similar to: R-estimators

Dear R-users, I'm working with the Rpart package and trying to understand how the procedure select the best split in the case the method "class" and the splitting index "Gini" are used. In particular I'd like to have look to the source code that works out the best split for un unordered predictor. Does anyone can suggest me which functions in the sources I should

Dear all, I'm trying to manage with user defined split function in rpart (file rpart\tests\usersplits.R in http://cran.r-project.org/src/contrib/rpart_3.1-34.tar.gz - see bottom of the email). Suppose to have the following data.frame (note that x's values are already sorted) > D y x 1 7 0.428 2 3 0.876 3 1 1.467 4 6 1.492 5 3 1.703 6 4 2.406 7 8 2.628 8 6 2.879 9 5 3.025 10 3 3.494

Dear R users, I'm working with the rpart package and want to evaluate the performance of user defined split functions. I have some problems in understanding the meaning of the xval argument in the two functions rpart.control and xpred.rpart. In the former it is defined as the number of cross-validations while in the latter it is defined as the number of cross-validation groups. If I am

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Dear List, quantile(x) function allows to obtain specified quantiles of a vector of observations x. Is there an analogous function to compute quantiles in the case one have the vector of the observations x and the correspondent vector f of relative frequencies ? Thank you Paolo Radaelli [[alternative HTML version deleted]]

how do I check the rank of a matrix ? say A= 1 0 0 0 1 0 then rank(A)=2 what is this function? thanks I did try help.search("rank"), but all the returned help information seem irrelevant to what I want. I would like to know how people search for help information like this. rank(base) Sample Ranks SignRank(stats) Distribution of the

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 <-

Hallo, is there any extension of the pairwise Wilcoxon test to a dependent samples layout with replicates (or, in other terms, a one-way layout with blocking and replicates)? The Wilcoxon method with matched pairs works for the case of dependent samples with one observation per block, while the Mann-Whitney test works for independent samples, thus one single block and replicated observations. Is

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

Hi, I recently ran into the problem that I needed a Siegel-Tukey test for equal variability based on ranks. Maybe there is a package that has it implemented, but I could not find it. So I programmed an R function to do it. The Siegel-Tukey test requires to recode the ranks so that they express variability rather than ascending order. This is essentially what the code further below does. After 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:

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

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

Given the following data, and hypothesized median M.0 I've found a method to implement the Wilcoxon signed-rank test. Data: (with one zero difference and tied ranks) x <- c(136, 103, 91, 122, 96, 145, 140, 138, 126, 120, 99, 125, 91,142, 119, 137) M.0 <- 119 > library(exactRankTests) Package ?exactRankTests? is no longer under development. Please consider using package ?coin?

Dear forum users, It's 3:35am and I am swamped with statistics homework lol I'm terrible with R and this time I have no idea what the prof wants. Here is the question: Consider the (two-??sample) Wilcoxon rank statistic T = ?rank(Xi). For n1=106 and n2=192, determine by simulation the ?=.05 critical point for testing H0: ?=0, H1:?<0. We can do this as follows: For m=10000 (no wimpy

Hello, Gurus: I tried to test if the sample mean of a dataset is zero. The data has 1500 numbers with a lot of zeros and some small positive numbers. The data range on [0,1] but the distribution is unknown. It is zero inflated anyway. I tried to use the Wilcoxon Signed Ranks test. But I read from this website that it does assume the population pdf is symmetric.

Dear All, is there a stratified version of the Wilcoxon test (also known as van Elteren test) available in R? I could find it in the survdiff function of the survival package for censored data. I think, it should be possible to use this function creating a dummy censoring indicator and setting it to not censored, but may be there is a better way to perform the test. Thanks, Heinz T??chler

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, 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

I am contemplating improving my ctest package as follows: * Add exact p,q,r,s functions for the Wilcoxon distribution, and change the test accordingly (make `exact' work). * Make Fisher's test work for tables larger than 2 by 2. * Perhaps add an exact unconditional test for 2 by 2 tables? * Perhaps add something on estimating/testing relative risk and odds? As clearly I'd like to