similar to: how to obtain p values from an ANOVA result

Hi, I was wondering why anova() and drop1() give different tail probabilities for F tests. I guess overdispersion is calculated differently in the following example, but why? Thanks for any advice, Tom For example: > x<-c(2,3,4,5,6) > y<-c(0,1,0,0,1) > b1<-glm(y~x,binomial) > b2<-glm(y~1,binomial) > drop1(b1,test="F") Single term deletions Model: y ~

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

Dear List, I have tried a stratified Cox Regression, it is working fine, except for the "Anova"-Tests: Here the commands (should work out of the box): library(survival) d = colon[colon$etype==2, ] m = coxph(Surv(time, status) ~ strata(sex) + rx, data=d) summary(m) # Printout ok anova(m, test='Chisq') This is the output of the anova command: > Analysis of Deviance Table

This is a follow-up to a query that was posted regarding some problems that emerge when running anova analyses for cox models, posted by Mathias Gondan: Matthias Gondan wrote: >* Dear List,*>**>* I have tried a stratified Cox Regression, it is working fine, except for*>* the "Anova"-Tests:*>**>* Here the commands (should work out of the box):*>**>*

Dear R users, I noticed a problem in the anova command when applied on a single coxph object if there are missing observations in the data: This example code was run on R-2.6.1: > library(survival) > data(colon) > colondeath = colon[colon$etype==2, ] > m = coxph(Surv(time, status) ~ rx + sex + age + perfor, data=colondeath) > m Call: coxph(formula = Surv(time, status) ~ rx +

Full_Name: Robert King Version: 1.5.0 OS: windows Submission from: (NULL) (134.148.4.19) Also occurs in 1.6.0 on linux anova.glm(fitted.object,test="Chisq") is giving strange answers in this situation > resptime sex task time 1 m s 210 2 m s 300 3 m s 420 4 f s 250 5 f s 310 6 f s 390 7 m c 310 8 m c 400 9 m c 600

Hi all, I have done a backward stepwise selection on a full binomial GLM where the response variable is gender. At the end of the selection I have found one model with only one explanatory variable (cohort, factor variable with 10 levels). I want to test the significance of the variable "cohort" that, I believe, is the same as the significance of this selected model: >

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

Dear useRs, I am using mgcv version 1.7-16. When I create a model with a few non-linear terms and a random intercept for (in my case) country using s(Country,bs="re"), the representative line in my model (i.e. approximate significance of smooth terms) for the random intercept reads: edf Ref.df F p-value s(Country) 36.127 58.551 0.644

Hi there - I am using R1.5.1 on WinNT and the latest MASS (Venables and Ripley) library. Running the following code: >minimod<-glm(miniSF~gtbt*f.batch+log(mxjd),data=gtbt,family="binomial") >summary(minimod,cor=F) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 0.91561 0.32655 2.804 0.005049 ** gtbtgt 0.47171

Hallo all, I have the following glm model: f1 <- as.formula(paste("factor(y.fondi)~", "flgsess + segmeta2 + udm + zona.geo + ultimo.prod.", "+flg.a2 + flg.d.na2 + flg.v2 + flg.cc2", " +(flg.a1 + flg.d.na1 + flg.v1 + flg.cc1)^2", " + flg.a2:flg.d.na2 + flg.a2:flg.v2 +

System: R 2.3.1 on Windows XP machine. I am building a logistic regression model for a sample of 100 cases in dataframe "d", in which there are 3 binary covariates: x1, x2 and x3. ---------------- > summary(d) y x1 x2 x3 0:54 0:50 0:64 0:78 1:46 1:50 1:36 1:22 > fit <- glm(y ~ x1 + x2 + x3, data=d, family=binomial(link=logit)) >

Dears useRs, I have 2 factors, (for the sake of explanation - A and B), with 4 levels each. I've already fitted a negative binomial generalized linear model to my data, and now I need to split the factors in two distinct analysis of deviance table:  - A within B1, A within B2, A within B3 and A within B4  - B within A1, B within A2, B within A3 and B within A4 Here is a code that illustrates

Hi, I have a problem with the df. I read in a big csv file. Tabelle <- read.csv("C:\\Users\\Public\\Documents\\Bachelorarbeit\\eingabe8_durchnummeriert.csv" , header = T , sep=";") then I try this: > ygamma <- glm(Tabelle$sb_ek_ber ~1+ Tabelle$FAHRL_C + Tabelle$NUTZKREIS + Tabelle$schw_drittel_c   , family = Gamma) >  anova(ygamma, test="Chisq")

Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared

Dear list, I want to perform a logistic regression analysis with multiple categorical predictors (i.e., a logit) on some data where there is a very definite relationship between one predicator and the response/independent variable. The problem I have is that in such a case the p value goes very high (while I as a naive newbie would expect it to crash towards 0). I'll illustrate my problem

We are using GAM in mgcv (Wood), relatively new users, and wonder if anyone can advise us on a problem we are encountering as we analyze many short time series datasets. For each dataset, we have four models, each with intercept, predictor x (trend), z (treatment), and int (interaction between x and z). Our models are Model 1: gama1.1 <- gam(y~x+z+int, family=quasipoisson) ##no smooths Model

Hi Peter and others, If it helps, I wrote a small function glm.scoretest() for the statmod package on CRAN to compute score tests from glm fits. The score test for adding a covariate, or any set of covariates, can be extracted very neatly from the standard glm output, although you probably already know that. Regards Gordon --------------------------------------------- Professor Gordon K

Hi, I?ve got this model > model<-glm(prevalence~agesex+agesex:month,binomial) and the output of anova is like that > anova(model,test="Chisq") Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 524 206.97 agesex 2 9.9165 522 197.05 0.007025 ** agesex:month 9 18.0899

Hi, I have a dichotomous variable (Q1) whose answers are Yes or No. Also I have 2 categorical explanatory variables (V1 and V2) with two levels each. I used logistic regression to determine whether there is an effect of V1, V2 or an interaction between them. I used the R and SAS, just for the conference. It happens that there is disagreement about the effect of the explanatory variables