similar to: help on ks.test and 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

Hello, I'm trying to test my data for normality. I enter the data (95ish species counts) run >ks.test (data,pnorm) and get a p- value <2.2e-16 But this seems to be the p-value no matter what the data I enter. (I have multiple datasets and am testing them all for normality). [Actually, I just entered a vector of 1's and the p-value changed.] When I use the >Shapiro.test command,

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, 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

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",

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

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, I am not so sure about an error note I got when using shapiro.test. I imported some data into R by wrinting it into a .txt file via > tab1<-read.table("etc....txt",header=T) > attach(tab1) The following object(s) are masked _by_ .GlobalEnv : ozon > ozon$V1 [1] 2.5 3.0 5.6 4.7 6.5 6.7 1.7 5.3 4.6 7.4 5.4 4.1 5.1 5.6 5.4 6.1 7.6[18] 6.2 6.0 5.5 5.8 8.2 3.1 5.8 2.6 Now

Hello, I was wandering if it is possible to perform on a dataframe called 'all' a shapiro wilk normality test for COUNTS by variable Group ACTIVITY? Could it be done using plyer? I saw an eg that applies to an array but not to a dataframe: lapply(split(dataset1$Height,dataset1$Group),shapiro.test) Any thoughts would be much appreciated. My dataframe is in shape: dat ACTIVIT

Hi all! I tried to perform Shapiro-Wilk test for my sample of 243 values. > Us [1] -10.4 -13.1 -12.2 38.1 -18.8 -13.3 -11.7 29.3 49.7 6.8 12.7 16.3 [13] 5.8 -0.7 -29.4 4.1 38.8 -1.4 8.8 15.6 32.9 -5.3 19.1 35.8 [25] 4.0 -1.5 0.6 -4.2 -10.0 -4.0 1.1 48.9 -21.0 -5.3 5.8 -10.8 [37] 21.9 8.2 -3.2 -3.9 -2.3 12.6 -4.7 -8.0 11.8 27.4 -9.5 -20.8 [49]

Dear all, I am trying to create beanplots from a dataset for which boxplot works fine. (MACOS, R 2.8.1 GUI 1.27 Tiger build 32-bit (5301)) I am getting the following error message: Error in shapiro.test(x) : sample size must be between 3 and 5000 I am not even sure why the shapiro.test is being used, but is there any workaround ? Thanks ! Markus [[alternative HTML version deleted]]

Can anybody tell me the exact meaning of the $statistic and $p.value calculated by shapiro.test? Unfortunately it is not covered in my few text books, and I cannot find the explanation in the R documentatiom or on-line. If I have a test statistic, T, which is Normally distributed with mean=m and sd=s under the null hypothesis, then I can convert T to a p-value (one-sided) using: p <- pnorm(T,

Hello, I have tested a distribution for normality using the Shapiro-Welch statistic. The result of this is the following: Shapiro-Wilk normality test data: mydata W = 0.9989, p-value = 0.8791 I know that the p-value > 0.05 (for my purposes) means that the data IS normally distributed but what I am not sure is with the W value, what values tell me that the data is normally

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

Hi The shapiro.test function outputs a value of the W statistic, which should be 1 if the distribution is normal, and a p-value for the test (as the documentation states). I'm a bit confused with some results. I'm getting a W=0.9977 and a p-value=0.1889. I was expecting that a W of 0.9977 would tell me that the distribution is normal so p-value should be small ... What am I missing ?

hey all, I'm just getting used to R and i'm having issues when it comes to reading my data in rows rather than columns. any good advice would be much appreciated ! here is the error: > data1 <- read.table(file.choose(),header=T) > x1 <- c(data1[1,1:5]) > shapiro.test(x1) Error in sort.int(x, na.last = na.last, decreasing = decreasing, ...) : 'x' must be atomic

"An Introduction to R", Venables and Smith, Version 1.6.2 (2003-01-10) http://cran.r-project.org/doc/manuals/R-intro.pdf has in its "Appendix A: A sample session", page 81, file.show("morley.tab") I get the response NO FILE morley.tab The following "Introduction to the R Project for Statistical Computing" www.itc.nl/~rossiter/teach/sstat14/

Hi all, I am newbie in using R software and also doing statistical test. I want to know if my data in in normal distribution. I have 2 groups of data and I did calculate Shapiro Wilks using R software. Here is the results: Group 1: W = 0.9206, p-value = 0.01683 Group 2: W = 0.9626, p-value = 0.4694 I am not quite sure what default confidence level (CF) is used in calculating Shapiro Wilks.