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josuemarquezz27
01.12.2021 •
Mathematics
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?
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Ответ:
(15/2) / (3/4) =
15/2 * 4/3 =
60/6 = 10 pieces