similar to: interpreting Shapiro-Wilks test result

Hello, I have a question about multivariate Shapiro-Wilks test. I tried to analyze if the data I have are multivariate normal, or how far they are from being multivariate normal. However, any time I did >mshapiro.test(mydata) I get the message: Error in solve.default(R %*% t(R), tol = 1e-18) : system is computationally singular: reciprocal condition number = 5.38814e-021 I tried

Hi all, I can't figure out how to compute Wilks Lambda in a one way repeated measure design. My matrix looks like: > t2.m Blank ECR ENC UEA UED 1 -0.15 0.14 0.16 0.09 0.14 2 0.30 0.08 0.14 0.14 0.14 [...] where each row is a case and the columns are levels of one factor (named trial): > t2.fit <- manova(t2.m ~ 1) > summary(t2.fit, intercept=T,

I don't understand which is the problem Could you help me please? Thanks in advance Marta > str(data_indiciN2) 'data.frame': 200 obs. of 36 variables: $ gruppo: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ... $ I001N2: num 19.32 8.22 28.35 7.24 14.7 ... $ I002N2: num 2.92 2.54 0.11 1.6 7.12 ... $ I003N2: num -22.362 -0.222 -19.291

Dear All, I would be very appreciative of your help with the following 1). I am running multivariate multiple regression through the manova() function (kindly suggested by Professor Venables) and getting two different answers for test=c("Wilks","Roy","Pillai") and tests=c("Wilks","Roy",'"Pillai") as shown below. In the

Hi everybody, I was wondering if you can help me about a module. In fact, I'm looking for a package or module about Wilks' lambda criterion in R environment. I didn't find it in R website ( http://cran.cict.fr/web/packages/index.html#available-packages-W or http://search.cpan.org/faq.html). If this module exists, could you show me the command line. Thank you very much for your help,

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

>>> 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

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.

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,

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, using the lda-method from MASS-package I was wondering whether there is a built-in method for figuring out Wilks' Lambda? Searching the Web I found in the r-help archive a thread form june 2002, but it didn't help me. I understand I can use manova and its summary-method to get Wilks' Lambda on the screen, but I don't see the connection to lda from MASS. Or does

Dear R-users, I'm wondering how to obtain Wilks-lambda values when discriminant analyses have only one discriminant function (i.e. 2 categories to discriminate between). The use of manova(predictions~groups, test="Wilks") asks for multiple response and the use of anova(lm(predictions~groups), test="Wilks") simply does not consider the last term. (The alternative

I have used the greedy.wilks to stepwise discriminant analysis, but it doesn't work with my dataset. I don't understand which is the problem Could you help me please? Thanks in advance Marta > str(data_indiciN2) 'data.frame': 200 obs. of 36 variables: $ gruppo: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ... $ I001N2: num 19.32 8.22 28.35 7.24

Hello! As I remember there were questions concerning wilks' lambda lately. But I lost those e-mails. So I dare ask again. Is there a way of calculating wilks' lambda for lda? Christoph. -- Christoph Lange clange at epost.de Verhaltensbiologie, FU Berlin 838-55068 Haderslebener Str. 9, 12163 Berlin

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 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, 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 helpeRs, the following data set comes from Johnson/Wichern: Applied Multivariate Statistical Analysis, 6th ed, pp. 304-306. /X <- structure(c(9, 6, 9, 3, 2, 7), .Dim = as.integer(c(3, 2))) Y <- structure(c(0, 2, 4, 0), .Dim = as.integer(c(2, 2))) Z <- structure(c(3, 1, 2, 8, 9, 7), .Dim = as.integer(c(3, 2)))/ I would like to compute Wilk's Lambda in R, which I know is 0.0385.

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

Full_Name: Tao Shi Version: 1.7.0 OS: Windows XP Professional Submission from: (NULL) (149.142.163.65) > x [,1] [,2] [1,] 149 151 [2,] 1 8 > c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value > for(i in (1:20)){c2x<-c(c2x,chisq.test(x, simulate.p.value=T,B=100000)$p.value)} > c2tx<-chisq.test(t(x), simulate.p.value=T, B=100000)$p.value > for(i in