Chap 15
15.8
At least two means differ.
Rejection region:
IBMDellHPOther
Mean13.3311.009.6717.00
Variance12.3379.0022.3339.00
Grand mean = 12.75
SST = = 3(13.33 – 12.75)+ 3(11.00 – 12.75)+ 3(9.67 – 12.75) + 3(17.00 – 12.75) = 92.92
SSE = (3 –1)(12.33) + (3 – 1)(79.00) + (3 – 1)(22.33) + (3 – 1)(39.00) = 305.33
ANOVA table
Source Degrees of FreedomSum of Squares Mean Squares F .
TreatmentsSST = 92.92
ErrorSSE = 305.33
F = .81, p-value = .5224. There is not enough evidence to conclude that there are differences in age between the computer brands.
15.10
a
At least two means differ.
Rejection region:
Grand mean = 101.0
SST =
= 30(90.17 – 101.0)+ 30(95.77 – 101.0) + 30(106.8 – 101.0) + 30(111.17 – 101.0) = 8,464
SSE =
= (30 –1)(991.52) + (30 – 1)(900.87) + (30 – 1)(928.70) + (30 – 1)(1,023.04) = 111,480
ANOVA table
Source Degrees of FreedomSum of Squares Mean Squares F .
TreatmentsSST = 8,464
ErrorSSE = 111,480
F = 2.94, p-value = .0363. There is enough evidence to infer that there are differences between the completion times of the four income tax forms.
b The times for each form must be normally distributed with the same variance.
c The histograms are approximately bell-shaped and the sample variances are similar.
15.12
At least two means differ.
Rejection region:
Grand mean = 173.3
SST = = 25(164.6 – 173.3)+ 25(185.6 – 173.3) + 25(154.8 – 173.3)
+ 25(182.6 – 173.3) + 25(178.9 – 173.3) = 17,251
SSE = = (25 –1)(1,164) + (25 – 1)(1,720) + (25 – 1)(1,114) + (25 – 1)(1,658)
+ (25 – 1)(841.8) = 155,941
ANOVA table
Source Degrees of FreedomSum of Squares Mean Squares F .
TreatmentsSST = 17,251
ErrorSSE = 155,941
F = 3.32, p-value = .0129. There is not enough evidence to allow the manufacturer to conclude that differences exist between the five lacquers.
b The times until first sign of corrosion for each lacquer must be normally distributed with a common variance.
c The histograms are approximately bell-shaped with similar sample variances.
15.16
At least two means differ.
Rejection region:
Grand mean = 77.39
SST = = 30(74.10 – 77.39)+ 30(75.67 – 77.39) + 30(78.50 – 77.39)
+ 30(81.30 – 77.39) = 909.42
SSE = = (30 –1)(249.96) + (30 – 1)(184.23) + (30 – 1)(233.36) + (30 – 1)(242.91) = 26,403
ANOVA table
Source Degrees of FreedomSum of Squares Mean Squares F .
TreatmentsSST = 909.4
ErrorSSE = 26,403
F = 1.33, p-value = .2675. There is not enough evidence of a difference between the four groups of companies.
15.60a =
At least two means differ.
F = 7.67, p-value = .0001. There is sufficient evidence to infer that differences in productivity exist between the four groups of companies.
b
Using either the Bonferroni adjustment or Tukey’s method we conclude that differs from , and . Companies that offered extensive training have productivity levels different from the other companies.
15.70=
At least two means differ.
F = 10.26, p-value = 0. There is enough evidence of differences between the four groups of investors.
15.72=
At least two means differ.
F = 9.17, p-value = 0. There is sufficient evidence to infer that there are differences in changes to the TSE depending on the loss the previous day.
15.74
At least two means differ.
Data file for this problems is "stacked"; see pp. 426-427 of text.
To prepare it for the Anova tool, follow the procedure on p. 474 of text.
The idea is to sort the data first.
F = 45.49, p-value = 0. There is enough evidence to infer that family incomes differ between the three market segments.
15.76 =
At least two means differ.
As in 15.74, you need to “unstack” the data first.
F = 211.61, p-value = 0. There is enough evidence to conclude that there are differences in the age of the car between the four market segments.
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