# Is K-W test statistics a U statistics?

The K-W H test statistic is given by:

$$H=(N-1) \frac{\sum_{i=1}^{g} n_{i}\left(\bar{r}_{i \cdot}-\bar{r}\right)^{2}}{\sum_{i=1}^{g} \sum_{j=1}^{n_{i}}\left(r_{i j}-\bar{r}\right)^{2}}, \text { where: }$$ If the data contain no ties $$H=\frac{12}{N(N+1)} \sum_{i=1}^{g} n_{i}\left(\bar{r}_{i}-\frac{N+1}{2}\right)^{2}$$$$=\frac{12}{N(N+1)} \sum_{i=1}^{g} n_{i} \bar{r}_{i .}^{2}-3(N+1)$$

It extends the Mann–Whitney U test (also called Wilcoxon rank sum test) , which is used for comparing only two groups. I learned that Mann–Whitney U statistic is U statistic. I want to know is K-W test statistics a U statistics?

The Kruskal–Wallis test is also called one-way ANOVA on ranks. Is one-way ANOVA (not on rank) a U statistics?

• Neither neither of them seem to be U-statistics if there are more than 2 groups. – John L Mar 22 at 15:01
• See Wikipedia on K-W tests. Often the test statistic $H$ has approximately a chi-squared distribution. – BruceET Mar 22 at 19:11
• @BruceET Yes，it is also why I think that's U statistics. – Happy Superman Mar 23 at 1:19