similar to: how to create matrix in a loop

Hi all, I have 205 rows with measurements for three categories of people. I want to generate stripplots for each of these rows. How can I do it without having to do them one by one. I am giving a sample dataset:- A B C A B C A B C A B C 10.34822 10.18426 9.837874 9.65047 8.020482 9.17312 6.349595 13.55664 5.286697 11.85409 2.827027 7.002696 11.54984 12.14591

Dear R-users, We expect to form N homogeneous groups of n features from an experimentation including N*n data. The aim is to prevent group effects. How to do that with R functionalitites ? Does anyone know any way enabling this ? Example : 100 patients are observed. 3 biochemical parameters are mesured for each one (Red and white globules ans glycemia). Patient RG RW

I was wondering if it was possible to do a normality test on a very large sample dataset. It is approx. 160,000 residual estimates from meshes modelling the brain surfaces of 50 subjects (25 patients). shapiro.test only works with at most 5000 points. Are there issues with very large samples sizes that I should be aware of? Cheers, -Morgan

R Users: My question is probably more about elementary statistics than the mechanics of using R, but I've been dabbling in R (version 2.2.0) and used it recently to test some data . I have a relatively small set of observations (n = 12) of arsenic concentrations in background groundwater and wanted to test my assumption of normality. I used the Shapiro-Wilk test (by calling shapiro.test()

I am having a hard time interpreting the results of the 'shapiro.test' for normality. If I do ?shapiro.test I see two examples using rnorm and runif. When I run the test using rnorm I get a wide variation of results. Most of this may be from variability of rnorm, samll sample size (limited to 5000 for the test), etc but if I repeat the test multiple times I can get: >

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

Hi. I'm going to introduce the R-package for a group of medical doctors later this week and is a little confused about there use of a test named "willcoxon-pratt" for testing if the clinical and biochemical markers has decreased significantly after the use of some drugs for a group of patients. Looking into the R-functions I would in R recommand using a matched-pairs Wilcoxon

Hi, I am looking for a normality test in R to see if a vector of data I have can be assumed to be normally distributed and hence used in a linear regression. > help.search("normality test") suggests the Shapiro test, ?shapiro.test. Now maybe I am interpreting things incorrectly (as is usually the case), am I right in assuming that this is a composite test for normality, and hence a

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

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

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

Dear R-users Idea: Analysing tree height frequency with hist(), normal distribution (ks.test & shapiro.test) and skewness (package e1071 - thanks a lot for this useful package)as an indication of possible self-thinning in an experimental tree stand. Problem: Results from the ks.test and the shapiro.test are not comparable (see example of both tests). Tree height is a nice continuous

Dear friends. I wanted to reproduce a sheet of paper useful for plotting the reciprocal of some biochemical variable (creatinine) over time (Bleyer Am J Kidney Dis 34:576-578,1999). Time on X axis with units years and months. On Y axis, 1/creatinine, but with the original creatinine value indicated - to make it easier for people to use it. When using the graph, the linear decrease in 1/crea over

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]

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 all T gurus, I would like to test if my dataset is indeed from N(0, 0.011908969). K.S. test gives following result: > ks.test(data, "pnorm", 0, 0.011908969) One-sample Kolmogorov-Smirnov test data: data D = 0.1092, p-value = 1.318e-05 alternative hypothesis: two-sided How ever "Shapiro-Wilk" test give following : >

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

In the course of applying Shapiro-Wilk to 100,000 samples of 60 items from 100,000 different distributions, I encountered a fatal error in apply(). This can be reconstructed as follows, using the attached data file distr.dat containing 2 lines of my original 100,000-line file: > version _ platform Windows arch x86 os Win32 system x86, Win32 status

Hi all, I have two very large samples of data (10000+ data points) and would like to perform normality tests on it. I know that p < .05 means that a data set is considered as not normal with any of the two tests. I am also aware that large samples tend to lead more likely to normal results (Andy Field, 2005). I have a few questions to ensure that I am using them right. 1) The Shapiro-Wilk