*Bounty: 100*

*Bounty: 100*

Dunn test for two groups is equivalent to Kruskal-Wallis. The null is the absence of stochastic dominance, but if we impose some more restrictions we can assume that the null is the equality of medians, $H_0: M1 = M2$.

Apparently, the Dunn test statistic is assumed Normal, but it is computed in rank terms, not in terms of the original observations. Therefore, we can obtain p-value for $H_0$, but no explicit CI for $(M1 – M2)$.

1) Is it ok to assume that $(M1 – M2)$ is also Normal? If yes, we can use the point estimate and p-value to derive the s.e. and construct a confidence interval.

2) This question is not related to 1). I would like to get p-value for $H_0 : M1/M2 = 1$. Let’s assume no ties. Then, using the fact that the Dunn p-value is invariant to a monotone transformation of the original response, $Y$, and that $log(median(Y)) = median(log(Y))$, it looks like the p-value for $M1/M2$ is exactly the same as for $M1 – M2$. Did I get that right?