similar to: Wilcoxon Mann-Whitney Rank Sum Test in R

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

Hi, I am running the following example which can be found on page 12 of the pdf file of COIN package wt<-wilcox_test(pd~age,data=water_transfer,distribution="exact", conf.int=TRUE) "wt" actually contains the estimate of difference in location and the confidence interval of it. I am just wondering how can I extract these values? From the examples, I understand that the

Hello, I'm having some issues grouping cases for some Mann-Whitney U tests I'm attempting to run. I'm willing to use wilcox.test if it'll work; I've also tried wilcox_test() from the "coin" package. Here's the deal: for each column (A through H), I would like to run the two-sample independent test, comparing Group 5 (CD8.14 through CD8.17) to Group 6 (CD8.18

Dear Sir, I am comparing two samples using wilcox.test in R. Literature appears to describe mann whitney u test as the most appropriate test to use on my data. is the wilcox.test function equivalent to mann-whitney u? Is there a way to gain the U-value as apposed to the W-value in R? Thank you

Hello, can someone please help me with coding a function for a bootstrap WMW test (package boot, R under Windows, version 1.6.2)?

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 <-

Dear R-forumites, I want to apply a Wilcoxon test on subsets of the data frame mydata, splitted using the myindice variable. When I send : wilcoxtest <- by(mydata, mydata$myindice, function(x) {wilcox_test(x$value~x$fact)}) I get : Error in eval(expr, envir, enclos) : object "x" not found whereas wilcoxtest <- by(mydata, mydata$myindice, function(x)

Hello everyone, I'm having some trouble interpreting the results of a Wilcoxon (Mann-Whitney U) test. Hope you can help. This is the R script that I am running: a <- c(1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 5, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 3, 1, 1) b <- c(1, 2, 1, 1, 2, 3, 2, 2, 1, 2, 1, 1, 1, 2) wilcox.test(a, b, alternative="t", mu=0, exact=FALSE, paired=FALSE) #1st

Dear Stefan I need to apply a Mann Whitney test to my data, by applying a bootstrap, in R. Loonkig for an answer in internet I found your question to the R mailing list but I have not found the answer to your question. Please, could you help me? thank you very much in advance. Best regards Luciana

I need to perform the Exact Wilcoxon Mann-Whitney on a subset of my database. Assuming that IPPO is my data frame and IPPOBIS is the subset my variable still have 3 different levels and the function wilcox_test (package "coin") does not accept it. I do not know how to overcome this problem. ippo <- c(rep("A",10),rep("B",10),rep("C",10)) ippo2 <-

Hello List, I'm trying to prepare some lecture notes on non parametric methods, and I can't manually reproduce the results of the wilcox.test function for ordinal data. The data I'm using are from David Howell's website, available here http://www.uvm.edu/~dhowell/StatPages/More_Stuff/OrdinalChisq/OrdinalChiSq.html If I run the wilcox.test function on the data I get a p-value of

Hi, I have a statistical question. To apply the Mann Whitney test to compare a pair of vectors, they must to have the same length ? Thanks, Rodrigo [[alternative HTML version deleted]]

Dear All, I have two simple data samples (no groups or factors, etc.) and would just like to compute the two-sample Wilcoxon Rank Sum test using the wilcox_test function contained in the coin package, which is reportedly better than the regular wilcox.test function because it performs some adjustment for ties. Would anyone know how to craft a script to perform this task? Much appreciated. Janh

Hello, I use t.test for normal distributed and wilcox.test for non-normal distributed samples. It is easy to write a function for t.test that calculates the effect size, because all parts of the formula are available from the t.test result: r = sqrt(t*t / (t*t + df)) However, for Wilcoxon tests, the formula for effect sizes is: r = Z / sqrt(N) I wonder how I can calculate the Z-score in R for

Does anyone have a piece of code or know how I can use R to generate a table of critical values for the Mann-Whitney (aka Wilcoxon Rank Sum) test. Ideally, I'd like a table that contains the critical values for any two samples of size 3 through 30. I could use Monte Carlo simulation or the normal approximation when n1 and n2 are greater than, 10, but I figured someone may know how to

Hi, Could someone please tell me how to perform a Mann-Whitney U test on a dataset with 2 groups where one group has more data values than another? I have split up my 2 groups into 2 columns in my .txt file i'm using with R. Here is the code i have so far... group1 <- c(LeafArea2) group2 <- c(LeafArea1) wilcox.test(group1, group2) This code works for datasets with the same number

Hi, I know that I can perform a Mann-Whitney U test with wilcox.test(x,y) But instead of an U-value I get a W-Value. Is it possible to change this?

Dear R-users, If you use the exact Wilcoxon test in the coin package, I would like make you aware of that SPSS/StatXact MAY perform a round-off before doing their exact Wilcoxon-Mann-Whitney test (if you ever are unlucky enough not to use R). I have data from two treatments and was surprised to find that SPSS (15 under Windows) and R differed in their p-values (0.167 resp. 0.172). It turns out

Hi, I became a little bit confused when working with the Wilcoxon test in R. As far as I understood, there are mainly two versions: 1) wilcox.test{stats}, which is the default and an approximation, especially, when ties are involved 2) wilcox_test{coin}, which does calculate the distribution _exactly_ even, with ties. I have the following scenario: #---BeginCode--- # big example size = 60

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