Question

As part of an ANOVA that compares three treatments, you carry out Tukey

pairwise tests at the overall 5% significance level. The Tukey tests find that p11
is significantly different from uz but that the other two comparisons are not
significant. You can be 95% confident that
(a) H17 H2 and p1 = P3 and 42 = P13.
(b) just pi + H2; there is not enough evidence to draw conclusions about the
other pairs of means.
(C) H1 = uz and u2 = 73, and this implies that it must also be true that p1 = u2.
Which one?

Answer 1

In an ANOVA with Tukey pairwise tests, if only one pair of means (μ1 and μ2) is significantly different, we can only conclude that those two means differ; we cannot make any conclusions about other pairwise comparisons without sufficient evidence.

Step-by-step explanation:

When conducting a Tukey pairwise test as part of an ANOVA to compare three treatments, finding that μ1 is significantly different from μ2 but no significant difference is observed for the other pairs at the 5% significance level suggests that you cannot confidently conclude that all means are equal. However, you can be 95% confident that μ1 is different from μ2. Since there is not enough evidence to draw conclusions about the other pairs (μ1 with μ3 and μ2 with μ3), it would be incorrect to assume they are equal. Thus, the correct answer is (b) just μ1 ≠ μ2; there is not enough evidence to draw conclusions about the other pairs of means.