similar to: wilcox.test; data type conversion?

I still need to do some repetitive statistical analysis on some outcomes from a dataset. Take the following as an example; id sex hiv age famsize bmi resprate 1 M Pos 23 2 16 15 2 F Neg 24 5 18 14 3 F Pos 56 14 23 24 4 F Pos 67 3 33 31 5 M Neg 34 2 21 23 I want to know if there are statistically detectable differences in all of the continuous variables in

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)) >

Hi All, I have data with 400 individuals and the following information Grade: pass or fail coded as 1 for pass and 0 for fail Sex: male or female ( coded as 1 for male and 2 for female ) Age Teaching.method : can be 1,2,3 I want to fit a logistic regression where the outcome if (1=pass or 0 for fail) and the rest of the variables are the regressors. My question is that I am not sure

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)

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

Hi there, I am trying to do multiple pairwise Wilcoxon signed rank tests in a manner similar to: a <- c(runif(1000, min=1,max=50), rnorm(1000, 50), rnorm(1000, 49.9, 0.5), rgeom(1000, 0.5)) b <- c(rep("group_a", 1000), rep("group_b", 1000), rep("group_c", 1000), rep("group_d", 1000)) pairwise.wilcox.test(a, b, alternative="two.sided",

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

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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

This is my first post to the mailing list and I guess it's a pretty stupid question but I can't figure it out. I hope this is the right forum for these kind of questions. Before I started using R I was using STATA to run a Wilcoxon signed-rank test on two variables. See data below:

I've got two set of data : 22.45 21.56 20.48 19.59 21.52 = A and 22.15 21.98 20.42 20.58 19.61 = B I perform a wilcox.test on this two set wilcox.test(A, B) and I'd this answer: Wilcoxon rank sum test data: A and B W = 12, p-value = 1 alternative hypothesis: true mu is not equal to 0 Should I interpret that there is no difference between the two sets ? -- Cordialement

Hi, Trying to setup a logistic regression model. (Something new to me. I usually use SVM.) The person explaining the concept explained to me that I can include a "group" variable so that the probabilities predicted by the model will be "per group" Does this make sense to anyone? If so, how would I implement this? Using the glm or lrm function? Thanks! -N

>>>>> Karolis Koncevi?ius >>>>> on Mon, 9 Dec 2019 23:43:36 +0200 writes: > So I tried adding Infinity support for all cases. > And it is (as could be expected) more complicated than I thought. "Of course !" Thank you, Karolis, in any case! > It is easy to add Inf support for the test. The problems start with conf.int=TRUE.

does anybody know why there are the two warnings in the example above? Regards Knut > day_4 [1] 540 1 1 1 1 1 1 300 720 480 > day_1 [1] 438 343 1 475 1 562 500 435 1045 890 > is.vector (day_1) [1] TRUE > is.vector (day_4) [1] TRUE > wilcox.test(day_4 ,day_1,paired=TRUE,alternative="two.sided",exact=TRUE,conf.int=TRUE) Wilcoxon

This was filed as a bug report on the Debian r-base package. It is more properly a bug report on the ctest package in R. The default method for wilcox.test manipulates x and y without checking the class or data.class of these objects. Possible solutions are - create wilcox.test.factor (if appropriate) - check the class and/or data.class of x and y in wilcox.test.default and produce error

Dear all, When we perform a Wilcoxon rank sum test (on two samples with different sizes) we get a test statistic. My question is, as the value of test statistic increases the difference between the distributions of the two samples also increase, right? Thanks in advance, João Fadista [[alternative HTML version deleted]]

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,

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

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)

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