similar to: shapiro wilk normality test

Hi all, though i know this is a simple question, i really hope someone could. I am just trying to compare percents respectively the difference in percents by a simple test. t.test usually compares means and is working like it is supposed to be. Now i wonder how i can use a test in R to test for significant difference between two percentages.

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 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]

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 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

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, 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 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

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.

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 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, Do you know a R implemented normality test like the shapiro test but more suitable for large data set ? Thanks, _________________________________________________________________ Découvrez toutes les possibilités de communication avec vos proches [[alternative HTML version deleted]]

Hi there, I am trying to perform different normality tests in R. I have no problem when I use Shapiro.test( ). However, when I try to use any normality tests from the Nortest package, I kept getting this error " Error in read.table(source, header = T, fill = T) : 'file' must be a character string or connection". I have no idea why because there was no problem with

Hi there, is there a possibility to count the number of appearances of an element in a vector ? i mean of any given element.. deliver all elements which are exactly xtimes in this vector ? thx in advance !!

>>> 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

Hi, this is my first help request so please bear with me. I've been running some normality tests using the nortest package. For some of my datasets the Cramer-von Mises normality test generates an extremely high probability (e.g., 1.637e+31) and indicates normality when the other tests do not. Is there something I'm misunderstanding or potentially a bug in the code? Below are the

Dear all, i have a problem with processing dataframes within a function using the "$". Here´s my code: recode_items = function(dataframe,number,medium=2){ # this works q<-paste("columna",number,sep="") # this does not work, particularly because "dataframe" is not processed # dataframe should be: givenframe$columnagivennumber

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

Dear all, i am stuck with a syntax problem. i have a matrix which has about 500 rows and 6 columns. now i want to kick some data out. i want create a new matrix which is basically the old one except for all entries which have a 4 in the 5 column AND a 1 in the 6th column. i tried the following but couldn´t get a new matrix, just some wierd errors:

I currently have a program that automates 2-way ANOVA on a series of endpoints, but before the ANOVA is carried out I want the code to test the assumptions of normality and equal variance and report along with each anova result in the output file.  How can I do this? I have pasted below the code that I currently use.   library(car) numFiles = x #