similar to: ks.test - continuous vs discrete

Dear All I am trying to replicate a numerical application (not computed on R) from an article. Using, ks.test() I computed the exact D value shown in the article but the p-values I obtain are quite different from the one shown in the article. The tests are performed on a sample of 37 values (please see "[0] DATA" below) for truncated Exponential, Pareto and truncated LogNormal

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 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)

Hello, I have a bunch of files containing 300 data points each with values from 0 to 1 which also sum to 1 (I don't think the last element is relevant though). In addition, each data point is annotated as an "a" or a "b". I would like to know in which files (if any) the data is uniformly distributed. I used Google and found out that a Kolmogorov-Smirnov or a Chi-square

> x <- runif(100) > y <- runif(100) > ks.test(x,y) Two-sample Kolmogorov-Smirnov test data: x and y D = 0.11, p-value = 0.5806 alternative hypothesis: two-sided ok I expected that, but: > ks.test(runif(100), "runif") One-sample Kolmogorov-Smirnov test data: runif(100) D = 0.9106, p-value < 2.2e-16 alternative hypothesis: two-sided How

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 .... >

Hello, While doing test of normality under R and SAS, in order to prove the efficiency of R to my company, I notice that Anderson Darling, Cramer Van Mises and Shapiro-Wilk tests results are quite the same under the two environnements, but the Kolmogorov-smirnov p-value really is different. Here is what I do: > ks.test(w,pnorm,mean(w),sd(w)) One-sample Kolmogorov-Smirnov test data: w D

Hi, I'm using the ks.test() on two vectors. I looked up the reference and also coded up a version of the two sample Smirnov test. My question is that how can I decide from the output of R that the two vectors x & y come from the same distribution? Am I correct in assuming that smaller D values indicate that they come from the same distribution? In addition how can I use the p value that

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 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 am interested in a statistical comparison of multiple (5) time series' generated from modeling software (Hydrologic Simulation Program Fortran). The model output simulates daily bacteria concentration in a stream. The multiple time series' are a result of varying our representation of the stream within the model. Our main question is: Do the different methods used to represent a

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]]

Dear all, One can use ks.test(x,y) for a one-sample kolmogorov-smirnov test: x being the data sample y being a string specifying a distribution I notice the help on ks.test does not tell you how to get such a list. Is this a hole in my R knowledge? Where can I get a list of the strings specifying the possible distributions? and more specifically What would be the string and following

Hi! how can I do the Kolmogorov Smirnov test for discrepancy between the estimated and empirical tails? Regards Anuradha [[alternative HTML version deleted]]

Hello all, Would somebody be kind enough to show me how to do a KS test in R for a lognormal distribution with ESTIMATED parameters. The R function ks.test()says "the parameters specified must be prespecified and not estimated from the data" Is there a way to correct this when one uses estimated data? Regards, Kwabena. -------------------------------------------- Kwabena Adusei-Poku

Based on a report to the Windows maintainers from Richard Rowe <Richard.Rowe@jcu.edu.au>: NEWS for 1.3.0 says o Exact p-values are available for the two-sided two-sample Kolmogorov-Smirnov test. I think the (new) p-values are computed but are backwards: > set.seed(123) > x <- rnorm(50) > y <- runif(50) > ks.test(x,y, exact=T)$p [1] 1 > 1 - ks.test(x,y,

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

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

The note to 1004 says "fixed for 1.3.1" Uh. No. It ain't. The problem was more serious than guessed as even the simplest testing would show. For example, Example 5.4 in Hollander and Wolfe (Nonparametric Statistical, Methods, 2nd ed., Wiley, 1999, pp. 180-181) R Version 1.3.1 (SuSE Linux 7.1) > X <-

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