similar to: p value in MANOVA

R users, I got a problem to analyze with probit model. What package contains the algorithm to do probit model. Lawrence N.M Kazembe Mathematical Sciences Department Chancellor College University of Malawi P.O. Box 280 Zomba Malawi Tel: (265) 524 222 ext 284 Fax: (265) 524 046 e-mail: lkazembe at chirunga.sdnp.org.mw url: kazembe.cjb.net kazembe.tsx.org

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 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

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

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

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 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 =

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

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.

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

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 All, I have questions about MANOVA which I am still not sure if appropriately I should use it. For example I have a data set like this: BloodPressure (BP) Weight Height 120 115 165 125 145 198 156 99 176 I know that BloodPressure is correlated with both Weight and Height, however colinearity exists between Weight and Height. When I use BP = Weight + Height