similar to: confidence intervals in Manova and Mancova in Splus

Dear all, I'm trying to compare two sets of variables, the first set is composed exclusively of numerical variables and the second regroups factors and numerical variables. I can't use a Manova because of this inclusion of numerical variables in the second set. The solution should be to perform a Mancova, but I didn't find any package that allow this type of test. I've already

Hello, I have subjects in 4 groups: X1, X2, X3, X4. There are 33 subjects in group X1, 35 in X2, 31 in X3, and 46 in group X4. I have 7 continuous response variables (actually integers, approximately normal) measured for each subject: Y1 to Y7, and two continuous covariates C1, C2 (they are both integers). I want to perform all pairwise comparisons for each response variable between groups.

Hi everybidy Does anybody knows the code in R relate to MANCOVA???I saw somewhere lm(Y`...) but ma not so sure.. Thanks in advance!! Paola -- View this message in context: http://r.789695.n4.nabble.com/MANCOVA-tp4647735.html Sent from the R help mailing list archive at Nabble.com.

Hello, I need to perform a mixed-model (with nesting) MANCOVA, using Type III sums of squares. I know how to perform each of these types of tests individually, but I am not sure if performing a mixed-model MANCOVA is possible. Please let me know. Erika <>< <>< <>< <>< <>< <>< <>< Erika Crispo, PhD candidate

Dear List, I am trying to analyze a dataset where I have 1 continuous between-item variable (C), and 2 factorial within-item variables (3- and 2-level: F3, F2). I'm interested in whether slope of C is different from 0 at different combinations of F3 and F2, and whether it varies between these combinations. And unfortunately I need a decent anova-like table with p-values. The reason is that

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

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 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 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 =

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

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'

Hi, I have a nested unbalanced data set of four correlated variables. When I do univariate analyses, my factor of interest is significant or marginally significant with all of the variables. Small effect size but always in the same direction. If I do a MANOVA instead (because the variables are not independent!) then my factor is far from being significant. How does that come about? I have

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