similar to: Shapiro-Wilk cpoefficients: 2 Qs

# Why does expressing one function require(ctest) t.test # return only function (x, ...) UseMethod("t.test") <environment: namespace:ctest> # but expressing another function shapiro.test # returns more complete code? function (x) { DNAME <- deparse(substitute(x)) x <- sort(x[complete.cases(x)]) n <- length(x) if (n < 3 || n > 5000)

From: aguitatierra@hotmail.com Sent: Friday, April 05, 2013 11:47 PM To: r-help@r-project.org ; R Help Subject: Reversing data transformation Hi everybody, I would be very grateful if you could give me your thoughts on the following issue. I need to perform Box-Cox (bcPower€) transformation on my data. To do this, I calculated lambda using the function '€powerTransform'€.

Hi everybody, I would be very grateful if you could give me your thoughts on the following issue. I need to perform Box-Cox (bcPower€) transformation on my data. To do this, I calculated lambda using the function '€powerTransform'€. powerTransform(data) However, I got an error message when performing this function: Convergence failure: return code = 52 I was told by John Fox

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,

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.

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

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

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 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

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

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 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.

> library(Kendall) > Kendall(1:3,1:3) WARNING: Error exit, tauk2. IFAULT = 12 <<<<<< tau = 1, 2-sided pvalue =1 I believe Kendall tau is well-defined for this case and the reported value is correct; isn't it a bug to give a warning? (And if, e.g., the pvalue is not well-defined in this case, wouldn't it be better to return NA or NaN or something?) Also,

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]]

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

>>> Ted Harding <Ted.Harding at manchester.ac.uk> 14/07/2008 00:16 >>> >said: >What constitutes "reasonable doubt" can become a very interesting >question, but there are some crimes for which it has a definite >statistical interpretation Warning for potential courtgoers: "reasonable doubt" NEVER has a direct statistical interpretation in a

Hello, I am trying to call a FORTRAN subroutine from R. The Fortran code is @: http://lib.stat.cmu.edu/apstat/206 It performs a bivariate isotonic regression on a rectangular grid (m X n) matrix. I used the g77 compiler and successfully created a dll file and it also loads successfully from R. But somehow the programs fails to run properly. (I do get the correct result when I compile the

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

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",