similar to: p-values from multiple testing

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 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)

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,

Dear All, Suppose I have a categorical variable a=as.factor(sample(1:3,10,replace=T)) plot(a) and hist(as.numeric(a),freq=F) would give the histogram of it. But I do not know how to add the counts or percentage information for plot.factor(). hist() can do it but as a numeric variable, the x-axis is not 3 categories in this case. Thank you for any suggestion. Best wishes, Jie

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'

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

> 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

Hi, I have two sets of data, an observed data and generated data. The generated data is obtained from the model where the parameters is estimated from the observed data. So I'm not sure which to use either one-sample test ks.test(x+2, "pgamma", 3, 2) # two-sided, exact or two-sample test ks.test(x, x2, alternative="l") If I use the one-sample test I need to

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,

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

Although I saw this issue being discussed many times before, I still did not find the answer to: why does R can not calculate p-values for data with ties (i.e. - sample with two or more values the same)? Can anyone elaborate some details about how does R calculate the p- values for the Kolmogorov Smirnov test statistics? I can understand the theoretical problem that continuous distributions do

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, does the shapiro wilk test in R-1.3.0 work correctly? Maybe it does, but can anybody tell me why the following sample doesn't give "W = 1" and "p-value = 1": R> x<-1:9/10;x [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 R> shapiro.test(qnorm(x)) Shapiro-Wilk normality test data: qnorm(x) W = 0.9925, p-value = 0.9986 I can't imagine a sample being

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 list members, I'm quite new to R, and though I tried to find the answer to my probably very basic question through the available resources (website, mailing list archives, docs, google), I've not found it. If I try to use the "integrate" function from within my own functions, my functions seem to misbehave in some contexts. The following example is a bit silly, but

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 .... >

Dear All, I would like to store quote as part of an vector. For instance, I would like to get an character object as x = " 12"ab"34 " or y = c("1", "2", """, "a", "b", """, "3", "4") Is that possible? Thank you. Best wishes, Jie [[alternative HTML version deleted]]

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-users, I want to perform a One-Sample parametric bootstrapped Kolmogorov-Smirnov GoF test (note package "Matching" provides "ks.boot" which is a 2-sample non-parametric bootstrapped K-S version). So I wrote this code: ---[R Code] --- ks.test.bootnp <- function( x, dist, ..., alternative=c("two.sided", "less", "greater"), B = 1000 ) {

Dear All, I have a question, suppose X is a dataframe, with column names as "x1", "x2", "x3", ..... And I would like to use the i-th column by X[,'xi']. But it seems the single quote and double quote are different. So if I run X[, names(X)[i]], it has some error. Please use the below example code X = matrix(rnorm(50),ncol = 5) X = data.frame(X)