I have an ordinal variable which assumes only 3 values. I need to do a comparison between 2 treatments.

I used Mann-Whitney test, since is an ordinal. My "p-value" is quite high (0.9388) which is strange because the values distribution is quite different for each group.

Then I perform a Fisher a chi-squared test and my "p-value" decreased to 0.0885, which I believe, reflects a little bit better the behavior of the sample.

What should I do? Stick to Mann-Whitney, go to chi-squared or do another type of evaluation?

Thank you very much for your answer.

You are right, there is not a such thing as a Fisher chi-squared. It was a mistype error. In fact I perform both test (since one of the cells has the frequency of 2) and I had that in my mind. I apologize for the mistake.

Thank you very much for your advice of using the proportional odds ordinal logistic model's likelihood ratio chi-squared. I will follow that path.

Can I ask you one more thing? I also have to analyse the same variable but over time: baseline vs. follow-up. Which test can I apply? I believe that the Wilcoxon Signed Rank is not the most appropriate.


1 Answer 1


I don't know of a Fisher $\chi^2$ test for a contingency table. I assume you mean either Pearson's test or Fisher's "exact" test (the former being preferred as it is more accurate - Fisher's "exact" test is conservative). But to your point, with only three values the excessive number of ties present makes the $P$-value from the Wilcoxon-Mann-Whitney two-sample test questionable. I recommend that you use the proportional odds ordinal logistic model's likelihood ratio $\chi^2$ test, which handles arbitrarily heavy ties well. The proportional odds model is a generalization of the Wilcoxon/Kruskal-Wallis test.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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