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I've read a few sources that describe ANOVA on ranks, and I am still confused on how the calculation is done, and why.

Specifically:

  • Is ANOVA on ranks just applying a rank transformation to the data (combined across all groups) and then doing regular ANOVA on the transformed data? If not, how is it different from that?

  • Why is it valid to assume that the ranks are normally distributed? Wouldn't ranks be uniformly distributed under the null (no effect, no interactions) hypothesis?

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    $\begingroup$ Of course ranks aren't normally distributed: barring any ties, within any given group they act like a sample without replacement from a discrete uniform distribution. But AFAIK no aspect of ANOVA requires this assumption of normality--it only requires that its test statistics are reasonably well approximated by various distributions (such as the F-ratio or chi-squared distributions). $\endgroup$ – whuber Oct 2 '17 at 21:24
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    $\begingroup$ See the answer to this question: What is a one-way ANOVA on the ranks of the observations discussing the equivalence of the test statistics for a Kruskal-Wallis test and a one-way ANOVA performed on the ranks - really, the difference between the two is in the tables you look up to approximate the distribution of the test statistic under the null, as described there. The statistics themselves are monotonically related; indeed some people call the Kruskal-Wallis one-way ANOVA on the ranks $\endgroup$ – Glen_b -Reinstate Monica Oct 2 '17 at 21:38
  • $\begingroup$ @whuber Thanks for confirming the ranks are sampled from uniform. The wikipedia page says ANOVA assumes normality of the residuals, how does that apply here? $\endgroup$ – Alex I Oct 2 '17 at 23:12
  • $\begingroup$ It's not clear which Wikipedia page you refer to, so I cannot comment on it specifically. I can say that many people focus on the introductory paragraph in a Wikipedia article. Those paragraphs tend to be informal summaries rather than rigorous or completely correct accounts. That might be what you are referring to. Of course it's not a problem in ANOVA to assume normal residuals, provided that's appropriate for your data--but ANOVA can be carried out and rigorously justified in a much broader setting. $\endgroup$ – whuber Oct 3 '17 at 12:57

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