Question

1. The following ANOVA summary table was obtained for a comparison of six teachers. Of interest is whether students of those teachers display different mean levels of performance on a standardized language test. Class size is about 20 to 22 students per teacher, for a total number of 126 students included in the analysis.
a. Complete the missing calculations in the table below. (0.5 points each; 3 points total)
Sum of Squares df Mean Square F Between Groups 600.00 _____? _____? _____? Within Groups _____? _____? _____?
Total 4800.00
b. Given the F ratio calculated, would one reject or fail to reject Ho for α = .05? Explain have you arrived at this conclusion. (2 points)
c. If α = .01, would one reject or fail to reject Ho? Explain have you arrived at this conclusion. (2 points)

Answer 1

Number of treatment, k = 6

Total sample Size, N = 126

df(between) = k-1 = 5

df(within) = N-k = 120

df(total) = N-1 = 125

SS(between) = 600

SS(total) = 4800

SS(within) = SS(total) - SS(between) = 4800 -600 = 4200

MS(between) = SS(between)/df(between) = 120

MS(within) = SS(within)/df(within) = 35

F = MS(between)/MS(within) = 3.4286

p-value = F.DIST.RT(3.4286, 5, 120) = 0.0062

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	600	5	120	3.4286	0.0062
Within Groups	4200	120	35
Total	4800	125

b) As P-value < 0.05, Reject the null hypothesis.

There is enough evidence to conclude that not all means are equal.

c) As P-value < 0.01, Reject the null hypothesis.