[R-sig-ME] Welch's ANOVA

[Paul Miller]

Hi Rafael,

Not sure if what I just prposed  will actually work or not. I searched online and found some places suggesting that the Kruskal-Wallis test doesn't require an assumption of equal variances. As an equivalent test, this would seem to suggest that an ANOVA on ranks also would not require this assumption either.

The Kruskal-Wallis test is also an extension of the Wilcoxon Rank-Sum test though. I've found information suggesting that this test assumes that distribution have the same dispersion and that this is analogous to the assumption of variance homogeneity in the two-sample t-test.

So now I'm confused.

Perhaps a more expert statistician than myself could lend some insight into this matter.

Thanks,

Paul