similar to: glm questions

> -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of David Firth > Sent: Tuesday, March 16, 2004 1:12 PM > To: Paul Johnson > Cc: r-help at r-project.org > Subject: Re: [R] glm questions > > > Dear Paul > > Here are some attempts at your questions. I hope it's of some help.

Following p.206 of "Statistical Models in S", I wish to change the code for summary.glm() so that it estimates the dispersion for binomial & poisson models when the parameter dispersion is set to zero. The following changes [insertion of ||dispersion==0 at one point; and !is.null(dispersion) at another] will do the trick: "summary.glm" <- function(object, dispersion =

I've noticed that glm and lrm give extremely different results if you attempt to fit a saturated model to a dataset with zero cells. Consider, for instance the data from, Agresti's Death Penalty example [0]. The crosstab table is: , , PENALTY = NO VIC DEF BLACK WHITE BLACK 97 52 WHITE 9 132 , , PENALTY = YES VIC DEF BLACK WHITE BLACK 6 11

I am running a binomial glm with response variable the no of mites of two species y->cbind(mitea,miteb) against two continuous variables (temperature and predatory mites) - see below. My model shows overdispersion as the residual deviance is 48.81 on 5 degrees of freedom. If I use quasibinomial to account for overdispersion the dispersion parameter estimate is 2501139, which seems

Hello, I am working on my thesis and can't really figure out how to produce a reasonable graph from the output from my glm., I could just give the R-output in my results and then discuss them, but it would be more interesting if I could visualise what is going on. My research is how bees react to different fieldmargins, for this I have 4 different types of field margin (A,B,C & D) and

Hi I am using R version 2.7.2. on a windows XP OS and have a question concerning an analysis of covariance with count data I am trying to do, I will give details of a scaled down version of the analysis (as I have more covariates and need to take account of over-dispersion etc etc) but as I am sure it is only a simple problem but I just can't see how to fix it. I have a data set with count

Dear R users, I am having problems trying to fit quasipoisson and negative binomials glm. My data set contains abundance (counts) of a species under different management regimens. First, I tried to fit a poisson glm: > summary(model.p<-glm(abund~mgmtcat,poisson)) Call: glm(formula = abund ~ mgmtcat, family = poisson) . . . (Dispersion parameter

Dear all, I am new to R and my question may be trivial to you... I am doing a GLM with binomial errors to compare proportions of species in different categories of seed sizes (4 categories) between 2 sites. In the model summary the residual deviance is much higher than the degree of freedom (Residual deviance: 153.74 on 4 degrees of freedom) and even after correcting for overdispersion by

List: I'm helping a colleague with some Poisson regression modeling. He uses SAS proc GENMOD and I'm using glm() in R. Note on the SAS and R output below that our estimates, standard errors, and deviances are identical but what we get for likelihoods differs considerably. I'm assuming that these must differ just by some constant but it would be nice to have some confirmation

There may be a bug in the anova.glm function. deathstar[32] R R : Copyright 2004, The R Foundation for Statistical Computing Version 2.0.1 (2004-11-15), ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project

Ted Harding said: > I can get the estimated RRs from > RRs <- exp(summary(GLM)$coef[,1]) > but do not see how to implement confidence intervals based > on "robust error variances" using the output in GLM. Thanks for the link to the data. Here's my best guess. If you use the following approach, with the HC0 type of robust standard errors in the

THANKS so very much for your help (previous and future!). I have a two follow-up questions. 1) You say that dispersion = 1 by definition ....dispersion changes from 1 to 13.5 when I go from binomial to quasibinomial....does this suggest that I should use the binomial? i.e., is the dispersion factor more important that the 2) Is there a cutoff for too much overdispersion - mine seems to be

I have been implementing a proposal of Jim Lindsey for glm(.) to return AIC values, and print.glm(.) and print.summary.glm(.) printing them.... however: >>>>> "Jim" == Jim Lindsey <jlindsey@luc.ac.be> writes: Jim> The problem still remains of getting the correct AIC when the user Jim> wants the scale parameter to be fixed. (The calculation should

I'm confused as to the trustworthiness of the dispersion parameters reported by glm. Any help or advice would be greatly appreciated. Context: I'm interested in using a fitted GLM to make some predictions. Along with the predicted values, I'd also like to have estimates of variance for each of those predictions. For a Gamma-family model, I believe this can be done as Var[y] =

Hello, The estimate of glm dispersion can be based on the deviance or on the Pearson statistic. I have compared output from R glm() to another statastical package and it appears that R uses the Pearson statistic. I was wondering if it is possible to make use R the deviance instead by modifying the glm(...) function? Thanks for your attention. Kind regards, Robin Smit This e-mail and its

Dear R users, I've notice that there are two ways to conduct a binomial GLM with binomial counts using R. The first way is outlined by Michael Crawley in his "Statistical Computing book" (p 520-521): >dose=c(1,3,10,30,100) >dead = c(2,10,40,96,98) >batch=c(100,90,98,100,100) >response = cbind(dead,batch-dead) >model1=glm(y~log(dose),binomial)

Hello R users I like to extract z values for x1 and x2. I know how to extract coefficents using model$coef but I don't know how to extract z values for each of independent variable. I looked around using names(model) but I couldn't find how to extract z values. Any help would be appreciated. Thanks TM ######################################################### >summary(model) Call:

I am doing a reanalysis of results that have previously been published. My hope was to demonstrate the value of adoption of more modern regression methods in preference to the traditional approach of univariate stratification. I have encountered a puzzle regarding differences between I thought would be two equivalent analyses. Using a single factor, I compare poisson models with and without

My data has three factors, a discrete response and an offset column. >From the summary of a glm object derived using three way interactions, I get this deviance information: (Dispersion parameter for poisson family taken to be 1) Null deviance: 39244 on 896 degrees of freedom Residual deviance: 11913 on 795 degrees of freedom AIC: 13905 Number of Fisher Scoring iterations: 6 The

glm(S ~ -1 + Mdif, family=quasipoisson(link=identity), start=strt, sdat) gives error: > Error in glm.fit(x = X, y = Y, weights = weights, start = start, etastart > = > etastart, : > cannot find valid starting values: please specify some strt is set to be the coefficient for a similar fit glm(S ~ -1 + I(Mdif + 1),... i.e. (Mdif + 1) is a vector similar to Mdif. The error