After fixing the parentheses in your code so it does run, it seems that the difference is that wilcox.test defaults to using a continuity correction and your manual calculation does not. Use wilcox.test(big1, big2, correct=FALSE). -thomas On Tue, 17 Aug 2010, Cedric Laczny wrote:
Hi, I became a little bit confused when working with the Wilcoxon test in R. As far as I understood, there are mainly two versions: 1) wilcox.test{stats}, which is the default and an approximation, especially, when ties are involved 2) wilcox_test{coin}, which does calculate the distribution _exactly_ even, with ties. I have the following scenario: #---BeginCode--- # big example size = 60 big1 = rnorm(size, 0, 1) big2 = rnorm(size, 0.5, 1 g1f = rep(1, size) g2f = rep(2, size) big = c(big1, big2) data_frame = data.frame(big, gr=as.factor(c(g1f, g2f))) wilcox_approx = wilcox.test(big1, big2) wilcox_exact = wilcox_test(big ~ gr, data=data_frame, distribution="exact") #---EndCode--- I found here http://www-stat.stanford.edu/~susan/courses/s141/hononpara.pdf that wilcox.test (at least for the signed rank test) relies on exact (p-)values until N = n1 + n2 = 50. I can reproduce this, when using e.g. size = 15. The p-values then are the same, as I would expect it, having read the info from the link. #---BeginCode--- print(paste("Wilcox approx p-value:", wilcox_approx$p.value), quote=F) print(paste("Wilcox exact p-value:", pvalue(wilcox_exact)), quote=F) #---EndCode--- That said, if I set e.g. size = 60, then the p-values of wilcox.test and wilcox_test differ, as expected. What's puzzling me particularly is the differing results when wanting to calculate the p-value manually, for bigger sample sizes. So, if we get the W-score from wilcox.test: #---BeginCode--- W_big = wilcox.test(big1, big2))$statistic #---EndCode--- and "convert" it to a Z-score, like this: #---BeginCode--- mu_big = (size^2)/2 sd_big = sqrt(size*size*(size + size + 1)/12) N = size + size sd_big_corr = sqrt( (size * size) / (N * (N - 1)) * (N^3 - N) / 12 ) Z_big = (((W_big - mu_big)/sd_big) #---EndCode--- The Z-Score (Z_big) is equal to the statistic of wilcox_test. So far so good. And now comes the main problem. When I follow the documentation correctly, the p-value for a given W-score/- statistic ist calculated using the normal-approximation with the Z-score. However, when I do that, I get a different result than what I would expect. Because I would expect the p-value of wilcox.test to be equal to 2*pnorm(Z_big), which is in fact _not_ equal. Please see: #---BeginCode--- p_value_manual = 2 * pnorm(Z_big) print("--- Resulting pvalues --- ", quote=F) print(paste("Wilcox approx p-value:", wilcox_approx$p.value), quote=F) print(paste("Wilcox exact p-value:", pvalue(wilcox_exact)), quote=F) print(paste("P-value manual:", p_value_manual), quote=F) #---EndCode--- So how is the calculation of the p-value performed in wilcox.test, when the sample sizes are big? Because this might explain why the value differs from that being calculated manually. Best regards, Cedric ______________________________________________ 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.
Thomas Lumley Professor of Biostatistics University of Washington, Seattle ______________________________________________ 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.