# Urgent-Statistical Test comparing means of unknown distributions

#### moudatsos

##### New Member
Is there a non-parametric test with which one can reject the null hypothesis of an equal mean (μ1 = μ2) between two sample distributions, over an alternative μ1 < μ2 or μ1!= μ2 or μ1 > μ2, without assuming anything about the sample distributions and their variances?

If there's no answer, is it at least true that for a very large sample size the normality assumption is relaxed for the t-test? My question lies in the fact that I ve seen there's a two sample distribution t-test which assumes unequal variances and can perform a tail test to reject the null hypothesis (μ1 = μ2) over one of the pre-mentioned alternative hypotheses, only that it assumes the two sample distributions follow the Normal distribution.

If the normality assumption can be ommited, for large samples I am done. There's one point however. I've read somewhere that this is true only if sample sizes do not differ significantly. Well my samples differ: I got two such pairs. The first wo samples have sizes 209826 and 11588 (which unfortunately differ by an order of magnitude) and I also have two samples made of averages of previous samples, which, sadly and contrary to the central limit theorem, fail to pass normality tests and have sizes 14958 and 1070.

Finally if there's nothing concerning the above, is there a test that proves that one distribution has more lower values than another distribution, which has more higher values?
Thank you all in advance! -If there's anyone out there...-

Michael

#### TheAnalysisFactor

##### New Member
Is there a non-parametric test with which one can reject the null hypothesis of an equal mean (μ1 = μ2) between two sample distributions, over an alternative μ1 < μ2 or μ1!= μ2 or μ1 > μ2, without assuming anything about the sample distributions and their variances?
There is a nonparametric equivalent to a t-test, but it doesn't test the means, since those are parameters.

If there's no answer, is it at least true that for a very large sample size the normality assumption is relaxed for the t-test? My question lies in the fact that I ve seen there's a two sample distribution t-test which assumes unequal variances and can perform a tail test to reject the null hypothesis (μ1 = μ2) over one of the pre-mentioned alternative hypotheses, only that it assumes the two sample distributions follow the Normal distribution.
Yes, true. I can't remember what it's called--SPSS does it automatically in their independent samples t-test.

If the normality assumption can be ommited, for large samples I am done. There's one point however. I've read somewhere that this is true only if sample sizes do not differ significantly. Well my samples differ: I got two such pairs. The first wo samples have sizes 209826 and 11588 (which unfortunately differ by an order of magnitude) and I also have two samples made of averages of previous samples, which, sadly and contrary to the central limit theorem, fail to pass normality tests and have sizes 14958 and 1070.
The normality assumption can't be omitted, but it can be relaxed. Meaning, your distribution can be somewhat skewed if it's still unimodal, but you can't just use any distribution, such as a bimodal or if half your observations are 0 or something.

It is also true that unequal sample sizes mess up that relaxation--yours are pretty different. I don't know how far is too far apart--I think you need JohnM to answer this one. It would also depend on how non-normal your data are. Slight skew or kurtosis? You're probably fine.

Finally if there's nothing concerning the above, is there a test that proves that one distribution has more lower values than another distribution, which has more higher values?
Yup. And this would be your safest bet. Mann Whitney U and Wilcoxan Rank Sum are equavalent. Whichever your software has will work.

Thank you all in advance! -If there's anyone out there...-

Michael
Karen