# Inconsistent information about Mood's median test assumptions

I'm a bit confused as to what assumptions my data needs to satisfy in order to use Mood's median test. Some resources state that you need the samples to come from population distributions with the same shape, while some sources, including wikipedia and the comments to this question:

seem not to mention any required assumptions at all. So my question is mainly this: What assumptions does Mood's Median test require (it would be great if some reliable references can be provided)? Does it require that the samples come from population distributions with the same shape?

If I have two sets of samples and I want to compare their group medians on some discrete scores, I can observe the sample distribution with a plot, but how does one check whether the population distributions of each sample have the same shape?

If the population distributions of each sample don't have the same shape and this is indeed required to use Mood's median test, then what statistical test is available to compare medians or means?

Any advice or help would be greatly appreciated!

The "shape" assumption is rather vague. However, if two different samples have the same median, then there is no implication that Mood's median test will be unbiased. But you can guarantee that it will be if the distributions are the same. I believe that when you see references to this shape assumption, that's really all it's saying.

One can heuristically see the effect of the distribution through a simple example: Let $X$ take values $-2$, $0$ and $1$ with probabilities $1-r$, $2r$ and $1-r$. Additionally, let $Y$ take values $-1$, $0$ and $2$ with probabilities $1-r$, $2r$ and $1-r$. The median for both distributions is zero. Now, assume that $r$ is very close to $0$ and that $n$ independent samples have been taken from both distributions. To make each variable continuous, you can assume that a small amount of white noise has been added to each.

The statistic for this test is $T = \sum_{i=1}^n I(X_i > Med([\boldsymbol X, \boldsymbol Y]))$ and it should have a Binomial(n, 1/2) distribution if X and Y are identically distributed. But we know there's a 50% chance that more than half of the $\boldsymbol Y$ sample will be comprised of 2's. Even if less than 50% of $\boldsymbol Y$ are 2's, then there's still a 50% chance that more than half of the $\boldsymbol X$ sample will be comprised of -2's. In either situation, there's a negligible probability that $T$ will exceed $n/2$. In other words, there's a 75% chance that $T$ can't exceed it's expected median. Thus, Mood's statistic can't follow a Binomial(n, 1/2) distribution for the given samples. This would lead you to reject the null hypothesis more often than whatever significance threshold you set.

I'm not aware of a nonparametric test for medians besides Mood's test. But if all you're interested in is a robust test of central tendency, then you could use the Mann-Whitney U test. That test skirts the shape assumption by testing for a different measure of centrality. You can think of it as a generalized median test where it actually tests for all quantiles but in weighted manner.

• The Mood test has such low power that I never recommend it. Dec 19 '16 at 19:07
• @jjet - Thank you for your answer. I'm a bit confused about your opening comment: "...if two different samples have the same median, then there is no implication that Mood's median test will be unbiased." So would it be correct to say that if the sample's have different medians, then Mood's test is unbiased? Dec 20 '16 at 2:25
• Maybe I should've clarified what I meant by that. A statistical test is unbiased if it satisfies two conditions: The probability of rejecting $H_0$ given $H_0$ is never greater than $\alpha$ and the probability of rejecting $H_0$ given $H_1$ is never less than $\alpha$. If the "shape" assumption is Mood's test is violated, then you can concoct distributions which would violate either assumption. In other words, if you assume that the distributions of the samples have different medians, then there's no guarantee that the probability of rejecting $H_0$ is greater than $\alpha$.
– jjet
Dec 20 '16 at 2:54