similar to: MANOVA permutation testing

Hi, I have a function for doing permutation tests for Manovas, written with the help of folks here on the list. It seems to work ok, and I've found that there is indeed a significant difference among groups in my analysis. I want to follow up on this by testing for which pairs of groups are significantly different. Reasoning that since Wilks Lambda is a generalization of the Hotelling T test,

Hi all, I am experimenting the function "manova" in R. I tried it on a few data sets, but I did not understand the result: I used "summary(manova_result)" and "summary(manova_result, test='Wilks')" and they gave a bunch of numbers... But I need the Sum-of-Squares of BETWEEN and WITHIN matrices... How do I read off from the R's manova results? Any

Dear all, I need to do a Manova but I have an unbalanced design. I have morphological measurements similar to the iris dataset, but I don't have the same number of measurements for all species. Does anyone know a procedure to do Manova with this kind of input in R? Thank you very much, Naiara. -------------------------------------------- Naiara S. Pinto Ecology, Evolution and Behavior 1

Hello, I had a couple questions about manova modeling in R. I have calculated a manova model, and generated a summary.manova output using both the Wilks test and Pillai test. The output is essentially the same, except that the Wilks lambda = 1 - Pillai. Is this normal? (The output from both is appended below.) My other question is about the use of MANOVA. If I have one variable which has a

Dear R-users; Previously I posted a question about the problem of rank deficiency in summary.manova. As somebody suggested, I'm attaching a small part of the data set. #*************************************************** "test" <- structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)), levels = c("1", "2", "3"), class =

While I was helping a SAS-using friend with an analysis I noticed some differences in the multivariate test statistics, approximate F statistics, and p-values in the manova function using R and proc GLM using SAS. The univariate coefficients are identical. Is there a reason to expect R and SAS to give different results? Thanks, Bill Kristan.

Hi R-users I'm trying to do multivariate analysis of variance of a experiment with 3 treatments, 2 variables and 5 replicates. The procedure adopted in SAS is as follow, but I'm having difficulty in to implement the contrasts for comparison of all treatments in R. I have already read manuals and other materials about manova in R, but nothing about specific contrasts were found in them,

Hi.? There appears to be a bug in R function manova.? My friend and I both ran it the same way as shown below (his run) with the shown data set. His results are shown below. we both got the same results.? I was running with R 2.3.1. I'm not sure what version he used. Thanks very much, David Booth Kent State University -----Original Message----- From: dvdbooth at cs.com To: kberk at

Dear R users, Can anyone tell me how to get the p value out of the output from summary.manova? I tried all the methods I can think of, but failed. Many thanks Yu-Kang _________________________________________________________________ ¥ß§Y¥Ó½Ð MSN Mobile ªA°È¡G¦b±zªº¤â¾÷¤W¦¬µo MSN Hotmail http://msn.com.tw/msnmobile

Hi, I would like to conduct a MANOVA. I know that there 's the manova() funciton and the summary.manova() function to get the appropriate summary of test statistics. I just don't manage to specify my model in the manova() call. How to specify a model with multiple responses and one explanatory factor? If I type:

Hi there, Does anyone know how to extract elements from the table returned by Manova()? Using the univariate equivalent, Anova(), it's easy: a.an<-Anova(lm(y~x1*x2)) a.an$F This will return a vector of the F-values in order of the terms of the model. However, a similar application using Manova(): m.an<-Manova(lm(Y~x1~x2)) m.an$F Returns NULL. So does any attempt at calling the

Hi, Suppose I have: > summary(manova(plank.man)) Df Pillai approx F num Df den Df Pr(>F) plankton.new[, 1] 1 0.5267 9.8316 6 53 2.849e-07 *** Residuals 58 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 My understanding is the MANOVA summary returns a list.

Hello everybody After doing a MANOVA on a bunch of data, I want to be able to make some comment on the amount of variation in the data that is explained by the factor of interest. I want to say this in the following way: XX% of the data is explained by A. I can acheive something like what I want by doing the following: X <- structure(c(9, 6, 9, 3, 2, 7), .Dim = as.integer(c(3,

Dear friends, Sorry for this somewhat generically titled posting but I had a question with using contrasts in a manova context. So here is my question: Suppose I am interested in doing inference on \beta in the case of the model given by: Y = X %*% \beta + e where Y is a n x p matrix of observations, X is a n x m design matrix, \beta is m x p matrix of parameters, and e is a

I would like to test the hypothesis that the difference between pairs, for several variables, is zero. This is easily done separately for each variable with: lm(Y ~ rep(0, nrow(Y))) where Y is a matrix whose columns are the differences for each variable between pair members. However, I would like to get an overall probability across all variables from a Wilks or Pillai-Bartlett statistic as in

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

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.

I'm trying to learn to use manova(), and don't understand why none of the following work: > data(iris) > fit <- manova(~ Species, data=iris) Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : incompatible dimensions > fit <- manova(iris[,1:4] ~ Species, data=iris) Error in model.frame(formula, rownames, variables, varnames, extras,

Can R do a repeated measures MANOVA and tell what dimensionality the statistical variance occupies? I have been using MATLAB and SPSS to do my statistics. MATLAB can do ANOVAs and MANOVAs. When it performs a MANOVA, it returns a parameter d that estimates the dimensionality in which the means lie. It also returns a vector of p-values, where each p_n tests the null hypothesis that the mean

Hi all, I have received the following error from summary.manova: Error in summary.manova(manova.test, test = "Pillai") : residuals have rank 36 < 64 The data is simulated data for 64 variables. The design is a 2*2 factorial with 10 replicates per treatment. Looking at the code for summary.manova, the error involves a problem with qr(). Does anyone have a suggestion as to how to