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I have two arrays of data and I wish to test whether or not they are from some distribution with the same variance (whether or not they have the same mean is not important). I don't know what distribution the data are from and certainly have no particular reason to assume it is normal. I was thinking of using a permutation test.

My question is; can I simply re-sample without replacement to create a distribution of difference in variance and then compare my true difference to this distribution? This seems almost too intuitive. Are there any other conditions that apply.

I think this question is probably related to Permutation tests: criteria to choose a test statistic .

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While you could certainly use a statistic of the difference in variance to test equality of variance, some people might feel it would make more sense to have a statistic of the ratio of variance (where equality implies a ratio of 1).

However, this resampling scheme (randomly dividing the whole of the data randomly into two groups) relies on an assumption that the means are identical. If that's not the case, you're not really testing for a difference in variance.

[I'm not sure that "being intuitive" is really something that you can have too much of, but yes, permutation tests tend to be quite intuitive in simple cases. ]

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  • $\begingroup$ Re your second paragraph: probably one can explicitly center both groups separately & then do the permutation test as described? $\endgroup$ – amoeba Nov 4 '15 at 22:12
  • $\begingroup$ @amoeba I think the test will no longer be distribution free. It should still perform fairly well though. $\endgroup$ – Glen_b Nov 4 '15 at 22:16
  • $\begingroup$ Thanks, I think in reality assuming equal means is reasonable. It would be satisfying, if unnecessary pragmatically speaking, to relax it. $\endgroup$ – Bowler Nov 5 '15 at 13:41

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