Question

Part B

3. In a completely randomized design, 7 experimental units were used for each of

the three levels of the factor. (2 points each; 6 points total)

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatment
Error	432076.5
Total	675643.3

1. Complete the ANOVA table.

2. Find the critical value at the 0.05 level of significance from the F table for testing whether the population means for the three levels of the factors are different.
3. Use the critical value approach and α = 0.05 to test whether the population

means for the three levels of the factors are the same.

Answer 1

a) Number of treatment, k = 3

Total sample Size, N = 6*3 = 18

df(between) = k-1 = 2

df(within) = N-k = 15

df(total) = N-1 = 17

SS(within) = 432076.5

SS(total) = 675643.3

SS(between) = SS(total) - SS(within) = 675643.3 -  432076.5 = 243566.8

MS(between) = SS(between)/df(between) =121783.4

MS(within) = SS(within)/df(within) =28805.1

F = MS(between)/MS(within) = 4.2278

Source of variation	SS	df	MS	F
Treatment	243566.8	2	121783.4	4.2278
Error	432076.5	15	28805.1
Total	675643.3	17

b) At α = 0.05, df1=2, df2 =15 Critical value, Fc = F.INV.RT(0.05, 2, 15) = 3.682

c) As F = 4.2278 > Fc = 3.682, we reject the null hypothesis.

There is not enough evidence to conclude that the population means for the three levels of the factors are the same at 0.05 significance level.