search for: shapiro

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, 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 dat...

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 shapi...

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 f...

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...

...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() in R) and I'm not sure how to interpret the output. Here's the input/output from the R console: >As = c(13, 17, 23, 9.5, 20, 15, 11, 17, 21, 14, 22, 13) >shapiro.test(As) Shapiro-Wilk normality test data: As W = 0.9513, p-value =...

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...

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 (i in 1:17) { j = 1:(i+3); pvalues[i]=shapiro.test(j)$p; } plot(pvalues); print(pv...

Hi, I am not so sure about an error note I got when using shapiro.test. I imported some data into R by wrinting it into a .txt file via > tab1<-read.table("etc....txt",header=T) > attach(tab1) The following object(s) are masked _by_ .GlobalEnv : ozon > ozon$V1 [1] 2.5 3.0 5.6 4.7 6.5 6.7 1.7 5.3 4.6 7.4 5.4 4.1 5.1 5.6 5.4 6.1 7.6[18] 6...

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...

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....

Hi, In ?shapiro.test, there seems to be no option to pass mean and sd information of the Normal distribution which I want to compare my sample data to. For example in the code below, I want to test my sample to N(0, 10). shapiro.test(rnorm(100, mean = 5, sd = 3)) Is there any way to pass the information of the...

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 a...

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 <...

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 columns selected" ks.test is "Error in...

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 norma...

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]]

Why can't this function be used with the 'by' command? > x = array(runif(16), dim=c(8,2)) > x = data.frame(x) > x$group = rep(c('wt', 'app'), each=4) > shapiro.p = function(x) shapiro.test(x)[[2]] > apply(x[,1:2], 2, shapiro.p) X1 X2 0.4126345 0.2208781 > by(x[,1:2], x$group, shapiro.p) Error in `[.data.frame`(x, complete.cases(x)) : undefined columns selected [[alternative HTML version deleted]]

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 th...

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...