2
$\begingroup$

I randomized samples into two experiments and I ended up with about 190 samples in one group and 80 in the other. The distributions of the two samples are not normal and I want to use Wilcoxon test to see if the treatments are identical. Should I pick equal samples from both groups or is this requirement waived off for non-parametric tests?

$\endgroup$
5
  • 3
    $\begingroup$ It's not even a requirement for parametric tests. There are situations where tests are more robust to violation of assumptions when the groups are equal sizes. $\endgroup$ Jun 5, 2015 at 20:54
  • $\begingroup$ @Scortchi - "it's not even a requirement for parametric tests" - could you elaborate more on this? $\endgroup$
    – rk567
    Jun 5, 2015 at 21:04
  • 1
    $\begingroup$ Did all 190 in one group receive one treatment while the other group of 80 receive the other treatment? Or did each group get divided in half, with a control and a treatment group in each group? If the latter, were the treatments the same in each group? $\endgroup$
    – EdM
    Jun 5, 2015 at 21:06
  • 1
    $\begingroup$ @rk567: I can give an example: the two sample t-test. $\endgroup$ Jun 5, 2015 at 21:15
  • $\begingroup$ @EdM yes, of all the 270 samples that came, each sample was randomly assigned to one of the two treatments and the result we have is 190 of them belonging to trt 1 (for the outcome I am interested in measuring) and the other 80 belonging to trt 2. $\endgroup$
    – rk567
    Jun 5, 2015 at 21:28

1 Answer 1

2
$\begingroup$

It depends on which Wilcoxon test you are refering to. The Wilcoxon signed rank test is for paired data. From what is sounds like you do not have paired data so this would not be the appropriate test. The Wilcoxon rank-sum test is for two independent samples and does not require equal sample size for each group. However, there are assumptions that come along with this test (as any non parametric test does-- just less than the parametric version) so I would suggest looking those up beforehand.

$\endgroup$

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.