus the Fitted Values
(response is Days)
5
4
3
Residual
2
1
0
-1
-2
-3
-4
3.5
4.0
4.5
Fitted Value
5.0
5.5
There is nothing unusual about the residuals.
(c) Notice that the response variable in this experiment is a count. Should this cause any potential
concerns about the validity of the analysis of variance?
Because the data is count data, a square root transformation could be applied. The analysis is shown below.
It does not change the interpretation of the data.
Minitab Output
One-way ANOVA: Sqrt Days versus Car Type
Analysis of Variance for Sqrt Day
Source
DF
SS
MS
Car Type
3
1.087
0.362
Error
36
11.807
0.328
Total
39
12.893
F
1.10
P
0.360
3-13
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.14. I belong to a golf club in my neighborhood. I divide the year into three golf seasons: summer (JuneSeptember), winter (November-March) and shoulder (October, April and May). I believe that I play my
best golf during the summer (because I have more time and the course isn’t crowded) and shoulder
(because the course isn’t crowded) seasons, and my worst golf during the winter (because all of the partyear residents show up, and the course is crowded, play is slow, and I get frustrated). Data from the last
year are shown in the following table.
Season
Summer
Shoulder
W inter
83
91
94
85
87
91
85
84
87
87
87
85
Observations
90
88
85
86
87
91
88
83
92
84
91
86
(a) Do the data indicate that my opinion is correct? Use α = 0.05.
Minitab Output
One-way ANOVA: Score versus Season
Analysis of Variance for Score
Source
DF
SS
MS
Season
2
35.61
17.80
Error
22
184.63
8.39
Total
24
220.24
F
2.12
P
0.144
The data do not support the author’s opinion.
(b) Analyze the residuals from this experiment and comment on model adequacy.
Normal Probability Plot of the Residuals
(response is Score)
99
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-5.0
-2.5
0.0
Residual
3-14
2.5
5.0
90
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals Versus the Fitted Values
(response is Score)
5.0
Residual
2.5
0.0
-2.5
-5.0
86.0
86.5
87.0
87.5
88.0
Fitted Value
88.5
89.0
89.5
There is nothing unusual about the residuals.
3.15. A regional opera company has tried three approaches to solicit donations from 24 potential sponsors.
The 24 potential sponsors were randomly divided into three groups of eight, and one approach was used for
each group. The dollar amounts of the resulting contributions are shown in the following table.
Approach
1
2
3
1000
1500
900
1500
1800
1000
Contributions (in $)
1200
1800
1600
1100
2000
1200
2000
1700
1200
1500
1200
1550
1000
1800
1000
1250
1900
1100
(a) Do the data indicate that there is a difference in results obtained from the three different approaches?
Use α = 0.05.
Minitab Output
One-way ANOVA: Contribution versus Approach
Analysis of Variance for Contribution
Source
DF
SS
MS
F
Approach
2
1362708
681354
9.41
Error
21
1520625
72411
Total
23
2883333
P
0.001
There is a difference between the approaches. The Tukey test will indicate which are different. Approach
2 is different than approach 1 and approach 3.
Minitab Output
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0200
Critical value = 3.56
Intervals for (column level mean) - (row level mean)
1
2
3-15
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
2
-770
-93
3
-214
464
218
895
(b) Analyze the residuals from this experiment and comment on the model adequacy.
Normal Probability Plot of the Residuals
(response is Contribution)
99
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-500
-250
0
Residual
250
500
Residuals Versus the Fitted Values
(response is Contribution)
500
Residual
250
0
-250
-500
1200
1300
1400
1500
Fitted Value
There is nothing unusual about the residuals.
3-16
1600
1700
1800
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.16. An experiment was run to determine whether four specific firing temperatures affect the density of a
certain type of brick. A completely randomized experiment led to the following data:
Temperature
100
125
150
175
21.8
21.7
21.9
21.9
Density
21.7
21.5
21.8
21.8
21.9
21.4
21.8
21.7
21.6
21.4
21.6
21.4
21.7
21.5
(a) Does the firing temperature affect the density of the bricks? Use α = 0.05.
No, firing temperature does not affect the density of the bricks. Refer to the Design-Expert output below.
Design Expert Output
Response:
Density
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
0.16
3
0.052
A
0.16
3
0.052
Residual
0.36
14
0.026
Lack of Fit
0.000
0
Pure Error
0.36
14
0.026
Cor Total
0.52
17
F
Value
2.02
2.02
Prob > F
0.1569
0.1569
not significant
The "Model F-value" of 2.02 implies the model is not significant relative to the noise. There is a
15.69 % chance that a "Model F-value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-100
21.74
0.072
2-125
21.50
0.080
3-150
21.72
0.072
4-175
21.70
0.080
Treatment
1 vs 2
1 vs 3
1 vs 4
2 vs 3
2 vs 4
3 vs 4
Mean
Difference
0.24
0.020
0.040
-0.22
-0.20
0.020
DF
1
1
1
1
1
1
Standard
Error
0.11
0.10
0.11
0.11
0.11
0.11
t for H0
Coeff=0
2.23
0.20
0.37
-2.05
-1.76
0.19
Prob > |t|
0.0425
0.8465
0.7156
0.0601
0.0996
0.8552
(b) Is it appropriate to compare the means using the Fisher LSD method in this experiment?
The analysis of variance tells us that there is no difference in the treatments. There is no need to proceed
with Fisher’s LSD method to decide which mean is difference.
(c) Analyze the residuals from this experiment. Are the analysis of variance assumptions satisfied? There
is nothing unusual about the residual plots.
3-17
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Normal plot of residuals
Residuals vs. Predicted
0.2
99
2
0.075
90
80
70
Res iduals
Norm al % probability
95
50
30
20
10
2
-0.05
2
-0.175
5
1
-0.3
-0.3
-0.175
-0.05
0.075
0.2
21.50
21.56
R es idual
21.62
21.68
21.74
Predicted
(d) Construct a graphical display of the treatments as described in Section 3.5.3. Does this graph
adequately summarize the results of the analysis of variance in part (b). Yes.
Scaled t Distribution
(125)
21.2
21.3
21.4
(175,150,100)
21.5
21.6
21.7
21.8
Mean Density
3.17. Rework Part (d) of Problem 3.16 using the Tukey method. What conclusions can you draw?
Explain carefully how you modified the procedure to account for unequal sample sizes.
When sample sizes are unequal, the appropriate formula for the Tukey method is
Tα =
Treatment 1
Treatment 1
Treatment 1
Treatment 3
Treatment 4
Treatment 3
qα (a, f )
2
1 1
MS E +
n n
j
i
vs. Treatment 2 = 21.74 – 21.50 = 0.24 < 0.314
vs. Treatment 3 = 21.74 – 21.72 = 0.02 < 0.296
vs. Treatment 4 = 21.74 – 21.70 = 0.04 < 0.314
vs. Treatment 2 = 21.72 – 21.50 = 0.22 < 0.314
vs. Treatment 2 = 21.70 – 21.50 = 0.20 < 0.331
vs. Treatment 4 = 21.72 – 21.70 = 0.02 < 0.314
3-18
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
All pairwise comparisons do not identify differences. Notice that there are different critical values for the
comparisons depending on the sample sizes of the two groups being compared.
Because we could not reject the hypothesis of equal means using the analysis of variance, we should never
have performed the Tukey test (or any other multiple comparison procedure, for that matter). If you ignore
the analysis of variance results and run multiple comparisons, you will likely make type I errors.
3.18. A manufacturer of television sets is interested in the effect of tube conductivity of four different
types of coating for color picture tubes. A completely randomized experiment is conducted and the
following conductivity data are obtained:
Coating Type
1
2
3
4
143
152
134
129
141
149
136
127
Conductivity
150
137
132
132
146
143
127
129
(a) Is there a difference in conductivity due to coating type? Use α = 0.05.
Yes, there is a difference in means. Refer to the Design-Expert output below..
Design Expert Output
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
844.69
3
281.56
A
844.69
3
281.56
Residual
236.25
12
19.69
Lack of Fit
0.000
0
Pure Error
236.25
12
19.69
Cor Total
1080.94
15
F
Value
14.30
14.30
Prob > F
0.0003
0.0003
The Model F-value of 14.30 implies the model is significant. There is only
a 0.03% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
145.00
2.22
2-2
145.25
2.22
3-3
132.25
2.22
4-4
129.25
2.22
Treatment
1 vs 2
1 vs 3
1 vs 4
2 vs 3
2 vs 4
3 vs 4
Mean
Difference
-0.25
12.75
15.75
13.00
16.00
3.00
DF
1
1
1
1
1
1
Standard
Error
3.14
3.14
3.14
3.14
3.14
3.14
t for H0
Coeff=0
-0.080
4.06
5.02
4.14
5.10
0.96
(b) Estimate the overall mean and the treatment effects.
3-19
Prob > |t|
0.9378
0.0016
0.0003
0.0014
0.0003
0.3578
significant
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
ö
µ = 2207 / 16 = 137.9375
ö1 = y1. − y .. = 145.00 − 137.9375 = 7.0625
τ
ö
τ 2 = y 2. − y .. = 145.25 − 137.9375 = 7.3125
ö
τ 3 = y 3. − y .. = 132.25 − 137.9375 = −5.6875
ö
τ 4 = y 4. − y .. = 129.25 − 137.9375 = −8.6875
(c) Compute a 95 percent interval estimate of the mean of coating type 4. Compute a 99 percent interval
estimate of the mean difference between coating types 1 and 4.
19.69
4
124.4155 ≤ µ 4 ≤ 134.0845
Treatment 4: 129.25 ± 2.179
Treatment 1 - Treatment 4: ( − 129.25)± 3.055
145
(2) .69
19
4
6.164 ≤ µ1 − µ 4 ≤ 25.336
(d) Test all pairs of means using the Fisher LSD method with α=0.05.
Refer to the Design-Expert output above. The Fisher LSD procedure is automatically included in the
output.
The means of Coating Type 2 and Coating Type 1 are not different. The means of Coating Type 3 and
Coating Type 4 are not different. However, Coating Types 1 and 2 produce higher mean conductivity than
does Coating Types 3 and 4.
(e) Use the graphical method discussed in Section 3.5.3 to compare the means. Which coating produces
the highest conductivity?
S yi . =
MS E
19.96
=
= 2.219 Coating types 1 and 2 produce the highest conductivity.
n
4
Scaled t Distribution
(4)
(3)
130
(1)
(2)
135
140
145
150
Conductivity
(f) Assuming that coating type 4 is currently in use, what are your recommendations to the manufacturer?
We wish to minimize conductivity.
3-20
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Since coatings 3 and 4 do not differ, and as they both produce the lowest mean values of conductivity, use
either coating 3 or 4. As type 4 is currently being used, there is probably no need to change.
3.19. Reconsider the experiment in Problem 3.18. Analyze the residuals and draw conclusions about
model adequacy.
There is nothing unusual in the normal probability plot. A funnel shape is seen in the plot of residuals
versus predicted conductivity indicating a possible non-constant variance.
Residuals vs. Predicted
Normal plot of residuals
6.75
99
3
90
80
70
Res iduals
Norm al % probability
95
50
30
20
10
-0.75
2
-4.5
5
1
-8.25
-8.25
3
-0.75
-4.5
129.25
6.75
133.25
137.25
141.25
145.25
Predicted
R es idual
Residuals vs. Coating Type
6.75
Res iduals
3
2
-0.75
-4.5
-8.25
1
2
3
4
Coating Type
3.20. An article in the ACI Materials Journal (Vol. 84, 1987. pp. 213-216) describes several experiments
investigating the rodding of concrete to remove entrapped air. A 3” x 6” cylinder was used, and the
number of times this rod was used is the design variable. The resulting compressive strength of the
concrete specimen is the response. The data are shown in the following table.
3-21
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Rodding Level
10
15
20
25
Compressive Strength
1530
1530
1610
1650
1560
1730
1500
1490
1440
1500
1530
1510
(a) Is there any difference in compressive strength due to the rodding level? Use α = 0.05.
There are no differences.
Design Expert Output
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
28633.33
3
9544.44
A
28633.33
3
9544.44
Residual
40933.33
8
5116.67
Lack of Fit
0.000
0
Pure Error
40933.33
8
5116.67
Cor Total
69566.67
11
F
Value
1.87
1.87
Prob > F
0.2138
0.2138
not significant
The "Model F-value" of 1.87 implies the model is not significant relative to the noise. There is a
21.38 % chance that a "Model F-value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated Standard
Mean
Error
1-10
1500.00 41.30
2-15
1586.67 41.30
3-20
1606.67 41.30
4-25
1500.00 41.30
Treatment
1 vs 2
1 vs 3
1 vs 4
2 vs 3
2 vs 4
3 vs 4
Mean
Difference
-86.67
-106.67
0.000
-20.00
86.67
106.67
DF
1
1
1
1
1
1
Standard
Error
58.40
58.40
58.40
58.40
58.40
58.40
t for H0
Coeff=0
-1.48
-1.83
0.000
-0.34
1.48
1.83
Prob > |t|
0.1761
0.1052
1.0000
0.7408
0.1761
0.1052
(b) Find the P-value for the F statistic in part (a). From computer output, P=0.2138.
(c) Analyze the residuals from this experiment. What conclusions can you draw about the underlying
model assumptions?
Slight inequality of variance can be observed in the residual plots below; however, not enough to be
concerned about the assumptions.
3-22
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Predicted
Normal plot of residuals
123.333
99
70.8333
90
80
70
Res iduals
Norm al % probability
95
50
2
18.3333
30
20
10
-34.1667
5
1
-86.6667
-86.6667
-34.1667
70.8333
18.3333
1500.00
123.333
1526.67
1553.33
1580.00
Predicted
R es idual
Residuals vs. Rodding Level
123.333
Res iduals
70.8333
2
18.3333
-34.1667
-86.6667
1
2
3
4
Rodding Level
(d) Construct a graphical display to compare the treatment means as describe in Section 3.5.3.
3-23
1606.67
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Scaled t Distribution
(10, 25)
(15)
(20)
1418 1459 1500 1541 1582 1623 1664
Mean Compressive Strength
3.21. An article in Environment International (Vol. 18, No. 4, 1992) describes an experiment in which the
amount of radon released in showers was investigated. Radon enriched water was used in the experiment
and six different orifice diameters were tested in shower heads. The data from the experiment are shown in
the following table.
Orifice Dia.
0.37
0.51
0.71
1.02
1.40
1.99
80
75
74
67
62
60
Radon Released (%)
83
83
75
79
73
76
72
74
62
67
61
64
85
79
77
74
69
66
(a) Does the size of the orifice affect the mean percentage of radon released? Use α = 0.05.
Yes. There is at least one treatment mean that is different.
Design Expert Output
Response: Radon Released in %
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
1133.38
5
226.68
A
1133.38
5
226.68
Residual
132.25
18
7.35
Lack of Fit
0.000
0
Pure Error
132.25
18
7.35
Cor Total
1265.63
23
F
Value
30.85
30.85
The Model F-value of 30.85 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
EstimatedStandard
Mean Error
1-0.37
82.75
1.36
2-0.51
77.00
1.36
3-0.71
75.00
1.36
4-1.02
71.75
1.36
5-1.40
65.00
1.36
6-1.99
62.75
1.36
3-24
Prob > F
< 0.0001
< 0.0001
significant
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Treatment
1 vs 2
1 vs 3
1 vs 4
1 vs 5
1 vs 6
2 vs 3
2 vs 4
2 vs 5
2 vs 6
3 vs 4
3 vs 5
3 vs 6
4 vs 5
4 vs 6
5 vs 6
Mean
Difference
5.75
7.75
11.00
17.75
20.00
2.00
5.25
12.00
14.25
3.25
10.00
12.25
6.75
9.00
2.25
DF
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Standard
Error
1.92
1.92
1.92
1.92
1.92
1.92
1.92
1.92
1.92
1.92
1.92
1.92
1.92
1.92
1.92
t for H0
Coeff=0
3.00
4.04
5.74
9.26
10.43
1.04
2.74
6.26
7.43
1.70
5.22
6.39
3.52
4.70
1.17
Prob > |t|
0.0077
0.0008
< 0.0001
< 0.0001
< 0.0001
0.3105
0.0135
< 0.0001
< 0.0001
0.1072
< 0.0001
< 0.0001
0.0024
0.0002
0.2557
(b) Find the P-value for the F statistic in part (a).
P=3.161 x 10-8
(c) Analyze the residuals from this experiment.
There is nothing unusual about the residuals.
Residuals vs. Predicted
Normal plot of residuals
4
99
2
2
1.8125
90
80
70
Res iduals
Norm al % probability
95
50
30
20
2
-0.375
2
10
-2.5625
5
2
1
-4.75
-4.75
-2.5625
-0.375
1.8125
62.75
4
67.75
72.75
Predicted
R es idual
3-25
77.75
82.75
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Orifice Diameter
4
Res iduals
2
2
1.8125
2
-0.375
2
-2.5625
2
-4.75
1
2
3
6
5
4
Orifice Diam eter
(d) Find a 95 percent confidence interval on the mean percent radon released when the orifice diameter is
1.40.
Treatment 5 (Orifice =1.40): 65 ± 2.101
62.152 ≤ µ ≤ 67.848
7.35
4
(e) Construct a graphical display to compare the treatment means as describe in Section 3.5.3. What
conclusions can you draw?
Scaled t Distribution
(6)
(5)
60
(5)
65
(4)
70
(3)
75
(2)
(1)
80
Conductivity
Treatments 5 and 6 as a group differ from the other means; 2, 3, and 4 as a group differ from the other
means, 1 differs from the others.
3-26
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.22. The response time in milliseconds was determined for three different types of circuits that could be
used in an automatic valve shutoff mechanism. The results are shown in the following table.
Circuit Type
1
2
3
9
20
6
Response Time
10
23
8
12
21
5
8
17
16
15
30
7
(a) Test the hypothesis that the three circuit types have the same response time. Use α = 0.01.
From the computer printout, F=16.08, so there is at least one circuit type that is different.
Design Expert Output
Response: Response Time in ms
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
543.60
2
271.80
A
543.60
2
271.80
Residual
202.80
12
16.90
Lack of Fit
0.000
0
Pure Error
202.80
12
16.90
Cor Total
746.40
14
F
Value
16.08
16.08
Prob > F
0.0004
0.0004
significant
The Model F-value of 16.08 implies the model is significant. There is only
a 0.04% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
10.80
1.84
2-2
22.20
1.84
3-3
8.40
1.84
Treatment
1 vs 2
1 vs 3
2 vs 3
Mean
Difference
-11.40
2.40
13.80
DF
1
1
1
Standard
Error
2.60
2.60
2.60
t for H0
Coeff=0
-4.38
0.92
5.31
Prob > |t|
0.0009
0.3742
0.0002
(b) Use Tukey’s test to compare pairs of treatment means. Use α = 0.01.
S yi . =
MS E
16.90
=
= 1.8385
5
n
q0.01,(3,12 ) = 5.04
t0 = 1.8385(5.04) = 9.266
1 vs. 2: 10.8-22.2=11.4 > 9.266
1 vs. 3: 10.8-8.4=2.4 < 9.266
2 vs. 3: 22.2-8.4=13.8 > 9.266
1 and 2 are different. 2 and 3 are different.
Notice that the results indicate that the mean of treatment 2 differs from the means of both treatments 1 and
3, and that the means for treatments 1 and 3 are the same. Notice also that the Fisher LSD procedure (see
the computer output) gives the same results.
(c) Use the graphical procedure in Section 3.5.3 to compare the treatment means. What conclusions can
you draw? How do they compare with the conclusions from part (a).
3-27
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
The scaled-t plot agrees with part (b). In this case, the large difference between the mean of treatment 2
and the other two treatments is very obvious.
Scaled t Distribution
(3)
5
(1)
10
(2)
15
20
25
Tensile Strength
(d) Construct a set of orthogonal contrasts, assuming that at the outset of the experiment you suspected the
response time of circuit type 2 to be different from the other two.
H 0 = µ1 − 2 µ 2 + µ3 = 0
H1 = µ1 − 2 µ 2 + µ3 ≠ 0
C1 = y1. − 2 y2. + y3.
C1 = 54 − 2 (111) + 42 = −126
(−126 )
5 (6 )
2
SSC1 =
FC1 =
= 529.2
529.2
= 31.31
16.9
Type 2 differs from the average of type 1 and type 3.
(e) If you were a design engineer and you wished to minimize the response time, which circuit type would
you select?
Either type 1 or type 3 as they are not different from each other and have the lowest response time.
(f) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?
The normal probability plot has some points that do not lie along the line in the upper region. This may
indicate potential outliers in the data.
3-28
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Predicted
Normal plot of residuals
7.8
99
4.55
90
80
70
Res iduals
Norm al % probability
95
50
30
20
10
1.3
-1.95
5
1
-5.2
-5.2
-1.95
1.3
8.40
7.8
4.55
11.85
15.30
18.75
22.20
Predicted
R es idual
Residuals vs. Circuit Type
7.8
Res iduals
4.55
1.3
-1.95
-5.2
1
3
2
Circuit Type
3.23. The effective life of insulating fluids at an accelerated load of 35 kV is being studied. Test data have
been obtained for four types of fluids. The results from a completely randomized experiment were as
follows:
Fluid Type
1
2
3
4
17.6
16.9
21.4
19.3
18.9
15.3
23.6
21.1
Life (in h) at 35 kV Load
16.3
17.4
18.6
17.1
19.4
18.5
16.9
17.5
20.1
19.5
20.5
18.3
21.6
20.3
22.3
19.8
(a) Is there any indication that the fluids differ? Use α = 0.05.
At α = 0.05 there is no difference, but since the P-value is just slightly above 0.05, there is probably a
difference in means.
3-29
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Design Expert Output
Response:
Life
in in h
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
30.17
3
10.06
A
30.16
3
10.05
Residual
65.99
20
3.30
Lack of Fit
0.000
0
Pure Error
65.99
20
3.30
Cor Total
96.16
23
F
Value
3.05
3.05
Prob > F
0.0525
0.0525
not significant
The Model F-value of 3.05 implies there is a 5.25% chance that a "Model F-Value"
this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
18.65
0.74
2-2
17.95
0.74
3-3
20.95
0.74
4-4
18.82
0.74
Treatment
1 vs 2
1 vs 3
1 vs 4
2 vs 3
2 vs 4
3 vs 4
Mean
Difference
0.70
-2.30
-0.17
-3.00
-0.87
2.13
DF
1
1
1
1
1
1
Standard
Error
1.05
1.05
1.05
1.05
1.05
1.05
t for H0
Coeff=0
0.67
-2.19
-0.16
-2.86
-0.83
2.03
Prob > |t|
0.5121
0.0403
0.8753
0.0097
0.4183
0.0554
(b) Which fluid would you select, given that the objective is long life?
Treatment 3. The Fisher LSD procedure in the computer output indicates that the fluid 3 is different from
the others, and it’s average life also exceeds the average lives of the other three fluids.
(c) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?
There is nothing unusual in the residual plots.
Residuals vs. Predicted
Normal plot of residuals
2.95
99
1.55
90
80
70
Res iduals
Norm al % probability
95
50
30
20
10
0.15
-1.25
5
1
-2.65
-2.65
-1.25
0.15
1.55
17.95
2.95
18.70
19.45
Predicted
R es idual
3-30
20.20
20.95
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Fluid Type
2.95
Res iduals
1.55
0.15
-1.25
-2.65
1
4
3
2
Fluid Type
3.24. Four different designs for a digital computer circuit are being studied in order to compare the amount
of noise present. The following data have been obtained:
Circuit Design
1
2
3
4
19
80
47
95
20
61
26
46
Noise Observed
19
73
25
83
30
56
35
78
8
80
50
97
(a) Is the amount of noise present the same for all four designs? Use α = 0.05.
No, at least one treatment mean is different.
Design Expert Output
Response:
Noise
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
12042.00
3
4014.00
A
12042.00
3
4014.00
Residual
2948.80
16
184.30
Lack of Fit
0.000
0
Pure Error
2948.80
16
184.30
Cor Total
14990.80
19
F
Value
21.78
21.78
The Model F-value of 21.78 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
19.20
6.07
2-2
70.00
6.07
3-3
36.60
6.07
4-4
79.80
6.07
3-31
Prob > F
< 0.0001
< 0.0001
significant
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Mean
Difference
-50.80
-17.40
-60.60
33.40
-9.80
-43.20
Treatment
1 vs 2
1 vs 3
1 vs 4
2 vs 3
2 vs 4
3 vs 4
Standard
Error
8.59
8.59
8.59
8.59
8.59
8.59
DF
1
1
1
1
1
1
t for H0
Coeff=0
-5.92
-2.03
-7.06
3.89
-1.14
-5.03
Prob > |t|
< 0.0001
0.0597
< 0.0001
0.0013
0.2705
0.0001
(b) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?
There is nothing too unusual about the residual plots, although there is a mild outlier present.
Residuals vs. Predicted
Normal plot of residuals
17.2
99
2
4.45
90
2
80
70
Res iduals
Norm al % probability
95
50
30
20
10
-8.3
-21.05
5
1
-33.8
-33.8
-21.05
4.45
-8.3
19.20
17.2
34.35
49.50
64.65
79.80
Predicted
R es idual
Residuals vs. Circuit Design
17.2
2
4.45
Res iduals
2
-8.3
-21.05
-33.8
1
2
3
4
Circuit Des ign
(c) Which circuit design would you select for use? Low noise is best.
From the Design Expert Output, the Fisher LSD procedure comparing the difference in means identifies
Type 1 as having lower noise than Types 2 and 4. Although the LSD procedure comparing Types 1 and 3
has a P-value greater than 0.05, it is less than 0.10. Unless there are other reasons for choosing Type 3,
Type 1 would be selected.
3-32
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.25. Four chemists are asked to determine the percentage of methyl alcohol in a certain chemical
compound. Each chemist makes three determinations, and the results are the following:
Chemist
1
2
3
4
Percentage of Methyl Alcohol
84.99
84.04
85.15
85.13
84.72
84.48
84.20
84.10
84.38
84.88
85.16
84.55
(a) Do chemists differ significantly? Use α = 0.05.
There is no significant difference at the 5% level, but chemists differ significantly at the 10% level.
Design Expert Output
Response: Methyl Alcohol in %
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
1.04
3
0.35
A
1.04
3
0.35
Residual
0.86
8
0.11
Lack of Fit
0.000
0
Pure Error
0.86
8
0.11
Cor Total
1.90
11
F
Value
3.25
3.25
Prob > F
0.0813
0.0813
The Model F-value of 3.25 implies there is a 8.13% chance that a "Model F-Value"
this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
84.47
0.19
2-2
85.05
0.19
3-3
84.79
0.19
4-4
84.28
0.19
Treatment
1 vs 2
1 vs 3
1 vs 4
2 vs 3
2 vs 4
3 vs 4
Mean
Difference
-0.58
-0.32
0.19
0.27
0.77
0.50
DF
1
1
1
1
1
1
Standard
Error
0.27
0.27
0.27
0.27
0.27
0.27
t for H0
Coeff=0
-2.18
-1.18
0.70
1.00
2.88
1.88
(b) Analyze the residuals from this experiment.
There is nothing unusual about the residual plots.
3-33
Prob > |t|
0.0607
0.2703
0.5049
0.3479
0.0205
0.0966
not significant
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Predicted
Normal plot of residuals
0.52
99
0.2825
90
80
70
Res iduals
Norm al % probability
95
50
30
20
10
0.045
-0.1925
5
1
-0.43
-0.43
-0.1925
0.2825
0.045
84.28
0.52
84.48
84.67
84.86
85.05
Predicted
R es idual
Residuals vs. Chemist
0.52
Res iduals
0.2825
0.045
-0.1925
-0.43
1
3
2
4
Chem is t
(c) If chemist 2 is a new employee, construct a meaningful set of orthogonal contrasts that might have
been useful at the start of the experiment.
Chemists
1
2
3
4
Total
253.41
255.16
254.36
252.85
Contrast Totals:
3-34
C1
1
-3
1
1
-4.86
C2
-2
0
1
1
0.39
C3
0
0
-1
1
-1.51
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(− 4.86)2 = 0.656
3( )
12
(0.39)2 = 0.008
SS C 2 =
3(6 )
(− 1.51)2 = 0.380
SS C 3 =
3(2 )
SS C1 =
0.656
= 6.115*
0.10727
0.008
FC 2 =
= 0.075
0.10727
0.380
FC 3 =
= 3.54
0.10727
FC1 =
Only contrast 1 is significant at 5%.
3.26. Three brands of batteries are under study. It is s suspected that the lives (in weeks) of the three
brands are different. Five randomly selected batteries of each brand are tested with the following results:
Brand 1
100
96
92
96
92
Weeks of Life
Brand 2
Brand 3
76
108
80
100
75
96
84
98
82
100
(a) Are the lives of these brands of batteries different?
Yes, at least one of the brands is different.
Design Expert Output
Response:
Life
in Weeks
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
1196.13
2
598.07
A
1196.13
2
598.07
Residual
187.20
12
15.60
Lack of Fit
0.000
0
Pure Error
187.20
12
15.60
Cor Total
1383.33
14
F
Value
38.34
38.34
Prob > F
< 0.0001
< 0.0001
The Model F-value of 38.34 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
95.20
1.77
2-2
79.40
1.77
3-3
100.40
1.77
Mean
Standard
Treatment Difference
DF
Error
1 vs 2
15.80
1
2.50
1 vs 3
-5.20
1
2.50
2 vs 3
-21.00
1
2.50
t for H0
Coeff=0
6.33
-2.08
-8.41
(b) Analyze the residuals from this experiment.
There is nothing unusual about the residuals.
3-35
Prob > |t|
< 0.0001
0.0594
< from 0.0001
significant
Solutions Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Predicted
Normal plot of residuals
7.6
99
4.6
90
80
70
Res iduals
Norm al % probability
95
50
30
20
1.6
2
2
10
-1.4
5
1
2
-4.4
-4.4
-1.4
1.6
4.6
79.40
7.6
84.65
89.90
95.15
100.40
Predicted
R es idual
Residuals vs. Brand
7.6
Res iduals
4.6
1.6
2
2
-1.4
2
-4.4
1
3
2
Brand
(c) Construct a 95 percent interval estimate on the mean life of battery brand 2. Construct a 99 percent
interval estimate on the mean difference between the lives of battery brands 2 and 3.
y i . ± tα
2
,N − a
MS E
n
Brand 2: 79.4 ± 2.179
15.60
5
79.40 ± 3.849
75.551 ≤ µ 2 ≤ 83.249
Brand 2 - Brand 3: y i . − y j . ± tα
2
,N − a
2( .60 )
15
5
−28.631 ≤ µ 2 − µ3 ≤ −13.369
79.4 − 100.4 ± 3.055
3-36
2 MS E
n
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(d) Which brand would you select for use? If the manufacturer will replace without charge any battery
that fails in less than 85 weeks, what percentage would the company expect to replace?
Chose brand 3 for longest life. Mean life of this brand in 100.4 weeks, and the variance of life is estimated
by 15.60 (MSE). Assuming normality, then the probability of failure before 85 weeks is:
85 − 100.4
= Φ (− 3.90 ) = 0.00005
15.60
Φ
That is, about 5 out of 100,000 batteries will fail before 85 week.
3.27. Four catalysts that may affect the concentration of one component in a three component liquid
mixture are being investigated. The following concentrations are obtained from a completely randomized
experiment:
Catalyst
2
3
56.3
50.1
54.5
54.2
57.0
55.4
55.3
1
58.2
57.2
58.4
55.8
54.9
4
52.9
49.9
50.0
51.7
(a) Do the four catalysts have the same effect on concentration?
No, their means are different.
Design Expert Output
Response:
Concentration
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
85.68
3
28.56
A
85.68
3
28.56
Residual
34.56
12
2.88
Lack of Fit
0.000
0
Pure Error
34.56
12
2.88
Cor Total
120.24
15
F
Value
9.92
9.92
Prob > F
0.0014
0.0014
The Model F-value of 9.92 implies the model is significant. There is only
a 0.14% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
56.90
0.76
2-2
55.77
0.85
3-3
53.23
0.98
4-4
51.13
0.85
Treatment
1 vs 2
1 vs 3
1 vs 4
2 vs 3
2 vs 4
3 vs 4
Mean
Difference
1.13
3.67
5.77
2.54
4.65
2.11
DF
1
1
1
1
1
1
Standard
Error
1.14
1.24
1.14
1.30
1.20
1.30
t for H0
Coeff=0
0.99
2.96
5.07
1.96
3.87
1.63
3-37
Prob > |t|
0.3426
0.0120
0.0003
0.0735
0.0022
0.1298
significant
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(b) Analyze the residuals from this experiment.
There is nothing unusual about the residual plots.
Residuals vs. Predicted
Normal plot of residuals
2.16667
99
0.841667
90
80
70
Res iduals
Norm al % probability
95
-0.483333
50
30
20
10
-1.80833
5
1
-3.13333
-3.13333
-1.80833
-0.483333
0.841667
51.13
2.16667
52.57
54.01
55.46
Predicted
R es idual
Residuals vs. Catalyst
2.16667
Res iduals
0.841667
-0.483333
-1.80833
-3.13333
1
3
2
4
Catalys t
(c) Construct a 99 percent confidence interval estimate of the mean response for catalyst 1.
y i . ± tα
2
,N − a
MS E
n
Catalyst 1: 56.9 ± 3.055
2.88
5
56.9 ± 2.3186
54.5814 ≤ µ1 ≤ 59.2186
3-38
56.90
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.28. An experiment was performed to investigate the effectiveness of five insulating materials. Four
samples of each material were tested at an elevated voltage level to accelerate the time to failure. The
failure times (in minutes) is shown below:
Material
1
2
3
4
5
110
1
880
495
7
Failure Time (minutes)
157
194
2
4
1256
5276
7040
5307
5
29
178
18
4355
10050
2
(a) Do all five materials have the same effect on mean failure time?
No, at least one material is different.
Design Expert Output
Response:
Failure Timein Minutes
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Model
1.032E+008 4
2.580E+007
6.19
A
1.032E+008 4
2.580E+007
6.19
Residual
6.251E+007 15
4.167E+006
Lack of Fit
0.000
0
Pure Error
6.251E+007 15
4.167E+006
Cor Total
1.657E+008 19
Prob > F
0.0038
0.0038
significant
The Model F-value of 6.19 implies the model is significant. There is only
a 0.38% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
159.75
1020.67
2-2
6.25
1020.67
3-3
2941.75
1020.67
4-4
5723.00
1020.67
5-5
10.75
1020.67
Treatment
1 vs 2
1 vs 3
1 vs 4
1 vs 5
2 vs 3
2 vs 4
2 vs 5
3 vs 4
3 vs 5
4 vs 5
Mean
Difference
153.50
-2782.00
-5563.25
149.00
-2935.50
-5716.75
-4.50
-2781.25
2931.00
5712.25
DF
1
1
1
1
1
1
1
1
1
1
Standard
Error
1443.44
1443.44
1443.44
1443.44
1443.44
1443.44
1443.44
1443.44
1443.44
1443.44
t for H0
Coeff=0
0.11
-1.93
-3.85
0.10
-2.03
-3.96
-3.118E-003
-1.93
2.03
3.96
Prob > |t|
0.9167
0.0731
0.0016
0.9192
0.0601
0.0013
0.9976
0.0732
0.0604
0.0013
(b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals.
What information do these plots convey?
3-39
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Predicted
Normal plot of residuals
4327
99
95
Norm al % probability
Res iduals
1938.25
-450.5
-2839.25
90
80
70
50
30
20
10
5
1
-5228
6.25
1435.44
2864.62
4293.81
5723.00
-5228
-2839.25
Predicted
-450.5
1938.25
4327
Res idual
The plot of residuals versus predicted has a strong outward-opening funnel shape, which indicates the
variance of the original observations is not constant. The normal probability plot also indicates that the
normality assumption is not valid. A data transformation is recommended.
(c) Based on your answer to part (b) conduct another analysis of the failure time data and draw appropriate
conclusions.
A natural log transformation was applied to the failure time data. The analysis in the log scale identifies
that there exists at least one difference in treatment means.
Design Expert Output
Response:
Failure Timein Minutes Transform:
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
165.06
4
41.26
A
165.06
4
41.26
Residual
16.44
15
1.10
Lack of Fit
0.000
0
Pure Error
16.44
15
1.10
Cor Total
181.49
19
Natural log
Constant:
0.000
F
Value
37.66
37.66
Prob > F
< 0.0001
< 0.0001
significant
The Model F-value of 37.66 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
5.05
0.52
2-2
1.24
0.52
3-3
7.72
0.52
4-4
8.21
0.52
5-5
1.90
0.52
Treatment
1 vs 2
1 vs 3
1 vs 4
1 vs 5
2 vs 3
2 vs 4
Mean
Difference
3.81
-2.66
-3.16
3.15
-6.47
-6.97
DF
1
1
1
1
1
1
Standard
Error
0.74
0.74
0.74
0.74
0.74
0.74
t for H0
Coeff=0
5.15
-3.60
-4.27
4.25
-8.75
-9.42
3-40
Prob > |t|
0.0001
0.0026
0.0007
0.0007
< 0.0001
< 0.0001
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
2 vs
3 vs
3 vs
4 vs
5
4
5
5
-0.66
-0.50
5.81
6.31
1
1
1
1
0.74
0.74
0.74
0.74
-0.89
-0.67
7.85
8.52
0.3856
0.5116
< 0.0001
< 0.0001
There is nothing unusual about the residual plots when the natural log transformation is applied.
Residuals vs. Predicted
Normal plot of residuals
1.64792
99
0.733576
90
80
70
Res iduals
Norm al % probability
95
-0.180766
50
30
20
10
-1.09511
5
1
-2.00945
-2.00945
-1.09511
-0.180766
0.733576
1.24
1.64792
4.73
2.99
6.47
8.21
Predicted
R es idual
Residuals vs. Material
1.64792
Res iduals
0.733576
-0.180766
-1.09511
-2.00945
1
2
3
5
4
Material
3.29. A semiconductor manufacturer has developed three different methods for reducing particle counts
on wafers. All three methods are tested on five wafers and the after-treatment particle counts obtained.
The data are shown below:
Method
1
2
3
31
62
58
10
40
27
Count
21
24
120
(a) Do all methods have the same effect on mean particle count?
3-41
4
30
97
1
35
68
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
No, at least one method has a different effect on mean particle count.
Design Expert Output
Response:
Count
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
8963.73
2
4481.87
A
8963.73
2
4481.87
Residual
6796.00
12
566.33
Lack of Fit
0.000
0
Pure Error
6796.00
12
566.33
Cor Total
15759.73
14
F
Value
7.91
7.91
Prob > F
0.0064
0.0064
significant
The Model F-value of 7.91 implies the model is significant. There is only
a 0.64% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
13.40
10.64
2-2
38.20
10.64
3-3
73.00
10.64
Treatment
1 vs 2
1 vs 3
2 vs 3
Mean
Difference
-24.80
-59.60
-34.80
DF
1
1
1
Standard
Error
15.05
15.05
15.05
t for H0
Coeff=0
-1.65
-3.96
-2.31
Prob > |t|
0.1253
0.0019
0.0393
(b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals.
Are there potential concerns about the validity of the assumptions?
The plot of residuals versus predicted appears to be funnel shaped. This indicates the variance of the
original observations is not constant. The residuals plotted in the normal probability plot do not fall along a
straight line, which suggests that the normality assumption is not valid. A data transformation is
recommended.
Residuals vs. Predicted
Normal plot of residuals
47
99
95
Norm al % probability
Res iduals
23.75
0.5
-22.75
90
80
70
50
30
20
10
5
1
-46
13.40
28.30
43.20
58.10
73.00
-46
Predicted
-22.75
0.5
23.75
Res idual
(c) Based on your answer to part (b) conduct another analysis of the particle count data and draw
appropriate conclusions.
3-42
47
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
For count data, a square root transformation is often very effective in resolving problems with inequality of
variance. The analysis of variance for the transformed response is shown below. The difference between
methods is much more apparent after applying the square root transformation.
Design Expert Output
Response:
Count Transform: Square root
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
63.90
2
31.95
A
63.90
2
31.95
Residual
38.96
12
3.25
Lack of Fit
0.000
0
Pure Error
38.96
12
3.25
Cor Total
102.86
14
Constant:
0.000
F
Value
9.84
9.84
Prob > F
0.0030
0.0030
significant
The Model F-value of 9.84 implies the model is significant. There is only
a 0.30% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
3.26
0.81
2-2
6.10
0.81
3-3
8.31
0.81
Treatment
1 vs 2
1 vs 3
2 vs 3
Mean
Difference
-2.84
-5.04
-2.21
DF
1
1
1
Standard
Error
1.14
1.14
1.14
t for H0
Coeff=0
-2.49
-4.42
-1.94
Prob > |t|
0.0285
0.0008
0.0767
3.30 A manufacturer suspects that the batches of raw material furnished by his supplier differ
significantly in calcium content. There are a large number of batches currently in the warehouse. Five of
these are randomly selected for study. A chemist makes five determinations on each batch as obtains the
following data:
Batch 1
23.46
23.48
23.56
23.39
23.40
Batch 2
23.59
23.46
23.42
23.49
23.50
Batch 3
23.51
23.64
23.46
23.52
23.49
Batch 4
23.28
23.40
23.37
23.46
23.39
Batch 5
23.29
23.46
23.37
23.32
23.38
(a) Is there significant variation in the calcium content from batch to batch? Use α=0.05. The computer
output below shows that for the random effects model there is batch to batch variation.
Based on the ANOVA in the JMP output below, the batches differ significantly.
JMP Output
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
0.525399
0.430479
0.066182
23.4436
25
3-43
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Analysis of Variance
Source
DF
Model
4
Error
20
C. Total
24
Sum of Squares
0.09697600
0.08760000
0.18457600
Effect Tests
Source
Batch
DF
4
Nparm
4
Mean Square
0.024244
0.004380
Sum of Squares
0.09697600
F Ratio
5.5352
F Ratio
5.5352
Prob > F
0.0036*
Prob > F
0.0036*
(b) Estimate the components of variance.
ˆ
= MS E 0.004380
σ2 =
ˆ
=
σ τ2
MSTreatments − MS E 0.024244 − 0.004380
=
= 0.003973
n
5
This is verified in the JMP REML analysis shown below.
JMP Output
Parameter Estimates
Term
Intercept
Estimate
23.4436
REML Variance Component Estimates
Random
Var Ratio
Var Component
Effect
Batch
0.907032
0.0039728
Residual
0.00438
Total
0.0083528
Std Error
0.031141
DFDen
4
t Ratio
752.82
Prob>|t|
<.0001*
Std Error
95% Lower
95% Upper
Pct of Total
0.0034398
0.0013851
-0.002769
0.0025637
0.0107147
0.0091338
47.562
52.438
100.000
Covariance Matrix of Variance Component Estimates
Random Effect
Batch
Residual
Batch
1.1832e-5
-3.837e-7
Residual
-3.837e-7
1.9184e-6
(c) Find a 95 percent confidence interval for σ τ2
(σ
2
τ
+σ 2 )
1 0.024244 1
1 MSTreatments
1
=
− 1
= 0.1154
−1
5 0.004380 3.51
n MS E
Fα 2,a −1, N − a
1 0.024244 1
1 MSTreatments
1
U
=
=
− 1
= 9.2780
−1
MS E
5 0.004380 0.1168
n
F1−α 2,a −1, N − a
2
ˆ
σ
L
U
≤ 2 τ 2≤
ˆ
ˆ
1 + L στ + σ
1+U
=
L
ˆ
σ2
9.2780
0.1154
≤ 2 τ 2≤
ˆ
ˆ
1 + 0.1154 σ τ + σ
1 + 9.2780
0.1035 ≤
ˆ
σ τ2
≤ 0.9027
ˆ
ˆ
σ τ2 + σ 2
(d) Analyze the residuals from this experiment. Are the analysis of variance assumptions are satisfied?
3-44
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
The plot of residuals vs. predicted show no concerns.
0.1
Residual
Calcium content
0.0
0.0
-0.
-0.
23.25
23.35
23.45
23.55
2
Calcium content Predicted
The residuals used in the plot below are based on the REML analysis and shows no concerns. Note,
normality is not a concern for this analysis.
0.2
-1.64 -1.28
0.0
-0.67
1.28 1.64
0.67
0.1
0.1
0.0
0
-0.
-0.
-0.
-0.
0.1
0.2
0.5
0.8
0.9 0.95
Normal Quantile Plot
3.31. Several ovens in a metal working shop are used to heat metal specimens. All ovens are supposed to
operate at the same temperature, although it is suspected that this may not be true. Three ovens selected at
random, and their temperatures on successive heats are noted. The data collected are as follows:
Oven
Temperature
1
491.50
498.30
498.10
493.50
2
488.50
484.65
479.90
477.35
3
480.10
484.80
488.25
473.00
493.60
471.85
(a) Is there significant variation in temperature between ovens? Use α=0.05.
The computer output below shows that there is oven to oven variation.
3-45
478.65
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Minitab Output
ANOVA: Temp versus Oven
Factor
Oven
Type
random
Levels
3
Values
1, 2, 3
Analysis of Variance for Temp
Source
Oven
Error
Total
DF
2
12
14
S = 5.14224
SS
705.10
317.31
1022.41
R-Sq = 68.96%
MS
352.55
26.44
F
13.33
P
0.001
R-Sq(adj) = 63.79%
(b) Estimate the components of variation for this model.
a
∑ ni2 1 77
1 a
∑ ni − i =1 =
=
15 − = 4.9333
n0
a
3 −1
15
a − 1 i =1
ni
∑
i =1
ˆ
= MS E 26.44
σ2 =
ˆ
=
σ τ2
MSTreatments − MS E 352.55 − 26.44
= = 66.10
4.9333
n0
(c) Analyze the residuals from this experiment and draw conclusions about model adequacy.
There are no concerns with the residual plots below.
Residual Plots for Temp
Normal Probability Plot
Versus Fits
5
Residual
10
90
Percent
99
50
10
1
-10
-5
0
Residual
5
0
-5
-10
10
480
484
Histogram
496
Versus Order
2
Residual
Frequency
492
10
3
1
0
488
Fitted Value
-8
-4
0
Residual
4
5
0
-5
-10
8
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15
Observation Order
3-46
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.32. An article in the Journal of the Electrochemical Society (Vol. 139, No. 2, 1992, pp. 524-532)
describes an experiment to investigate low-pressure vapor deposition of polysilicon. The experiment was
carried out in a large capacity reactor at Sematech in Austin, Texas. The reactor has several wafer
positions, and four of these positions are selected at random. The response variable is film thickness
uniformity. Three replicates of the experiment were run, and the data are as follows:
Wafer
Positions
Uniformity
1
2.76
5.67
4.49
2
1.43
1.70
2.19
3
2.34
1.97
1.47
4
0.94
1.36
1.65
(a) Is there a difference in the wafer positions? Use Use α=0.05.
The JMP output below identifies a difference in the wafer positions.
JMP Output
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
Analysis of Variance
Source
DF
Model
3
Error
8
C. Total
11
Effect Tests
Source
Wafer Position
Nparm
3
0.756617
0.665349
0.807579
2.330833
12
Sum of Squares
16.219825
5.217467
21.437292
DF
3
Mean Square
5.40661
0.65218
Sum of Squares
16.219825
F Ratio
8.2900
Prob > F
0.0077*
F Ratio
8.2900
Prob > F
0.0077*
(b) Estimate the variability due to wafer position.
The JMP REML output below identifies the variance component for the wafer position as 1.5848.
JMP Output
Parameter Estimates
Term
Intercept
Estimate
2.3308333
REML Variance Component Estimates
Random Effect
Var Ratio
Wafer Position
2.4300043
Residual
Total
Std Error
0.671231
DFDen
3
Var Component
1.5848083
0.6521833
2.2369917
Std Error
1.4755016
0.3260917
Covariance Matrix of Variance Component Estimates
Random Effect
Wafer Position
Wafer Position
2.177105
Residual
-0.035445
Residual
-0.035445
0.1063358
3-47
t Ratio
3.47
95% Lower
-1.307122
0.2975536
Prob>|t|
0.0403*
95% Upper
4.4767383
2.393629
Pct of Total
70.846
29.154
100.000
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(c) Estimate the random error component.
The JMP REML output above identifies the random error variance component as 0.6522..
(d) Analyze the residuals from this experiment and comment on model adequacy.
The plot of residuals vs. predicted shows some uniformity concerns.
1.5
1.0
Residual
Uniformity
0.5
0.0
-0.
-1.
-1.
-2.
1
2
3
4
5
6
Uniformity Predicted
The residuals used in the plot below are based on the REML analysis. The normal plot shows some
concerns with the normality assumption; however, the normality is not important for this analysis.
-1.28
0.0
-0.67
1.28
0.67
3
2
1
0
-1
0.1
0.2
0.3 0.4 0.5 0.6 0.7
0.8
0.9
Normal Quantile Plot
Uniformity data often requires a transformation, such as a log transformation, and should be
considered for this experiment.
3.33. Consider the vapor-deposition experiment described in Problem 3.32.
(a) Estimate the total variability in the uniformity response.
The JMP REML output shown in part (b) of Problem 3.32 identifies the total variability as 2.2370.
3-48
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(b) How much of the total variability in the uniformity response is due to the difference between positions
in the reactor?
From the JMP REML output shown in part (b) of Problem 3.32, the differences between positions
represents 70.846% of the total variability.
(c) To what level could the variability in the uniformity response be reduced if position-to-position
variability in the reactor could be eliminated? Do you believe this is a significant reduction?
The variability could be reduced to 29.154% of the current total variability. Based on the 95%
confidence intervals calculated below, this is not significant. An increase in sample size might reverse
this decision.
=
L
1 5.40661
1 MSTreatments
1
1
=
− 1
= 0.073623
−1
4 0.65218 6.059467
n MS E
Fα 2,a −1, N − a
U
=
1 5.40661
1 MSTreatments
1
1
=
− 1
= 65.08093
−1
4 0.65218 0.025398
n MS E
F1−α 2,a −1, N − a
ˆ
σ2
L
U
≤ 2 τ 2≤
ˆ
ˆ
1 + L στ + σ
1+U
ˆ
σ2
65.08093
0.073623
≤ 2 τ 2≤
ˆ
ˆ
1 + 0.073623 σ τ + σ
1 + 65.08093
0.068575 ≤
ˆ
σ τ2
≤ 0.984867
2
ˆ
ˆ
στ + σ 2
3.34. An article in the Journal of Quality Technology (Vol. 13, No. 2, 1981, pp. 111-114) describes and
experiment that investigates the effects of four bleaching chemicals on pulp brightness. These four
chemicals were selected at random from a large population of potential bleaching agents. The data are as
follows:
Chemicals
Brightness
1
77.199
74.466
92.746
76.208
82.876
2
80.522
79.306
81.914
80.346
73.385
3
79.417
78.017
91.596
80.802
80.626
4
78.001
78.358
77.544
77.364
77.386
(a) Is there a difference in the chemical types? Use Use α=0.05.
From the analysis below, there does not appear to be a difference in chemical types.
JMP Output
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
Analysis of Variance
Source
0.123254
-0.04114
4.898921
79.90395
20
DF
Sum of Squares
Mean Square
3-49
F Ratio
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Source
Model
Error
C. Total
DF
3
16
19
Effect Tests
Source
Chemical
Sum of Squares
53.98207
383.99085
437.97292
Nparm
3
DF
3
Mean Square
17.9940
23.9994
Sum of Squares
53.982073
F Ratio
0.7498
Prob > F
0.5383
F Ratio
0.7498
Prob > F
0.5383
(b) Estimate the variability due to chemical types.
The JMP REML output below identifies the variance component for chemical types as -1.201081.
This negative value is a concern. One solution would be to convert this to zero, but this has concerns
as identified in Section 3.9.3 of the textbook. Another course of action is to re-estimate this variance
component using a method that always provides a non-negative value. Another alternative is to
assume that the underlying model is non-linear and re-examine the problem.
JMP Output
Parameter Estimates
Term
Intercept
Estimate
79.90395
Std Error
0.948526
REML Variance Component Estimates
Random Effect
Var Ratio
Var Component
Chemical
-0.050046
-1.201081
Residual
23.999428
Total
22.798347
DFDen
3
Std Error
3.3932473
8.4850792
95% Lower
-7.851723
13.312053
t Ratio
84.24
95% Upper
5.4495617
55.589101
Prob>|t|
<.0001*
Pct of Total
-5.268
105.268
100.000
Covariance Matrix of Variance Component Estimates
Random Effect
Chemical
Residual
Chemical
11.514127
-14.39931
Residual
-14.39931
71.996569
(c) Estimate the variability due to random error.
From the JMP REML output shown above, the variance component due to random error is 23.999428.
(d) Analyze the residuals from this experiment and comment on model adequacy.
Examination of the residuals identified two outliers. These outliers correspond to the Brightness
values of 92.746 and 91.596. The experimenter should resolve these outliers.
Residual
Brightness
10
5
0
-5
75
80
85
90
95
Brightness Predicted
3-50
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
15
-1.64 -1.28
-0.67
0.0
0.67
1.28 1.64
10
5
0
-5
0.1
0.2
0.5
0.8
0.9 0.95
Normal Quantile Plot
3.35. Consider the single-factor random effects model discussed in this chapter. Develop a procedure for
finding a 100(1-α)% confidence interval on the ratio σ τ2
(σ
2
τ
+ σ 2 ) . Assume that the experiment is
balanced.
The procedure shown below is based on the guidelines presented in Section 1.4 of the textbook. Rather
than repeat the details of the seven steps, only additional information is provided below that is specific to
the single-factor random effects and the confidence interval.
1. Recognition of and statement of the problem.
2. Selection of the response variable.
3. Choice of factors, levels, and range. For this case, one factor is chosen. However, the number of
levels chosen and the number of replicates determines the degrees of freedom for the F value used
in the confidence interval calculations. Because the levels are random, it is important to choose an
adequate representation of this effect.
4. Choice of experimental design. For this case, the experimental design is a single factor
experiment. As mentioned above, the number of replicates is important in the estimation of the
confidence intervals. The value for α should also be identified as this could influence the number
of replicates chosen.
5. Performing the experiment.
6. Statistical analysis of the data. Perform the analysis of variance in the same manner as a fixed
effects case. Identify the MS τ and MS E from the ANOVA. Select the F α /2,a-1,N-a and F 1-α /2,a-1,N-a .
Perform the calculations as identified in Equations 3.59a, 3,59b, and 3.60.
7. Conclusions and recommendations.
3.36. Consider testing the equality of the means of two normal populations, where the variances are
unknown but are assumed to be equal. The appropriate test procedure is the pooled t test. Show that the
pooled t test is equivalent to the single factor analysis of variance.
t0 =
y1. − y 2.
Sp
2
n
~ t 2 n − 2 assuming n 1 = n 2 = n
3-51
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
∑ (y
) ∑ (y2 j − y2. )2 ∑∑ (yij − y1. )2
n
n
1j
Sp =
j =1
j =1
(y1. − y2. )2 n = ∑ yi . − y..
2
n
2
i =1
i =1 j =1
=
2n − 2
2
Furthermore,
2
− y1. 2 +
2n
≡ MS E for a=2
2
n
2n − 2
, which is exactly the same as SS Treatments in a one-way
2
classification with a=2. Thus we have shown that t 0 =
SS Treatments
2
. In general, we know that t u = F1,u so
MS E
2
that t 0 ~ F1,2 n − 2 . Thus the square of the test statistic from the pooled t-test is the same test statistic that
results from a single-factor analysis of variance with a=2.
a
3.37. Show that the variance of the linear combination
∑
c i y i . is σ 2
i =1
a
V ci yi . =
i =1
∑
a
∑
V (ci yi . ) =
i =1
a
∑
i =1
=
∑
∑c
∑n c
2
ii
.
i =1
ni
ci2V
yij =
j =1
a
a
ni
∑ ∑ V (y ), V (y )= σ
a
ci2
i =1
2
ij
ij .
j =1
2
2
i ni σ
i =1
3.38. In a fixed effects experiment, suppose that there are n observations for each of four treatments. Let
2
2
Prove that
Q12 , Q2 , Q3 be single-degree-of-freedom components for the orthogonal contrasts.
2
2
SS Treatments = Q12 + Q 2 + Q3 .
C1 = 3 y1. − y 2. − y 3. − y 4.
C 2 = 2 y 2. − y 3. − y 4.
C 3 = y 3. − y 4.
SS C1 = Q12
2
SS C 2 = Q 2
2
SS C 3 = Q3
( 3 y1. − y 2. − y 3. − y 4. ) 2
12n
( 2 y 2. − y 3. − y 4. ) 2
2
Q2 =
6n
( y − y 4. ) 2
2
Q 3 = 3.
2n
4
y i2. − 6
yi. y j.
9
i =1
i< j
2
2
and since
Q12 + Q 2 + Q3 =
12n
Q12 =
∑
∑∑
4
∑∑
i< j
1 2
yi . y j . = y.. −
2
4
∑
i =1
2
2
yi2. , we have Q12 + Q 2 + Q3 =
for a=4.
3-52
12
∑y
2
i.
2
− 3 y ..
i =1
12n
4
=
∑
i =1
2
y i2. y ..
−
= SS Treatments
n
4n
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.39. Use Bartlett's test to determine if the assumption of equal variances is satisfied in Problem 3.24. Use
α = 0.05. Did you reach the same conclusion regarding the equality of variance by examining the residual
plots?
2
χ 0 = 2.3026
2
q = (N − a )log 10 S p −
q
, where
c
a
∑ (n
i
− 1)log 10 S i2
i =1
1
(ni − 1)−1 − (N − a )−1
3(a − 1) i =1
a
∑
c = 1+
a
2
Sp =
S12
2
S2
2
S3
∑ (n
− 1)S i2
i =1
N −a
= 11.2
= 14.8
i
2
Sp =
(5 − 1)11.2 + (5 − 1)14.8 + (5 − 1)20.8
15 − 3
= 20.8
c = 1+
a
1
(5 − 1)−1 − (15 − 3)−1
3(3 − 1) i =1
c = 1+
= 15.6
1 3 1
+ = 1.1389
3(3 − 1) 4 12
∑
a
2
q = (N − a )log10 S p − ∑ (ni − 1)log10 Si2
i =1
q = (15 − 3)log10 15.6 − 4 (log10 11.2 + log10 14.8 + log10 20.8 )
q = 14.3175 − 14.150 = 0.1673
q
c
χ 02 = 2.3026 = 2.3026
0.1673
= 0.3382
1.1389
2
χ 0.05,2 = 5.99
Cannot reject null hypothesis; conclude that the variance are equal. This agrees with the residual plots in
Problem 3.24.
3.40. Use the modified Levene test to determine if the assumption of equal variances is satisfied on
Problem 3.24. Use α = 0.05. Did you reach the same conclusion regarding the equality of variances by
examining the residual plots?
The absolute value of Battery Life – brand median is:
yij −
yi
Brand 1
4
0
4
0
4
Brand 2
4
0
5
4
2
3-53
Brand 3
8
0
4
2
0
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
The analysis of variance indicates that there is not a difference between the different brands and therefore
the assumption of equal variances is satisfied.
Design Expert Output
Response: Mod Levine
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
0.93
2
0.47
A
0.93
2
0.47
Pure Error
80.00
12
6.67
Cor Total
80.93
14
F
Value
0.070
0.070
Prob > F
0.9328
0.9328
3.41. Refer to Problem 3.22. If we wish to detect a maximum difference in mean response times of 10
milliseconds with a probability of at least 0.90, what sample size should be used? How would you obtain a
preliminary estimate of σ 2 ?
Φ2 =
nD 2
2 aσ 2
, use MS E from Problem 3.20 to estimate σ 2 .
Φ2 =
n( )2
10
= 0.986n
2(3)( .9 )
16
Letting α = 0.05 , P(accept) = 0.1 , υ1 = a − 1 = 2
Trial and Error yields:
n
υ2
5
6
7
12
15
18
Φ
2.22
2.43
2.62
P(accept)
0.17
0.09
0.04
Choose n ≥ 6, therefore N ≥ 18
Notice that we have used an estimate of the variance obtained from the present experiment. This indicates
that we probably didn’t use a large enough sample (n was 5 in problem 3.20) to satisfy the criteria specified
in this problem. However, the sample size was adequate to detect differences in one of the circuit types.
When we have no prior estimate of variability, sometimes we will generate sample sizes for a range of
possible variances to see what effect this has on the size of the experiment. Often a knowledgeable expert
will be able to bound the variability in the response, by statements such as “the standard deviation is going
to be at least…” or “the standard deviation shouldn’t be larger than…”.
3.42. Refer to Problem 3.26.
(a) If we wish to detect a maximum difference in mean battery life of 10 hours with a probability of at
least 0.90, what sample size should be used? Discuss how you would obtain a preliminary estimate of
σ2 for answering this question.
Use the MS E from Problem 3.26.
3-54
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
nD 2
n (10 )
2
= 1.0684n
2 (3)(15.60 )
Letting α = 0.05 , P(accept) = 0.1 , υ1 = a − 1 = 2
Φ=
2
2 aσ 2
Φ=
2
Trial and Error yields:
n
υ2
4
5
6
9
12
15
Φ
2.067
2.311
2.532
P(accept)
0.25
0.12
0.05
Choose n ≥ 6, therefore N ≥ 18
See the discussion from the previous problem about the estimate of variance.
(b) If the difference between brands is great enough so that the standard deviation of an observation is
increased by 25 percent, what sample size should be used if we wish to detect this with a probability of
at least 0.90?
υ1 = a − 1 = 2
[
υ 2 = N − a = 15 − 3 = 12
]
[
α = 0.05
]
P( accept ) ≤ 0.1
1
1
λ = 1 + n ( + 0.01P )2 − 1 = 1 + n ( + 0.01(25))2 − 1 = 1 + 0.5625n
Trial and Error yields:
n
υ2
8
9
10
21
24
27
Φ
2.12
2.25
2.37
P(accept)
0.16
0.13
0.09
Choose n ≥ 10, therefore N ≥ 30
3.43. Consider the experiment in Problem 3.26. If we wish to construct a 95 percent confidence interval
on the difference in two mean battery lives that has an accuracy of ±2 weeks, how many batteries of each
brand must be tested?
α = 0.05
MS E = 15.6
width = t 0.025 ,N − a
2MS E
n
Trial and Error yields:
n
υ2
t
width
5
10
31
32
12
27
90
93
2.179
2.05
1.99
1.99
5.44
3.62
1.996
1.96
Choose n ≥ 31, therefore N ≥ 93
3-55
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.44. Suppose that four normal populations have means of µ 1 =50, µ 2 =60, µ 3 =50, and µ 4 =60. How many
observations should be taken from each population so that the probability or rejecting the null hypothesis of
equal population means is at least 0.90? Assume that α=0.05 and that a reasonable estimate of the error
variance is σ 2 =25.
µ i = µ + τ i ,i = 1,2 ,3,4
4
∑µ
i
220
µ=
=
= 55
4
4
τ 1 = −5,τ 2 = 5,τ 3 = −5,τ 4 = 5
i =1
Φ=
2
n
∑τ
aσ
2
2
i
=
100n
=n
4(25)
Φ= n
4
∑τ
2
i
= 100
i =1
υ1 = 3,υ 2 = 4(n − 1),α = 0.05 , From the O.C. curves we can construct the following:
n
4
5
υ2
12
16
Φ
2.00
2.24
β
0.18
0.08
1-β
0.82
0.92
Therefore, select n=5
3.45. Refer to Problem 3.44.
(a) How would your answer change if a reasonable estimate of the experimental error variance were σ 2 =
36?
Φ2 =
n
∑τ
2
i
=
aσ 2
100n
= 0.6944n
4(36 )
Φ = 0.6944n
υ1 = 3,υ 2 = 4(n − 1),α = 0.05 , From the O.C. curves we can construct the following:
n
5
6
7
υ2
16
20
24
Φ
1.863
2.041
2.205
β
0.24
0.15
0.09
1-β
0.76
0.85
0.91
Therefore, select n=7
(b) How would your answer change if a reasonable estimate of the experimental error variance were σ 2 =
49?
Φ2 =
n
∑τ
aσ
2
2
i
=
100n
= 0.5102n
4(49 )
Φ = 0.5102n
3-56
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
υ1 = 3,υ 2 = 4(n − 1),α = 0.05 , From the O.C. curves we can construct the following:
n
7
8
9
Φ
1.890
2.020
2.142
υ2
24
28
32
β
0.16
0.11
0.09
1-β
0.84
0.89
0.91
Therefore, select n=9
(c) Can you draw any conclusions about the sensitivity of your answer in the particular situation about
how your estimate of σ affects the decision about sample size?
As our estimate of variability increases the sample size must increase to ensure the same power of the test.
(d) Can you make any recommendations about how we should use this general approach to choosing n in
practice?
When we have no prior estimate of variability, sometimes we will generate sample sizes for a range of
possible variances to see what effect this has on the size of the experiment. Often a knowledgeable expert
will be able to bound the variability in the response, by statements such as “the standard deviation is going
to be at least…” or “the standard deviation shouldn’t be larger than…”.
3.46. Refer to the aluminum smelting experiment described in Section 3.8.3. Verify that ratio control
methods do not affect average cell voltage. Construct a normal probability plot of residuals. Plot the
residuals versus the predicted values. Is there an indication that any underlying assumptions are violated?
Design Expert Output
Response:
Cell Average
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
2.746E-003 3
9.153E-004
A
2.746E-003 3
9.153E-004
Residual
0.090
20
4.481E-003
Lack of Fit
0.000
0
Pure Error
0.090
20
4.481E-003
Cor Total
0.092
23
F
Value
0.20
0.20
Prob > F
0.8922
0.8922
not significant
The "Model F-value" of 0.20 implies the model is not significant relative to the noise. There is a
89.22 % chance that a "Model F-value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-1
4.86
0.027
2-2
4.83
0.027
3-3
4.85
0.027
4-4
4.84
0.027
Mean
Treatment Difference
1 vs 2
0.027
1 vs 3
0.013
1 vs 4
0.025
2 vs 3
-0.013
2 vs 4
-1.667E-003
3 vs 4
0.012
DF
1
1
1
1
1
1
Standard
Error
0.039
0.039
0.039
0.039
0.039
0.039
t for H0
Coeff=0
0.69
0.35
0.65
-0.35
-0.043
0.30
The following residual plots are satisfactory.
3-57
Prob > |t|
0.4981
0.7337
0.5251
0.7337
0.9660
0.7659
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Predicted
Normal plot of residuals
0.105
99
2
0.05125
90
80
70
Res iduals
Norm al % probability
95
50
30
20
10
3
-0.0025
-0.05625
5
1
-0.11
-0.11
-0.05625
0.05125
-0.0025
4.833
0.105
4.840
4.847
4.853
4.860
Predicted
R es idual
Residuals vs. Algorithm
0.105
2
Res iduals
0.05125
3
-0.0025
-0.05625
-0.11
1
2
3
4
Algorithm
3.47. Refer to the aluminum smelting experiment in Section 3.8.3. Verify the ANOVA for pot noise
summarized in Table 3.16. Examine the usual residual plots and comment on the experimental validity.
Design Expert Output
Response:
Cell StDev Transform: Natural log
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
6.17
3
2.06
A
6.17
3
2.06
Residual
1.87
20
0.094
Lack of Fit
0.000
0
Pure Error
1.87
20
0.094
Cor Total
8.04
23
Constant:
0.000
F
Value
21.96
21.96
Prob > F
< 0.0001
< 0.0001
The Model F-value of 21.96 implies the model is significant. There is only
a 0.01% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
3-58
significant
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
1-1
2-2
3-3
4-4
Mean
-3.09
-3.51
-2.20
-3.36
Treatment
1 vs 2
1 vs 3
1 vs 4
2 vs 3
2 vs 4
3 vs 4
Error
0.12
0.12
0.12
0.12
Mean
Difference
0.42
-0.89
0.27
-1.31
-0.15
1.16
DF
1
1
1
1
1
1
Standard
Error
0.18
0.18
0.18
0.18
0.18
0.18
t for H0
Coeff=0
2.38
-5.03
1.52
-7.41
-0.86
6.55
Prob > |t|
0.0272
< 0.0001
0.1445
< 0.0001
0.3975
< 0.0001
The following residual plots identify the residuals to be normally distributed, randomly distributed through
the range of prediction, and uniformly distributed across the different algorithms. This validates the
assumptions for the experiment.
Residuals vs. Predicted
Normal plot of residuals
0.512896
99
2
0.245645
90
80
70
Res iduals
Norm al % probability
95
-0.0216069
50
30
20
10
3
2
2
-0.288858
5
2
1
-0.55611
-0.55611
-0.288858 -0.0216069
0.245645
-3.51
0.512896
Residuals vs. Algorithm
0.512896
2
Res iduals
0.245645
3
2
2
-0.288858
2
-0.55611
1
3
2
-2.85
Predicted
R es idual
-0.0216069
-3.18
4
Algorithm
3-59
-2.53
-2.20
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
3.48. Four different feed rates were investigated in an experiment on a CNC machine producing a
component part used in an aircraft auxiliary power unit. The manufacturing engineer in charge of the
experiment knows that a critical part dimension of interest may be affected by the feed rate. However,
prior experience has indicated that only dispersion effects are likely to be present. That is, changing the
feed rate does not affect the average dimension, but it could affect dimensional variability. The engineer
makes five production runs at each feed rate and obtains the standard deviation of the critical dimension (in
10-3 mm). The data are shown below. Assume that all runs were made in random order.
Feed Rate
(in/min)
10
12
14
16
1
0.09
0.06
0.11
0.19
Production
2
0.10
0.09
0.08
0.13
Run
3
0.13
0.12
0.08
0.15
4
0.08
0.07
0.05
0.20
5
0.07
0.12
0.06
0.11
(a) Does feed rate have any effect on the standard deviation of this critical dimension?
Because the residual plots were not acceptable for the non-transformed data, a square root transformation
was applied to the standard deviations of the critical dimension. Based on the computer output below, the
feed rate has an effect on the standard deviation of the critical dimension.
Design Expert Output
Response:
Run StDev Transform: Square root
Constant:
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
F
Source
Squares
DF
Square
Value
Model
0.040
3
0.013
7.05
A
0.040
3
0.013
7.05
Residual
0.030
16
1.903E-003
Lack of Fit
0.000
0
Pure Error
0.030
16
1.903E-003
Cor Total
0.071
19
0.000
Prob > F
0.0031
0.0031
significant
The Model F-value of 7.05 implies the model is significant. There is only
a 0.31% chance that a "Model F-Value" this large could occur due to noise.
Treatment Means (Adjusted, If Necessary)
Estimated
Standard
Mean
Error
1-10
0.30
0.020
2-12
0.30
0.020
3-14
0.27
0.020
4-16
0.39
0.020
Mean
Treatment Difference
1 vs 2
4.371E-003
1 vs 3
0.032
1 vs 4
-0.088
2 vs 3
0.027
2 vs 4
-0.092
3 vs 4
-0.12
DF
1
1
1
1
1
1
Standard
Error
0.028
0.028
0.028
0.028
0.028
0.028
t for H0
Coeff=0
0.16
1.15
-3.18
0.99
-3.34
-4.33
Prob > |t|
0.8761
0.2680
0.0058
0.3373
0.0042
0.0005
(b) Use the residuals from this experiment of investigate model adequacy. Are there any problems with
experimental validity?
The residual plots are satisfactory.
3-60
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
Residuals vs. Predicted
Normal plot of residuals
0.0584817
99
2
0.028646
90
80
70
Res iduals
Norm al % probability
95
2
-0.00118983
50
30
20
10
-0.0310256
5
1
-0.0608614
-0.0608614 -0.0310256 -0.00118983 0.028646
0.27
0.0584817
0.30
0.33
0.36
0.39
Predicted
R es idual
Residuals vs. Feed Rate
0.0584817
2
Res iduals
0.028646
2
-0.00118983
-0.0310256
-0.0608614
1
4
3
2
Feed Rate
3.49. Consider the data shown in Problem 3.22.
(a) Write out the least squares normal equations for this problem, and solve them for µ and τ using the
i,
usual constraint
∑
3
ö
τ
i =1 i
= 0 . Estimate τ 1 − τ 2 .
ö
15µ
ö
5µ
ö
5µ
ö
+5τ 1
ö
+5τ 1
ö
+5τ 2
3
∑ τö
i
=207
=54
ö
+5τ 2
ö
5µ
Imposing
ö
+5τ 3
=111
ö
+5τ 3
=42
ö
ö
ö
ö
= 0 , therefore µ = 13.80 , τ 1 = −3.00 , τ 2 = 8.40 , τ 3 = −5.40
i =1
3-61
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
öö
τ1 − τ 2 = −3.00 − 8.40 = −11.40
ö
ö
ö
(b) Solve the equations in (a) using the constraint τ 3 = 0 . Are the estimators τ i and µ the same as you
found in (a)? Why? Now estimate τ 1 − τ 2 and compare your answer with that for (a). What statement
can you make about estimating contrasts in the τ i ?
ö
ö
ö
Imposing the constraint, τ 3 = 0 we get the following solution to the normal equations: µ = 8.40 , τ 1 = 2.40
ö
ö
, τ 2 = 13.8 , and τ 3 = 0 .
These estimators are not the same as in part (a).
However,
öö
τ1 − τ 2 = 2.40 − 13.80 = −11.40 , is the same as in part (a). The contrasts are estimable.
(c) Estimate µ + τ 1 , 2τ 1 − τ 2 − τ 3 and µ + τ 1 + τ 2 using the two solutions to the normal equations.
Compare the results obtained in each case.
1
Contrast
µ + τ1
Estimated from Part (a)
10.80
Estimated from Part (b)
10.80
2
2τ 1 − τ 2 − τ 3
-9.00
-9.00
3
µ +τ1 +τ 2
19.20
24.60
Contrasts 1 and 2 are estimable, 3 is not estimable.
3.50. Apply the general regression significance test to the experiment in Example 3.5. Show that the
procedure yields the same results as the usual analysis of variance.
From the etch rate table:
y.. = 12355
from Example 3.5, we have:
ö
µ = 617.75 τö = −66.55 τö2 = −30.35
1
τö = 7.65 τö4 = 89.25
3
4
5
∑∑ y
i =1 j =1
2
ij
= 7, 704,511 , with 20 degrees of freedom.
5
ö
R (µ ,τ ) = µ y.. + ∑τöyi.
i =1
= (617.75 )(12355 ) + (−66.55 )(2756 ) + (−30.35 )(2937 ) + (7.65 )(3127 ) + (89.25 )(3535 )
= 7, 632,301.25 + 66, 870.55 = 7, 699,171.80
with 4 degrees of freedom.
4
5
2
SS E = ∑∑ yij − R (µ ,τ ) = 7, 704,511 − 7, 699,171.80 = 5339.2
i =1 j =1
with 20-4 degrees of freedom.
This is identical to the SS E found in Example 3.5.
The reduced model:
3-62
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
ö
R (µ ) = µ y.. = (617.75 )(12355 ) = 7, 632,301.25 , with 1 degree of freedom.
R (τ µ ) = R (µ ,τ ) − R (µ ) = 7, 699,171.80 − 7, 632,301.25 = 66,870.55 , with 4-1=3 degrees of
freedom.
Note: R( µ )= SS Treatment from Example 3.1.
τ
Finally,
R (τ µ )
F0 =
3
SS E
16
66,870.55
22290.8
3
=
=
= 66.8
5339.2
333.7
16
which is the same as computed in Example 3.5.
3.51. Use the Kruskal-Wallis test for the experiment in Problem 3.23. Are the results comparable to those
found by the usual analysis of variance?
From Design Expert Output of Problem 3.21
Response:
Life
in in h
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
30.17
3
10.06
A
30.16
3
10.05
Residual
65.99
20
3.30
Lack of Fit
0.000
0
Pure Error
65.99
20
3.30
Cor Total
96.16
23
H=
F
Value
3.05
3.05
Prob > F
0.0525
0.0525
not significant
a Ri2
12
12
.
[4060.75]− 3 (24 + 1) = 6.22
∑ − 3 (N + 1) =
N (N + 1) i =1 ni
24 (24 + 1)
2
χ 0.05 ,3 = 7.81
Accept the null hypothesis; the treatments are not different. This agrees with the analysis of variance.
3.52. Use the Kruskal-Wallis test for the experiment in Problem 3.24. Compare conclusions obtained with
those from the usual analysis of variance?
From Design Expert Output of Problem 3.22
Response:
Noise
ANOVA for Selected Factorial Model
Analysis of variance table [Partial sum of squares]
Sum of
Mean
Source
Squares
DF
Square
Model
12042.00
3
4014.00
A
12042.00
3
4014.00
Residual
2948.80
16
184.30
Lack of Fit
0.000
0
Pure Error
2948.80
16
184.30
Cor Total
14990.80
19
F
Value
21.78
21.78
3-63
Prob > F
< 0.0001
< 0.0001
significant
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
H=
a Ri2
12
12
.
[2726.8]− 3 (20 + 1) = 14.91
∑ − 3 (N + 1) =
N (N + 1) i =1 ni
20 (20 + 1)
2
χ 0.05,3 = 7.81
Reject the null hypothesis because the treatments are different. This agrees with the analysis of variance.
3.53. Consider the experiment in Example 3.5. Suppose that the largest observation on etch rate is
incorrectly recorded as 250A/min. What effect does this have on the usual analysis of variance? What
effect does it have on the Kruskal-Wallis test?
The incorrect observation reduces the analysis of variance F 0 from 66.8 to 0.50. It does change the value
of the Kruskal-Wallis test statistic but not the result.
Minitab Output
One-way ANOVA: Etch Rate 2 versus Power
Analysis of Variance for Etch Rat
Source
DF
SS
MS
Power
3
15927
5309
Error
16
168739
10546
Total
19
184666
F
0.50
P
0.685
3.54 A textile mill has a large number of looms. Each loom is supposed to provide the same output of
cloth per minute. To investigate this assumption, five looms are chosen at random, and their output is
noted at different times. The following data are obtained:
Loom
1
2
3
4
5
14.0
13.9
14.1
13.6
13.8
14.1
13.8
14.2
13.8
13.6
Output
14.2
13.9
14.1
14.0
13.9
14.0
14.0
14.0
13.9
13.8
14.1
14.0
13.9
13.7
14.0
(a) Explain why this is a random effects experiment. Are the looms equal in output? Use α=0.05.
(b) Estimate the variability between looms.
(c) Estimate the experimental error variance.
(d) Find a 95 percent confidence interval for sdfdsfseererrw(need equation typed here)
(e) Analyze the residuals from this experiment.
assumptions are satisfied?
Do you think that the analysis of variance
(f) Use the REML method to analyze this data. Compare the 95 percent confidence interval on the
error variance from REML with the exact chi-square confidence interval.
3.55 A manufacturer suspects that the batches of raw material furnished by his supplier differ significantly
in calcium content. There are a large number of batches currently in the warehouse. Five of these are
3-64
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
randomly selected for study. A chemist makes five determinations on each batch as obtains the following
data:
Batch 1
23.46
23.48
23.56
23.39
23.40
Batch 2
23.59
23.46
23.42
23.49
23.50
Batch 3
23.51
23.64
23.46
23.52
23.49
Batch 4
23.28
23.40
23.37
23.46
23.39
Batch 5
23.29
23.46
23.37
23.32
23.38
This is the same as question 3.30 except for (e).
(a) Is there significant variation in the calcium content from batch to batch? Use α=0.05.
(b) Estimate the components of variance.
(c) Find a 95 percent confidence interval for sdfdsfseererrw(need equation typed here)
(d) Analyze the residuals from this experiment.
satisfied?
Are the analysis of variance assumptions are
(e) Use the REML method to analyze this data. Compare the 95 percent confidence interval on the
error variance from REML with the exact chi-square confidence interval.
3-65

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