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.
Best Answer
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.