similar to: Repeated measures MANOVA

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

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

Hello all, Is there an r function that exists that will perform repeated measures MANOVAs? For example, let's say I have 3 DVs, one between-subjects IV, and one within-subjects IV. Based on the documentation for the manova command, a function like that below is not appropriate because it cannot take Error arguments. manova(cbind(DV1,DV2,DV3) ~ BetweenSubjectsIV * WithinSubjectsIV +

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 <-

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

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.? 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, 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 ! I try to obtain in R eigenvectors and canonical analysis on MANOVA results, but I don't find how to process? In particular, I would be interesting to obtain "standardized canonical coefficients" of the canonical variates. There analysis give some information on the correlation between response variates. My data are organised in 2 terms (one is continu, one is a

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 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

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,

Dear r-helpers, I have a query regarding use of contrasts in MANOVA. summary(manova(model)) gives me only result of test for overall difference. Would you be so kind and give me a hint how to get the same test statistics (e.g.Pillai's) and P values for the predefined contrasts? Best regards Ondrej Mikula -- Institute of Animal Physiology and Genetics Academy of Sciences of the Czech

I just finished a MANOVA test and got the following output: > summary(M, test="Pillai") Df Pillai approx F num Df den Df Pr(>F) as.factor(X) 3 1.1922 6.5948 36 360 < 2.2e-16 *** Residuals 129 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' '

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

Full_Name: Yves Rosseel Version: 2.2.1 OS: Submission from: (NULL) (157.193.116.152) The code below sketches a possible implementation of a function 'mlh.mlm' which I think would be a good complement to the 'anova.mlm' function in the stats package. It tests a single linear hypothesis of the form H_0: LBT'= 0 where B is the matrix of regression coefficients; L is a matrix

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

> > 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

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

Dear R-helpers, In a data set I got from a medical doctor there are six treatment groups and (about) 5 bivariate responses in each group. Using 'manova', it is easy to see significant differences in treatment effects, but the doctor is more interested in the correlation between the two responses (within groups). I'm willing to assume a common value over groups, and one way of