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