I have two independent groups with skewed distribution of continuous data. I want to compare the medians of two groups. Should i use Mann-Whitney or Mood's median test?
Hello Harmandeep,
The Mann-Whitney U test is a test of equality of distributions for ordinal data. Only if you assume that the distribution is symmetric does it equate to a test of equality of medians (as well, it assumes homogeneity of variance).
Because Mood's test is based on the fractions of cases in the two groups above the (combined groups) median, it is a direct test of equality of medians. Be aware that, the direct competitors of similarly structured methods (of a 2 x 2 table; group by above/below combined median) would be the chi-square test or Fisher's exact test.
Either way, larger sample sizes will enhance statistical power.
You could also use exact/resampling methods to determine the solution.
Good luck with your work!
- Badges
- Science topic
- Similar topics
- Statistics
- Survey