similar to: wilcox.test returned estimates

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 recently ran into the problem that I needed a Siegel-Tukey test for equal variability based on ranks. Maybe there is a package that has it implemented, but I could not find it. So I programmed an R function to do it. The Siegel-Tukey test requires to recode the ranks so that they express variability rather than ascending order. This is essentially what the code further below does. After the

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

Hi, Could anyone provide the formula of the statistics which the wilcox.test used for the two-sample rank-sum test? I got some statistics of 0 values, but it is impossible to have 0 "rank-sum". Does the function use the Mann-Whitney U test statistics? Thanks. Ting-Yuan Liu

Greetings, I'm working on some analyses where I need to calculate wilcox tests for paired samples. In my current literature search I've found a few papers on sample size determination for the wilcox test notably: Sample Size Determination for Some Common Nonparametric Tests Gottfried E. Noether Journal of the American Statistical Association

Good day all, My first posting to this list. It looked like the best place to post this question. When running the wilcox.test(), I noticed that the output values change if you change the ordering of the levels (example below which includes a t.test for comparison). I think this has something to do with the change in ranking order, but this doesn't make much sense as I would expect the

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:

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

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)

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.

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

>>>>> 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.

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

The Partek package (www.partek.com) allows only two selections for Multiple Test Correction: Bonferroni and Dunn-Sidak. Can anyone suggest why Partek implemented Dunn-Sidak and not the other methods that R has? Is there any particular advantage to the Dunn-Sidak method? R knows about these methods (in R 2.1.1): > p.adjust.methods [1] "holm" "hochberg" "hommel"

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)

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

Hi, Two functions wilcox.exact and wilcox_test give slightly different confidence intervals of the difference of the medians: for example y<-c(0,0,1.081,0.594,0,0.769,0,0.009,0,0,0.798,0.405,0.498,0.946,1.35,1.149,0.528) x<-c(rep(1,10),rep(2,7)) aa<-wilcox.exact(y~x,conf.int=TRUE) bb<-wilcox_test(y~factor(x),distribution="exact",conf.int=TRUE) aa bb Does anyone know

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Dear All, I am using wilcox.test to test two samples, data_a and data_b, earch sample has 3 replicates, suppose data_a and data_b are 20*3 matrix. Then I used the following to test the null hypothesis (they are from same distribution.): wilcox.test(x=data_a, y=data_b, alternative="g") I got pvalue = 1.90806170863311e-09. When I switched data_a and data_b by doing the following:

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,