Wilcoxon Test and Mean Ratios
Mohamed Radhouane Aniba <aradwen <at> gmail.com> writes:
Thank you Thomas, So you think a t-test is more adequate to use in this case ? Rad
No, because a t-test makes even stronger parametric assumptions. You were given more specific advice on stackoverflow http://stackoverflow.com/questions/12499687/wilcoxon-test-and-mean-ratios If you want to prove that there is *some* difference between the distributions, you're done. If you want to test for some specific difference, you need to think more about what kind of test you want to do. Permutation tests with various test statistics are a way to approach that. Ben Bolker
On Sep 19, 2012, at 8:43 PM, Thomas Lumley <tlumley <at> uw.edu> wrote:
On Thu, Sep 20, 2012 at 5:46 AM, Mohamed Radhouane Aniba <aradwen <at> gmail.com> wrote:
Hello All, I am writing to ask your opinion on how to interpret this case. I have two
vectors "a" and "b" that I am trying
to compare.
[snip]
There's nothing conceptually strange about the Wilcoxon test showing a difference in the opposite direction to the difference in means. It's probably easiest to think about this in terms of the Mann-Whitney version of the same test, which is based on the proportion of pairs of one observation from each group where the `a' observation is higher. Your 'c' vector has a lot more zeros, so a randomly chosen observation from 'c' is likely to be smaller than one from 'a', but the non-zero observations seem to be larger, so the mean of 'c' is higher. The Wilcoxon test probably isn't very useful in a setting like this, since its results really make sense only under 'stochastic ordering', where the shift is in the same direction across the whole distribution. -thomas -- Thomas Lumley Professor of Biostatistics University of Auckland