similar to: MANOVA

R (Windows version) users/developers: While I have checked R-help/-devel mailing lists and R-bug-tracking system, this problem has not been reported so far. I would like to know whether this problem is specific to my machine. Lately I downloaded the version 1.5.0 patched [for windows] from http://www.stats.ox.ac.uk/pub/bdr/RWin/Rpre/ [patched: 2002-06-05] and

Guido has fixed Rterm hanging on Windows 2000 in interactive and --ess modes. The problem was that 'PulseEvent' doesn't work under 2000 as documented (and I can find nothing about this in MicroSoft's Knowledge Base except that it is known to happen under a debugger). The cross-compiled pre-release at http://www.stats.ox.ac.uk/pub/bdr/RWin/Rpre now incoporates the fix, and we

Dear list, (I've not upgraded to R1.4 yet) I have the following $class component in a predict.qda object: > unique(mod23S.qda.pred$class) [1] 12 17 8 10 4 9 5 13 14 19 20 15 6 3 7 1 23 11 18 21 16 2 22 NA Levels: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Nevertheless, when I try to identify the individual(s) with NA, I get: >

I've put up a build of the current R-patched (and updated recommended packages) for Windows at http://www.stats.ox.ac.uk/pub/bdr/RWin/Rpre/SetupRpatched.exe This should be able to rename files on Win 9x (unlike the mid-May build) and can build packages accessing Rblas.dll. As ever, this is less tested than an official release, in particular only on Windows XP, but ought to have fewer

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

The planned release of R 1.1.0 will be next Thursday. The development version is going into feature freeze tomorrow, meaning that we'll only do testing and fix simple bugs during the next week. Those of you who prefer to find bugs before releases instead of afterwards may want to try it out and tell us what crawls out. -- O__ ---- Peter Dalgaard Blegdamsvej 3 c/ /'_

Hi there, I found a small problem after installing the recent version of R (1.5.0) for Windows. The link on its documentation (.PDF) does not work. Am I just the one who has this problem ? I am using Windows ME. Thank you. Baju Indradjaja Dept. of Astronomy, ITB Jl. Ganesha 10, Bandung INDONESIA -----------------------

Hi, I tried running Rcmd.exe to no avail. It doesn't even start because it tries to read some memory address that NT says it can't and terminate. I tried downloading and reinstalling rw1020sp.zip with the installer twice just in case it got corrupted somehow in transport but it resulted in the same problem. What should I do? platform i386-pc-mingw32 arch x86 os Win32 system

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'