Usually it is equal to "n-1". These methods compare only one parameter of 2 samples - median. You can also compare other parameters like distribution function, homoscedasticity, etc.

Do not forget on testing of normality (normal distribution of both samples).

In Wilcoxon rank test you just calculate the direction of the difference (d) between matched pairs 1 snd 2 and you put + and -. The smaller sum of + or - is your answer. You just give N of pairs no other df is needed. Then you can look up the p in a Table or on the net. Small samples are N less than 25 pairs.

here are web calculators:

Wilcoxon Signed-Rank Test Online Calculator -EasyMedStat

In Mann-Whitney you give the value of U. The size of the U determines if something is sign or not.

Mann-Whitney U Test Calculator - Statology

Roman Vavrek is incorrect in stating that the Wilcoxon Mann-Whitney calculates or compares medians. It calculates the probability that an observation from one sample will be greater than an observation from the other. Indeed, the title of Mann and Whitney's paper describes the test perfectly :

Mann, H. B. & Whitney, D. R. On a Test of Whether one of Two Random Variables is Stochastically Larger than the Other. Ann. Math. Statist. 18, 50–60 (1947).

It does not assume anything about the distribution of the data.

2 parameters are used n, which corresponds to the size of group 1 and m which is the group of size 2.

Hello all,

The M-W-W (Mann-Whitney-Wilcoxon) test doesn't have degrees of freedom associated with a sampling distribution, as is the case for tests based on distributions such as the t, the F, or chi-square. The appropriate determination of critical values is via exact enumeration of all possible assortments of score differences. As this was computationally intensive, back in the day (e.g., 1947, when M&W published their explication), exact critical values are usually furnished in texts only for sample sizes up to about 20 per group. The "large-sample approximation," via an estimated z-statistic, is typically used in statistical programs, as it is easier to implement from a programming perspective. This is consistent with S. Béatrice Marianne Ewalds-Kvist 's response.

As Ronán Michael Conroy correctly points out, the test is of equality of distributions. However, if and when one is willing to tack on an additional assumption, symmetry of the respective distributions, the M-W-W becomes a test of equality of medians.

Best regards

Sorry, I cannot agree with you, see: https://statistics.laerd.com/spss-tutorials/mann-whitney-u-test-using-spss-statistics.php

Roman Vavrek the Mann whitney is a test of equality of medians only under some unlikely assumptions of identical distribution. You are not disagreeing with me, however, you are disagreeing with Drs Mann and Whitney. I'd take it up with them. Have a read of their original paper.

It's a really common myth, though, and I'm not surprised you encountered it – you find it all over the internet. If you want to test equality of medians, then quantile regression is there to do exactly that. Why use a test for stochastic superiority and hope for the best when you can actually use a method designed to do the job?