similar to: How to run Shapiro-Wilk test for each grouped variable?

Hi, I use two similar scripts to put p-values of shapiro.test and, respectively, of wilcox.test in a table: a) d <- data.frame(dataset$GroupFactor, dataset[2:11]) # p-values for the shapiro test (by levels of GroupFactor) with(d, aggregate(d[,-1], list(d[,1]), FUN = function(x) shapiro.test(x)$p.value)) b) d <- data.frame(dataset$GroupFactor, dataset[2:11]) with(d, aggregate(d[,-1],

Hi! I searched some time ago a way to get the Wilcoxon test results as a table more or less formatted. Nobody told me any solution and I found nothing on the Internet. Recently I came across this link ( http://myowelt.blogspot.com/2008/04/beautiful-correlation-tables-in-r.html), which helped me to find a solution. Here's the solution (I'm using R Commander): W <-

Dear R users, I have a dataset with two variables: $esan - a grouping factor with 8 levels and $reus. I'd like to do wilcox.test on this dataset as sugested Weiwei here: https://stat.ethz.ch/pipermail/r-help/2007-July/136627.html. I tried to adapt his recommendation but no succes. Can anyone help me? Regards, Iurie Malai, Senior Lecturer Department of Psychology Faculty of Psychology and

Hi, I have a data set: > Dataset X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 1 user1 m 22 19 28 24 12 18 9 7 4 5 4 7 5 7 9 2 user2 f 25 19 23 18 18 15 6 8 6 6 7 10 7 7 7 3 user3 f 28 21 24 18 15 12 10 6 7 9 5 10 5 9 5 4 user4 f 26 19 26 21 12 18 6 6 5 1 3 8 6 5 6 5 user5 m 21 22 26 18 9 6 4 6 1

Hi, All! To calculate Levene's Test for Homogeneity of Variance I use R Commander, and this is the output: > levene.test(Dataset$age, Dataset$sex) Levene's Test for Homogeneity of Variance Df F value Pr(>F) group 1 0.8739 0.3567 33 I am not sure what means "Pr(>F)"? Can anyone explain/translate this? Regards, Iurie Malai Department of Psychology and

How to make a launcher for Rattle? Regards, Iurie Malai Moldova Pedagogical State University [[alternative HTML version deleted]]

In the "Introduction and preliminaries" the "An Introduction to R" manual says about R: "... Among other things it has ... a well developed, simple and effective programming language (Called 'S') ... ". Now I'm a little confused. This means that language S is a component part of R? And S is not free? But R is free? Or the mentioned S is only "a free

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

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,

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

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'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.

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

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]]

I want to start JGR as a full sized/maximized window automatically, but how to do this? For other applications I can add option "--geometry 1440x900" to the launch command, but for JGR this do not work. Sugestions? Iurie Malai Senior Lecturer Psychology Department Ion Creanga Moldova Pedagogical State University [[alternative HTML version deleted]]

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.

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