A Fisher LSD test is basically a series of pairwise t-test except that the pooled variance is computed on all groups (and therefore does not vary from one comparison to another) while the normal t-tests calculate the pooled variance on only the two groups considered in the comparison.

On wikipedia, I read

In each t-test, a pooled standard deviation is computed from only the two groups being compared, while the Fisher's LSD test computes the pooled standard deviation from all groups - thus increasing power.

I failed to have an intuition why Fisher's LSD would have more power than a series of t.tests. Why does it?

Also, what should be considered when having to chose between pairwise t.test and Fisher LSD test after an ANOVA?

Note by the way, that for my particular case of interest, I am considering bootstrapping (getting the F-values for the ANOVA and the t-values for the t-tests) to calculate the p-values.


1 Answer 1


The LSD test has more power because the df denominator are greater. That's the gain you get by assuming homogeneity of variance. In general, the Fisher LSD test does not control the Type I error rate. Better would be the Tukey hsd. The Tukey hsd controls the Type I error rate even if you don't do an ANOVA so I would skip the ANOVA.


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