similar to: Adding parameters for Benchmark normal distribution in shapiro.test

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, I am trying to create a table with information from Shapiro Wilk's test of normality. However, it fails due to lack of sample size, it says, but the way I see it, this is not a problem. (See the table of sample sizes (almost) at the bottom). Applying a different function using a similar ftable call is not a problem (See the bottom table). This is R 2.1.0 on Linux (Gentoo). /Fredrik

Hi, from what you're writing: "The logaritmic transformation "shapiro.test(log10(y))" says: W=0.9773, p-value= 2.512e-05." it seems the log-values are not distributed normally and so original data are not distributed like a log-normal: the p-value is extremally small! Other tests for normality are available in package: nortest compare the log-transformation of your ecdf

Hi, I couldn't figure out how to use the functions from the ctest library. I'm using the r-base package that comes with debian potato. library("ctest") told me that no such package existed. I checked the CRAN, but no such package was availiable, instead I was told that it would be part of the standard installation. But functions from ctest like shapiro-wilk don't work. The

I have a large amount of data which I break down into a collection of vectors of 100-125 values each. I would like to test the normality of the vectors and compare them. In the interactive mode I can test any one vector using the Shapiro-Wilk test or the Kolmogorov-Smirnov test. My problem is that when I try to write out the results to a file, the term output is a mixture of alpha characters

Hi all, I have two very large samples of data (10000+ data points) and would like to perform normality tests on it. I know that p < .05 means that a data set is considered as not normal with any of the two tests. I am also aware that large samples tend to lead more likely to normal results (Andy Field, 2005). I have a few questions to ensure that I am using them right. 1) The Shapiro-Wilk

Hello: Yes I know that sort of questions comes up quite often. But with all due respect I din't find how to perform what I want. I am searching archives and bowsing manuals but it isn't there, though, it is a ridiculous simple task for the experienced R user. I have data and can do the following with them: == hist(y, prob=TRUE) lines(density(y,bw=0.03) == The result actually is a

Hi all T gurus, I would like to test if my dataset is indeed from N(0, 0.011908969). K.S. test gives following result: > ks.test(data, "pnorm", 0, 0.011908969) One-sample Kolmogorov-Smirnov test data: data D = 0.1092, p-value = 1.318e-05 alternative hypothesis: two-sided How ever "Shapiro-Wilk" test give following : >

Hi everybody, somehow i dont get the shapiro wilk test for normality. i just can?t find what the H0 is . i tried : shapiro.test(rnorm(5000)) Shapiro-Wilk normality test data: rnorm(5000) W = 0.9997, p-value = 0.6205 If normality is the H0, the test says it?s probably not normal, doesn ?t it ? 5000 is the biggest n allowed by the test... are there any other test ? ( i know qqnorm

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,

Greetings! I want to have the coefficients that R uses in shapiro.test() for the Shapiro-Wilk test for a prticular sample size, i.e. the a[i] in W = Sum(a[i]*x[i])/(Sum(x[i] - mean(x))^2) (where the x[i] are sorted). Two questions: Q1: Is there a readymade R function from which I can extract these? Q2: I was wondering if I might be able to modify the code for the function shapiro.test() so

Full_Name: Jason Polak Version: R version 2.5.0 (2007-04-23) OS: Xubuntu 7.04 Submission from: (NULL) (137.122.144.35) Dear R group, I have noticed a strange anomaly with the shapiro.test() function. Unfortunately I do not know how to calculate the shapiro test P values manually so I don't know if this is an actual bug. So, to produce the results, run the following code: pvalues = 0; for

Hello, I have applied the Shapiro test to a matrix with 26.925 rows of data using the following F1.norm<-apply(F1.n.mat,1,shapiro.test) I would now like to view and export a table of the p and W values from the Shapiro test, but I am not sure how to approach this. I have tried the following with errors. > write.table(x=F1.norm,file="I:/R_Work/F1/Shapiro.csv",

@Eric - with this approach I am getting below error : Error in FUN(X[[i]], ...) : unused argument (list()) On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote: > Hi Christofer, > You cannot assign to list(...). You can do the following > > myList <- list(...)[!names(list(...)) %in% 'mc.cores'] > > HTH, > Eric > > On Sun, Mar

R Users: My question is probably more about elementary statistics than the mechanics of using R, but I've been dabbling in R (version 2.2.0) and used it recently to test some data . I have a relatively small set of observations (n = 12) of arsenic concentrations in background groundwater and wanted to test my assumption of normality. I used the Shapiro-Wilk test (by calling shapiro.test()

Dear R-users Idea: Analysing tree height frequency with hist(), normal distribution (ks.test & shapiro.test) and skewness (package e1071 - thanks a lot for this useful package)as an indication of possible self-thinning in an experimental tree stand. Problem: Results from the ks.test and the shapiro.test are not comparable (see example of both tests). Tree height is a nice continuous

Estimados todos: Me es grato escribir a esta lista de ayuda para R, ya que comparto 100% la filosofía de Software libre en especial software en Estadística ya que es la carrera que estoy siguiendo. El motivo de este mensaje es por un par de dudas que no pude resolver: 1. He utilizado las funciones para realizar pruebas de normalidad (kolmogorov-smirnov, cuando n>50) y (Shapiro, cuando