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Bootstrap statistical testing is a way to compare two populations. I asked a question before on whether bootstrap distributions are always Gaussian or not. The answer was that no, they are not always Gaussian.

The bootstrap statistical test involves computing the bootstrap distribution for two populations, then applying Student's t-test to determine whether they are equivalent or not. However Student's t-test assumes that they both come from a Gaussian.

How does that make sense? If the bootstrap distribution can be non-gaussian, is the Bootstrap t-test flawed?

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    Where do these statements come from about "the bootstrap statistical test"? – Nick Cox Mar 21 '16 at 09:57
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    stat.ucla has a nice example of a one-sample bootstrap test of a mean when the sampling distribution of the mean is not normal. See http://www.stat.ucla.edu/~rgould/110as02/bshypothesis.pdf – Robert Alan Greevy Jr PhD Mar 21 '16 at 10:40
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    Continuing from the one-sample case linked above, apply this to the two-sample case. Adjust the data so Ho is true and bootstrap. The test statistic is no longer the mean, but the unequal variance t-test statistic. The bootstrapping is done to get the distribution of the t-test statistic under Ho. I think your confusion comes from the bootstrap test using the t-test statistic but not the t-test distribution under Ho and normally distributed data. – Robert Alan Greevy Jr PhD Mar 21 '16 at 10:48
  • I understand that bootstrap distributions do not have to be normal. But, the whole point of the non-parametric bootstrap test to have zero distribution assumptions. –  Mar 21 '16 at 11:25
  • Thanks Robert. Unequal variance = unpooled t-test. So we need to apply Welches test on the bootstrap distributions. –  Mar 21 '16 at 11:32

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