similar to: Finding the Degrees of Freedom in a Wilcoxon Test

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

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

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

Hello there, I would like to do a Wilcoxon matched pairs signed rank sum test in R and tried the function wilcoxon.test. Is it in the "base" library? If not, please let me know which library it is in. Thank you in advance for your help Lisa Wang Biostatistician Princess Margaret Hospital Toronto, Ca

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 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

Dear all, I am using the ROCR library to compute the AUC and also the Hmisc library to compute the C-index of a predictor and a group variable. The results of AUC and C-index are similar and give a value of about 0.57. The Wilcoxon p-value is <0.001! Why the AUC is showing small value and the p-value is high significant? The AUC is based on Wilcoxon calculation? Many thanks, Lina

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

Dear R-users, I?m trying to perform a non-parametric statistical pairwise comparison of two samples "x" and "y" using the Wilcoxon test in R, but each of the pairs have a weight associated. An example of my data is the following one: set.seed(9) x <- sample(x = c(1:100), size = 20, replace = TRUE) y <- sample(x = c(1:100), size = 20, replace = TRUE) weight <-

Hi! I need to run a wilcoxon (Mann-whitly, in fact) test with bonferroni correction, as I am running 10 consecutive wilcoxon test not independent, and I know that bonferroni will partially correct for this problem, but I have no idea how to do it with R, I have been looking in the archive but couldn't understand how to do it. The format I am using at the moment is r4_o <- [1]

Hi, I have a data set: > Dataset X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 1 user1 m 22 19 28 24 12 18 9 7 4 5 4 7 5 7 9 2 user2 f 25 19 23 18 18 15 6 8 6 6 7 10 7 7 7 3 user3 f 28 21 24 18 15 12 10 6 7 9 5 10 5 9 5 4 user4 f 26 19 26 21 12 18 6 6 5 1 3 8 6 5 6 5 user5 m 21 22 26 18 9 6 4 6 1

I am sure I am opening myself up to looking stupid, but I have two samples with medians of 613.5 and 189 (difference in location of 424 compared to the difference suggested from the wilcoxon of 291.5) > wilcox.test(pipwtCount,pipwdCount, conf.int=TRUE, na.rm=TRUE) Wilcoxon rank sum test data: pipwtCount and pipwdCount W = 822, p-value = 0.01227 alternative hypothesis: true location

Hello All, I am writing to ask your opinion on how to interpret this case. I have two vectors "a" and "b" that I am trying to compare. The wilcoxon test is giving me a pvalue of 5.139217e-303 of a over b with the alternative "greater". Now if I make a summary on each of them I have the following > summary(a) Min. 1st Qu. Median Mean 3rd Qu.

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:

Hallo, is there any extension of the pairwise Wilcoxon test to a dependent samples layout with replicates (or, in other terms, a one-way layout with blocking and replicates)? The Wilcoxon method with matched pairs works for the case of dependent samples with one observation per block, while the Mann-Whitney test works for independent samples, thus one single block and replicated observations. Is

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

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

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

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 =

Hello All, I posted a similar question before, but the direction was driven to whether my case is suitable for a wilcoxon test. After research about the appropriateness, I am pretty sure that a wilcoxon test is the right tool for my case. But how to compute the power of the test is still an unanswered question bothering me. The basic stats of my two paired samples are: mean1 = 0.0032, sd1 =