similar to: BIZARRE results from wilcox.test()

The point estimates produced by wilcox.test are perverse (not wrong, just brain damaged). The Hodges-Lehmann estimator that goes with the signed rank test is the median of the Walsh averages. The Hodges-Lehmann estimator that goes with the rank sum test is the median of the pairwise differences. wilcox.test agrees except that it uses the following very peculiar definition of "sample

Thank you for a fast response. Nice to see this mailing list being so alive. Regarding Inf issue: I agree with your assessment that Inf should not be removed. The code gave me an impression that Inf values were intentionally removed (since is.finite() was used everywhere, except for paired case). I will try to adjust my patch according to your feedback. One more thing: it seems like you

Hello, Writing to share some things I've found about wilcox.test() that seem a a bit inconsistent. 1. Inf values are not removed if paired=TRUE # returns different results (Inf is removed): wilcox.test(c(1,2,3,4), c(0,9,8,7)) wilcox.test(c(1,2,3,4), c(0,9,8,Inf)) # returns the same result (Inf is left as value with highest rank): wilcox.test(c(1,2,3,4), c(0,9,8,7), paired=TRUE)

>>>>> Karolis Koncevi?ius >>>>> on Mon, 9 Dec 2019 23:43:36 +0200 writes: > So I tried adding Infinity support for all cases. > And it is (as could be expected) more complicated than I thought. "Of course !" Thank you, Karolis, in any case! > It is easy to add Inf support for the test. The problems start with conf.int=TRUE.

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

Full_Name: Patrick Hodgson Version: 2.0 OS: solaris 2.9 Submission from: (NULL) (65.94.128.161) The value reported for the parameter W in the function wilcox.exact appears to be incorrect. I have checked the reference in the help file for this function (Myles & Hollander 1973, as well as 2nd ed. 1999 by same authors) and it is clear that W is the sum of the ranks of the data set with the

This was filed as a bug report on the Debian r-base package. It is more properly a bug report on the ctest package in R. The default method for wilcox.test manipulates x and y without checking the class or data.class of these objects. Possible solutions are - create wilcox.test.factor (if appropriate) - check the class and/or data.class of x and y in wilcox.test.default and produce error

Hi there, I am trying to do multiple pairwise Wilcoxon signed rank tests in a manner similar to: a <- c(runif(1000, min=1,max=50), rnorm(1000, 50), rnorm(1000, 49.9, 0.5), rgeom(1000, 0.5)) b <- c(rep("group_a", 1000), rep("group_b", 1000), rep("group_c", 1000), rep("group_d", 1000)) pairwise.wilcox.test(a, b, alternative="two.sided",

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

I've got two set of data : 22.45 21.56 20.48 19.59 21.52 = A and 22.15 21.98 20.42 20.58 19.61 = B I perform a wilcox.test on this two set wilcox.test(A, B) and I'd this answer: Wilcoxon rank sum test data: A and B W = 12, p-value = 1 alternative hypothesis: true mu is not equal to 0 Should I interpret that there is no difference between the two sets ? -- Cordialement

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,

This is my first post to the mailing list and I guess it's a pretty stupid question but I can't figure it out. I hope this is the right forum for these kind of questions. Before I started using R I was using STATA to run a Wilcoxon signed-rank test on two variables. See data below:

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

I'm working on a quick tutorial for my students, and was planning on using Mann-Whitney U as one of the tests. I have the following (fake) data grade <- c("MVG", "VG", "VG", "G", "MVG", "G", "VG", "G", "VG") sex <- c( "male", "male", "female", "male",

Dear R people, I notice that the confidence intervals of a very small sample (e.g. n=6) derived from the one-sample wilcox.test are just the maximum and minimum values of the sample. This only occurs when the required confidence level is higher than 0.93. Example: > sample <- c(1.22, 0.89, 1.14, 0.98, 1.37, 1.06) > summary(sample) Min. 1st Qu. Median Mean 3rd Qu. Max.

I'm having trouble with the wilcox.test command in R. To demonstrate the anomalous behavior of wilcox.test, consider > wilcox.test(c(1.5,5.5), c(1:10000), exact = F)$p.value [1] 0.01438390 > wilcox.test(c(1.5,5.5), c(1:10000), exact = T)$p.value [1] 6.39808e-07 (this calculation takes noticeably longer). > wilcox.test(c(1.5,5.5), c(1:20000), exact = T)$p.value (R closes/crashes)

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

Hi, I am very confused with constructing the wilcox.test in R. I have two populations 'original' and 'test'. I want to know if the 'test' is generally 'lower' than original. I use alpha of 0.05. So do I write the function as wilcox.test(original, test, alternative="l")? or wlcox.test(original, test, alternative = "g")? or wilcox.test(test,

Hello, I think I am right in saying that a 2 sample wilcox.test is equal to a 2 sample kruskal.test and a 2 sample t.test is equal to a 2 sample anova. This is also stated in the ?kruskal.test man page: The Wilcoxon rank sum test (wilcox.test) as the special case for two samples; lm together with anova for performing one-way location analysis under normality assumptions; with Student's t

Hello, I'm trying to compare the allozyme data from two taxa. I have several columns of data (19 loci) for each species. I want to do a Mann-Whitney U-test or the wilcox.test (two sample Wilcoxon). When I try to run my code (the first two columns are 1:name of the species, 2:name of individual) I get the error message: "Error in wilcox.test.default(CaScSc, CaScCo, alternative