similar to: Orders of levels affecting wilcox.test() output

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

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,

Recently, I ran a series of Kruskal-Wallace tests [kruskal.test()] using by() to group by site Output is a list: >Herb.KW Herb.df$ID: 10-1 Kruskal-Wallis rank sum test data: Indicator_Rating by Year Kruskal-Wallis chi-squared = 15.24, df = 7, p-value = 0.03302 ----------------------------------------------------------------------------------------------------- Herb.df$ID: 18-1

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)

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,

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

>>>>> 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 all, I want to compare the efficiency of 2 methods in extracting proteins from algal samples. I collected 6 independant algal samples and I extracted 3 by the method 1 and 3 others by the method 2. So I have 2 groups of 3 samples, that are not paired. I would like to know if the results obtained by these 2 methods are significantly different, I hope method 2 to be more efficient than method

Dear Developers Team, I have two items: 1. wilcox.test with the paired=T option appears to delete zeros before ranking absolute differences. Would it be possible to add the feature of removing zeros after ranking, which is given in Lehmann's Nonparametrics as the preferred choice. See also Pratt (1959), JASA 54, 655-667. It is given in wilcoxsign_test of the coin package as an option

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

I get these BIZARRE results from wilcox.test() When INCREASING the number of samples i get INCREASED p-values. When increasing the number of samples further, the p-values goes down again. This seems really bizarre! Can anyone explain why this is so?! Example: > w <- wilcox.test(c(1:40),(c(1:40)+100)) > w$p.value [1] 1.860340e-23 > w <- wilcox.test(c(1:50),(c(1:50)+100)) >

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

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

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,

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

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