similar to: MANOVA / Hotelling's 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

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,

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

I am using rw0641 with Windows 98. In the help for aov it states that the formula can sepecify multiple responses for a "maov". The help doesn't give an example, Venables and Ripley 1997:381 doesn't either. It isn't mentioned in the scripts nor in the section R Complements nor in Rnotes. I tried aov(c(y1,y2,y3)~x) which seemed reasonable to me but doesn't work. How

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 R fans, I have got a difficult sounding problem. For fitting a linear model using continuous response and then for re-fitting the model after excluding every single variable, the following functions can be used. library(MASS) model = lm(perf ~ syct + mmin + mmax + cach + chmin + chmax, data = cpus) dropterm(model, test = "F") But I am not sure whether any similar functions is

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, 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

# Peter B. Mandeville kindly offered to send me code for Hotelling's T^2 # Test. Unfortunately there seems to be no route to his machine. # So i'm trying to reach him via the Mailing List. ------------------------------------------------------------------------ Sir, this morning i recieved your message about the availability of the code for Hotelling's Test. I hurried to find out

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

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.

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,

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 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, 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 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",

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

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

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 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 =