similar to: MANOVA power, degrees of freedom, and RAO's paradox

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

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 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

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.? 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

> > After running a MANOVA test, I used univariate Fs in order to see > which IV(s) was contributing the difference. > However, I have been critised about it, > and the reason I was given is: > > "MANOVA allows you to explore multiple associations but > does not excuse you from tracing out the causal relationships > for each variable. This does

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 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 everybody ... I have to perform Hotteling's T^2 Test - more generally a MANOVA - on a set of data. Is there a (simple?) possibility to do it in R? The somewhat obvious way to do it would be > summary(aov(Y ~ x1 + x2 ... + xn)) where Y would be a two-column matrix. But this does not work in any possible combination of matrix or factor dimensions! Is it principally not possible

Hello, I've been playing with the manova() function to do some pretty straightforward multivariate analyses, and I can't for the life of me figure out how to get at the discriminant functions used. When predicting several variables simultaneously, it's important to be able to gauge how much each variable is contributing to the analysis...a simple p-value isn't really enough. I

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,

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

Hi, Does anyone know what the error message mean? > Boot2.Pillai <- function(x, ind) { + x <- as.matrix(x[,2:ncol(x)]) + boot.x <- as.factor(x[ind, 1]) + boot.man <- manova(x ~ boot.x) + summary(manova(boot.man))[[4]][[3]] + } > > man.res <- manova(as.matrix(pl.nosite) ~ + as.factor(plankton.new[,1]))$residuals > boot2.plank <-

Dear colleagues, I had a question wrt the car package. How do I evaluate whether a simpler multivariate regression model is adequate? For instance, I do the following: ami <- read.table(file = "http://www.public.iastate.edu/~maitra/stat501/datasets/amitriptyline.dat", col.names=c("TCAD", "drug", "gender", "antidepressant","PR",

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

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 All, I am trying to perform MANOVA. I have table with 504 columns(species) and 36 rows) with two grouping (season and location) Zx <- Z[c(4:504)] Zxm <- as.matrix(Z) m<- manova(Zxm~Season*location, data=Z) when I do summary.aov, I get respond for each species but summary.manova summary.manova(m) :" residuals have rank" 24<501. What can it be the reason for this error

Hello all, I have a question on manova in R: I'm using the function "manova()" from the stats package. Is there anything like a stepwise (backward or forward) manova in R (like there is for regression and anova). When I enter: step(Model1, data=Mydata) R returns the message: Error in drop1.mlm(fit, scope$drop, scale = scale, trace = trace, k = k, : no 'drop1'