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

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

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

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

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,

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

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,

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,

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)) >

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

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)

Dear All, I am using wilcox.test to test two samples, data_a and data_b, earch sample has 3 replicates, suppose data_a and data_b are 20*3 matrix. Then I used the following to test the null hypothesis (they are from same distribution.): wilcox.test(x=data_a, y=data_b, alternative="g") I got pvalue = 1.90806170863311e-09. When I switched data_a and data_b by doing the following:

Dear all, I'm a newbie to R and I would really apperciate any help with the following: I have two lists, l1 and l2: l1: $"A*0101" [1] 0.076 0.109 0.155 0.077 0.09 0 0 0.073 [9] 0.33 0.0034 0.0053 $"A*0247" [1] 0 0 0.5 .004 0 0 0 $"A*0248" [1] 0 0 0.3 0 0.06 .... l2: $"A*1101" [1] 0.17 0.24 0.097 0.075 0.067 $"A*0247" numeric(0)

Full_Name: Patrick Hodgson Version: 2.0 OS: solaris 2.9 Submission from: (NULL) (65.94.128.161) The value reported for the parameter W in the function wilcox.exact appears to be incorrect. I have checked the reference in the help file for this function (Myles & Hollander 1973, as well as 2nd ed. 1999 by same authors) and it is clear that W is the sum of the ranks of the data set with the

Dear R friends, here I write again about the wilcox.exact() problem. I want to compare two sets of categorical data, and in one case it says "negative length vectors not allowed", and in the other one I get the error "cannot allocate vector of length ...". On http://rapidshare.com/files/298621893/wilcox.exact_trouble.Rdata.html you can download the data that cause the

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

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

Greetings, I'm working on some analyses where I need to calculate wilcox tests for paired samples. In my current literature search I've found a few papers on sample size determination for the wilcox test notably: Sample Size Determination for Some Common Nonparametric Tests Gottfried E. Noether Journal of the American Statistical Association

Dear Contributors I have a problem with the collection of data from the results of a test. I need to perform a comparative test over groups of data , recall the value of the pvalue and create a table. My problem is in the way to replicate the analysis over and over again over subsets of data according to a condition. I have this database, called y: gg t1 t2 d 40 1 1