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Post hoc tests, like Bonferroni, allow us to detect which of the groups is the different one after ANOVA.

To use it, we calculate effectively what is like the independent t-test for each of the pairwise groups. This value is then compared to the $t$ critical value.

The $t$ critical value is found using Degrees of Freedom (DoF) $= N-k$ where $k$ is number of groups.

My question is why do we use $N-k$ when comparing to a $t$ value made up from 2 groups only. Why isn't DoF equal to total sum of items in the 2 groups, minus 2 like in independent $t$ test?

John Hon
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1 Answers1

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In your ANOVA I suppose you were making the assumption that all $k$ groups have the same population variance $\sigma^2.$ MS(Resid) is an estimate of that variance with $\mathrm{DF} = N-k.$ Appropriate ad hoc comparisons would use that estimate.

Because the Bonferroni method of guarding against false discovery is based on an inequality, it is sometimes too conservative (requiring P-values to be smaller than really necessary in order to declare ad hoc differences). In this situation, you might do better to use Tukey's HSD method to make all comparisons within a fixed error probability.

By contrast, if you are using the Welch ANOVA (implemented in the R procedure oneway.test), which does not assume equal variances, you might do Welch t tests for ad hoc comparisons along with Bonferroni. In that case each Welch t test would use a variance estimate derived from the two groups being compared. Then the (smaller) DF would be decided by Welch's two-sample formula.

BruceET
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