similar to: feature and bug in wilcox.test

I have a question concerning the Wilcoxon signed-rank test, and specifically, which R subroutine I should use for my particular dataset. There are three different commands in R (that I'm aware of) that calculate the Wilcoxon signed-rank test; wilcox.test, wilcox.exact, and wilcoxsign_test. When I run the three commands on the same dataset, I get different p-values. I'm hoping that

Hi, R allows me to run a one sample Wilcoxon test like this: wilcox.test(c(1,3.5,2.1,4,1.5,5), mu=2, exact=TRUE) The function 'wilcoxsign_test' from the package 'coin' should (I suppose) be able to calculate exact p values even if there are ties in the ranks. However, I couldn't find information on how to run a one sample test using 'wilcoxsign_test' like in the

hi, when performing a wilcox.test or a wilcox.exact i get results that looks like this: wilcox.exact(x, mu=.5) Exact Wilcoxon signed rank test data: x V = 207, p-value = 0.0006905 alternative hypothesis: true mu is not equal to 0.5 the way i understand the wilcox.test (or wilcox.exact) the V-value represents the summed up ranks of either the positive or negative differences,

hi, i am looking for a way to correct for ties in the wilcoxon signed rank test -> e.g. wilcox.test(x,mu=.5) one way i have heard of is to look up the p value in a table that has been produced by Buck (1975). obviously i need to know the T-value to do that -> how do i get the T-value from the wilcox.test() function. is there any other (already implemented) way to correct for ties in

Dear all, I have two samples that I need to compare. Running Shapiro Wilk normality test does not allow me to reject the normality assumption neither for the first sample (p-value = 0.8938) not for the second one (p-value = 0.07905). So the t.test seems to be applicable. However, the p-value for the second sample seems to be quite close to 0.05 threshold, so wouldn't it be better to use a

Hi, In the exactRankTest package, I've become aware that you can get anomalous p-values (i.e., above 1) from the wilcox.exact method, as in: > wilcox.exact(c(-0.6,0.8,-0.5)) Exact Wilcoxon signed rank test data: c(-0.6, 0.8, -0.5) V = 3, p-value = 1.25 alternative hypothesis: true mu is not equal to 0 This is disturbing. Has anyone encountered this before, and if so

Hi, When I execute a Wilcox u-test on two variables I receive a warning : 'cannot compute exact p-value with ties' - What are ties? What does this mean for my data? - Is that a problem for significance testing? - is there a way to overcome this problem? I have different threads in this forum but it hard to find what the exact meaning of this warning message is. -- View this message in

Full_Name: David Wooff Version: 1.7.0 OS: i686-pc-linux-gnu Submission from: (NULL) (129.234.4.10) wilcox.test exits with error message when confidence interval required, under some situations. I suspect this occurs when the data contain a zero and for some data lengths only: print(wilcox.test(c(2,1,4,3,6,-5,0),conf.int=T)) fails print(wilcox.test(c(2,1,4,3,6,-5,0,1),conf.int=T)) works

There is a fairly new bug in wilcox.test in R-2.2.1 (stable). It wasn't there when I last taught nonparametrics in fall 2003. Line 86 of wilcox.test.R achieved.alpha<-2*psignrank(trunc(qu),n) It should be achieved.alpha<-2*psignrank(trunc(qu)-1,n) If you don't see why, decode the cookbook instructions p. 56 in Hollander and Wolfe (2nd ed.) or see

Dear R Core, pairwise.wilcox.test does not handle "paired = TRUE" correctly; e.g. set.seed(13) x <- rnorm(20) g <- c(rep(1, 10), rep(2, 10)) wilcox.test(x ~ g)$p.value # 0.075 pairwise.wilcox.test(x, g)$p.value # 0.075, o.k wilcox.test(x ~ g, paired = TRUE)$p.value # 0.105 pairwise.wilcox.test(x, g, paired = TRUE)$p.value # 0.075, wrong The line wilcox.test(xi, xj,

Your second issue seems like a more or less unavoidable floating-point computation issue. The paired test operates by computing differences between corresponding values of x and y. It's not impossible to try to detect "almost-ties" (by testing for differences less than, say, sqrt(.Machine$double.eps)), but it's delicate and somewhat subjective/problem-dependent. Example:

does anybody know why there are the two warnings in the example above? Regards Knut > day_4 [1] 540 1 1 1 1 1 1 300 720 480 > day_1 [1] 438 343 1 475 1 562 500 435 1045 890 > is.vector (day_1) [1] TRUE > is.vector (day_4) [1] TRUE > wilcox.test(day_4 ,day_1,paired=TRUE,alternative="two.sided",exact=TRUE,conf.int=TRUE) Wilcoxon

Does anyone know why wilcox.exact gives W-statistic 6 instead of 12 as indicated below. 12 is the rank sum of group 0 of x, which is the linear statistic computed by wilcox_test. y<-c(1,2,3,4,5) x<-c(1,1,0,0,0) (a) wilcox.exact wilcox.exact(y~x) Exact Wilcoxon rank sum test data: y by x W = 6, p-value = 0.2 alternative hypothesis: true mu is not equal to 0 (b) wilcox_test

Dear all, When we perform a Wilcoxon rank sum test (on two samples with different sizes) we get a test statistic. My question is, as the value of test statistic increases the difference between the distributions of the two samples also increase, right? Thanks in advance, João Fadista [[alternative HTML version deleted]]

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I have a stupid question on how to get the real p-values for wilcox.test and correlation. the minmun can be reached is 2.2E-16 using the R version 2.6.2. I do not think it is the R version causing this but other issues. Any help is highly appreciated. Charles -- View this message in context: http://www.nabble.com/minmun-p-value-for-wilcox.test-and-correlation-tp23637568p23637568.html Sent

Dear R friends, I want to compare two datasets and I get the message Error in .Call("cpermdist2", ma = as.integer(m), mb = as.integer(col), : negative length vectors are not allowed after specifying the exact test. I'm using the exactRankTests package. Do you suggest me using the coin library, or is there anything "wrong" with my data? Kind regards, David --

I have been examining the Mann-Whitney test closely. And there are two features of the R implementation that puzzles me. The test statistic is reported as "W" and depends on the order of the arguments to the function. > x <- c(1,3,5,7,9) > y <- x-1 > x [1] 1 3 5 7 9 > y [1] 0 2 4 6 8 > wilcox.test(x, y)$statistic W 15 > wilcox.test(y,x)$statistic W 10

Hello, I have used Wilcox test to find the p-value for hgua95 spiken data. It did not give the required result, and instead gave me warnings like the values are duplicated. Kindly suggest me how to overcome this problem as I am new to R. Thank you Deepthi BM

Hi folks, sorry for this beginner question but what means a p-value = NA on a menn-whitney test? v1 <- c(0.022, 0.022, 0.022, 0.022, 0.022, 0.022) v2 <- c(0.022, 0.022, 0.022, 0.022, 0.022) wilcox.test(v1, v2, PAIRED=FALSE) W = 15, p-value = NA I know that there's no statistical difference between v1 and v2, so why my p-value is not ONE ?? Can I consider p-value= NA as ONE ?? what i