# Determine which mean is smaller over two non-normal distributions

Let's say I have a non-normal distribution A and another non-normal distribution B, the mean and std deviations of each distribution are different.

I then randomly sample 100 values from A, SampleA, and 50 values from B, SampleB.

Given only SampleA and SampleB, what is the equation to determine the probability that the mean of A is less than the mean of B.

I'm not a statistician, if this problem is underspecified leave a comment and I'll update it with any relevant information.

I've looked into t-tests, but things I've read have made it sound like it is inapplicable to non-normal distributions. I'm also unsure why I care about things like the null hypothesis and 95% confidence intervals when I'm only concerned with the specific probability that one distribution has a lower mean than the other.

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You want the Mann-Whitney U test: en.wikipedia.org/wiki/Mann%E2%80%93Whitney_U –  Michael Lugo Jul 16 '13 at 23:02
I would do a quick sampling to estimate the probability. Do you need theoretical values? –  Memming Jul 16 '13 at 23:09

## 1 Answer

Why not resample?

Without getting into the difficulties of what you describe as "the probability that the mean of A is less than the mean of B," suppose you take 10,000 resamples (with replacement) of size 100 from sampleA and size 50 from sampleB. Now construct the distribution of the differences of their sample means.

As an example, suppose 97% of the differences have the sample mean from sampleA being greater than the sample mean from sampleB. What could you conclude? How confident are you of this conclusion?

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