similar to: Help with Wilcoxon Test

Hi I have managed to do a paired t-test with a data set i have 5 colums of data im dealing with GENE              SampA              SampB              SampC                   SampVehicle ctcc                   859                     na                     145                           24 gtcg                   45                         5                       54                           69

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

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

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

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.

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 =

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 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

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 list, I am facing a problem in my statistical analyses on R. My experiments are about plants, I record there growth after each cutting (every 3 weeks). 'BC' is for the plant, and '1' to '5' is the time of cutting and recording. The data and R script are : "" BourdCoup <- c(21, 7.2, 9.2, 0, 8.52, 14.7, 8.31, 6.2, 127.05, 115.2, 100.7, 24, 162.64, 136.8,

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 <-

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, I am using wilcox.test function to test the difference between the means of two samples. The data points are paired, so I am using a paired test. There is one strange case. Sample A has a higher mean than a sample B. However, wilcox.test function says that sample B has a significantly higher "mean rank" than sample A. How is it possible? Here is the code (data file is attached):

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, 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

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, 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

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 =

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