similar to: Stepwise regression for multivariate case in R?

Hello R list, I'm looking to do some stepwise discriminant function analysis (DFA) based on the minimization of Wilks' lambda in R to end up with a composite signature (of metals "Al","Sb","Bi","Cr","Ba") capable of discriminating 100% of the source factors (LANDUSE: "A","B","C"). The Wilks' lambda

I don't understand which is the problem Could you help me please? Thanks in advance Marta > str(data_indiciN2) 'data.frame': 200 obs. of 36 variables: $ gruppo: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ... $ I001N2: num 19.32 8.22 28.35 7.24 14.7 ... $ I002N2: num 2.92 2.54 0.11 1.6 7.12 ... $ I003N2: num -22.362 -0.222 -19.291

Hello, I have a question about multivariate Shapiro-Wilks test. I tried to analyze if the data I have are multivariate normal, or how far they are from being multivariate normal. However, any time I did >mshapiro.test(mydata) I get the message: Error in solve.default(R %*% t(R), tol = 1e-18) : system is computationally singular: reciprocal condition number = 5.38814e-021 I tried

Hi Since a stepwise procedure for variable selection (as e.g. in SPSS) for a LDA is not implemented in R and anyway I cannot be sure, that all the required assumptions for e.g. a procedure using a statistic based on wilks' lambda, hold (such as normality and variance homogeneity) I would like to ask you, what you would recommend me: shall I e.g. define a criterion such as the error-rate

Dear helpeRs, the following data set comes from Johnson/Wichern: Applied Multivariate Statistical Analysis, 6th ed, pp. 304-306. /X <- structure(c(9, 6, 9, 3, 2, 7), .Dim = as.integer(c(3, 2))) Y <- structure(c(0, 2, 4, 0), .Dim = as.integer(c(2, 2))) Z <- structure(c(3, 1, 2, 8, 9, 7), .Dim = as.integer(c(3, 2)))/ I would like to compute Wilk's Lambda in R, which I know is 0.0385.

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

Dear list member, My question is related to input file format to an Anova from car package. Here is an example of what I did: My file format is like this (and I dislike the idea that I will need to recode it): Hormone day Block Treatment Plant Diameter High N.Leaves SH 23 1 1 1 3.19 25.3 2 SH 23 1 1 2 3.42 5.5 1 SH 23 1 2 1 2.19 5.2 2 SH 23 1 2 2 2.17 7.6 2 CH 23 1 1 1 3.64 6.5 2 CH 23 1 1 2

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,

Hi, I wish to compute multivariate test statistics for a within-subjects repeated measures design with anova.mlm. This works great if I only have two factors, but I don't know how to compute interactions with more than two factors. I suspect, I have to create a new "grouping" factor and then test with this factor to get these interactions (as it is hinted in R News 2007/2), but

Good Day, I fit a multivariate linear regression model with 3 dependent variables and several predictors using the lm function. I would like to use stepwise variable selection to produce a set of candidate models. However, when I pass the fitted lm object to step() I get the following error: Error from R: Error in drop1.mlm(fit, scope$drop, scale = scale, trace = trace, k = k, : no

Hi, I'm doing anova on a matrix of multivariate data where I want to assess the effect of each column (element). My matrix is 86 rows x 31 columns. I've created a grouping factor of length 86 containing group assignments of 6 types. Then I run: x<- aov(matrix~grouping.factor) summary(aov.fit.raw, test="Wilks") This is working fine enough, but I'm getting different

In the meantime I found out that it should be possible to compute wilks' lambda from the singular values that are returned by "lda". But since I'm only a biologist I got totally stuck in formulas I don't understand ... Can anybody help me to get equivalent outputs of wilks' lambda in R as in SPSS? Yours, Christoph. -- Christoph Lange

Hi, I've got a dataset with 7 variables for 8 different species. I'd like to test the null hypothesis of no difference among species for these variables. MANOVA seems like the appropriate test, but since I'm unsure of how well the data fit the assumptions of equal variance/covariance and multivariate normality, I want to use a permutation test. I've been through CRAN looking at

Dear all, is there anyone who can help me extracting p-values from an Anova object from the car library? I can't seem to locate the p-values using str(result) or str(summary(result)) in the example below > A <- factor( rep(1:2,each=3) ) > B <- factor( rep(1:3,times=2) ) > idata <- data.frame(A,B) > fit <- lm( cbind(a1_b1,a1_b2,a1_b3,a2_b1,a2_b2,a2_b3) ? sex,

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

Hello, I am switching from SPSS-Windows to R-Linux. My university is very SPSS-oriented so maybe that's the cause of my problems. I am a beginner in R and my assignments are SPSS-oriented, so I hope I don't annoy anyone with my questions... Right now I've got 2 problems: -I have to evaluate discriminant functions I have calculated with lda(MASS). I can't find a measure that

Hi all, I am trying to fit a logistic regression for a bivariate response using five independent variables in a stepwise procedure. My outputs look okay but does any one know (or is there any literature on) how the confidence intervals are calculated for the reported odds ratios..? Thanks! [[alternative HTML version deleted]]

The test is manova. I tried to use manova() function, I used the code below:fit <- manova(Y ~ X)summary(fit, test="Wilks")but I get p values for intercept and regression coefficient as in anova() function, not for the hull model. Date: Mon, 7 Feb 2011 00:57:43 -0800 Subject: Re: [R] FW: multivariate regression From: djmuser@gmail.com To: denizsigirli@hotmail.com CC:

I have used the greedy.wilks to stepwise discriminant analysis, but it doesn't work with my dataset. I don't understand which is the problem Could you help me please? Thanks in advance Marta > str(data_indiciN2) 'data.frame': 200 obs. of 36 variables: $ gruppo: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ... $ I001N2: num 19.32 8.22 28.35 7.24

My stat textbook tells me that using Shapiro-Wilk test for each variable one by one is not equal to a test for multivariate normality as a whole. Does R have a function of testing for multivariate normality? Thanks. Hongsheng (Hank) Liao, Ph.D. Lab Manager Center for Quantitative Fisheries Ecology 800 West 46th Street Old Dominion University Norfolk, Virginia 23508 Phone:757.683.4571