We calculate BSS = 10 * (540.3 - 543.6)2 + 16 * (602.9 - 543.6)2 + 17 * (489.8 - 543.6)2 = 105578.2 WSS = (10 - 1) * 46.42 + (16 - 1) * 47.42 + (17 - 1) * 48.92 = 91337.4 F = (BSS / (k-1)) / (WSS / (n-k)) = (105578.2 / (3 - 1)) / (91337.4 / (43 - 3)) = 23.1 The critical F value is F.05,2,40 = 3.23. The observed value vastly exceeds this, so we clearly reject the null hypothesis and conclude that the means differ across the groups.

Question

Answer 1

We can calculate the F statistic as follows
BSS = 5 * (7.6 - 6.7)2 + 5 * (6.0 - 6.7)2 + 5 * (6.64 - 6.7)2 = 6.5
WSS = (5 - 1)*1.12 + (5 - 1)*0.82 + (5 - 1)*0.62 = 8.8
F = (BSS / (k-1)) / (WSS / (n-k)) = (6.5 / 2) / (8.8 / 12) = 4.43
Similarly, the KW statistic is given by summing the ranks of the different groups, so that we get
H = (12/(n(n+1)) sum{i=1->k} Ri
2 / ni) - 3(n+1)
= (12/(15(15+1)) sum{i=1->k} Ri
2 / ni) - 3(15+1)
= 0.05 (572/5 + 232/5 + 402/5) - 48
= 5.78
Thus, comparing F with F.05,2,12
= 3.89, the critical value leads us to reject the null hypothesis.
Comparing H with the chi-squared2 = 5.991, so that we fail to reject the null hypothesis.
Sometimes the lack of power associated with a nonparametric test prevents us from rejecting the
null hypothesis.