Question

The following data represent a simple random sample of n = 4 from three populations that are known to be normally distributed. Verify that the F-test statistic is 2.04.

Sample 1	Sample 2	Sample 3
28	22	25
23	25	24
30	17	19
27	23	30

Answer 1

The Data from these 3 samples are as below

	Sample 1	Sample 2	Sample 3
Mean(x bar)	27	21.75	24.5
SD	2.9439	3.4034	4.5092
Variance	8.6667	11.5833	20.3333
n	4	4	4
columns	3

The overall mean = (27 + 21.75 + 24.5)/3 = 73.25/3 = 24.4167

SS treatment = SUM n* ( x̅_i - overall mean)² = 4 * (24.4167 - 27)² + 4 * (24.4167 - 21.75)² + 4 * (24.4167 - 24.5)² = 55.17

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 55.17 / 2 = 27.58

SSerror = SUM [(n - 1) * Variance] = 3 * 8.6667 + 3 * 11.5833 + 3 * 20.3333 = 121.75

df2 = N - k = 12 - 3 = 9

Therefore MS error = SSerror/df2 = 121.75/9 = 13.53

F = MSTR/MSE = 27.58/13.53 = 2.0384     2.04

The Complete ANOVA table is as below

Source	SS	DF	Mean Square	F
Between	55.17	2	27.58	2.04
Within/Error	121.75	9	13.53
Total	176.92	11