similar to: How to apply the wilcox_test function to subsets ?

Dear R helpers, I try to use the 'delaunayn' function in the 'geometry' package for Delaunay triangulation in 2 dimensions. For the four following points, I get a warning message : > coord=matrix(ncol=2,byrow=TRUE,c(622633,7073452, + 621228,7073517, + 621879,7071762, +

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

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

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

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, 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

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 <-

An earlier post had posed the question: "Does anybody know what is relation between 'T' value calculated by 'wilcox_test' function (coin package) and more common 'W' value?" I found the question interesting and ran the commands in R and SPSS. The W reported by R did not seem to correspond to either Mann-Whitney U, Wilcoxon W or the Z which I have more

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

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

LO HICE DE ESTA MANERA Y FUNCIONÓ a_HTA <- (estudiantes1[,c(1)]) IMC <- (estudiantes1[,c(2)]) estudiantes4 <- data.frame( IMC, a_HTA = factor(c(rep("si", 29), rep("no",30)))) wt <- wilcox_test(IMC ~ a_HTA, data = estudiantes4, distribution = "exact", conf.int = TRUE) print(wt) De: Jorge I Velez

Dear Sir, I am doing some work on a population of patients. About half of them are admitted into hospital with albumin levels less than 33. The other half have albumin levels greater than 33, so I stratify them into 2 groups, x and y respectively. I suspect that the average length of stay in hospital for the group of patients (x) with albumin levels less than 33 is greater than those

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

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

R 2.1.1 on Windows 2K I hope one and all will allow both an R question and a general stats question: (1) Is there any function that will perform non-parametric tests such as a sign test or a signed rank test? I know I could program both but I would prefer not re-inventing the wheel. (2) When performing a sign test, the usual practice is to drop zero values. Is there any theoretical reason

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 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

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

De la manera que me aconsejas no funciona y da este mansaje Warning message: In wilcox.test.default(x = c(19L, 18L, 17L, 16L, 18L, 19L, 18L, : cannot compute exact p-value with ties como aconsejaste tmp <- read.table("estudiantes1.csv", header=TRUE, sep=",", dec=".", na.strings = "NA") wt <- wilcox.test(IMC ~ a_HTA, data=tmp, paired = FALSE,