I have a numerical response variable A which depends on a categorical explanatory variable B. I also have another variable C that I'd like to check for confounding effects. So far I've been using ANOVA, but I've realised my response variable A is not normally distributed, so I need a non-parametric test. Thus I thought about Kruskal-Wallis, but I am not aware it's possible to perform such test (I'm an R user). Do you know if there's any equivalent to ANOVA but non-parametric for such task?

  • $\begingroup$ "Non normal response" does not necessarily imply you need a nonparametric test. Firstly, it's not the marginal distribution of the response that has a distributional assumption, but the conditional distribution. Secondly, there are other parametric assumptions than normal for which ANOVA-like models can be used. What is A measuring? $\endgroup$
    – Glen_b
    Commented Nov 18, 2015 at 15:47

2 Answers 2


First, the response variable does not have to be normally distributed to use ANOVA. The errors (as estimated by the residuals) do.

Second, if you want a method that does not require the assumption that the residuals are normally distributed, you can use robust regression or quantile regression.


The generalization of the Wilcoxon and Kruskal-Wallis tests is the proportional odds ordinal logistic model, which allows for covariate adjustment.

  • $\begingroup$ Frank, do by generalization you mean that the ordinal regression is the direct extension of KW and KW is a particular case of it? If yes, can you show it, please? - the demonstration that both analyses produce, in cases when KW is used, the same test statistic value or the same p-value? $\endgroup$
    – ttnphns
    Commented Nov 18, 2015 at 14:02
  • $\begingroup$ Yes; yes. This has been shown by either Agresti or Whitehead. The numerator of score statistic for the global test that all regression coefficients (other than the intercepts) are zero is exactly the Wilcoxon/K-W test statistic in $U$-statistic form. In non $U$-statistic form (e.g., $z$-statistic) the different approaches will note give exactly the same values, and the $P$-values are not identical, but they are all very close. In some ways the proportional odds model gives $\endgroup$ Commented Nov 18, 2015 at 17:22
  • $\begingroup$ ... more accurate $P$-values by better taking heavy ties into account. $\endgroup$ Commented Nov 18, 2015 at 17:52

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