similar to: Power curve for a wilcoxon test

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 =

Hi, I' am doing a stats project using R to work out the size of a t-test and wilcoxon test depending on the distribution and sample size. I just can't get it to work - I want to put my results from the function size() into an array.At the moment I keep getting the error message:Error in res[distribution, test, samplesize] <- results : subscript out of boundsCan anyone tell me where

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

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, 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

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

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

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

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 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

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.

Hi I have 2 sets of data that I want to do a Wilcoxon test on. They are of the same dimension. One has 4 zero values and the other has 5. > dim(SampA) [1] 1 10 > dim(SampV) [1] 1 10 I get the folowing error Error in wilcox.test.default(SampA, SampV, na.rm = TRUE, paired = FALSE, : 'x' must be numeric I am using the function wilcox.test(SampA, SampV, na.rm=TRUE,

When I use wilcox.test, I get vastly different p-values than the problems from Statistics textbooks. For example: The following problem comes from "Applied Statistics and Probability for Engineers", 2nd Edition, by D. C. Montgomery. Page736, problem 14.7. The problem is to compare the sample data with a population median of 8.5. The book answer is p = 0.25, wilcox.test answer is p =

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

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 =

Dear list, Let's say I have data a=c(37.961,38.214,57.68) b=c(77.56,61.875,67.683) wilcox.test(a,b) the wilcoxon test only gives me a p value with one decimal place. Is this normal? Thanks. Jun [[alternative HTML version deleted]]

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 everyone, I am trying to do a Wilcoxon one-sided test on my gene expression data. These are the data i have in R: data.matrix (matrix, numeric) containing all gene expression data (42 rows=genes, 42 columns=tumors), no column header or row names data.cl (vector, numeric) consisting of 42 0's and 1's to indicate class 0 or class 1 for each column in data.matrix I want to do a

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

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