I would like to examine the equality of the 25th and 75th percentile between two groups.
One way would be to apply Mood's test adapted to the 25th percentile (and then to the 75th):
Construct the 2x2 contingency table counting the number of instances above versus below the 25th percentile for each of the two groups (similarly a contingency table for the 75th percentile)
Then compute the chi-square test-statistic as the sum of square differences between observed versus expected frequencies.
Then accept or reject the null hypothesis of homogeneity if the chi-square test-statistic falls beyond the critical (e.g. 95%) chi-square value.
Another way would be to:
- Split the data in the middle (by the median value).
- For each of the two resulting datasets apply Mood's test as explained above but testing for equality of the median value (the median of the median would become the 25th and 75th percentile).
My questions are:
- Do you think these two approaches are equivalent?
- Is any of these approaches violating any of the chi-square-test assumptions ?