similar to: shapiro.test() output

R users: I'd like to get some insight on an error I encounter when attempting to work with two moderately sized sets of time series data. FYI - I'm using the following versions of R and supporting packages on a Windows 2000 OS: - R version 2.4.1 (2006-12-18) - zoo version 1.2-2 - chron version 2.3-10 The two time series I'm working with are from the summer of 2004 and are: 1.)

R Users: I'm wondering: Why does my logical vector becomes a numeric vector when stuffed into a data frame? How do I change this so that it's retained as a logical data type? I've tried a couple of things but to no avail. Here's my example code: # Exercise 4-1 in Non-Detects and Data Analysis. Dennis Helsel. 2005. # "Create two new variables in the Interval Endpoints

Hi, I try to boxplot following data on the subset of (V1,V3,V5,V7) and (V2,V4,V6,V8) V1 V2 V3 V4 V5 V6 V7 V8 2 4 6 7 12 33 43 53 how can I use boxplot function to plot it? thanks, William

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",

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 ?

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

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

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, 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 benchmark normal distribution?

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, 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] 6.2 6.0 5.5 5.8 8.2 3.1 5.8 2.6 Now

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

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,

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 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 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]

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

hey all, I'm just getting used to R and i'm having issues when it comes to reading my data in rows rather than columns. any good advice would be much appreciated ! here is the error: > data1 <- read.table(file.choose(),header=T) > x1 <- c(data1[1,1:5]) > shapiro.test(x1) Error in sort.int(x, na.last = na.last, decreasing = decreasing, ...) : 'x' must be atomic

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