similar to: Use of "ks.test()" function

> -----Original Message----- > From: Peter Dalgaard > Sent: Thursday, May 06, 2004 4:32 PM > To: Janete Borges > Cc: r-help at stat.math.ethz.ch > Subject: Re: [R] help on ks.test > > "Janete Borges" <janeteborges at gmx.net> writes: > > > Dear All > > > > I need to test the goodness-of-fit of a (Negative) Exponential >

Hello r-group I have a question to the ks.test. I would expect different values for less and greater between data1 and data2. Does anybody could explain my point of misunderstanding the function? data1<-c(8,12,43,70) data2<- c(70,43,12,8) ks.test(data1,"pnorm") ks.test(data1,"pnorm",alternative ="less") #expected < 0.001

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

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 am testing if my data are distributed under Laplace's distribution, which I managed to do with: > library(rmutil) > ks.test(jitter(x), "plaplace", mean(x), sd(x)) Nevertheless, I am trying to bootstrap ks.test without any success. Something is wrong in my commands: > data = read.table("data.dat",T) > library(rmutil) > library(boot) > ks.laplace =

Hi all, How can we use ks.test() to evaluate goodness of fit of mixtures of distributions? For example I have the following dataset: > x [1] 176.1 176.8 259.6 171.6 90.0 234.3 145.7 113.7 105.9 176.2 168.9 136.1 [13] 109.2 110.3 164.3 117.7 131.3 163.7 200.4 196.4 196.2 168.6 190.4 127.5 [25] 136.0 114.2 112.0 91.9 333.4 295.5 172.0 293.3 91.7 289.7 118.8 55.1 [37] 161.9 233.9 197.7

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

I am trying to do a ks.test in R 2.6.2 (running on Mac OS X 10.4.11). In the help guides it specifies that the y variable can be a character string for the type of distribution I want. I am doing this on the residuals of a regression model, but I continue to get an error. This is some of the code I have tried: > ks.test(res,"Norm") Error in get(y, mode =

Hi, I have a dataframe with multiple (appr. 20) columns containing vectors of different values (different distributions). Now I'd like to create a crosstable where I compare the distribution of each vector (df-column) with each other. For the comparison I want to use the ks.test(). The result should contain as row and column names the column names of the input dataframe and the cells should

I am not sure whether I'm doing something wrong or there is a bug in the documentation of ks.test. Following the posting guide, as I'm not sure, I haven't found this in the bug tracker, and the R FAQ says that stats is an Add-on package of R, I think this is the place to send it. ?ks.test provides the example <QUOTE> # Does x come from a shifted gamma distribution with shape 3

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

I have three files of data which are available at http://zhangw.com/ R/, varied at the number of data. I tried to use R to analyze using shapiro.test, ks.test, and t.test. t.test ran as expected, however, when I run shapiro.test and ks.test commands, error message always occurred. Error message for shapiro.test is "Error in "[.data.frame"(x, complete.cases(x)) : undefined

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

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

Couldn't you do this by subtracting 0.5 + x from your y values and checking for normality with mean 0 and sd = 1 (using ks.test or another test of normality). If you fail, you'll have to do additional work to find out whether pairs with some particular x value (or range of x values) is causing the problem, but I think this fits the question as stated. Of course, if you have discrete x

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 =

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

Hi, I am performing GEV analysis on temperature/precipitation data and want to use the ks.boot function but I am unsure of how to implement it using the pgev distribution. For example: ks.test(data,pgev,shape,location,scale) ks.boot(data,(distribution parameters?),alternative="pgev",nboots=1000) Any advice? Apologies in advance if I have used the wrong email address. Regards, Louise

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 the first example you are performing a one-sample test against a continuous cumulative distribution (in this case a normal distribution). In the second case you are performing a two-sample test. You drew your values for x non-randomly by specifying fixed intervals along a normal distribution, but ks.test() just sees that you have provided two samples, not one sample and values along a