On May 5, 2011, at 10:58 , JP wrote: > On 4 May 2011 15:32, peter dalgaard <pda...@gmail.com> wrote: > >> >> On May 4, 2011, at 15:11 , JP wrote: >> >>> Peter thanks for the fantastically simple and understandable explanation... >>> >>> To sum it up... to find the z values of a number of pairwise wilcox >>> tests do the following: >>> >>> # pairwise tests with bonferroni correction >>> x <- pairwise.wilcox.test(a, b, alternative="two.sided", >>> p.adj="bonferroni", exact=F, paired=T) >> >> >> You probably don't want the bonferroni correction there. Rather >> p.adj="none". You generally correct the p values for multiple testing, not >> the test statistics. >> > > Oh, I see thanks... of course since I have 5 groups (samples) and 10 > comparisons I still have to correct when quoting p values... > > >> (My sentiment would be to pick apart the stats:::wilcox.test.default >> function and clone the computation of Z from it, but presumably backtracking >> from the p value is a useful expedient.) >> > > Should this be so onerous for the user [read non-statistician] ?
My main reservation is that if the p value gets very small or close to 1 (for 1-sided tests), then it might not be possible to reconstruct the underlying Z (qnorm(pnorm(Z)) breaks down around Z = -37 and Z=+8). If you get sensible values, there's probably not a problem. > >>> # what is the data structure we got back >>> is.matrix(x$p.value) >>> # p vals >>> x$p.value >>> # z.scores for each >>> z.score <- qnorm(x$p.value / 2) >>> >> >> Hmm, you're not actually getting a signed z out of this, you might want to >> try alternative="greater" and drop the division by 2 inside qnorm(). (If the >> signs come out inverted, I meant "less" not "greater"...) >> > > But I need a two sided test (changing the alternative would change the > hypothesis!)... do I still do this? For the z's, yes. Hmm, the asymmetry of qnorm(pnorm(...)) indicates that you might want to be a bit more careful. > All my z values are negative.... > > Is this correct? No, that's the problem. The sign of Z should depend on whether the signed ranks are predominantly positive or negative. If you do a two-sided p-value and divide by two, then by definition you get something less than .5 and qnorm of that will be negative. So if the 2-sided p is 0.04, the one-sided p will be either 0.02 or 0.98, depending on the alternative and on whether the V statistic is above or below its expectation. Notice that this corresponds to Z statistics of opposite sign: > qnorm(.02) [1] -2.053749 > qnorm(.98) [1] 2.053749 -- Peter Dalgaard 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.
