While reading this link: http://www.statsdirect.com/help/Default.htm#nonparametric_methods/kruskal_wallis.htm, I came across the following:

"An alternative to Kruskal-Wallis is to perform a one way ANOVA on the ranks of the observations."

What precisely is a one-way ANOVA on the ranks in this context? I thought that is the K-W? If not, how does it differ from K-W exactly?

  • $\begingroup$ stats.stackexchange.com/q/115519/3277 or probably elswhere: this question has been already asked here. $\endgroup$ – ttnphns Nov 5 '14 at 10:29
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    $\begingroup$ It's almost the same as one-way ANOVA -- they no doubt mean to do it using an F-test exactly as in ANOVA; if instead you work out the permutation distribution of the statistic under the null, you should get an equivalent test to the exact K-W. $\endgroup$ – Glen_b -Reinstate Monica Nov 5 '14 at 11:11

I believe that the intent of 'one way ANOVA on the ranks' in that quote is to do an actual F-test just as in ANOVA; if instead you work out the permutation distribution of the statistic under the null, you'll get an equivalent test to the Kruskal-Wallis.

The F-statistic based on the ranks and the Kruskal-Wallis statistic are monotonically related:


Let $\text{SSB}=\sum_{i=1}^g n_i(r_{i.}-\bar{r}_{..})^2$

$\quad\:\: \text{SSE} = \sum_{i=1}^{g}\sum_{j=1}^{n_i} (r_{ij}-\bar{r}_{i\cdot})^2$

$\quad\:\: \text{SST} = \sum_{i=1}^g\sum_{j=1}^{n_i}(r_{ij} - \bar{r})^2=\text{SSE}+\text{SSB}$

The usual ANOVA F statistic is $\frac{\text{SSB}/(g-1)}{\text{SSE}/(n-g)}$


The Kruskal-Wallis statistic is

$$K = (n-1)\frac{\sum_{i=1}^g n_i(\bar{r}_{i\cdot} - \bar{r})^2}{\sum_{i=1}^g\sum_{j=1}^{n_i}(r_{ij} - \bar{r})^2} = \frac{\text{SSB}/(g-1)}{SST/(n-1)}$$


We can easily see that the reciprocals of the two statistics are linear functions of $\frac{\text{SSE}}{\text{SSB}}$, so the two statistics themselves are monotonically related.

As a result, if we used tests with exactly the same type I error rate with either statistic, it would be an equivalent test.

In practice the difference is we don't usually use the exact distribution of either statistic for either test; except for very small samples, the Kruskal Wallis statistic is typically compared with a chi-square distribution, and one-way ANOVA on the ranks would be compared with an F. Both are approximate and because of those approximations, they don't quite reject exactly the same cases* because they're not conducted at identical significance levels.

* (but they nearly do)


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