similar to: Manova in R vs. SAS

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

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 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 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.

Dear all; working with a 'fat' data set (700 variables / 50 samples) and trying to run a manova test on it (I'm aware that it's not the best option for this kind of data set) I got the error in the summary.manova function about the rank of the residuals (rank < # variables). Ok. The thing that I don't understand is why I don't get the same type of error in SAS. There

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

I'm running R 1.2.2. The help information for manova says that the result is "A list with components SS: A names list of sums of squares and product matrices. Eigenvalues: A matrix of eigenvalues, stats: A matrix of the statistics, approximate F value and degrees of freedom." However, when I run the example, with fit <- manova(Y ~ rate*additive), I find that fit$SS is NULL.

Hi, I've got a dataset with 7 variables for 8 different species. I'd like to test the null hypothesis of no difference among species for these variables. MANOVA seems like the appropriate test, but since I'm unsure of how well the data fit the assumptions of equal variance/covariance and multivariate normality, I want to use a permutation test. I've been through CRAN looking at

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 =

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

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

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, 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:

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,

Hi, I'm doing anova on a matrix of multivariate data where I want to assess the effect of each column (element). My matrix is 86 rows x 31 columns. I've created a grouping factor of length 86 containing group assignments of 6 types. Then I run: x<- aov(matrix~grouping.factor) summary(aov.fit.raw, test="Wilks") This is working fine enough, but I'm getting different

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

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, I have problem getting the standard deviation from the manova output. I have used the manova function: myfit <- manova(cbind(y1, y2) ~ x1 + x2 + x3, data=mydata) . I tried to get the predicted values and their standard deviation by using: predict(myfit, type="response", se.fit=TRUE) But the problem is that I don't get the standard deviation values, I only