Chapter 11: Analysis of Variance and Design of Experiments
Chapter 11
Analysis of Variance and
Design of Experiments
LEARNING OBJECTIVES
The focus of this chapter is learning about the design of experiments and the analysis of
variance thereby enabling you to:
1.
2.
3.
4.
5.
6.
Understand the differences between various experiment designs and when to use
them.
Compute and interpret the results of a one-way ANOVA.
Compute and interpret the results of a random block design.
Compute and interpret the results of a two-way ANOVA.
Understand and interpret interaction.
Know when and how to use multiple comparison techniques.
CHAPTER TEACHING STRATEGY
This important chapter opens the door for students to a broader view of statistics
than they have seen to this time. Through the topic of experimental designs, the student
begins to understand how they can scientifically set up controlled experiments in which
to test certain hypotheses. They learn about independent and dependent variables. With
the completely randomized design, the student can see how the t test for two independent
samples can be expanded to include three or more samples by using analysis of variance.
This is something that some of the more curious students were probably wondering about
in chapter 10. Through the randomized block design and the factorial designs, the
student can understand how we can analyze not only multiple categories of one variable,
but we can simultaneously analyze multiple variables with several categories each. Thus,
this chapter affords the instructor an opportunity to help the student develop a structure
for statistical analysis.
In this chapter, we emphasize that the total sum of squares in a given problem do
not change. In the completely randomized design, the total sums of squares are parceled
into between treatments sum of squares and error sum of squares. By using a blocking
design when there is significant blocking, the blocking effects are removed from the error
effects which reduces the size of the mean square error and can potentially create a more
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Chapter 11: Analysis of Variance and Design of Experiments
powerful test of the treatment. A similar thing happens in the two-way factorial design
when one significant treatment variable siphons off sum of squares from the error term
that reduces the mean square error and creates potential for a more powerful test of the
other treatment variable.
In presenting the random block design in this chapter, the emphasis is on
determining if the F value for the treatment variable is significant or not. There is a deemphasis on examining the F value of the blocking effects. However, if the blocking
effects are not significant, the random block design may be a less powerful analysis of the
treatment effects. If the blocking effects are not significant, even though the error sum of
squares is reduced, the mean square error might increase because the blocking effects
may reduce the degrees of freedom error in a proportional greater amount. This might
result in a smaller treatment F value than would occur in a completely randomized
design. The repeated-measures design is shown in the chapter as a special case of the
random block design.
In factorial designs, if there are multiple values in the cells, it is possible to
analyze interaction effects. Random block designs do not have multiple values in cells
and therefore interaction effects cannot be calculated. It is emphasized in this chapter
that if significant interaction occurs, then the main effects analysis are confounded and
should not be analyzed in the usual manner. There are various philosophies about how to
handle significant interaction but are beyond the scope of this chapter. The main factorial
example problem in the chapter was created to have no significant interaction so that the
student can learn how to analyze main effects. The demonstration problem has
significant interaction and these interactions are displayed graphically for the student to
see. You might consider taking this same problem and graphing the interactions using
row effects along the x axis and graphing the column means for the student to see.
There are a number of multiple comparison tests available. In this text, one of the
more well-known tests, Tukey's HSD, is featured in the case of equal sample sizes.
When sample sizes are unequal, a variation on Tukey’s HSD, the Tukey-Kramer test, is
used. MINITAB uses the Tukey test as one of its options under multiple comparisons
and uses the Tukey-Kramer test for unequal sample sizes. Tukey's HSD is one of the
more powerful multiple comparison tests but protects less against Type I errors than
some of the other tests.
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Chapter 11: Analysis of Variance and Design of Experiments
CHAPTER OUTLINE
11.1 Introduction to Design of Experiments
11.2 The Completely Randomized Design (One-Way ANOVA)
One-Way ANOVA
Reading the F Distribution Table
Using the Computer for One-Way ANOVA
Comparison of F and t Values
11.3 Multiple Comparison Tests
Tukey's Honestly Significant Difference (HSD) Test: The Case of Equal Sample
Sizes
Using the Computer to Do Multiple Comparisons
Tukey-Kramer Procedure: The Case of Unequal Sample Sizes
11.4 The Randomized Block Design
Using the Computer to Analyze Randomized Block Designs
11.5 A Factorial Design (Two-Way ANOVA)
Advantages of the Factorial Design
Factorial Designs with Two Treatments
Applications
Statistically Testing the Factorial Design
Interaction
Using a Computer to Do a Two-Way ANOVA
KEY TERMS
a posteriori
a priori
Analysis of Variance (ANOVA)
Blocking Variable
Classification Variables
Classifications
Completely Randomized Design
Concomitant Variables
Confounding Variables
Dependent Variable
Experimental Design
F Distribution
F Value
Factorial Design
Factors
Independent Variable
Interaction
Levels
Multiple Comparisons
One-way Analysis of Variance
Post-hoc
Randomized Block Design
Repeated Measures Design
Treatment Variable
Tukey-Kramer Procedure
Tukey’s HSD Test
Two-way ANOVA
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Chapter 11: Analysis of Variance and Design of Experiments
SOLUTIONS TO PROBLEMS IN CHAPTER 11
11.1
a) Time Period, Market Condition, Day of the Week, Season of the Year
b) Time Period - 4 P.M. to 5 P.M. and 5 P.M. to 6 P.M.
Market Condition - Bull Market and Bear Market
Day of the Week - Monday, Tuesday, Wednesday, Thursday, Friday
Season of the Year – Summer, Winter, Fall, Spring
c) Volume, Value of the Dow Jones Average, Earnings of Investment Houses
11.2
a) Type of 737, Age of the plane, Number of Landings per Week of the plane,
City that the plane is based
b) Type of 737 - Type I, Type II, Type III
Age of plane - 0-2 y, 3-5 y, 6-10 y, over 10 y
Number of Flights per Week - 0-5, 6-10, over 10
City - Toronto, Montreal, Calgary, Vancouver
c) Average annual maintenance costs, Number of annual hours spent on
maintenance
11.3
a) Type of Card, Age of User, Economic Class of Cardholder, Geographic Region
b) Type of Card - MasterCard, Visa, Discover, American Express
Age of User - 21-25 y, 26-32 y, 33-40 y, 41-50 y, over 50
Economic Class - Lower, Middle, Upper
Geographic Region – Atlantic Provinces, Central Canada, the Prairies, the West
Coast, the North
c) Average number of card usages per person per month,
Average balance due on the card, Average per expenditure per person,
Number of cards possessed per person
11.4
Average dollar expenditure per day/night, Age of adult registering the family,
Number of days stay (consecutive)
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Chapter 11: Analysis of Variance and Design of Experiments
11.5
Source
Treatment
Error
Total
df
2
14
16
 = .05
SS
MS
F__
22.20 11.10 11.07
14.03 1.00______
36.24
Critical F.05,2,14 = 3.74
Since the observed F = 11.07 > F.05,2,14 = 3.74, the decision is to reject the null
hypothesis.
11.6
Source
Treatment
Error
Total
df
4
18
22
SS
MS
F__
93.77 23.44 15.82
26.67 1.48______
120.43
 = .01
Critical F.01,4,18 = 4.58
Since the observed F = 15.82 > F.01,4,18 = 4.58, the decision is to reject the null
hypothesis.
11.7
Source
Treatment
Error
Total
df
3
12
15
SS
MS
F_
544.2 181.4 13.00
167.5 14.0______
711.8
 = .01
Critical F.01,3,12 = 5.95
Since the observed F = 13.00 > F.01,3,12 = 5.95, the decision is to reject the null
hypothesis.
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Chapter 11: Analysis of Variance and Design of Experiments
11.8
Source
Treatment
Error
Total
df
1
12
13
SS
64.29
43.43
107.71
MS
F__
64.29 17.76
3.62______
 = .05
Critical F.05,1,12 = 4.75
Since the observed F = 17.76 > F.05,1,12 = 4.75, the decision is to reject the null
hypothesis.
Observed t value using t test:
1
n1 = 7
x 1 = 29
s1 2 = 3
t =
(29  24.71429)  (0)
3(6)  (4.238095)(6) 1 1

772
7 7
Also, t =
11.9
2
n2 = 7
x 2 = 24.71429
s22 = 4.238095
= 4.21
F  17.76 = 4.21
Source
SS
Treatment
583.39
Error
972.18
Total
1,555.57
11.10 Source
Treatment
Error
Total
SS
29.64
68.42
98.06
df
MS
F__
4 145.8475 7.50
50
19.4436______
54
df
2
14
16
MS
F _
14.820 3.03
4.887___ __
F.05,2,14 = 3.74
Since the observed F = 3.03 < F.05,2,14 = 3.74, the decision is to fail to reject
the null hypothesis
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Chapter 11: Analysis of Variance and Design of Experiments
11.11 Source
Treatment
Error
Total
 = .01
df
3
15
18
SS
.007076
.003503
.010579
MS
F__
.002359
10.10
.000234________
Critical F.01,3,15 = 5.42
Since the observed F = 10.10 > F.01,3,15 = 5.42, the decision is to reject the null
hypothesis.
11.12 Source
Treatment
Error
Total
 = .01
df
2
12
14
SS
180700000
11700000
192400000
MS
F__
90350000
92.67
975000_________
Critical F.01,2,12 = 6.93
Since the observed F = 92.67 > F.01,2,12 = 6.93, the decision is to reject the null
hypothesis.
11.13 Source
Treatment
Error
Total
 = .05
df
2
15
17
SS
29.61
18.89
48.50
MS
F___
14.80
11.76
1.26________
Critical F.05,2,15 = 3.68
Since the observed F = 11.76 > F.05,2,15 = 3.68, the decision is to reject the null
hypothesis.
11.14 Source
Treatment
Error
Total
 = .05
df
3
16
19
SS
456630
220770
677400
MS
F__
152210 11.03
13798_______
Critical F.05,3,16 = 3.24
Since the observed F = 11.03 > F.05,3,16 = 3.24, the decision is to reject the null
hypothesis.
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Chapter 11: Analysis of Variance and Design of Experiments
11.15 There are 4 treatment levels. The sample sizes are 18, 15, 21, and 11. The F
value is 2.95 with a p-value of .04. Fcrit = 2.7555 is approximately equal to
F.05, 3, 60 . So,   0.05 . Since the observed F = 2.95 > Fcrit = 2.7555, the decision is
to reject the null hypothesis. There is an overall significant difference at alpha of
.05. The means are 226.73, 238.79, 232.58, and 239.82.
11.16 The independent variable for this study was plant with five classification levels
(the five plants). There were a total of 43 workers who participated in the study.
The dependent variable was number of hours worked per week. An observed F
value of 3.10 was obtained with an associated p-value of .026595. With an alpha
of .05, there was a significant overall difference in the average number of hours
worked per week by plant. A cursory glance at the plant averages revealed that
workers at plant 3 averaged 61.47 hours per week (highest number) while workers
at plant 4 averaged 49.20 (lowest number).
11.17 C = 6
MSE = .3352
q.05,6,40 = 4.23
n3 = 8
 = .05
n6 = 7
N = 46
x 3 = 15.85
x 6 = 17.21
.3352  1 1 
   = 0.896
2 8 7
HSD = 4.23
x 3  x 6  15.85  17.21 = 1.36
Since 1.36 > 0.896, there is a significant difference between the means of
groups 3 and 6.
11.18 C = 4
n=6
MSE = 2.389
HSD = q
N = 24
dfE = N – C = 24 – 4 = 20
 = .05
q.05,4,20 = 3.96
MSE
2.389
= (3.96)
= 2.50
n
6
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Chapter 11: Analysis of Variance and Design of Experiments
11.19 C = 3
 = .05
MSE = 1.002381
q.05,3,14 = 3.70
HSD = 3.70
n1 = 6
N – C = 14
N = 17
n2 = 5
x1 =2
x 2 = 4.6
1.002381  1 1 
   = 1.586
2
 6 5
x1  x 2  2  4.6 = 2.6
Since 2.6 > 1.586, there is a significant difference between the means of
groups 1 and 2.
11.20 From problem 11.6,
n2 = 5
MSE = 1.481481
 = .01
n5 = 6
C=5
N = 23 N – C = 18
q.01,5,18 = 5.38
HSD = 5.38
1.481481  1 1 
   = 2.80
2
5 6
x 2 = 10
x 5 = 13
x 2  x 5  10  13 = 3
Since 3 > 2.80, there is a significant difference in the means of
groups 2 and 5.
11.21 N = 16
HSD = q
n=4
C=4
N – C = 12
MSE = 13.95833
q.01,4,12 = 5.50
MSE
13.95833
= 5.50
= 10.27
4
n
x 1 = 115.25
x 2 = 125.25
x 3 = 131.5
x 4 = 122.5
x 1 and x 3 are the only pair that are significantly different using the
HSD test.
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Chapter 11: Analysis of Variance and Design of Experiments
11.22 n = 7
C=2
MSE = 3.619048
N – C = 14 – 2 = 12
N = 14
 = .05
q.05,2,12 = 3.08
HSD = q
MSE
3.619048
= 3.08
= 2.215
7
n
x 1 = 29 and x 2 = 24.71429
Since x 1 – x 2 = 4.28571 > HSD = 2.215, there is a significant difference in
means.
11.23 C = 4
MSE = .000234
q.01,4,15 = 5.25
n1 = 4
 = .01
n2 = 6
n3 = 5
N = 19
N – C = 15
n4 = 4
x 1 = 4.03, x 2 = 4.001667, x 3 = 3.974, x 4 = 4.005
HSD1,2 = 5.25
.000234  1 1 
   = .0367
2 4 6
HSD1,3 = 5.25
.000234  1 1 
   = .0381
2  4 5
HSD1,4 = 5.25
.000234
= .0402
4
HSD2,3 = 5.25
.000234  1 1 
   = .0344
2 6 5
HSD2,4 = 5.25
.000234  1 1 
   = .0367
2 6 4
HSD3,4 = 5.25
.000234  1 1 
   = .0381
2 5 4
x 1  x 3 = .056
This is the only pair of means that are significantly different.
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Chapter 11: Analysis of Variance and Design of Experiments
11.24  = .01
C=3
MSE = 975,000
MSE
n
HSD = q
x 1 = 40,900
N – C = 12
n=5
N = 15
q.01,3,12 = 5.04
975,000
5
= 5.04
x 2 = 49,400
= 2,225.6
x 3 = 45,300
x1  x 2 = 8,500
x 1  x 3 = 4,400
x 2  x 3 = 4,100
Using Tukey's HSD, all three pairwise comparisons are significantly
different.
11.25  = .05
C=3
q.05,3,15 = 3.67
N – C = 15
N = 18
n1 = 5
n2 = 7
MSE = 1.259365
n3 = 6
x 1 = 7.6
x 2 = 8.8571
x 3 = 5.8333
HSD1,2 = 3.67
1.259365  1 1 
   = 1.705
2
5 7
HSD1,3 = 3.67
1.259365  1 1 
   = 1.763
2
5 6
HSD2,3 = 3.67
1.259365  1 1 
   = 1.620
2
7 6
x 1  x 3 = 1.767 (is significant)
x 2  x 3 = 3.024 (is significant)
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Chapter 11: Analysis of Variance and Design of Experiments
11.26  = .05
n=5
q.05,4,16 = 4.05
HSD = q
C=4
x 1 = 591
N = 20
N – C = 16
x 2 = 350
MSE = 13,798.13
x 3 = 776
x 4 = 563
MSE
13,798.13
= 4.05
= 212.76
5
n
x1  x 2 = 241
x 1  x 3 = 185
x1  x 4 = 28
x 2  x 3 = 426
x 2  x 4 = 213
x 3  x 4 = 213
Using Tukey's HSD = 212.76, means 1 and 2, means 2 and 3, means 2 and 4,
and means 3 and 4 are significantly different.
11.27  = .05. There were five plants and ten pairwise comparisons. The MINITAB
output reveals that the only significant pairwise difference is between plant 2 and
plant 3 where the reported confidence interval (0.180 to 22.460) contains the same
sign throughout indicating that 0 is not in the interval. Since the confidence
interval does not contain zero, then we are 95% confident that there is a
significant difference in the pair of means. The lower and upper values for all
other confidence intervals have different signs which indicates that zero is
included in the interval, and there is no difference in the means for these pairs.
11.28 H0: µ1 = µ2 = µ3 = µ4
Ha: At least one treatment mean is different from the others
Source
Treatment
Blocks
Error
Total
df
3
4
12
19
SS
62.95
257.50
45.30
365.75
MS
F__
20.9833 5.56
64.3750 17.05
3.7750______
 = .05
Critical F.05,3,12 = 3.49 for treatments
For treatments, the observed F = 5.56 > F.05,3,12 = 3.49, the decision is to
reject the null hypothesis.
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Chapter 11: Analysis of Variance and Design of Experiments
11.29 H0: µ1 = µ2 = µ3
Ha: At least one treatment mean is different from the others
Source
df
Treatment 2
Blocks
3
Error
6
Total
11
 = .01
SS
MS
F_
.001717 .000858
1.48
.076867 .025622 44.13
.003483 .000581_______
.082067
Critical F.01,2,6 = 10.92 for treatments
For treatments, the observed F = 1.48 < F.01,2,6 = 10.92 and the decision is to
fail to reject the null hypothesis.
11.30 Source
Treatment
Blocks
Error
Total
 = .05
df
5
9
45
59
SS
2477.53
3180.48
11661.38
17319.39
MS
F__
495.506 1.91
353.387 1.36
259.142______
Critical F.05,5,45 = 2.45 for treatments
For treatments, the observed F = 1.91 < F.05,5,45 = 2.45 and decision is to fail to
reject the null hypothesis.
11.31 Source
Treatment
Blocks
Error
Total
 = .01
df
3
6
18
27
SS
199.48
265.24
306.59
771.31
MS
F__
66.493 3.90
44.207 2.60
17.033______
Critical F.01,3,18 = 5.09 for treatments
For treatments, the observed F = 3.90 < F.01,3,18 = 5.09 and the decision is to
fail to reject the null hypothesis.
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Chapter 11: Analysis of Variance and Design of Experiments
11.32 Source
Treatment
Blocks
Error
Total
 = .05
df
3
9
27
39
SS
2302.5
5402.5
1322.5
9027.5
MS
F__
767.5000 15.67
600.2778 12.26
48.9815____ __
Critical F.05,3,27 = 2.96 for treatments
For treatments, the observed F = 15.67 > F.05,3,27 = 2.96 and the decision is to
reject the null hypothesis.
11.33 Source
Treatment
Blocks
Error
Total
 = .01
df
2
4
8
14
SS
64.5333
137.6000
16.8000
218.9333
MS
F
32.2667
15.37
34.4000
16.38
2.1000_ _____
Critical F.01,2,8 = 8.65 for treatments
For treatments, the observed F = 15.37 > F.01,2,8 = 8.65 and the decision is to
reject the null hypothesis.
11.34 This is a randomized block design with 3 treatments (machines) and 5 block
levels (operators). The F for treatments is 6.72 with a p-value of .019. There is a
significant difference in machines at  = .05. The F for blocking effects is 0.22
with a p-value of .807. There are no significant blocking effects. The blocking
effects reduced the power of the treatment effects since the blocking effects were
not significant.
11.35 The p value for Phone Type, .00018, indicates that there is an overall significant
difference in treatment means at alpha .001. The lengths of calls differ according
to type of telephone used. The p-value for managers, .00028, indicates that there
is an overall difference in block means at alpha .001. The lengths of calls differ
according to Manager. The significant blocking effects have improved the power
of the F test for treatments.
11.36 This is a two-way factorial design with two independent variables and one
dependent variable. It is 2x4 in that there are two row treatment levels and four
column treatment levels. Since there are three measurements per cell, interaction
can be analyzed.
dfrow treatment = 1
dfcolumn treatment = 3
dfinteraction = 3
© 2010 John Wiley & Sons Canada, Ltd.
dferror = 16
dftotal = 23
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Chapter 11: Analysis of Variance and Design of Experiments
11.37 This is a two-way factorial design with two independent variables and one
dependent variable. It is 4x3 in that there are four row treatment levels and three
column treatment levels. Since there are two measurements per cell, interaction
can be analyzed.
dfrow treatment = 3
11.38 Source
Row
Column
Interaction
Error
Total
df
3
4
12
60
79
dfcolumn treatment = 2
SS
126.98
37.49
380.82
733.65
1278.94
dfinteraction = 6
dferror = 12
dftotal = 23
MS
F__
42.327 3.46
9.373 0.77
31.735 2.60
12.228______
 = .05
Critical F.05,3,60 = 2.76 for rows. For rows, the observed F = 3.46 > F.05,3,60 = 2.76
and the decision is to reject the null hypothesis.
Critical F.05,4,60 = 2.53 for columns. For columns, the observed F = 0.77 <
F.05,4,60 = 2.53 and the decision is to fail to reject the null hypothesis.
Critical F.05,12,60 = 1.92 for interaction. For interaction, the observed F = 2.60 >
F.05,12,60 = 1.92 and the decision is to reject the null hypothesis.
Since there is significant interaction, the researcher should exercise extreme
caution in analyzing the "significant" row effects.
11.39 Source
Row
Column
Interaction
Error
Total
df
1
3
3
16
23
SS
1.047
3.844
0.773
6.968
12.632
MS
F__
1.047 2.40
1.281 2.94
0.258 0.59
0.436______
 = .05
Critical F.05,1,16 = 4.49 for rows. For rows, the observed F = 2.40 < F.05,1,16 = 4.49
and decision is to fail to reject the null hypothesis.
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Chapter 11: Analysis of Variance and Design of Experiments
Critical F.05,3,16 = 3.24 for columns. For columns, the observed F = 2.94 <
F.05,3,16 = 3.24 and the decision is to fail to reject the null hypothesis.
Critical F.05,3,16 = 3.24 for interaction. For interaction, the observed F = 0.59 <
F.05,3,16 = 3.24 and the decision is to fail to reject the null hypothesis.
11.40 Source
Row
Column
Interaction
Error
Total
df
1
2
2
6
11
SS
60.750
14.000
2.000
9.500
86.250
MS
F___
60.750
38.37
7.000
4.42
1.000
0.63
1.583________
 = .01
Critical F.01,1,6 = 13.75 for rows. For rows, the observed F = 38.37 >
F.01,1,6 = 13.75 and the decision is to reject the null hypothesis.
Critical F.01,2,6 = 10.92 for columns. For columns, the observed F = 4.42 <
F.01,2,6 = 10.92 and the decision is to fail to reject the null hypothesis.
Critical F.01,2,6 = 10.92 for interaction. For interaction, the observed F = 0.63 <
F.01,2,6 = 10.92 and the decision is to fail to reject the null hypothesis.
11.41 Source
Treatment 1
Treatment 2
Interaction
Error
Total
df
1
3
3
24
31
SS
1.24031
5.09844
0.12094
0.46750
6.92719
MS
F__
1.24031 63.67
1.69948 87.25
0.04031
2.07
0.01948______
 = .05
Critical F.05,1,24 = 4.26 for treatment 1. For treatment 1, the observed F = 63.67 >
F.05,1,24 = 4.26 and the decision is to reject the null hypothesis.
Critical F.05,3,24 = 3.01 for treatment 2. For treatment 2, the observed F = 87.25 >
F.05,3,24 = 3.01 and the decision is to reject the null hypothesis.
Critical F.05,3,24 = 3.01 for interaction. For interaction, the observed F = 2.07 <
F.05,3,24 = 3.01 and the decision is to fail to reject the null hypothesis.
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Chapter 11: Analysis of Variance and Design of Experiments
11.42 Source
df
Age
3
No. Children 2
Interaction
6
Error
12
Total
23
SS
42.4583
49.0833
4.9167
11.5000
107.9583
MS
F__
14.1528 14.77
24.5417 25.61
0.8194
0.86
0.9583______
 = .05
Critical F.05,3,12 = 3.49 for Age. For Age, the observed F = 14.77 >
F.05,3,12 = 3.49 and the decision is to reject the null hypothesis.
Critical F.05,2,12 = 3.89 for No. Children. For No. Children, the observed
F = 25.61 > F.05,2,12 = 3.89 and the decision is to reject the null hypothesis.
Critical F.05,6,12 = 3.00 for interaction. For interaction, the observed F = 0.86 <
F.05,6,12 = 3.00 and fail to reject the null hypothesis.
11.43 Source
Location
Competitors
Interaction
Error
Total
df
2
3
6
24
35
SS
1736.22
1078.33
503.33
607.33
3925.22
MS
F__
868.11 34.31
359.44 14.20
83.89
3.32
25.31_______
 = .05
Critical F.05,2,24 = 3.40 for rows. For rows, the observed F = 34.31 >
F.05,2,24 = 3.40 and the decision is to reject the null hypothesis.
Critical F.05,3,24 = 3.01 for columns. For columns, the observed F = 14.20 >
F.05,3,24 = 3.01 and decision is to reject the null hypothesis.
Critical F.05,6,24 = 2.51 for interaction. For interaction, the observed F = 3.32 >
F.05,6,24 = 2.51 and the decision is to reject the null hypothesis.
Note: There is a significant interaction in this study. This may confound the
interpretation of the main effects, Location and Number of Competitors.
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Chapter 11: Analysis of Variance and Design of Experiments
11.44 This two-way design has 3 row treatments and 5 column treatments. There are 45
total observations with 3 in each cell.
FR =
MSR 46.16

= 13.23
MSE
3.49
p-value = .000 and the decision is to reject the null hypothesis for rows.
FC =
MSC 249.70

= 71.55
MSE
3.49
p-value = .000 and the decision is to reject the null hypothesis for columns.
MSI 55.27

FI =
= 15.84
MSE
3.49
p-value = .000 and the decision is to reject the null hypothesis for interaction.
Because there is a significant interaction, the analysis of main effects is
confounded. The graph of means displays the crossing patterns of the line
segments indicating the presence of interaction.
11.45 The null hypotheses are that there are no interaction effects, that there are no
significant differences in the means of the valve openings by machine, and that
there are no significant differences in the means of the valve openings by shift.
Since the p-value for interaction effects is .876, there are no significant interaction
effects and that is good since significant interaction effects would confound that
study. The p-value for columns (shifts) is .008 indicating that column effects are
significant at alpha of .01. There is a significant difference in the mean valve
opening according to shift. No multiple comparisons are given in the output.
However, an examination of the shift means indicates that the mean valve
opening on shift one was the largest at 6.47 followed by shift three with 6.3 and
shift two with 6.25. The p-value for rows (machines) is .937 and that is not
significant.
11.46 This two-way factorial design has 3 rows and 5 columns with three observations
per cell. The observed F value for rows is 0.19, for columns is 1.19, and for
interaction is 1.40. Using an alpha of .05, the critical F value for rows and
columns (same df) is F2,18,.05 = 3.55. Neither the observed F value for rows nor
the observed F value for columns is significant. The critical F value for
interaction is F4,18,.05 = 2.93. There is no significant interaction.
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Chapter 11: Analysis of Variance and Design of Experiments
11.47 Source
Treatment
Error
Total
df
3
12
15
 = .05
SS
66.69
30.25
96.94
MS
F__
22.23 8.82
2.52______
Critical F.05,3,12 = 3.49
Since the treatment F = 8.82 > F.05,3,12 = 3.49, the decision is to reject the null
hypothesis.
For Tukey's HSD:
MSE = 2.52
n=4
N = 16
N – C = 12
C=4
q.05,4,12 = 4.20
HSD = q
x 1 = 12
MSE
2.52
= (4.20)
= 3.33
n
4
x 2 = 7.75
x 3 = 13.25
x 4 = 11.25
Using HSD of 3.33, there are significant pairwise differences between
means 1 and 2, means 2 and 3, and means 2 and 4.
11.48 Source
Treatment
Error
Total
df
6
19
25
SS
68.19
249.61
317.80
11.49 Source
Treatment
Error
Total
df
5
36
41
SS
210
655
865
11.50 Source
Treatment
Error
Total
 = .01
df
2
22
24
SS
150.91
102.53
253.44
MS
F__
11.365 0.87
13.137______
MS
F__
42.000 2.31
18.194______
MS
F__
75.46 16.19
4.66________
Critical F.01,2,22 = 5.72
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Chapter 11: Analysis of Variance and Design of Experiments
Since the observed F = 16.19 > F.01,2,22 = 5.72, the decision is to reject the null
hypothesis.
x 1 = 9.200
x 2 = 14.250
x 3 = 8.714286
n1 = 10
n2 = 8
n3 = 7
MSE = 4.66
 = .01
C=3
N = 25
q.01,3,22 = 4.64
N – C = 22
HSD1,2 = 4.64
4.66  1 1 
   = 3.36
2  10 8 
HSD1,3 = 4.64
4.66  1 1 
   = 3.49
2  10 7 
HSD2,3 = 4.64
4.66  1 1 
   = 3.67
2 8 7
x1  x 2 = 5.05 and
x 2  x 3 = 5.5357 are significantly different at  = .01
11.51 This design is a repeated-measures type random block design. There is one
treatment variable with three levels. There is one blocking variable with six
people in it (six levels). The degrees of freedom treatment are two. The degrees
of freedom block are five. The error degrees of freedom are ten. The total
degrees of freedom are seventeen. There is one dependent variable.
11.52 Source
Treatment
Blocks
Error
Total
 = .05
df
3
9
27
39
SS
20,994
16,453
33,891
71,338
MS
F__
6998.00 5.58
1828.11 1.46
1255.22_____
Critical F.05,3,27 = 2.96 for treatments
Since the observed F = 5.58 > F.05,3,27 = 2.96 for treatments, the decision is to
reject the null hypothesis.
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Chapter 11: Analysis of Variance and Design of Experiments
11.53 Source
Treatment
Blocks
Error
Total
df
3
5
15
23
 = .05
SS
240.125
548.708
38.125
MS
F__
80.042 31.49
109.742 43.20
2.542_ _____
Critical F.05,3,15 = 3.29 for treatments
Since for treatments the observed F = 31.49 > F.05,3,15 = 3.29, the decision is to
reject the null hypothesis.
For Tukey's HSD:
Ignoring the blocking effects, the sum of squares blocking and sum of squares
error are combined together for a new SSerror = 548.708 + 38.125 = 586.833.
Combining the degrees of freedom error and blocking yields a new dferror = 20.
Using these new figures, we compute a new mean square error, MSE =
(586.833/20) = 29.34165.
n=6
C=4
HSD = q
N = 24
MSE
= (3.96)
n
x 1 = 16.667
x 2 = 12.333
N – C = 20
q.05,4,20 = 3.96
29.34165
= 8.757
6
x 3 = 12.333
x 4 = 19.833
None of the pairs of means are significantly different using Tukey's HSD = 8.757.
This may be due in part to the fact that we compared means by folding the
blocking effects back into error and the blocking effects were highly
significant.
11.54 Source
Treatment 1
Treatment 2
Interaction
Error
Total
df
4
1
4
30
39
SS
MS
F__
29.13
7.2825 1.98
12.67 12.6700 3.44
73.49 18.3725 4.99
110.38
3.6793______
225.67
 = .05
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Chapter 11: Analysis of Variance and Design of Experiments
Critical F.05,4,30 = 2.69 for treatment 1. For treatment 1, the observed F = 1.98 <
F.05,4,30 = 2.69 and the decision is to fail to reject the null hypothesis.
Critical F.05,1,30 = 4.17 for treatment 2. For treatment 2 observed F = 3.44 <
F.05,1,30 = 4.17 and the decision is to fail to reject the null hypothesis.
Critical F.05,4,30 = 2.69 for interaction. For interaction, the observed F = 4.99 >
F.05,4,30 = 2.69 and the decision is to reject the null hypothesis.
Since there are significant interaction effects, examination of the main effects
should not be done in the usual manner. However, in this case, there are no
significant treatment effects anyway.
11.55 Source
Treatment 2
Treatment 1
Interaction
Error
Total
df
3
2
6
24
35
SS
257.889
1.056
17.611
54.000
330.556
MS
F___
85.963 38.21
0.528
0.23
2.935
1.30
2.250________
 = .01
Critical F.01,3,24 = 4.72 for treatment 2. For the treatment 2 effects, the observed
F = 38.21 > F.01,3,24 = 4.72 and the decision is to reject the null hypothesis.
Critical F.01,2,24 = 5.61 for Treatment 1. For the treatment 1 effects, the observed
F = 0.23 < F.01,2,24 = 5.61 and the decision is to fail to reject the null hypothesis.
Critical F.01,6,24 = 3.67 for interaction. For the interaction effects, the observed
F = 1.30 < F.01,6,24 = 3.67 and the decision is to fail to reject the null hypothesis.
11.56 Source
Age
Column
Interaction
Error
Total
df
2
3
6
24
35
SS
49.3889
1.2222
1.2778
15.3333
67.2222
MS
F___
24.6944 38.65
0.4074 0.64
0.2130 0.33
0.6389_______
 = .05
Critical F.05,2,24 = 3.40 for Age. For the age effects, the observed F = 38.65 >
F.05,2,24 = 3.40 and the decision is to reject the null hypothesis.
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Chapter 11: Analysis of Variance and Design of Experiments
Critical F.05,3,24 = 3.01 for Region. For the region effects, the observed F = 0.64
< F.05,3,24 = 3.01 and the decision is to fail to reject the null hypothesis.
Critical F.05,6,24 = 2.51 for interaction. For interaction effects, the observed
F = 0.33 < F.05,6,24 = 2.51 and the decision is to fail to reject the null hypothesis.
There are no significant interaction effects. Only the Age effects are significant.
Computing Tukey's HSD for Age:
x 1 = 2.667
x 2 = 4.917
n = 12 C = 3
N = 36
x 3 = 2.250
N – C = 33
MSE is recomputed by folding together the interaction and column sum of
squares and degrees of freedom with previous error terms:
MSE = (1.2222 + 1.2778 + 15.3333)/(3 + 6 + 24) = 0.5404
q.05,3,33 = 3.49
HSD = q
MSE
= (3.49)
n
.5404
= 0.7406
12
Using HSD, there are significant pairwise differences between means 1 and 2 and
between means 2 and 3.
Shown below is a graph of the interaction using the cell means by Age.
© 2010 John Wiley & Sons Canada, Ltd.
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Chapter 11: Analysis of Variance and Design of Experiments
11.57 Source
Treatment
Error
Total
 = .05
df
3
20
23
SS
90477679
81761905
172239584
MS
F__
30159226 7.38
4088095_______
Critical F.05,3,20 = 3.10
The treatment F = 7.38 > F.05,3,20 = 3.10 and the decision is to reject the null
hypothesis.
11.58
Source
Treatment
Blocks
Error
Total
df
2
5
10
17
SS
460,353
33,524
22,197
516,074
MS
F__
230,176 103.70
6,705
3.02
2,220_______
 = .01
Critical F.01,2,10 = 7.56 for treatments
Since the treatment observed F = 103.70 > F.01,2,10 = 7.56, the decision is to
reject the null hypothesis.
11.59 Source
Treatment
Error
Total
 = .05
df
2
18
20
SS
9.555
185.1337
194.6885
MS
F__
4.777
0.46
10.285_______
Critical F.05,2,18 = 3.55
Since the treatment F = 0.46 < F.05,2,18 = 3.55, the decision is to fail to reject the
null hypothesis.
Since there are no significant treatment effects, it would make no sense to
compute Tukey-Kramer values and do pairwise comparisons.
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Chapter 11: Analysis of Variance and Design of Experiments
11.60 Source
Years
Size
Interaction
Error
Total
df
2
3
6
36
47
SS
4.875
17.083
2.292
17.000
41.250
MS
F___
2.437
5.16
5.694 12.06
0.382
0.81
0.472_______
 = .05
Critical F.05,2,36 = 3.32 for Years. For Years, the observed F = 5.16 >
F.05,2,36 = 3.32 and the decision is to reject the null hypothesis.
Critical F.05,3,36 = 2.92 for Size. For Size, the observed F = 12.06 > F.05,3,36 = 2.92
and the decision is to reject the null hypothesis.
Critical F.05,6,36 = 2.42 for interaction. For interaction, the observed F = 0.81 <
F.05,6,36 = 2.42 and the decision is to fail to reject the null hypothesis.
There are no significant interaction effects. There are significant row and column
effects at  = .05.
11.61 Source
Treatment
Blocks
Error
Total
 = .05
df
4
7
28
39
SS
53.400
17.100
27.400
97.900
MS
F___
13.350 13.64
2.443
2.50
0.979________
Critical F.05,4,28 = 2.71 for treatments
For treatments, the observed F = 13.64 > F.05,4,28 = 2.71 and the decision is to
reject the null hypothesis.
11.62 This is a one-way ANOVA with four treatment levels. There are 36 observations
in the study. The p-value of .045 indicates that there is a significant overall
difference in the means at  = .05. An examination of the mean analysis shows
that the sample sizes are different with sizes of 8, 7, 11, and 10, respectively. No
multiple comparison technique was used here to conduct pairwise comparisons.
However, a study of sample means shows that the two most extreme means are
from levels one and four. These two means would be the most likely candidates
for multiple comparison tests. Note that the confidence intervals for means one
and four (shown in the graphical output) are seemingly non-overlapping
indicating a potentially significant difference.
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Chapter 11: Analysis of Variance and Design of Experiments
11.63 Excel reports that this is a two-factor design without replication indicating that
this is a random block design. Neither the row nor the column p-values are less
than .05 indicating that there are no significant treatment or blocking effects in
this study. Also displayed in the output to underscore this conclusion are the
observed and critical F values for both treatments and blocking. In both
cases, the observed value is less than the critical value.
11.64 This is a two-way ANOVA with 5 rows and 2 columns. There are 2 observations
per cell. For rows, FR = 0.98 with a p-value of .461 which is not significant. For
columns, FC = 2.67 with a p-value of .134 which is not significant. For
interaction, FI = 4.65 with a p-value of .022 which is significant at  = .05. Thus,
there are significant interaction effects and the row and column effects are
confounded. An examination of the interaction plot reveals that most of the lines
cross verifying the finding of significant interaction.
11.65 This is a two-way ANOVA with 4 rows and 3 columns. There are 3 observations
per cell. FR = 4.30 with a p-value of .0146 is significant at  = .05. The null
hypothesis is rejected for rows. FC = 0.53 with a p-value of .594 is not
significant. We fail to reject the null hypothesis for columns. FI = 0.99 with a
p-value of .453 for interaction is not significant. We fail to reject the null
hypothesis for interaction effects.
11.66 This was a random block design with 5 treatment levels and 5 blocking levels.
For both treatment and blocking effects, the critical value is F.05,4,16 = 3.01. The
observed F value for treatment effects is MSC / MSE = 35.98 / 7.36 = 4.89 which
is greater than the critical value. The null hypothesis for treatments is rejected,
and we conclude that there is a significant different in treatment means. No
multiple comparisons have been computed in the output. The observed F value
for blocking effects is MSR / MSE = 10.36 /7.36 = 1.41 which is less than the
critical value. There are no significant blocking effects. Using random block
design on this experiment might have cost a loss of power.
11.67 This one-way ANOVA has 4 treatment levels and 24 observations. The F = 3.51
yields a p-value of .034 indicating significance at  = .05. Since the sample sizes
are equal, Tukey’s HSD is used to make multiple comparisons. The computer
output shows that means 1 and 3 are the only pairs that are significantly different
(same signs in confidence interval). Observe on the graph that the confidence
intervals for means 1 and 3 barely overlap.
© 2010 John Wiley & Sons Canada, Ltd.
387