CHAPTER 2 – ONE-FACTOR DESIGNS AND THE ANALYSIS OF VARIANCE
1 2 2.1
Treatment
1
2
3
6
6
11
Replications
3
5
10
(R = 4)
8
4
8
3
9
11
5
6
10
(C = 3)
Column means Grand Mean
7
SSBc = 4 5 − 7 2 + 6 − 7 2 + 10 − 7 2 = 4 4 + 1 + 9 = 56 SSW =
6 − 5 2 + 3 − 5 2 + 8 − 5 2 + ⋯ + 11 − 10 2 = 38
𝐹𝑐𝑎𝑙𝑐 = 56 2
38 9 = 6.632
Alternatively, using Excel, we have
ANOVA Source of Variation
SS
df
MS
F
Between Groups
56
2
28
Within Groups
38
9
4.222
Total
94
11
P-value 6.632
F crit
0.017
4.257
Yes, there is sufficient evidence of differences is sales due to the level of the treatment; Fcalc = 6.632 is well within the rejection region; thus, we reject H0. 16
3
1
2.2
2 ANOVA Source of Variation
SS
df
MS
Between Groups
112
2
56
Within Groups
76
21
3.619
Total
188
23
F
P-value
15.474
7.411E-05
F crit 3.467
3 4
Yes, there is evidence of differences is sales due to the level of the treatment; Fcalc = 15.474 is well
5
within the rejection region.
6 7
2.3
8 9
For the "same" data with twice as many replications, MSBc is doubled, MSW is reduced by 14%,
10
and Fcalc is more than doubled. Simultaneously, the change in denominator df reduces the critical
11
value from 4.256 to 3.466. The net result is a change in p-value from .017 to .0000741. In other
12
words, the same difference in means becomes more significant with increased replication because
13
the estimates of the means are better (have narrower confidence intervals).
14 15
2.4
16 ANOVA Source of Variation
SS
df
MS
Between Groups
6.108
4
1.527
Within Groups
3.37
15
0.225
Total
9.478
19
F
P-value 6.797
0.003
F crit 3.056
17 18
Yes, there is sufficient evidence of a difference due to dominant technology; p-value = .0025 <
19
.05.
20 21
4
1
2.5
2 3
n = 100 indicates total df = 99; therefore, error df = 96. Fcalc = 15 and MSW = 600 indicates SSBC
4
= 9000. SSQ values follow from these.
5 ANOVA Table Source of Variability
SSQ
df
MSQ
Fcalc
Diet
27000
3
9000
15
Error
57600
96
600
Total
84600
99
6 7
2.6
8 ANOVA Source of Variation
SS
Between Groups
313.333
3
104.444
Within Groups
442.667
20
22.133
756
23
Total
df
MS
F
P-value 4.719
0.012
F crit 3.098
9 10
Yes, there is sufficient evidence to conclude that the level of the room affects perception of the
11
degree of motion; p-value = .012 < .05.
12 13
2.7
14 15
Sum of Squares for each column = 𝑑𝑓 Standard Deviation 2
16 17 18
For Column 1: SS = 22 9.18 2 = 1853.993
19 20
SSW = 10056.19
21
df = 100
22
MSW = 100.56 5
1 Grand mean = 23 38.12 + 35 29.72 + 17 33.40 + 29 36.15
2
104 = 33.97
3 SSBc = 23 38.12 − 33.97 2 + 35 29.72 − 33.97 2 + 17 33.40 − 33.97 2 +
4
29 36.15 − 33.97 2 = 1171.647
5 6
df = 3
7
MSBc = 390.549
8 Species
1
2
3
4
Column Mean
38.12
29.72
33.40
36.15
Standard Deviation
9.18
10.42
11.34
9.36
Sample Size
23
35
17
29
104
df
22
34
16
28
100
1853.993
3691.598
2057.53
2453.069
10056.19
Sums of Squares
Total
9 10
𝐹𝑐𝑎𝑙𝑐 = 390.549 100.56 = 3.884; for α = .05 and df = (3, 100), c = 2.68; there is evidence of a
11
difference in cost per visit for the four dog species. Since, for α = .01 and df = (3, 100), c = 3.95,
12
.05 > p-value > .01.
13 14
2.8
15 ANOVA Source of Variation
SS
Between Groups
35.583
Within Groups Total
df
MS
F
2
17.792
6.188
60.375
21
2.875
95.958
23
P-value 0.008
F crit 3.467
16 17
The texts are not equally effective; p-value = .008.
18 19 20 21
6
1
2.9
2 ANOVA Source of Variation
SS
df
MS
Between Groups
0.2606
2
0.130
Within Groups
0.043
12
0.004
Total
0.304
14
F
P-value
36.024
8.47E-06
F crit 3.8853
3 4
At any traditional value of α, there is substantial evidence of a difference in price among the three
5
cities; p-value = .00000847, essentially zero.
6 7
2.10
8 9
Boynton Beach and Delray Beach ANOVA Source of Variation
SS
df
MS
Between Groups
0.021
1
0.021
Within Groups
0.021
8
0.003
Total
0.042
9
F 8.186
P-value 0.021
F crit 5.318
10 11
Delray Beach and Boca Raton ANOVA Source of Variation
SS
df
MS
Between Groups
0.123
1
0.123
Within Groups
0.030
8
0.004
Total
0.154
9
F 32.424
P-value 0.0005
F crit 5.318
12 13 14 15
7
1
Boynton Beach and Boca Raton ANOVA Source of Variation
SS
df
MS
Between Groups
0.247
1
0.247
Within Groups
0.036
8
0.005
Total
0.282
9
F
P-value
55.205
F crit 5.318
7.4E-05
2 3
There is evidence of significant difference in price for each pair of cities.
4 5
2.11
6 ANOVA Source of Variation
SS
df
MS
Between Groups
20.5
3
6.833
Within Groups
32.5
12
2.708
53
15
Total
F
P-value 2.523
F crit
0.107
3.490
7 8
No. There is insufficient evidence at α = .01 that there is a difference in rental duration among the
9
four sizes of cars; p-value = .1071 > .01.
10 11
2.12
12 ANOVA Source of Variation
SS
Between Groups
1007.475
Within Groups Total
df
MS
F
P-value
3
335.825
106.401
2.494E-39
555.497
176
3.156
1562.972
179
F crit 2.656
13 14
Yes, there is a difference in waiting time among the four offices; p-value is essentially zero.
15 16
8
1
2.13
2 ANOVA Source of Variation
SS
df
MS
Between Groups
75257.55
3
25085.850
Within Groups
10130.44
396
25.582
Total
85387.99
399
F
P-value
980.609
7.450E-183
F crit 2.627
3 4
Yes, there is a difference in average scores among the four courses; p-value is essentially zero.
5 6
2.14
7 ANOVA Source of Variation
SS
df
MS
Between Groups
2859.524
2
1429.762
Within Groups
75907.143
18
4217.063
Total
78766.667
20
F
P-value 0.339
0.717
F crit 3.555
8 9
No, there are not significant differences due to state; p-value = .717 > .05.
10 11
2.15
12 ANOVA Source of Variation
SS
df
MS
Between Groups
4905500
2
2452750
Within Groups
18137500
27
671759.259
Total
23043000
29
F
P-value 3.651
0.039
F crit 3.354
13 14
Yes, there are differences in amount of donations due to solicitation approach, though it is a close
15
call; p-value = .039 < .05.
16
9
1
2.16
2 ANOVA Source of Variation
SS
Between Groups
2144.724
Within Groups Total
df
MS
F
P-value
7
306.389
105.407
9.184E-109
2302.126
792
2.907
4446.850
799
F crit 2.021
3 4
Yes, device does affect battery lifetime; p-value is essentially zero.
5 6
2.17
7 ANOVA Source of Variation
SS
df
MS
Between Groups
38.663
1
38.663
Within Groups
8435.248
53
159.156
Total
8473.911
54
F
P-value 0.243
0.624
F crit 4.023
8 9
No, mean grade does not differ by evening; p-value = .624 > .05.
10 11
2.18
12 ANOVA Source of Variation
SS
df
MS
Between Groups
727.198
1
727.198
Within Groups
7746.713
53
146.164
Total
8473.911
54
F
P-value 4.975
0.030
F crit 4.023
13 14
Yes, mean grade differs by status; p-value = .030 < .05.
15 16
10
1
2.19
2 ANOVA Source of Variation
SS
df
MS
Between Groups
555.022
1
555.022
Within Groups
7918.888
53
149.413
Total
8473.911
54
F
P-value 3.715
0.059
F crit 4.023
3 4
No, we cannot conclude that mean grade does not differ by gender; p-value = .059 > .05.
5 6
2.20
7 8
Yes. Differences thought due to gender may be partially due to status (and vice versa). This is the
9
type of issue that is usually present when the design isn't "balanced."
10 11
2.21
12 ANOVA Source of Variation
SS
Between Groups
139.497
Within Groups Total
df
MS
F
P-value
3
46.499
64.867
1.750E-10
14.337
20
0.717
153.833
23
F crit 3.098
13 14
Yes, mean yield differs by concentration of potassium; p-value is essentially zero.
15 16 17 18 19 20 21
11
1
2.22
2 ANOVA Source of Variation
SS
Between Groups
1768.273
4
442.068
Within Groups
6.073
15
0.405
1774.346
19
Total
df
MS
F
P-value
1091.976
2.726E-18
F crit 3.056
3 4
Yes, mean yield of fish oil differs by extraction temperature; p-value is essentially zero.
5 6
12
1
CHAPTER 3 – SOME FURTHER ISSUES IN ONE-FACTOR DESIGNS AND ANOVA
2 3 4
3.1
5 6
The rank-order array is below: 1
2
3
4
8
8
23.5
19
15.5
20.5
11
23.5
4.5
4.5
22
11
1.5
4.5
14
8
13
11
17.5
15.5
1.5
4.5
17.5
20.5
S
44.0
53.0
105.5
97.5
n
6
6
6
6
7 8
𝐻 = 12 24 24 + 1
44 2 6 + 53 2 6 + 105.5 2 6 + 97.5 2 6 − 3 24 + 1
9
= . 02 322.67 + 468.17 + 1855.04 + 1584.38 − 75
10
= .02 4230.25 − 75 = 9.605
11 12
We have 5 two-way ties, 2 three-way ties, and 1 four-way tie.
13 14
𝐻𝑐 = 9.605 1 − 5 23 − 2 + 2 33 − 3 + 43 − 4
15
= 9.605 1 − 30 + 48 + 60
16
= 9.605 . 99 = 9.702
243 − 24
13800
17 18
From the c2 tables with df = 3, c = 7.815; we reject H0 and conclude that room level affects
19
perception of the degree of motion. Also from the tables, for α = .025, c = 9.348, and for α = .01,
20
c = 11.345; .01 < p-value < .025.
21 22 23 24 13
1
Alternatively, using JMP®, we have
2 Wilcoxon / Kruskal-Wallis Tests (Rank Sums) Level Column 1 Column 2 Column 3 Column 4
Count Score Sum 6 44.000 6 53.000 6 105.500 6 97.500
Expected Score Score Mean (Mean-Mean0)/Std0 75.000 7.3333 -2.044 75.000 8.8333 -1.441 75.000 17.5833 2.010 75.000 16.2500 1.474
1-Way Test, ChiSquare Approximation ChiSquare 9.7020
3
DF Prob>ChiSq 3 0.0213*
4 5
3.2
6 7
The rank-order array is below. California
Kansas
Connecticut
2
1
12
16
5
10.5
17
7.5
5
3
5
10.5
13
7.5
14
20.5
20.5
9
18
19
15
S
89.5
65.6
76.0
n
7
7
7
8 9 10
𝐻 = 12 21 21 + 1
89.5 2 7 + 65.5 2 7 + 76 2 7 − 3 21 + 1
= 1.0742
11 12
We have 3 two-way ties and 1 three-way tie.
13 14
𝐻𝑐 = 1.0742
1 − 3 23 − 2 + 33 − 3
15
= 1.0742 1 − 18 + 24
16
= 1.0742 . 9955 = 1.0791
213 − 21
9240
17
14
1
From the c2 tables with df = 2, c = 5.991; we accept H0 and conclude that amounts of life insurance
2
for senators does not vary by state. Also from the tables, for α = .25, c = 2.773, and for α = .75, c
3
= .575; .25 < p-value < .75.
4 Wilcoxon / Kruskal-Wallis Tests (Rank Sums) Level California Connecticut Kansas
Count Score Sum 7 89.500 7 76.000 7 65.500
Expected Score 77.000 77.000 77.000
Score Mean 12.7857 10.8571 9.3571
(Mean-Mean0)/Std0 0.897 -0.037 -0.823
1-Way Test, ChiSquare Approximation ChiSquare 1.0791
5
DF Prob>ChiSq 2 0.5830
6 7
3.3
8 9
We did this as Exercise 14 in Chapter 2; our results were "No, there are not significant differences
10
due to state; p-value = .717 > .05." The F-test results are confirmed by the Kruskal-Wallis test. (Of
11
course, in either case, it's not a close call.)
12 13
3.4
14 15
The rank-order array is below. 1
2
3
4
20
24.5
7.5
36.5
17
24.5
7.5
36.5
7.5
13
36.5
28
7.5
17
31.5
39.5
13
13
13
24.5
7.5
31.5
2.5
36.5
2.5
24.5
31.5
17
20
13
31.5
31.5
2.5
31.5
2.5
39.5
7.5
20
24.5
24.5
S
105.0
212.5
188.5
314.0
n
10
10
10
10
15
1 𝐻 = 12 40 40 + 1
2 3
3 40 + 1
4
= 16.251
105 2 10 + 212.5 2 10 + 188.5 2 10 + 314 2 10 −
5 6
We have 1 two-way tie, 2 three-way ties, 2 four-way ties, 1 five-way tie, and 3 six-way ties.
7 8
𝐻𝑐 = 16.251 1 − 23 − 2 + 2 33 − 3 + 2 43 − 4 + 53 − 5 + 3 63 − 6
9
= 16.251 1 − 6 + 48 + 120 + 120 + 630
10
= 16.251 . 9856 = 16.489
403 − 40
63960
11 12
From the c2 tables with df = 3, c = 7.815; we reject H0 and conclude that amounts of life insurance
13
for senators does not vary by state. Also from the tables, for α = .025, c = 9.348, and for α = .01, c
14
= 11.345; p-value < .01.
15 Wilcoxon / Kruskal-Wallis Tests (Rank Sums) Level 1 2 3 4
Count Score Sum 10 105.000 10 212.500 10 188.500 10 314.000
Expected Score 205.000 205.000 205.000 205.000
Score Mean 10.5000 21.2500 18.8500 31.4000
(Mean-Mean0)/Std0 -3.131 0.220 -0.503 3.414
1-Way Test, ChiSquare Approximation ChiSquare 16.4891
16
DF Prob>ChiSq 3 0.0009*
17 18
3.5
19
Analysis of Variance
20
Source Label Error C. Total
DF 3 36 39
Sum of Squares Mean Square 1.3147500 0.438250 1.8690000 0.051917 3.1837500
F Ratio Prob > F 8.4414 0.0002*
21 22
No, our conclusion is the same; p-value = .0002 < .05.
16
1
3.6
2 A
B
C
D
6
14
2
8
10
16
1
12
5
13
4
9
7
15
3
11
S
28
58
10
40
n
4
4
4
4
3 4 5
𝐻 = 12 16 16 + 1
28 2 4 + 58 2 4 + 10 2 4 + 40 2 4 − 3 16 + 1
= 13.500
6 7
From the c2 tables with df = 3, c = 7.815; we reject H0 and conclude that amounts of life insurance
8
for senators does not vary by state. Also from the tables, for α = .025, c = 9.348, and for α = .01, c
9
= 11.345; p-value < .01.
10 Wilcoxon / Kruskal-Wallis Tests (Rank Sums) Level A B C D
Count Score Sum 4 28.000 4 58.000 4 10.000 4 40.000
Expected Score 34.000 34.000 34.000 34.000
Score Mean 7.0000 14.5000 2.5000 10.0000
(Mean-Mean0)/Std0 -0.667 2.850 -2.850 0.667
1-Way Test, ChiSquare Approximation ChiSquare DF Prob>ChiSq 13.5000 3 0.0037* Small sample sizes. Refer to statistical tables for tests, rather than large-sample approximations.
11 12 13
Running an F-test:
14
Analysis of Variance
15
Source Label Error C. Total
DF 3 12 15
Sum of Squares Mean Square 8038.6875 2679.56 718.2500 59.85 8756.9375
F Ratio Prob > F 44.7682 <.0001*
17
1
Our conclusion is the same; p-value < .0001.
2 3
3.7
4 5
𝐶 = 4 and
𝑅=6
6
𝜈1 = 𝐶 − 1 = 3
7
α = .01
8
𝛟 = 2.5
9
From the tables power =.82
and
𝜈2 = 𝑅𝐶 − 𝐶 = 20
10 11
3.8
12 13
For 𝑅 = 9, 𝜈2 = 32, and power = .93; for 𝑅 = 3, 𝜈2 = 8, and power is .64. From the perspective
14
of increasing power, more replication yields greater power, but with decreasing benefit as the total
15
number of replicates grows. The change in power when R goes from 3 to 6 is much larger than the
16
change in power when R goes from 6 to 9.
17 18
3.9
19 20
The change to α = .05 corresponds to an increase in power to .97, while the change to 𝑅 = 9
21
yielded a (smaller) increase in power to .93.
22 23
3.10
24 25
From the power tables, with power = .80, ∆ σ = 2, α = .01, and 𝐶 = 4, we find that 𝑅 = 10.
26 27 28 29 30 31
18
1
3.11
2 3
We can use what's given to populate the ANOVA table.
4 ANOVA Table Source of Variability
SSQ
df
MSQ
Fcalc
Diet
300
3
100
4
Error
900
36
25
Total
1200
39
5 6
We have that E MSE = σ2; also, E MSBc = σ2 + 𝑉𝑐𝑜𝑙
7
Replacing E(MSE) by MSE and E(MSBc) by MSBc, and using values from the above ANOVA
8
table, we have
9 10
σ2 = 25
and
σ2 + 𝑉𝑐𝑜𝑙 = 100
11 12
and, from Chapter 2, 𝑉𝑐𝑜𝑙 = 𝑅 𝐶 − 1
j µj
−µ
2
= 75
13 14
With 𝑅 = 10, 𝐶 = 4, 𝑅 𝐶 − 1
= 10 3 and
j µj − µ
2
= 75 10 3 = 22.5
15 16 17 18 19 20 21 22 23 24 25 26
19
1
3.12
2 A
B
C
30
28.5
25
26.5
28.5
17
26.5
22
13.5
24
22
13.5
22
19
8
20
17
8
17
15
8
11.5
11.5
4
5
8
2.5
2.5
8
1
S
185.0
179.5
100.5
n
10
10
10
3 4 5
𝐻 = 12 30 30 + 1
185 2 10 + 179.5 2 10 + 100.5 2 10 − 3 30 + 1
= 5.768
6 7
We have 5 two-way ties, 2 three-way ties, and 1 five-way tie.
8 9
𝐻𝑐 = 5.768 1 − 5 23 − 2 + 2 33 − 3 + 53 − 5
10
= 5.768 1 − 30 + 48 + 120
11
= 5.768 . 9927 = 5.811
303 − 30
26970
12 13
From the c2 tables with df = 2 and α = .05, c = 5.991; we accept H0 and conclude that there are not
14
differences in solicitation approach with respect to amount of contributions. Note that this result
15
is at odds with the conventional F-test results; in Exercise 15 of Chapter 2 we said "Yes, there are
16
differences in amount of donations due to solicitation approach, though it's a close call; p-value =
17
.0395 < .05."
18
20
Wilcoxon / Kruskal-Wallis Tests (Rank Sums) Level A B C
Count Score Sum 10 185.000 10 179.500 10 100.500
Expected Score 155.000 155.000 155.000
Score Mean 18.5000 17.9500 10.0500
(Mean-Mean0)/Std0 1.303 1.060 -2.384
1-Way Test, ChiSquare Approximation ChiSquare 5.8110
1
DF Prob>ChiSq 2 0.0547
2 3
3.13
4 5 6
a. From Chapter 2, Exercise 15, MSW = 671,759. For each estimate of the mean, n = 10.
7
From the t-tables for α = .05 and df = 27, t = 2.0518. The half-width of the confidence
8
interval for the mean is
9 𝑒 = 𝑡 MSW 𝑛 1
10
2
= 2.0518 671175 10 1
2
= 531.79
11 12
The confidence intervals for the three means (estimates) are as follows.
13 Approach
Mean
Confidence Interval
A
2205
1673.21 to 2736.79
B
2125
1593.21 to 2656.79
C
1310
778.21 to 1841.79
14
Means for Oneway Anova Level Number Mean Std Error Lower 95% Upper 95% 10 2205.00 259.18 1673.2 2736.8 A B 10 2125.00 259.18 1593.2 2656.8 C 10 1310.00 259.18 778.2 1841.8 Std Error uses a pooled estimate of error variance
15 16 17 18
b. Values in the range 1673.21 to 1841.79 are common to all three confidence intervals.
19 20 21
1
c.
2
This suggests that the means may well not be different, and the data are consistent with a
3
common mean in the region of overlap.
4 5
22
CHAPTER 4 – MULTIPLE-COMPARISON TESTING
1 2 3 4
4.1
5 6
The means for columns 1, 2, and 3 are 5, 6, and 10, respectively.
7 8
MSW =
6 − 5 2 + 3 − 5 2 + 8 − 5 2 + ⋯ + 11 − 10 2
21
= 3.619
9 10 11
For a = .05, df = 21, t = 2.080
12 13
LSD = 𝑡 2MSW 𝑅 1 2
14
= 2.080 2 3.619 8 1 2
15
= 1.9785
16 17
Column means are already in ascending order. Columns 1 and 2 are the same and are different
18
from column three.
19 Column: Mean:
1 5
2 6
3 10
20 21
Conclusion:
1 2
3
22 23
Alternatively, using SPSS®, we have
24 ANOVA Sales Sum of Squares
df
Mean Square
Between Groups
112.000
2
56.000
Within Groups
76.000
21
3.619
Total
188.000
23
F 15.474
Sig. .000
25 23
Multiple Comparisons Dependent Variable: Sales LSD 95% Confidence Interval
Mean Difference (I) Treatment
(J) Treatment
1
2
2 3
(I-J)
Std. Error
Sig.
Lower Bound
Upper Bound
-1.000
.951
.305
-2.98
.98
3
*
-5.000
.951
.000
-6.98
-3.02
1
1.000
.951
.305
-.98
2.98
3
*
-4.000
.951
.000
-5.98
-2.02
1
5.000*
.951
.000
3.02
6.98
2
4.000*
.951
.000
2.02
5.98
1 *. The mean difference is significant at the 0.05 level.
2 3
4.2
4 5
Number of treatment means = 3, error df = 21, and a = .05; q = 3.57
6 7
HSD = tuk1–𝐚 2 2MSW 𝑅 1
8
= 3.57 3.619 8 1
9
= 2.40
2
= 𝑞 MSW 𝑅 1
2
2
10 11
Our conclusions are the same as in Exercise 4.1. Columns 1 and 2 are the same and are different
12
from column three.
13 14 15 16 17 18 19 20 21
24
Multiple Comparisons Dependent Variable: Sales Tukey HSD 95% Confidence Interval
Mean Difference (I-J)
(I) Treatment (J) Treatment 1
2 3
Std. Error
Sig.
Lower Bound
Upper Bound
2
-1.000
.951
.554
-3.40
1.40
3
*
-5.000
.951
.000
-7.40
-2.60
1
1.000
.951
.554
-1.40
3.40
3
*
-4.000
.951
.001
-6.40
-1.60
1
5.000*
.951
.000
2.60
7.40
2
4.000*
.951
.001
1.60
6.40
1 2
*. The mean difference is significant at the 0.05 level. Sales Tukey HSD
a
Subset for alpha = 0.05 Treatment
N
1
2
1
8
5.00
2
8
6.00
3
8
Sig.
10.00 .554
1.000
3 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 8.000.
4 5
4.3
6 7
s = 3, q(3, 21) = 3.57, NKD = 𝑞 MSW 𝑅 1
2
= 3.57 3.619 8 1
2
= 2.40
8
s = 2, q(2, 21) = 2.94, NKD = 𝑞 MSW 𝑅 1
2
= 2.94 3.619 8 1
2
= 1.98
Difference in Means 1 4 5
Difference vs. NKD < 1.98 > 1.98 > 2.40
9 Columns Compared 1 vs. 2 2 vs. 3 1 vs. 3
Reject Equality No Yes Yes
10
25
1
Our conclusions are the same as in Exercise 4.1. Columns 1 and 2 are the same and are different
2
from column three.
3 Sales a
Student-Newman-Keuls
Subset for alpha = 0.05 Treatment
N
1
2
1
8
5.00
2
8
6.00
3
8
10.00
Sig.
.305
1.000
4 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 8.000.
5 6
4.4
7 8
Column 1 is the control. With a = .05, one control and two treatments, and df = 21, Dut = 2.38.
9 10
Dut-D = Dut1–𝐚 2 2MSW 𝑅 1
11
= 2.38 2 3.619 8 1
12
= 2.26
2
2
13 Columns Compared 1 vs. 2 1 vs. 3
Difference in Means 1 5
Difference vs. 2.26
Reject Equality
< >
No Yes
14 15
We conclude that column 2 is not significantly different from the control and that column 3 is
16
significantly different from the control.
17 18
Conclusion:
1 2
1 3
19 20 21
26
Multiple Comparisons Dependent Variable: Sales Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I-J)
(I) Treatment (J) Treatment
Std. Error
Sig.
Lower Bound
Upper Bound
2
1
1.000
.951
.481
-1.25
3.25
3
1
5.000
.951
.000
2.75
7.25
*
1 *. The mean difference is significant at the 0.05 level. a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
2 3
4.5
4 5
𝐿′ = −1 2 𝑌1 + 𝑌2 − 1 2 𝑌3
6
= −1 2 5 + 6 − 1 2 10
7
= 1.5
8 9
𝑎j2 = 1 4 + 1 + 1 4 = 1.5
10 11
𝐜=
12
=
13 14 15 16
MSW 𝑅
𝑎 2j 𝐶 − 1 𝐹 𝐶 − 1 , 𝐶 𝑅 − 1
3.619 8 1.5 2 𝐹 2,21 1
1 2
2
where F(2, 21) = 3.47 for a = .05 c = 2.17
17 18
L' = 1.5 < 2.17 = c; we accept H0 and conclude that the mean of column two is not significantly
19
different from the average of the means of columns one and three.
20 21 22 23 24
27
1
4.6
2 3
The SPSS® output is below.
4 ANOVA Staffing Sum of Squares
df
Mean Square
Between Groups
6.108
4
1.527
Within Groups
3.370
15
.225
Total
9.478
19
F 6.797
Sig. .002
5 Staffing Student-Newman-Keulsa Subset for alpha = 0.05 Technology
N
1
2
1
4
1.100
2
4
1.350
3
4
1.550
4
4
1.850
5
4
Sig.
2.700 .158
1.000
6 Means for groups in homogeneous subsets are displayed.
7
a. Uses Harmonic Mean Sample Size = 4.000.
8
The SPSS® output indicates that staffing ratios are not significantly different for industries with
9
dominant technologies as represented by A, B, C, or D. The staffing ratio for industry with
10
dominant technology E, however, is statistically different from those of the first group.
11 12 13 14 15 16 17 18 28
1
4.7
2 3
The SPSS® output is below.
4 Multiple Comparisons Dependent Variable: Motion Tukey HSD 95% Confidence Interval
Mean Difference (I) Level
(J) Level
1
2
-2.333
2.716
3
*
-8.667
4 2
3
4
5 6
(I-J)
Std. Error
Lower Bound
Upper Bound
.826
-9.94
5.27
2.716
.022
-16.27
-1.06
-7.667*
2.716
.048
-15.27
-.06
1
2.333
2.716
.826
-5.27
9.94
3
-6.333
2.716
.124
-13.94
1.27
4
-5.333
2.716
.235
-12.94
2.27
1
8.667*
2.716
.022
1.06
16.27
2
6.333
2.716
.124
-1.27
13.94
4
1.000
2.716
.982
-6.60
8.60
1
7.667
*
2.716
.048
.06
15.27
2
5.333
2.716
.235
-2.27
12.94
2.716
.982
-8.60
6.60
3 -1.000 * The mean difference is significant at the .05 level.
Sig.
Motion Tukey HSD
a
Subset for alpha = 0.05 Level
N
1
2
1
6
14.33
2
6
16.67
4
6
22.00
3
6
23.00
Sig.
.826
16.67
.124
7 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 6.000.
8
29
1
With regard to amount of motion perceived as a function of room level, the SPSS® output shows
2
that, if we put level two aside for the moment, levels three and four are similar (not significantly
3
different) to one another and are both different from level one. Level two cannot be said to be
4
significantly different from any of the other levels. Diagrammatically,
5 6
Conclusion:
1 2
4
3
7 8
4.8
9 10 11
The SPSS® output is below. Motion a
Student-Newman-Keuls
Subset for alpha = 0.05 Level
N
1
2
1
6
14.33
2
6
16.67
4
6
22.00
3
6
23.00
Sig.
.400
16.67
.074
12 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 6.000.
13 14
With regard to amount of motion perceived as a function of room level, the SPSS® output shows
15
that, if we put level two aside for the moment, levels three and four are similar (not significantly
16
different) to one another and are both different from level one. Level two cannot be said to be
17
significantly different from any of the other levels.
18 19
4.9
20 21
Yes, the practical conclusions are the same.
22 23 24
30
1
4.10
2 3 4
The SPSS® output is below. Multiple Comparisons Dependent Variable: Insurance LSD 95% Confidence Interval
Mean Difference (I-J)
(I) State
(J) State
California
Kansas
13.571
34.711
Connecticut
28.571
California
Kansas Connecticut
Std. Error
Sig.
Lower Bound
Upper Bound
.700
-59.35
86.50
34.711
.421
-44.35
101.50
-13.571
34.711
.700
-86.50
59.35
Connecticut
15.000
34.711
.671
-57.93
87.93
California
-28.571
34.711
.421
-101.50
44.35
Kansas
-15.000
34.711
.671
-87.93
57.93
5 6
The ANOVA table shows that there is no significant difference between the states (there is no
7
state effect). The LSD results are consistent with this conclusion.
8 9
4.11
10 11
The SPSS® output is below.
12 Multiple Comparisons Dependent Variable: Insurance Tukey HSD 95% Confidence Interval
Mean Difference (I-J)
(I) State
(J) State
California
Kansas
13.571
34.711
.920
-75.02
102.16
Connecticut
28.571
34.711
.694
-60.02
117.16
California
-13.571
34.711
.920
-102.16
75.02
Connecticut
15.000
34.711
.903
-73.59
103.59
California
-28.571
34.711
.694
-117.16
60.02
Kansas
-15.000
34.711
.903
-103.59
73.59
Kansas Connecticut
Std. Error
Sig.
Lower Bound
Upper Bound
13
31
Insurance Tukey HSD
a
Subset for alpha = 0.05 State
1
N
Connecticut
7
162.14
Kansas
7
177.14
California
7
190.71
Sig.
.694
1 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 7.000.
2 3 4
Tukey's HSD test shows there is no significant difference between the states.
5
4.12
6 7
The SPSS® output is below.
8 Multiple Comparisons Dependent Variable: Insurance Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I-J)
(I) State
(J) State
Std. Error
Sig.
Connecticut
California
-28.571
34.711
Kansas
California
-13.571
34.711
Lower Bound
Upper Bound
.630
-111.83
54.69
.896
-96.83
69.69
9 a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
10 Multiple Comparisons Dependent Variable: Insurance Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I-J)
(I) State
(J) State
Std. Error
California
Kansas
13.571
34.711
Connecticut
Kansas
-15.000
34.711
Sig.
Lower Bound
Upper Bound
.896
-69.69
96.83
.874
-98.26
68.26
11 a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
32
1 Multiple Comparisons Dependent Variable: Insurance Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I) State
(J) State
Std. Error
California
Connecticut
28.571
34.711
Kansas
Connecticut
15.000
34.711
(I-J)
Sig.
Lower Bound
Upper Bound
.630
-54.69
111.83
.874
-68.26
98.26
2 a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
3 4
The first set of results is for column one as control; the second is for column two as control; and
5
the third is for column three as control. Irrespective of which column is control, no other column
6
is significantly different from the control. The practical conclusions are the same as the ones
7
obtained when using the HSD test.
8 9
4.13
10 11 12
𝐿′ = 1 17 − 1 3 12 − 1 3 5 − 1 3 14 = 6.67
13 14
𝑎j2 = 1 + −1 3 2 + −1 3 2 + −1 3 2 = 1.33
15 16
j𝑎
2
∙ 𝐶 − 1 ∙ 𝐹1–𝐚 𝑑𝑓1, 𝑑𝑓2
17
𝐜=
MSW 𝑅 ∙
18
=
21.2 6 ∙ 1.33 ∙ 4 ∙ 2.78
19
= 7.238
20 21
L' = 6.67 < 7.24 = c; we accept H0 and conclude that the mean of column five is not significantly
22
different from the average of the means of columns two, three, and four.
23 24
33
1
4.14
2 3
The SPSS® output is below.
4 Lifetime Tukey HSD
a
Subset for alpha = 0.05 Device
N
1
2
1
3
2.600
2
3
4.600
4.600
6
3
4.600
4.600
3
3
5.800
5.800
5
3
6.200
6.200
4
3
7.000
7.000
8
3
7.400
7.400
7
3
Sig.
8.200 .052
.232
5 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 3.000.
6 7
Device 1 and device 7 are statistically different from each other; the remaining six devices are
8
not significantly different from each other, device 1, or device 7.
9 10
Conclusion: 1 2 6 3 5 4 8 7
11 12 13 14 15 16 17 18 19 20
34
1
4.15
2 3
The SPSS® output is below.
4 Lifetime a
Student-Newman-Keuls
Subset for alpha = 0.05 Device
N
1
2
1
3
2.600
2
3
4.600
4.600
6
3
4.600
4.600
3
3
5.800
5.800
5
3
6.200
6.200
4
3
7.000
7.000
8
3
7
3
Sig.
7.400 8.200 .057
.196
5 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 3.000.
6 7
Devices 8 and 7 are similar to each other and both are statistically different from device 1; the
8
remaining five devices are not significantly different from each other, device 1, or devices 8 or 7.
9 10
Conclusion: 1 2 6 3 5 4 8 7
11 12
4.16
13 14
Yes, the results are slightly different. The Newman-Keuls test is able to establish that device 8 is
15
different from device 1, while Tukey's HSD was not.
16 17
4.17
18 19
First, we relabel device 3 as device 1 and vice versa, because SPSS® will allow only the first or
20
last column to be the control; we choose the first. The SPSS® output is below.
35
1 Multiple Comparisons Dependent Variable: Lifetime Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I) Device
(J) Device
Std. Error
2
1
-1.2000
1.3952
.917
-5.279
2.879
3
1
-3.2000
1.3952
.160
-7.279
.879
4
1
1.2000
1.3952
.917
-2.879
5.279
5
1
.4000
1.3952
1.000
-3.679
4.479
6
1
-1.2000
1.3952
.917
-5.279
2.879
7
1
2.4000
1.3952
.397
-1.679
6.479
8
1
1.6000
1.3952
.764
-2.479
5.679
(I-J)
Sig.
Lower Bound
Upper Bound
2 3
a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
4
Remembering here that devices 3 and 1 have been swapped, we see that no column is significantly
5
different from the control (device 3). This is consistent with earlier results.
6 7
4.18
8 9 10
𝐿′ = 1 4 7.0 + 6.2 + 8.2 + 7.4 − 2.6 + 4.6 + 5.8 + 4.6 = 2.8
11 12
𝑎j2 = 8 1 4 2
13
= .5
14 𝑎2 ∙ 𝐶 − 1 ∙ 𝐹1–𝐚 𝑑𝑓1, 𝑑𝑓2
15
𝐜=
MSW 𝑅 ∙
16
=
2.92 3 ∙ . 5 ∙ 7 ∙ 2.66
17
= 3.01
18 19
L' = 2.8 < 3.01 = c; we accept H0 and conclude that the average of the four lowest column means
20
is not significantly different from the average of the four largest column means.
36
1
4.19
2 3
The SPSS® output is below.
4 Multiple Comparisons Dependent Variable: Waiting LSD 95% Confidence Interval
Mean Difference (I-J)
(I) Location (J) Location Amesbury
Andover
Salem
Sig.
Lower Bound
Upper Bound
Andover
.05711
.37454
.879
-.6820
.7963
Methuen
*
-1.07311
.37454
.005
-1.8123
-.3340
Salem
-5.70244*
.37454
.000
-6.4416
-4.9633
-.05711
.37454
.879
-.7963
.6820
*
-1.13022
.37454
.003
-1.8694
-.3911
Salem
*
-5.75956
.37454
.000
-6.4987
-5.0204
Amesbury
1.07311*
.37454
.005
.3340
1.8123
Andover
1.13022
*
.37454
.003
.3911
1.8694
Salem
*
-4.62933
.37454
.000
-5.3685
-3.8902
Amesbury
5.70244*
.37454
.000
4.9633
6.4416
Andover
5.75956
*
.37454
.000
5.0204
6.4987
Methuen
4.62933
*
.37454
.000
3.8902
5.3685
Amesbury Methuen
Methuen
Std. Error
5 *. The mean difference is significant at the 0.05 level.
6 7
We find that Andover and Amesbury are similar to (not statistically different) from each other,
8
and that the pair is different from both Methuen and Salem. Methuen and Salem are significantly
9
different from each other and from Andover and Amesbury.
10 11 12
Conclusion:
1 2
3
4
13 14 15 16 17
37
1
4.20
2 3
The SPSS® output is below.
4 5 Waiting Tukey HSD
a
Subset for alpha = 0.05 Location
N
1
2
Andover
45
4.9507
Amesbury
45
5.0078
Methuen
45
Salem
45
3
6.0809 10.7102
Sig.
.999
1.000
1.000
6 Means for groups in homogeneous subsets are displayed.
7 8
a. Uses Harmonic Mean Sample Size = 45.000.
The conclusions for Tukey's HSD test are the same as for Fisher's LSD test above (Exercise 4.19)
9 10
4.21
11 12
The SPSS® output is below.
13 Waiting a
Student-Newman-Keuls
Subset for alpha = 0.05 Location
1
N
2
Andover
45
4.9507
Amesbury
45
5.0078
Methuen
45
Salem
45
Sig.
3
6.0809 10.7102 .879
1.000
1.000
14 Means for groups in homogeneous subsets are displayed.
15
a. Uses Harmonic Mean Sample Size = 45.000.
38
1
The conclusions for the Newman-Keuls test are the same as for Fisher's LSD test and Tukey's HSD
2
test above (Exercises 4.19 and 4.20).
3 4
4.22
5 6
The SPSS® output is below.
7 Multiple Comparisons Dependent Variable: Waiting Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I-J)
(I) Location
(J) Location
Std. Error
Sig.
Lower Bound
Upper Bound
Amesbury
Andover
.05711
.37454
.997
-.8303
.9446
Methuen
Andover
1.13022
*
.37454
.008
.2428
2.0177
Salem
Andover
5.75956*
.37454
.000
4.8721
6.6470
8 *. The mean difference is significant at the 0.05 level.
9
a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
10
Amesbury is not significantly different from Andover; Methuen and Salem are both significantly
11
different from Andover.
12 13 14 15 16 17 18 19 20 21 22 23 24 39
1
4.23
2 3
The SPSS® output is below.
4 Multiple Comparisons Dependent Variable: Score LSD 95% Confidence Interval
Mean Difference (I) Course
(J) Course
Near Corners
Meadow Brook
Meadow Brook
Fountainbleau
Std. Error
Sig.
Lower Bound
Upper Bound
7.380*
.715
.000
5.97
8.79
Birch Briar
*
-8.240
.715
.000
-9.65
-6.83
Fountainbleau
-29.280*
.715
.000
-30.69
-27.87
Near Corners
-7.380*
.715
.000
-8.79
-5.97
*
-15.620
.715
.000
-17.03
-14.21
Fountainbleau
*
-36.660
.715
.000
-38.07
-35.25
Near Corners
8.240*
.715
.000
6.83
9.65
Meadow Brook
15.620
*
.715
.000
14.21
17.03
Fountainbleau
*
-21.040
.715
.000
-22.45
-19.63
Near Corners
29.280*
.715
.000
27.87
30.69
Meadow Brook
36.660
*
.715
.000
35.25
38.07
Birch Briar
21.040
*
.715
.000
19.63
22.45
Birch Briar Birch Briar
(I-J)
5 6
*. The mean difference is significant at the 0.05 level.
7
Fisher's LSD test shows that all four golf courses are significantly different from each other with
8
respect to golf scores for players like those at Eastern Electric.
9 10 11 12 13 14 15 16 17 40
1
4.24
2 3
The SPSS® output is below.
4 5 Score Tukey HSD
a
Subset for alpha = 0.05 Course
N
1
Meadow Brook
100
Near Corners
100
Birch Briar
100
Fountainbleau
100
2
3
4
98.58 105.96 114.20 135.24
Sig.
1.000
1.000
1.000
1.000
6 Means for groups in homogeneous subsets are displayed.
7
a. Uses Harmonic Mean Sample Size = 100.000.
8
Tukey's HSD test shows that all four golf courses are significantly different from each other with
9
respect to golf scores for players like those at Eastern Electric.
10 11
4.25
12 13
The SPSS® output is below.
14 Score a
Student-Newman-Keuls
Subset for alpha = 0.05 Course
1
N
Meadow Brook
100
Near Corners
100
Birch Briar
100
Fountainbleau
100
Sig.
2
3
4
98.58 105.96 114.20 135.24 1.000
1.000
1.000
1.000
15 Means for groups in homogeneous subsets are displayed. a. Uses Harmonic Mean Sample Size = 100.000.
41
1
The Newman-Keuls test shows that all four golf courses are significantly different from each other
2
with respect to golf scores for players like those at Eastern Electric.
3 4
4.26
5 6
The SPSS® output is below.
7 Multiple Comparisons Dependent Variable: Score Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I) Course
(J) Course
Near Corners
(I-J)
Std. Error
Sig.
Lower Bound
Upper Bound
Meadow Brook
7.380
*
.715
.000
5.69
9.07
Birch Briar
Meadow Brook
15.620*
.715
.000
13.93
17.31
Fountainbleau
Meadow Brook
36.660*
.715
.000
34.97
38.35
8 *. The mean difference is significant at the 0.05 level.
9
a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
10
Dunnett's test shows that all of the other three golf courses are significantly different Meadow
11
Brook with respect to golf scores for players like those at Eastern Electric.
12 13 14 15 16 17 18 19 20 21 22 23
42
1
4.27
2 3
The SPSS® outputs are below.
4 Multiple Comparisons Dependent Variable: Yield LSD 95% Confidence Interval
Mean Difference (I) Potassium
(J) Potassium
0
60
60
120
180
(I-J)
Std. Error
Sig.
Lower Bound
Upper Bound
-3.8667*
.4888
.000
-4.886
-2.847
120
*
-4.9167
.4888
.000
-5.936
-3.897
180
-6.5500*
.4888
.000
-7.570
-5.530
0
3.8667*
.4888
.000
2.847
4.886
120
*
-1.0500
.4888
.044
-2.070
-.030
180
*
-2.6833
.4888
.000
-3.703
-1.664
0
4.9167*
.4888
.000
3.897
5.936
60
1.0500
*
.4888
.044
.030
2.070
*
180
-1.6333
.4888
.003
-2.653
-.614
0
6.5500
*
.4888
.000
5.530
7.570
60
2.6833*
.4888
.000
1.664
3.703
120
1.6333
.4888
.003
.614
2.653
*
5 6
*. The mean difference is significant at the 0.05 level. Yield Tukey HSD
a
Subset for alpha = 0.05 Potassium
N
1
2
3
0
6
60
6
18.117
120
6
19.167
180
6
Sig.
14.250
20.800 1.000
.172
1.000
7 Means for groups in homogeneous subsets are displayed.
8 9
a. Uses Harmonic Mean Sample Size = 6.000.
43
Multiple Comparisons Dependent Variable: Yield Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I-J)
(I) Potassium
(J) Potassium
Std. Error
Sig.
Lower Bound
Upper Bound
60
0
3.8667*
.4888
120
0
4.9167
*
.000
2.625
5.108
.4888
.000
3.675
6.158
180
0
6.5500*
.4888
.000
5.308
7.792
1 *. The mean difference is significant at the 0.05 level.
2
a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
3
Our practical conclusions are the same using the tests above, which is not in agreement with the
4
results obtained with the Newman-Keuls test (shown below).
5 Yield Student-Newman-Keulsa Subset for alpha = 0.05 Potassium
1
N
0
6
60
6
120
6
180
6
Sig.
2
3
4
14.250 18.117 19.167 20.800 1.000
1.000
1.000
1.000
6 Means for groups in homogeneous subsets are displayed.
7 8 9 10 11 12 13 14 15 16 17
a. Uses Harmonic Mean Sample Size = 6.000.
44
1
4.28
2 3
The SPSS® outputs are below.
4 Multiple Comparisons Dependent Variable: Yield LSD 95% Confidence Interval
Mean Difference (I-J)
(I) Temperature
(J) Temperature
77
87
10.1500*
.4499
97
16.1250*
107
87
97
107
117
Lower Bound
Upper Bound
.000
9.191
11.109
.4499
.000
15.166
17.084
23.3500*
.4499
.000
22.391
24.309
117
26.0750
*
.4499
.000
25.116
27.034
77
-10.1500*
.4499
.000
-11.109
-9.191
97
5.9750
*
.4499
.000
5.016
6.934
107
13.2000
*
.4499
.000
12.241
14.159
117
15.9250
*
.4499
.000
14.966
16.884
77
-16.1250*
.4499
.000
-17.084
-15.166
87
*
-5.9750
.4499
.000
-6.934
-5.016
107
7.2250
*
.4499
.000
6.266
8.184
117
9.9500
*
.4499
.000
8.991
10.909
77
-23.3500*
.4499
.000
-24.309
-22.391
87
*
-13.2000
.4499
.000
-14.159
-12.241
97
*
-7.2250
.4499
.000
-8.184
-6.266
117
2.7250
*
.4499
.000
1.766
3.684
77
-26.0750*
.4499
.000
-27.034
-25.116
87
*
-15.9250
.4499
.000
-16.884
-14.966
97
-9.9500*
.4499
.000
-10.909
-8.991
*
.4499
.000
-3.684
-1.766
107
Std. Error
-2.7250
Sig.
5 6 7 8 9 10 11 12 13
*. The mean difference is significant at the 0.05 level.
45
Yield Tukey HSD
a
Subset for alpha = 0.05 Temperature
1
N
117
4
107
4
97
4
87
4
77
4
2
3
4
5
51.850 54.575 61.800 67.775 77.925
Sig.
1.000
1.000
1.000
1.000
1.000
1 Means for groups in homogeneous subsets are displayed.
2 3
a. Uses Harmonic Mean Sample Size = 4.000.
Yield a
Student-Newman-Keuls
Subset for alpha = 0.05 Temperature
1
N
117
4
107
4
97
4
87
4
77
4
2
3
4
5
51.850 54.575 61.800 67.775 77.925
Sig.
1.000
1.000
1.000
1.000
1.000
4 Means for groups in homogeneous subsets are displayed.
5
a. Uses Harmonic Mean Sample Size = 4.000. Multiple Comparisons Dependent Variable: Yield Dunnett t (2-sided)a 95% Confidence Interval
Mean Difference (I-J)
(I) Temperature
(J) Temperature
Std. Error
Sig.
Lower Bound
Upper Bound
77
117
26.0750*
.4499
.000
24.848
27.302
87
117
15.9250
*
.4499
.000
14.698
17.152
97
117
9.9500
*
.4499
.000
8.723
11.177
107
117
2.7250*
.4499
.000
1.498
3.952
6
46
*. The mean difference is significant at the 0.05 level. a. Dunnett t-tests treat one group as a control, and compare all other groups against it.
1 2
All treatments are found to have a different mean than the mean of the control group.
3 4 5
47
1
CHAPTER 5 – ORTHOGONALITY, ORTHOGONAL DECOMPOSITION, AND THEIR
2
ROLE IN MODERN EXPERIMENTAL DESIGN
3 4 5
5.1
6 Aspirin (Brand 1)
Aspirin (Brand 2)
Non-aspirin (Brand 1)
Non-aspirin (Brand 2)
Placebo
8
12
7
7
5
7
10
6
5
4
7.5
11
6.5
6
4.5
7 8
Grand Mean = 7.1
9 10 11
SSBc = 2 7.5 − 7.1 2 + 11 − 7.1 2 + 6.5 − 7.1 2 + ⋯ + 4.5 − 7.1 2 = 47.4
12 13 14
SSW =
8 − 7.5 2 + 7 − 7.5 2 + 12 − 11 2 + ⋯ + 4 − 4.5 2
= 5.5
15 Source of Variability
SSQ
Drug
47.4
4
11.85
Error
5.5
5
1.1
Total
df
MS
Fcalc 10.77
9
16 17 18 19 20 21 22
48
1
The SPSS® output is below.
2 ANOVA Reduction Sum of Squares
df
Mean Square
Between Groups
47.400
4
11.850
Within Groups
5.500
5
1.100
Total
52.900
9
F 10.773
Sig. .011
3 4
From the F tables for α = .05 and df = (4, 5), c = 5.19; since Fcalc = 10.77 > c = 5.19, we reject H0
5
and conclude that the column effect is not zero. That is, we conclude that there is a difference
6
among analgesics in terms of reduction of headache pain.
7 8
5.2
9 ORTHOGONAL COEFFICIENT MATRIX A1 A2 N1 N2 P 1 4 1 4 1 4 1 4 −1 1 2 1 2 0 −1 0 1 3 1 3 −1 1 3 0 −1 1 0 0 0 10 11
If we divide the coefficients in each row by the square root of the sum of the squared coefficients
12
in that row, we generate an orthonormal matrix.
13 ORTHONORMAL COEFFICIENT MATRIX A1 A2 N1 N2 P .2236 . 2236 . 2236 . 2236 −.8944 . 4082 . 4082 0 −.8164 0 . 2887 . 2887 −.8660 . 2887 0 . 7071 −.7071 0 0 0 14 15 16
49
1
5.3
2 Y1
Y2
Y3
Y4
Y5
7.5
11
6.5
6
4.5
Z
Z2
Z1 =
. 2236
. 2236
. 2236
. 2236
−.8944
2.9068
8.4495
Z2 =
. 4082
. 4082
0
−.8164
0
2.6533
7.0400
Z3 =
. 2887
. 2887
−.8660
. 2887
0
1.4442
2.0856
Z4 =
. 7071
−.7071
0
0
0
−2.4749
6.1249
SSBc = 23.7𝑅 3 4 5
SSBc = 23.7𝑅 = 47.4 Source of Variability Drug
SSQ
df
47.4
MS
4
Fcalc
11.85
10.77
P vs. P’
16.90
1
16.90
15.36
A1, A2 vs. N2
14.08
1
14.08
12.80
A1, A2, N2 vs. N1
4.17
1
4.17
3.79
A1 vs. A2
12.25
1
12.25
11.14
Error
5.5
5
Total
52.9
9
1.1
6 7
The SPSS® output is below.
8 Contrast Coefficients Drug Contrast
Asp_1
Asp_2
N_1
N_2
P
1
.224
.224
.224
.224
-.89
2
.408
.408
0
-.82
0
3
.289
.289
-.87
.289
0
4
.71
-.71
0
0
0
9 10
50
Contrast Tests Contrast Reduction
Value of Contrast
Std. Error
t
df
Sig. (2-tailed)
Assume equal
1
2.94a
.739
3.977
5
.011
variances
2
2.63a
.744
3.534
5
.017
3
1.43a
.744
1.915
5
.114
4
-2.49
.745
-3.337
5
.021
Does not assume
1
2.94a
.569
5.168
2.348
.025
equal variances
2
2.63a
.938
2.801
1.610
.136
3
1.43a
.614
2.321
2.834
.108
4
-2.49
.794
-3.130
1.471
.129
1 a. The sum of the contrast coefficients is not zero.
2 3
From the F tables for α = .05 and df = (1, 5), c = 6.61. The average reduction of headache pain for
4
those using A1, A2, or N2 is not significantly different from that of those using N1. All other
5
comparisons are significantly different from zero.
6 7
5.4
8 ORTHOGONAL MATRIX C A D L 1 3 1 3 1 3 −1 1 2 1 2 −1 0 1 −1 0 0 9 ORTHONORMAL MATRIX A1 A2 N1 N2 .2887 .2887 .2887 −.8660 . 4082 . 4082 −.8164 0 . 7071 −.7071 0 0 10 11 12 13 14
51
1
5.5
2 ORTHOGONAL MATRIX Low Medium High 1 2 −1 1 2 −1 0 1 3 4
5.6
5 ORTHOGONAL MATRIX Amesbury Andover Methuen Salem 1 1 −1 −1 1 −1 1 −1 1 1 −1 −1 6 7
Yes, the three questions depicted in the above table are orthogonal.
8 9
5.7
10 11
We divide each coefficient (above) by 1 + 1 + 1 + 1 1
2
=2
12 ORTHONORMAL MATRIX Amesbury Andover Methuen Salem 1 2 1 2 −1 2 −1 2 1 2 −1 2 1 2 −1 2 1 2 −1 2 −1 2 1 2 13 14 15 16 17 18 19 20 52