How do you calculate degrees of freedom for numerator and denominator with ANOVA
ID: 3218164 • Letter: H
Question
How do you calculate degrees of freedom for numerator and denominator with ANOVA?
example:
A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are below. Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same. (Let 1 = Working-class, 2 = Professional (middle incomes), and 3 = Professional (wealthy).)
Working-class Professional (middle incomes) Professional (wealthy) 18.3 16.7 9.0 27.4 17.5 6.5 50.3 22.7 5.1 9.5 7.9 13.5 65.5 9.7 11.7 48.0 2.7 29.0 19.6 7.3 15.9 51.7 14.5 9.9Explanation / Answer
Answer:
How do you calculate degrees of freedom for numerator and denominator with ANOVA?
degrees of freedom for numerator = number of groups -1
degrees of freedom for denominator = total sample size- number of groups
example:
A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are below. Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same. (Let 1 = Working-class, 2 = Professional (middle incomes), and 3 = Professional (wealthy).)
One factor ANOVA
Mean
n
Std. Dev
36.29
8
20.082
Working-class
12.38
8
6.575
Professional (middle incomes)
12.58
8
7.514
Professional (wealthy)
20.41
24
16.869
Total
ANOVA table
Source
SS
df
MS
F
p-value
Treatment
3,024.348
2
1,512.1738
9.02
.0015
Error
3,520.839
21
167.6590
Total
6,545.186
23
degrees of freedom for numerator = 3 -1 =2
degrees of freedom for denominator = total sample size- number of groups =24-3=21
Calculated F=9.02, P=0.0015 which is < 0.05 level of significance.
Ho is rejected.
We reject the null hypothesis and conclude that the three mean commuting mileages are different.
One factor ANOVA
Mean
n
Std. Dev
36.29
8
20.082
Working-class
12.38
8
6.575
Professional (middle incomes)
12.58
8
7.514
Professional (wealthy)
20.41
24
16.869
Total
ANOVA table
Source
SS
df
MS
F
p-value
Treatment
3,024.348
2
1,512.1738
9.02
.0015
Error
3,520.839
21
167.6590
Total
6,545.186
23
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