similar to: Contrasts in ANOVA table

Hello R Users, I am investigating the basic use of the LME function, using the following example; Response is Weight, covariate is Age, random factor is Genotype model.lme <- lme (Weight~Age, random=~ 1|Genotype) After summary(model.lme), I find that the estimate of Age is 0.098 with p=0.758. I am comparing the above model with the AOV function; model.aov <- aov (Weight~Age + Genotype)

Dear R Users, I suppose this is a school boy question, but here it is anyway. I'm trying to re-create the residuals for a poisson GLM with simulated data; x<-rpois(1000,5) model<-glm(x~1,poisson) my.resids<-(log(x)- summary(model)$coefficients[1]) plot(my.resids,residuals(model)) This shows that my calculated residuals (my.resids) are not the same as residuals(model). p 65 of

Dear Greg, Thanks for the new release. The decomposition of the SSQ is just what I need! Regards, Martin. Martin Hoyle, School of Life and Environmental Sciences, University of Nottingham, University Park, Nottingham, NG7 2RD, UK Webpage: http://myprofile.cos.com/martinhoyle >>> gregory_r_warnes at groton.pfizer.com 10/30/02 07:16PM >>> Version 0.7.3 of the gregmisc package

Hi, I'm trying to use the command "glm.nb" in library(MASS) to test for a significant difference in the aggregation parameter "theta" between the three levels of a factor. Any help gratefully received! Martin. Martin Hoyle, School of Life and Environmental Sciences, University of Nottingham, University Park, Nottingham, NG7 2RD, UK Webpage:

Hi, Can someone tell me how to normalize a data set so that the mean of the set is 0 and the variance is 1. As I understand, when you calculate the principle components of a data set through correlation as < princomp( dataset, cor=T ) > then a similar calculation is performed. I would like to know how I can perform such a calulation directly. Any help would be greatly appreciated. Many

Dear All, I'm not sure of the interpretation of interactions with contrasts. Can anyone help? I do an ANCOVA, dryweight is covariate, block and treatment are factors, c4 the response variable. model<-aov(log(c4+1)~dryweight+treatment+block+treatment:block) summary(model); Df Sum Sq Mean Sq F value Pr(>F) dryweight 1 3.947 3.947 6.6268 0.01076 *

Hello all. I have a simple barplot with sixteen different segments. When I plot my data, only five or six of the labels are showing in the x-axis. How do go get them all to show? Can I set them at a 45.degree angle? Thank you. Jess -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send

Hi, Please can someone tell me if this is correct for significance tests on count data; 1: If overdispersed, use quasipoisson, drop1(model, test="Chi"), 2: If not overdispersed, use (poisson or quasipoisson), drop1(model, test="Chi"). Thanks for your time, Martin. Martin Hoyle, School of Life and Environmental Sciences, University of Nottingham, University Park, Nottingham,

I am new to R. I am going to by one of the following book: 1. William N. Venables and Brian D. Ripley. Modern Applied Statistics with S-Plus. Third Edition. Springer, 1999. ISBN 0-387-98825-4. 2. The Fourth Edition of the book from point 1. 3. `S Programming' by W. N. Venables and B. D. Ripley Springer. ISBN 0-387-98966-8, 2000. I can only by one of the above books. Q1. I have found

Dear All, I'm rather a beginner on nested ANOVAs, so here goes with my 2 questions; Qu 1: I'm modelling the number of galls on a leaf (the response variable) as a function of; the tree on which I find the leaf, the branch on which I find the leaf. Then, the tree and the branch are both random factors, and I'm quite happy that I should write; aov(galls~tree/branch +

Hi all, I'm having some troubles with the Quantreg package. I am using R version 2.10.0, and have downloaded the most recent version of Quantreg (4.44) and SparseM (0.83 - required package). However, when I try to run an analysis (e.g. fit1<-rq(y~x, tau=0.5)) I get an error message saying that the function "rq" could not be found. I get the same message when I try to search

Hi all, Has anyone tested for FA in R? I need to seperate out the variance due to measurement error from variation between individuals (following Palmer & Strobeck 1986). Andy Higginson Animal Behaviour and Ecology Research Group School of Biology University of Nottingham NG7 2RD U.K. This message has been checked for viruses but the contents of an attachment may still contain

I want to form a 3x3 crosstabulation for the signs of two vectors (i.e. Negative, Zero, Positive). The problem is that I am simulating the data so for some iterations one of the categories is absent. Thus the resulting table shrinks to 3x2. I want it to be 3x3 with zero column corresponding to the missing category. Moreover, I have tried but failed to give the dimension names. -- Sohail Chand

Using the "ergoStool" data cited in Mixed-Effects Models in S and S-PLUS by Pinheiro and Bates as an example, we have ======== > library(nlme) > fm <- lme(effort~Type-1, data=ergoStool, random=~1|Subject) > summary(fm) Linear mixed-effects model fit by REML Data: ergoStool AIC BIC logLik 133.1308 141.9252 -60.5654 Random effects: Formula: ~1 | Subject

I'm looking for a tutorial or notes on the use of contrasts factor in linear model in R, I've found some mails and infos about in various documents about R, but I've probably missed a good review on this subject. -- Robert Espesser Laboratoire Parole et Langage UMR 6057, CNRS 29 Av. Robert Schuman 13621 AIX (FRANCE) Tel: +33 (0)4 42 95 36 26 Fax: +33 (0)4 42 59 50

Dear All, I wonder if anyone can advise me as to whether there is a consensus as to how the effect size should be calculated from GLIM models in R for any specified significant main effect or interaction. In investigating the causes of variation in infection in wild animals, we have fitted 4-way GLIM models in R with negative binomial errors. These are then simplified using the STEP procedure,

Is there a function/package that will allow you to test the hypothesis beta1 = x in a simple linear regression, where x is a constant? The AOV tests if beta1 in different from 0 but what if I want to know if beta1 one is different from 1? With best wishes and kind regards I am Sincerely, Corey A. Moffet Instructor Department of Range, Wildlife, and Fisheries Management Mail Stop 2125 Texas

Removing rows with NAs, using na.omit(), doesn't seem to be working for me. Dataset: > str ( ex10s ) 'data.frame': 2189576 obs. of 5 variables: $ LOPNR : int 58 58 58 58 64 64 64 64 64 64 ... $ DIAGNOS: Factor w/ 173 levels "F20","F200","F2000",..: 128 128 128 128 105 105 105 160 105 105 ... $ X_DATE : int 20060821 20061207 20080102 20090904

Hi guys, I am updating a package because of data format in data folder. So I just change an extension of a file to .txt ... nothing more. I get this error on the R CMD check ** help *** installing help indices ** building package indices ... Error in read.table(zfile, header = TRUE, as.is = FALSE) : more columns than column names ERREUR : installing package indices failed Note that the R CMD

Hi, I have a student trying to run R on an Apple (OS 10.3.9). She tried to cut-and-paste the data via the code: data<-read.table(pipe("pbpaste")) But she keeps getting the error message: 'error in pipe("pbpaste"): pipe connections are not available on this system' I do not know much about using an Apple. Has anyone run into this before? Does anyone have any ideas