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?
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:
ANOVA:
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 $\frac{\text{SSB}/(g-1)}{\text{SSE}/(n-g)}$
Kruskal-Wallis:
The Kruskal-Wallis statistic is
$$K = (n-1)\frac{\sum_{i=1}^g n_i(\bar{r}_{i\cdot} - \bar{r})^2}{\sum_{i=1}^g\sum_{j=1}^{n_i}(r_{ij} - \bar{r})^2} = \frac{\text{SSB}/(g-1)}{SST/(n-1)}$$
$K=\frac{(n-1)}{(g-1)}\frac{\text{SSB}}{SSE+SSB}=\frac{(n-1)}{(g-1)}\frac{1}{1+\frac{\text{SSE}}{\text{SSB}}}$
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)
