Question

A one-factor fixed-effects ANOVA is performed on data from three groups of equal size (n=10), and the null hypothesis is rejected at the .01 level. The following values were computed: MSwithin=40 and the sample means are ȳ1=4.5, ȳ2=12.5, ȳ3=13.0. Use the Tukey HSD method to test all possible pairwise contrasts.

Answer 1

Tukey HSD is calculated using the below formula

where   is a critical value of the studentized range for , the number of treatments or samples r, and the within-groups degrees of freedom . We get this value from studentized range table.

is the within groups mean square from the ANOVA table and n is the sample size for each treatment.

Given,

n = 10, r = 3

= 40 , = number of observations - number of groups = 10 * 3 - 3 = 27

= 0.01

From studentized range table, = 4.495

So,

So, if the absolute mean difference between groups is greater than 8.99, the mean difference is significant at 0.01 significance level.

ȳ2 -  ȳ1 = 12.5 - 4.5 = 8

ȳ3 -  ȳ2 = 13.0 - 12.5 = 0.5

ȳ3 -  ȳ1 = 13.0 - 4.5 = 8.5

Since none of the groups have absolute mean difference greater than 8.99, there are no significant differences in any of the groups.