3
$\begingroup$

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 ?

Thank you.

$\endgroup$
1
$\begingroup$

Your first approach uses more of the data and so seems preferable. Why not split the table into a 2 by 3 table and test both at once?

$\endgroup$
  • 1
    $\begingroup$ I am concerned that neither test may be valid, because they involve dependent values that are assumed independent. In the first one the two $\chi^2$ statistics are dependent. In the second one, the two subsets are dependent because they are determined by the same median value. I am also suspicious that the $\chi^2$ distribution might be incorrect in the first place. Could you comment on those issues? $\endgroup$ – whuber May 20 '16 at 21:22
  • 1
    $\begingroup$ @whuber Good points, my suggestion of the 3 by 2 table is also guilty of using data dependent cut-offs which may mean the calculated $\chi^2$ may follow the theoretical distribution even more approximately than usual. $\endgroup$ – mdewey May 21 '16 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.