similar to: "Reasonable doubt" - was "Re: shapiro wilk normality test"

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

What is the formula used in Shapiro-Wilk Statistic? Thanks Eduardo (S?o Paulo/ Brazil)

I'm pretty new to R and are trying to do some reliable normality testing... but, can't find the Shapiro Wilk test in R Does some experienced user have such a function that will be wanting to share with me? Or there is maybe some other way to get hte Shapiro Wilk test done... I'll appreciate any hint on this, Thanks -- *********************** Horacio Samaniego Dep. Ecologia P.

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,

Hi everybody, Does anyone know what problem may be with this test. I am applying 5 different normality tests and use p-values for them, but for some reason S-W gives me NA, while sample size is 100. Any ideas? Thanks a lot! [[alternative HTML version deleted]]

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

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

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

Is that really sufficient? See X86ISelLowering.cpp, look for addLegalFPImmediate. Usually targets have to tell legalizer what fp immediates are legl. Evan On Jan 5, 2010, at 8:38 AM, Anton Korobeynikov wrote: > Hello > >> I was wondering if it is possible to stop floating-point constants being converted to use the constant pool? As for our back-end we would like to be able to treat

I want to run Shapiro-Wilk test for each variable in my dataset, each grouped by variable groupFactor. I have these working commands: > data.n<-names(data) # put names into a vector called data.n > by(eval(parse(text=(paste("data",data.n[3],sep="$")))), data$factor, shapiro.test) #run shapiro.test but I must to change the variable number manualy. How to automate

Hello, I have a question about multivariate Shapiro-Wilks test. I tried to analyze if the data I have are multivariate normal, or how far they are from being multivariate normal. However, any time I did >mshapiro.test(mydata) I get the message: Error in solve.default(R %*% t(R), tol = 1e-18) : system is computationally singular: reciprocal condition number = 5.38814e-021 I tried

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.

Good morning and I appreciate the availability of a help-list. I am a professional hydrologist, but not a professional statistician. Yet I find myself using statistical tools at least part of the time. My discovery of the R-project through a friend has been most helpful. Here is my problem: I'm tasked with fitting a dataset comprising correlated discharges from adjacent watersheds to

Hello all, I tried several experiments with the mshapiro.test package in R and compared it with the energy package to test for multivariate normality and find that the mshapiro.test is not consistent which is a bit concerning and has suspicious behavior. On the other hand the energy test seems to be a more appropriate test for testing multivariate normality in any dimension. I looked for the

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]]

Hi, I need help to perform a Shapiro.test on a data frame, I know that this test works only with vector but I guess there most be a way to permor it on a data frame instead of vactor by vector (i.e. I've got 40 variables to analyze and its kinda annoying to do it one by one) Thanks to anyone that can help me. Gonzalo Quiroga

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: 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

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