Hi, http://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test
We can clearly see that null hypothesis is median different or not. One way of proving non difference is P(X>Y) = P(X<Y) where X and Y are ordered. On 9/20/12, peter dalgaard <pda...@gmail.com> wrote: > > On Sep 20, 2012, at 02:43 , Thomas Lumley wrote: > >> On Thu, Sep 20, 2012 at 5:46 AM, Mohamed Radhouane Aniba >> <arad...@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. >>> >>> The wilcoxon test is giving me a pvalue of 5.139217e-303 of a over b with >>> the alternative "greater". Now if I make a summary on each of them I have >>> the following >>> >>>> summary(a) >>> Min. 1st Qu. Median Mean 3rd Qu. Max. >>> 0.0000000 0.0001411 0.0002381 0.0002671 0.0003623 0.0012910 >>>> summary(c) >>> Min. 1st Qu. Median Mean 3rd Qu. Max. >>> 0.0000000 0.0000000 0.0000000 0.0004947 0.0002972 1.0000000 >>> >>> The mean ratio is then around 0.5399031 which naively goes in opposite >>> direction of the wilcoxon test ( I was expecting to find a ratio >> 1) >>> >> >> 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 > > I was sure I had seen a definition where X was "larger than" Y if P(X>Y) > > P(Y<X), but that's obviously not the normal definition. Anyways, it is worth > emphasizing that that is what the Wilcoxon test tests for, not whether the > means differ, nor whether the medians do. As a counterexample of the latter, > try > > x <- rep(0:1, c(60,40)) > y <- rep(0:1, c(80,20)) > wilcox.test(x,y) > median(x) > median(y) > > (and the "location shift" reference in wilcox.test output is a bit of a red > herring.) > > -- > Peter Dalgaard, Professor, > Center for Statistics, Copenhagen Business School > Solbjerg Plads 3, 2000 Frederiksberg, Denmark > Phone: (+45)38153501 > Email: pd....@cbs.dk Priv: pda...@gmail.com > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Avinash Barnwal, M.Sc. Statistics and Informatics Department of Mathematics IIT Kharagpur ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.