similar to: question about shapiro.test()

Hi, I have plot a graphics by next way: par(mar=c(5,5,1,1), font.axis=3, ps=14, xaxs="i", yaxs="i", lab=c(18,18,7), font .lab=3); plot(l, b, type="n", xlim=c(0,360), ylim=c(-90,90), xlab="l, deg", ylab="b, deg"); points(l, b, cex=vel/60, col=1, pch=21); On this plot I have axis x (denoted by l) ranged in [0,360] with labeling step 20. But I

Hi, I am very new to R and I've been trying to work through the R book to gain a better idea of the code (which is also completely new to me). Initially I imputed my data from a text file and that seemed to work ok, but I'm trying to examine linear relationships between gdist and gair, gdist and gsub, m6dist and m6air, etc. This didn't work and I think it might have something to do

Hi R users, I'm here trying to understand correlated residuals in nonlinear estimation. I'm reading/studying the book Bates, D. M. and D. G. Watts, (1988), /Nonlinear regression analysis and its applications/, Wiley, NY. pages 92-94, trying to reproduce the figures and to find out the code in R to perform the necessary calculations. I also consulted Pinheiro and Bates, but without

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

Estimada comunidad tengo un problema del que no encuentro datos que me ayuden mucho en la web. Estoy haciendo calculos por grupos con data,table. Tengo un archivo (zp.res) con tres columnas que clasifican los datos (sol, con, dia) y una columna de datos numericos (media), de la siguiente forma: sol con dia media 1: con 0 1 -22.6 2: con 0 1 -36.6 3: con 0 1 -35.6 y

Hi, Try: set.seed(25) X<- matrix(sample(1:50,3*140,replace=TRUE),nrow=140) #either res1<-do.call(rbind,lapply(split(as.data.frame(X),((seq_len(nrow(X))-1)%/%10)+1),function(x) apply(x,2,mean))) #or ?res2<-t(sapply(split(as.data.frame(X),((seq_len(nrow(X))-1)%/%10)+1),colMeans)) ?identical(res1,res2) #[1] TRUE #orlibrary(itertools) it<-ihasNext(isplitRows(X,chunkSize=10))

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

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

Some day I may figure out how ggplot2 works. I am trying to plot 5 columns of data on a graph (similar to a simple matplot) =========================================================================== library(ggplot2) bmi <- structure(list(pct = 2:21, P10 = c(14.6, 14.5, 14.2, 13.9, 13.7, 13.7, 13.9, 14.2, 14.5, 14.8, 15.3, 15.9, 16.6, 17.2, 17.8, 18.1, 18.3, 18.4, 18.5, 18.6), P25 =

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

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 ?

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",

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]]

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 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

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, 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

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.

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.