similar to: KS test and theoretical distribution

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)

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

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 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 =

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

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

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 .... >

Is the kolmogorov-smirnov test valid on both continuous and discrete data? I don't think so, and the example below helped me understand why. A suggestion on testing the discrete data would be appreciated. Thanks, a <- rnorm(1000, 10, 1);a # normal distribution a b <- rnorm(1000, 12, 1.5);b # normal distribution b c <- rnorm(1000, 8, 1);c # normal distribution c d <- rnorm(1000,

Hi, I have a problem with Kolmogorov-Smirnov test fit. I try fit distribution to my data. Actualy I create two test: - # First Kolmogorov-Smirnov Tests fit - # Second Kolmogorov-Smirnov Tests fit see below. This two test return difrent result and i don't know which is properly. Which result is properly? The first test return lower D = 0.0234 and lower p-value = 0.00304. The lower '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

Hi, Interpretation problem ! so what i did is by using the: >fit1 <- fitdist(vectNorm,"beta") Warning messages: 1: In dbeta(x, shape1, shape2, log) : NaNs produced 2: In dbeta(x, shape1, shape2, log) : NaNs produced 3: In dbeta(x, shape1, shape2, log) : NaNs produced 4: In dbeta(x, shape1, shape2, log) : NaNs produced 5: In dbeta(x, shape1, shape2, log) : NaNs produced 6: In

Hi, why don't you try try ks.test(VeriSeti1, VeriSeti2)$p.value All the best Jenny >How can i print only the P-Value of the kolmogorov smirnov test? > > >> ks.test(VeriSeti1, VeriSeti2) > > Two-sample Kolmogorov-Smirnov test > >data: VeriSeti1 and VeriSeti2 >D = 0.5, p-value = 0.4413 >alternative hypothesis: two-sided > > >This expression

Hi, I want to plot the residuals of a least-squares regression. plot(lm(y~x), which=1) does this, but it plots the y-axis of my data on the x-axis of the residuals plot. That is, it plots the residual for each y-value in the data. Can I instead use the x-axis of my data as the x-axis of the residuals plot, showing the residual for a given x? Thanks! Jason Priem University of North

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 <-

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]]

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 r-users,   I would like to use Kolmogorov smirnov test but in my observed data(xobs) there are ties.  I got the warning message.  My question is can I do something about it?   ks.test(xobs, xsyn)           Two-sample Kolmogorov-Smirnov test data:  xobs and xsyn D = 0.0502, p-value = 0.924 alternative hypothesis: two-sided Warning message: In ks.test(xobs, xsyn) : cannot compute correct

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,

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

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