similar to: Confusion about ks.test() handling of ties and exact vs approximate results

HI, I've noticed what I think is an incorrect behavior of stats::psmirnov function and consequently of ks.test when run in an exact mode. For example: psmirnov(1, sizes=c(50, 50), z=1:100, two.sided = FALSE, lower.tail = F, exact=TRUE) produces 2.775558e-15 However, the exact value should be 1/combination(100, 50), which is 9.9e-30. While the absolute error is small, the relative error is

Dear R-help, Hope this email finds you well. My name is Ziyan. I am a graduate student in Zhejiang University. My subject research involves ks.test in stats-package {stats}. Based on the code, I have two main questions. Could you provide me some more information? I download different versions of the r language source code through r language website (https://www.r-project.org/). By reading

Dear All, I wonder if there is now an implementation of a modified KS test that can handle ties? Steve. [[alternative HTML version deleted]]

This question is mainly aimed at Kurt Hornik as author of the ctest package, but I'm cc'ing it to r-help as I suspect there will be other valuable opinions out there. I have been attempting 2 sample Kolmogorov-Smirnov tests using the ks.test function from the ctest package (ctest v.0.9-15, R v.0.63.3 win32). I am comparing fish length-frequency distributions. My main reference for the

Hi there! I have two little different data. One is a computer test on people, the other is a paper and pencil test. two boxplots show me that the data is almost the same. So now I'd like to know if I could handle all data as one, by testing with ks.test: ==== > ks.test(el$angststoer, fl$angststoer) Two-sample Kolmogorov-Smirnov test data: el$angststoer and fl$angststoer D =

I frequently want to test for differences between animal size frequency distributions. The obvious test (I think) to use is the Kolmogorov-Smirnov two sample test (provided in R as the function ks.test in package ctest). The KS test is for continuous variables and this obviously includes length, weight etc. However, limitations in measuring (e.g length to the nearest cm/mm, weight to the nearest

Dear R users I am using KS test to compare two different distribution for the same variable (temperature) for two different time periods. H0: the two distributions are equal H1: the two distributions are different ks.test (temp12, temp22) Two-sample Kolmogorov-Smirnov test data: temp12 and temp22 D = 0.2047, p-value < 2.2e-16 alternative hypothesis: two-sided Warning message: In

Hi, I am trying the following: library(ismev) library(evd) fit <- gev.fit(x,show=FALSE) ks.test(x,pgev,fit$mle[1],fit$mle[2],fit$mle[3]) but I am getting: Warning message: cannot compute correct p-values with ties in: ks.test(x, pgev, fit$mle[1], fit$mle[2], fit$mle[3]) where x is: [1] 239 38 1 43 22 1 5 9 15 6 1 9 156 25 3 100 6 [18] 5 100

Full_Name: Thomas Waterhouse Version: 2.9.1 OS: OS X 10.5.7 Submission from: (NULL) (216.239.45.4) ks.test uses a biased approximation to the p-value in the case of the two-sample test with ties in the pooled data. This has been justified in R in the past by the argument that the KS test assumes a continuous distribution. But the two-sample test can be extended to arbitrary distributions by a

Full_Name: Andrew Grant McDowell Version: R 1.1.1 (but source in 1.3.0 looks fishy as well) OS: Windows 2K Professional (Consumer) Submission from: (NULL) (194.222.243.209) In article <xeQ_6.1949$xd.353840@typhoon.snet.net>, johnt@tman.dnsalias.com writes >Can someone help? In R, I am generating a vector of 1000 samples from >Bin (1000, 0.25). I then do a Kolmogorov Smirnov test

All, Happy World Cup and Wimbledon. This morning finds me with the first of my many daily questions. I am running a ks.test on residuals obtained from a regression model. I use this code: > ks.test(Year5.lm$residuals,pnorm) and obtain this output One-sample Kolmogorov-Smirnov test data: Year5.lm$residuals D = 0.7196, p-value < 2.2e-16 alternative hypothesis: two.sided Warning

I have to start by saying that I am new to R, so I might miss something crucial here. It seems to me that the results of friedman.test and ks.test are "wrong". Now, obviously, the first thing which crossed my mind was "it can't be, this is a package used by so many, someone should have observed", but I can't figure out what it might be. Problem: let's start with

Full_Name: t. avery Version: 2.0.0 OS: windows xp / Linux Submission from: (NULL) (131.162.134.159) ks.test does not produce the correct output. If given the script: d1 <- c(53.63984674,0.383141762,1.915708812,0.383141762,10.72796935,6.896551724,20.30651341,5.747126437,0) d1 d2 <- c(76.43312102,15.2866242,3.821656051,1.27388535,0,0.636942675,1.27388535,0.636942675,0.636942675) d2

I'm using ks.test() to compare two different measurement methods. I don't really know how to interpret the output in the absence of critical value table of the D statistic. I guess I could use the p-value when available. But I also get the message "cannot compute correct p-values with ties ..." does it mean I can't use ks.test() for these data or I can still use the D

Hi everybody, while performing ks.test for a standard exponential distribution on samples of dimension 2500, generated everytime as new, i had this strange behaviour: >data<-rexp(2500,0.4) >ks.test(data,"pexp",0.4) One-sample Kolmogorov-Smirnov test data: data D = 0.0147, p-value = 0.6549 alternative hypothesis: two.sided >data<-rexp(2500,0.4)

In message <Pine.GSO.4.31.0107010731110.7616-100000@auk.stats>, Prof Brian D Ripley <ripley@stats.ox.ac.uk> writes > >You do realize that the Kolmogorov tests (and the Kolmogorov-Smirnov >extension) assume continuous distributions, so the distribution theory >is not valid in this case? > >S-PLUS does stop you doing this: > >> ks.gof(o,

Dear r-helpers, I have a small dataset (n<50), and I want to compute the Hodges Lehmann exact confidence interval. So far, I know that "pairwiseCI" has the function "HL.diff". The description is as follows : HL.diff calculates the Hodges-Lehmann confidence interval for the difference of locations by calling wilcox.exact in package exactRankTests ; But when I check

Hi R Users, I have two vectors, x and y, of equal length representing two types of data from two studies. I would like to test if they are similar enough to use them interchangeably. No assumptions about distributions can be made (initial tests clearly show that they are not normal). Here some result: Two-sample Kolmogorov-Smirnov test data: x and y D = 0.1091, p-value < 2.2e-16 alternative

Dear List, after updating the exactRanksumTests package I receive a warning that the package is not developed any further and that one should consider the coin package. I don't find the signed rank test in the coin package, only the Wilcoxon Mann Whitney U-Test. I only found a signed rank test in the stats package (wilcox.test) which is able to calculate the exact pvalues but unfortunately

Hello all, A friend just showed me how ks.test fails to work with pbinom for small "size". Example: x<-rbinom(10000,10,0.5) x2<-rbinom(10000,10,0.5) ks.test(x,pbinom,10,0.5) ks.test(x,pbinom,size = 10, prob= 0.5) ks.test(x,x2) The tests gives significant p values, while the x did come from binom with size = 10 prob = 0.5. What test should I use instead ? Thanks, Tal