similar to: ks.test help

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

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 ) {

Hi everyone, when I use the two sample Kolmogorov¨CSmirnov ks2Test like this: x=read.table("e:/x.txt") y=rstable(1000,alpha,beta,gamma,delta) I alway get results as follows: Warning messages: 1: In ks.test(x = x, y = y, alternative = "two.sided") : cannot compute correct p-values with ties 2: In ks.test(x = x, y = y, exact = TRUE, alternative = "two.sided")

Tu fichero tiene los decimales como puntos y no como comas como tu le indicas. Te dejo un ejemplo #--------------------------------------------------------------------------------------------------------------------- setwd(dir="c:/Users/usuario/Desktop/") library(outliers) filename<-"timediff.csv" time<-read.csv(filename, sep=";",header=TRUE,dec=".")

Hola Marta, si observas cualquier artículo de psicología esto es una práctica típica. Te digo psicología porque creo que tus datos son de ese tipo. Tienes bibiligrafía de artículos cientificos en las que se quitan valores siguiendo los criterios que te he dicho solo has de buscar bibliografía del campo en el que te mueces. Osea que nos quitarlos por que sí, aunque en cada area de ciencia tienes

I've recently been dealing with various CentOS images on AWS, and am being forcibly reminded that the "/root/anaconda-ks.cfg" has only a passing resemblance to whatever the kickstart configuration file actually contained. But getting a copy of the actual "ks.cfg" is invaluable for updating and testing variations of the kickstart setup, especially when manipulating disk

Hello! I need to compare 2 datasets whether they come from the same distribution. I use function ks.boot{Matching}. And what is the confidence level of the p-value, returned by ks.boot function? The code is: set=read.table("http://stella.sai.msu.ru:8080/~gala/data/testsets.csv", header=T,sep=',') set1=set[!is.na(set$set1),'set1']

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 made a file partition-ks.cfg and put it in our machine 192.168.5.17 under /var/www/html. When I install a new machine I just do a linux ks=http://192.168.5.17/partition-ks.cfg I have a couple questions though: 1) I have the following in it: clearpart --all part /boot --fstype ext3 --size=100 part /var --fstype ext3 --size=10000 part / --fstype ext3 --size=10000

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)

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, 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

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, 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, I am trying to load the library(ks), but I am getting the following error: Loading required package: KernSmooth KernSmooth 2.22 installed Copyright M. P. Wand 1997 Loading required package: mvtnorm Loading required package: rgl Loading required package: misc3d Error in lazyLoadDBfetch(key, datafile, compressed, envhook) : ReadItem: unknown type 241 In addition: Warning messages: 1:

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

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

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 =

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

> 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