similar to: Better idea than Poisson?

Good afternoon, I REALLY try to answer to my question as an autonomous student searching in the huge pile of papers on my desk and on the Internet but I can't find out the solution. Would you mind giving me some help? Please. ######################################### I'm trying to use glm with factors: > Pyr.1.glm<-glm(Pyrale~Trait,DataRav,family=poisson) If I have correctly

Hello, I have a question about modelling via glm. I have a dataset (see dput) that looks like as if it where poisson distributed (actually I would appreciate that) but it isnt because mean unequals var. > mean (x) [1] 901.7827 > var (x) [1] 132439.3 Anyway, I tried to model it via poisson and quasipoisson. Actually, just to get an impression how glm works. But I dont know how to

Hi, I wonder if anyone can answer the following or point me in the direction of how to obtain answers to the questions. Below is Output from R and further down are the questions raised and explanation of the study. Output from R: glm(formula = CB95TO00 ~ URB + INC, family = poisson) Deviance Residuals: Min 1Q Median 3Q Max -1.2272 -1.1290 0.2709 0.4272 2.1376

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

I have these questions: (1) Use Poisson regression to estimate the main effects of car, age, and dist (each treated as categorical and modelled using indicator variables) and interaction terms. (2) It was determined by one study that all the interactions were unimportant and decided that age and car could be treated as though they were continuous variables. Fit a model incorporating these

R gurus, I'm working on data analysis for a small project. My response variable is total vines per tree (median = 0, mean = 1.65, min = 0, max = 24). My predictors are two categorical variables (four sites and four species) and one continuous (tree diameter at breast height (DBH)). The main question I'm attempting to answer is whether or not the species identity of a tree has

Hi, I receive the following warning message when I run a poisson GLM in R: "glm.fit: fitted rates numerically 0 occurred" The model summary is shown below. The variable 'Species' consists of counts of different species ranging from 0 to 4. I suspect this may have something to do with the warning message but I'm not sure. Can anybody help? Thank you! Anna Call:

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 =

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

Hello, I am using the glm model with a poisson distribution. The model runs just fine but when I try to get the null deviance for the model of the null degrees of freedom I get the following errors: > null.deviance(pAmeir_1) Error: couldn't find function "null.deviance" > df.null(pAmeir_1) Error: couldn't find function "df.null" When I do: >

Greetings, I am running glm models for species counts using a poisson link function. Normal summary functions for this provide summary statistics in the form of the deviance, AIC, and p-values for individual predictors. I would like to obtain the p-value for the overall model. So far, I have been using an analysis of deviance table to check a model against the null model with the intercept as

Full_Name: saraux Version: 2.6.1 OS: Windows vista Submission from: (NULL) (193.157.180.37) I would like to compute a glm with a distribution of poisson, using a maximum of likelihood method. But it seems not to work with a distribution of poisson. The same code with another distrubution (binomial for example) works. Here is the command I typed:

Using anova of a glm with test = "Chisq", I get this: Analysis of Deviance Table Model: poisson, link: log Response: Days Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 373 370.56 Block 3 71.05 370 299.51 2.543e-15 Variety 1 94.04 369

Hi everyone, My background is not math and I am trying to figure out exactly what equation to use to map a response variable in GIS based on the coefficients obtained from the GLM and the values of the independent variables in each grid cell of my study area. Most specifically, I want to know how to incorporate the Poisson family and log link function in the equation. I would really appreciate if

Hi R-people, I performed controlled experiments to evaluated the seeds germination of two palms under four levels of water treatments. I conducted a generalized linear model (GLM) with a Poisson distribution to verify whether there were significant differences in the number of seed germination (NS-count variable) between treatments and species (explanatory variables). Thus, my model and output

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

It seems that MASS suggest to judge on the basis of sum(residuals(mode,type="pearson"))/df.residual(mode). My question: Is there any rule of thumb of the cutpoiont value? The paper "On the Use of Corrections for Overdispersion" suggests overdispersion exists if the deviance is at least twice the number of degrees of freedom. Are there any further hints? Thanks. -- Ronggui

Sorry about re-posting this, it never went out to the mailing list when I posted this to r-help forum on Nabble and was pending for a few days, now that I am subscribe to the mailing list I hope that this goes out: I've been a viewer of this forum for a while and it has helped out a lot, but this is my first time posting something. I am running glm models for richness and abundances. For

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

Greetings, everybody. Can I ask some glm questions? 1. How do you find out -2*lnL(saturated model)? In the output from glm, I find: Null deviance: which I think is -2[lnL(null) - lnL(saturated)] Residual deviance: -2[lnL(fitted) - lnL(saturated)] The Null model is the one that includes the constant only (plus offset if specified). Right? I can use the Null and Residual deviance to