similar to: Modified KS test to handle ties.

Hello, Today I was investigating ks.test() with two numerical arguments (x and y) and was left a bit confused about the policy behind handling ties. I might be missing something, so sorry in advance, but here is what confuses me: The documentation states: "The presence of ties always generates a warning, since continuous distributions do not generate them" But when I run a test with

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

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

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

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

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

Dear R helpers, I need to use KS and AD test for Generalized Pareto and Generalized extreme value. E.g. if I need to use KS for Weibull, I have teh syntax ks.test(x.wei,"pweibull", shape=2,scale=1) Similarly, for AD I use ad.test(x, distr.fun, ...) My problem is fir given data, I have estimated the parameters of GPD and GEV using lmom. But I am not able to find out the distribution

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

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

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

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)

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

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,

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

Hello, I want to do normality test on my data I write this but I don't understand the display of the results ks.test(data,"pnorm") In fact I want to know if my data is a normal distribution. I have to check the p-value or D? Thanks. _____________________________________________________________________________ l [[alternative HTML version deleted]]

Hello! I need to use Kruskal-Wallis test and post-hoc test (Dunn's test) for my data. But when I searched around, I only found this function: kruskal.test. But nothing for Dunn's test. So I started to write one myself. But I do not know how to count ties in the data frame. I can use for loops but it seems long and unnecessary since the rank function actually knows the ties. So

On Sun, 1 Jul 2001 mcdowella@mcdowella.demon.co.uk wrote: > 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) Please upgrade: we've found a number of Win2k bugs and worked around them since then, let alone teh bug fixes and improvements in R .... >