similar to: one (small) sample wilcox.test confidence intervals

Dear all, I am trying to create a vector name, for example tmax.195012 from tmax., 1950 and 12. Obviously I don't wish to simply type it because the 3 name components are changing in each iteration within a loop. Is there any way of concatenating those 3 components (which are a mixture of numbers and letters)? Thanks for reading, Panos

Dear R colleagues, I am trying to plot some geophysical data as a filled contour on a continent map and so far the guidance from the R-help archives has been invaluable. The only bit that still eludes me is the colour key (legend) coming with filled.contour: I prefer to generate my own colour palette, mainly based on the quantiles of tenths of the data in order to capture the whole range (of

Dear R-helpers, I am doing a multiple linear regression of an ozone time-series on time and other explanatory variables. I have been using the "lm" model but I am recently experimenting with "gls". With the "lm" model I was able to look at the residuals by $res in the "summary (lm(...))" and then check with "acf" for autocorrelation in these

#Here is a vector of IDs > cwaves [1] 86 90 185 196 197 209 210 215 216 217 218 #Here is a matrix. The rows and columns correspond to the IDs in cwaves, and the matrix is populated with a coefficient > mat 86 90 185 196 209 210 215 216 217 218 86 0 0 0 0 0 0.00000 0 0.000000 0.000000 0.000000 90 0 0 0 0 0 0.00000 0

Hi to everybody, i tried to find an asnwer before posting this, but most astcc billing issues i searched refer to the case when no billing occurs at all. In my case i get only initial charges and any subsequent minute does not count for billing. In my iax.conf i entered the "notransfer = yes" but nothing changed. My test card and test calls are the following TEST-CARD en N/A

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:

Hello all I wanted to forward all incoming requests in port 80 to a server in my LAN, and by using DNAT lines, it actually works. However, it is unstable, in the sense that in the beggining of each connection (one or two seconds) it is extremely fast, then it sometimes pauses and waits 30 seconds or so, then it starts again and so on. The line i used is : DNAT net loc:192.168.0.210 tcp 80 DNAT

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

>>>>> Karolis Koncevi?ius >>>>> on Sat, 7 Dec 2019 20:55:36 +0200 writes: > 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))

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 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

So I tried adding Infinity support for all cases. And it is (as could be expected) more complicated than I thought. It is easy to add Inf support for the test. The problems start with conf.int=TRUE. Currently confidence intervals are computed via `uniroot()` and, in the case of infinities, we are computationally looking for roots over infinite interval which results in an error. I suspect this

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",

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

>>>>> Martin Maechler >>>>> on Thu, 12 Dec 2019 17:20:47 +0100 writes: >>>>> 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

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]]

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

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 working on a quick tutorial for my students, and was planning on using Mann-Whitney U as one of the tests. I have the following (fake) data grade <- c("MVG", "VG", "VG", "G", "MVG", "G", "VG", "G", "VG") sex <- c( "male", "male", "female", "male",

We've seen the integer overflow problem in ks.test before, easily solved. The help page says x and y must be numeric, so this is user error. I've added tests to the code. Why do people file bug reports without reading the help/man page? On Tue, 14 Jan 2003 bates@stat.wisc.edu wrote: > This was filed as a bug report on the Debian r-base package. It is > more properly a bug