I have checked with various sources, but I'm still puzzled: 1) What are the exact assumptions of Mood's Median Test? 2) Is homogeneity of variance needed for both groups or is the same shape criterion sufficient?

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    $\begingroup$ Median test is a particular form of Pearson z-test for proportions. It tests H0 that the groups do not differ in respect to the proportion of cases lying above (or below) the value which is the overall median (or another quantile) of the combined sample. stats.stackexchange.com/a/16226/3277. So, the asymptotic test has no assumptions besides usual Chi-square test assumptions (such as >=5 expected count in a cell). The exact version of the test has no assumptions. $\endgroup$ – ttnphns Aug 8 '14 at 8:56
  • $\begingroup$ See also the relevant Wikipedia entry $\endgroup$ – Glen_b -Reinstate Monica Aug 9 '14 at 11:36
  • $\begingroup$ @ttnphns - How come I have seen in many sources that an assumption is that the samples come from population distributions that have the same shape? This is not mentioned at all in the wikipedia article. $\endgroup$ – Terrence J Dec 16 '16 at 0:03
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    $\begingroup$ @Terrance J probably because those are the assumptions required to draw inference regarding equality of medians from ranksum tests. Mood's median test and ranksum tests of median equivalence are often discussed together. $\endgroup$ – Lucilled Feb 2 '17 at 0:01

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