similar to: Wilcoxon signed-rank test

>>>>> 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.

Dear List, after updating the exactRanksumTests package I receive a warning that the package is not developed any further and that one should consider the coin package. I don't find the signed rank test in the coin package, only the Wilcoxon Mann Whitney U-Test. I only found a signed rank test in the stats package (wilcox.test) which is able to calculate the exact pvalues but unfortunately

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

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

Hi all, I have a distribution, and take a sample of it. Then I compare that sample with the mean of the population like here in "Wilcoxon signed rank test with continuity correction": > wilcox.test(Sample,mu=mean(All), alt="two.sided") Wilcoxon signed rank test with continuity correction data: AlphaNoteOnsetDists V = 63855, p-value = 0.0002093 alternative hypothesis:

Hi. I'm going to introduce the R-package for a group of medical doctors later this week and is a little confused about there use of a test named "willcoxon-pratt" for testing if the clinical and biochemical markers has decreased significantly after the use of some drugs for a group of patients. Looking into the R-functions I would in R recommand using a matched-pairs Wilcoxon

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)

Hello, I am hoping someone could shed some light into the Wilcoxon Rank Sum Test for me? In looking through Stats references, the Mann-Whitney U-test and the Wilcoxon Rank Sum Test are statistically equivalent. When using the following dataset: m <- c(2.0863,2.1340,2.1008,1.9565,2.0413,NA,NA) f <- c(1.8938,1.9709,1.8613,2.0836,1.9485,2.0630,1.9143) and the wilcox.test command as

Dear, I'm using R software to evaluate Wilcoxon Rank Sum Test and I' getting one Warning message as this: > C1dea_com [1] 1.000 0.345 0.200 0.208 0.508 0.480 0.545 0.563 0.451 0.683 0.380 0.913 1.000 0.506 > C1dea_sem [1] 1.000 0.665 0.284 0.394 0.509 0.721 0.545 0.898 0.744 0.683 0.382 0.913 1.000 0.970 > wilcox.test(C1dea_sem,C1dea_com, paired = TRUE, alternative =

Hi everyone! I need to perform a Wilcoxon rank sum test, but I have some ties and the groups have different size also. When I deal with ties I use the wilcox.exact function, how can I solve the different size problem using this function? thanks net -- View this message in context: http://r.789695.n4.nabble.com/Wilcoxon-rank-sum-in-unbalanced-design-tp3447400p3447400.html Sent from the R help

Hi there, I'm a total newbie to R. I'd like to use a Wilcoxon Rank Sum test to compare two populations of values. Further, I'd like to do this simultaneously for 114 sets of values. The two populations are C and N. The different sets of values have arbitrary names (I'll call them a, b, c etc). The set-up is as follows: a b c d .... C 2 C 3 C 5 C 9 C 4

Hi guys,   I have two data sets of prices: endprice0, endprice1   I use the Wilcox test:   wilcox.test(endprice0, endprice1, paired = TRUE, alternative = "two.sided",  conf.int = T, conf.level = 0.9)   The result is with V = 1819, p-value = 0.8812.   Then I calculated the z-value of the test: z-value = -2.661263. The corresponding p-value is: p-value = 0.003892, which is different from

Given the following data, and hypothesized median M.0 I've found a method to implement the Wilcoxon signed-rank test. Data: (with one zero difference and tied ranks) x <- c(136, 103, 91, 122, 96, 145, 140, 138, 126, 120, 99, 125, 91,142, 119, 137) M.0 <- 119 > library(exactRankTests) Package ?exactRankTests? is no longer under development. Please consider using package ?coin?

Dear all, I am trying to run Wilcoxon Rank-Sum Test with Bonferroni's correction. I have two lists: l0, l1: mapply(function(x,y)wilcox.test(x,y)$p.value, l0, l1) How do I run Bonferroni's correction on mapply? Any help is much apperciated. Thanks, -Raj

Given this example #start code a<-c(0,70,50,100,70,650,1300,6900,1780,4930,1120,700,190,940, 760,100,300,36270,5610,249680,1760,4040,164890,17230,75140,1870,22380,5890,2430) b<-c(0,0,10,30,50,440,1000,140,70,90,60,60,20,90,180,30,90, 3220,490,20790,290,740,5350,940,3910,0,640,850,260) wilcox.test(a, b, paired=FALSE) #sum of rank for first sample sum.rank.a <-

how do I check the rank of a matrix ? say A= 1 0 0 0 1 0 then rank(A)=2 what is this function? thanks I did try help.search("rank"), but all the returned help information seem irrelevant to what I want. I would like to know how people search for help information like this. rank(base) Sample Ranks SignRank(stats) Distribution of the

I am running a pairwise wilcoxon signed rank test, and I am not sure how to interpret the result. I would like to see if there is a difference between the values in conditions a and b. It doesn't seem possible to have a V = 0, but a significant p value. Am I doing something wrong? The command I used is this: wilcox.test(x=a$x,y=b$x,paired=TRUE) The output looks like this: Wilcoxon

Dear R-ers, I am currently running some Wilcoxon tests in R-64. How do I find the degrees of freedom in the output I am receiving? > wilcox.test(good$TRUE, good$x4a, paired=FALSE) Wilcoxon rank sum test with continuity correction data: good$TRUE and good$x4a W = 2455, p-value < 2.2e-16 alternative hypothesis: true location shift is not equal to 0 Thank you, Stephen.

Estimado José, por lo que estuve mirando aqui https://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test, en ?wilcox.test y en libros de papel ... los datos deben cumplir ciertas condiciones, quiza la mas restrictiva es que sean pareados ... si no lo son, quiza esta variante del test es mas apropiada: https://en.wikipedia.org/wiki/Mann-Whitney-Wilcoxon_test la que en R se aplica usando el mismo

How does R compute the p-value in the Wilcoxon Rank Sum Test? If I have the test statistics, can I get the p-value using the function pwilcox? Thanks -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the