Hi All, I am using the glm.nb function from the MASS package (current version) to fit and compare GLMs with negative binomial error distributions. My question is: what is the appropriate method to use in the anova function to compare models? If only one fitted object like
m<-glm.nb(number<-p+sal+temp,data=data) is specified in the anova function (anova(m)), a fixed theta is used to generate the analysis of deviance: Analysis of Deviance Table Model: Negative Binomial(0.2345), link: log Response: number Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 117 122.707 p 1 11.327 116 111.380 0.001 sal 1 2.286 115 109.094 0.131 tem 1 1.979 114 107.115 0.159 ph 1 2.567 113 104.549 0.109 Warning message: In anova.negbin(m) : tests made without re-estimating 'theta' If multiple fitted objects like m1<-glm.nb(number~1,data=data) m2<-glm.nb(number~p,data=data) m3<-glm.nb(number~p+sal,data=data) m4<-glm.nb(number~p+sal+temp,data=data) is specified (anova(m1,m2,m3,4)), the theta is assumed re-estimated in each case to calculate the likelihood ratio tests: Likelihood ratio tests of Negative Binomial Models Response: number Model theta Resid. df 2 x log-lik. Test df LR stat. Pr(Chi) 1 1 0.1892056 117 -527.7463 2 p 0.2153105 116 -517.9349 1 vs 2 1 9.811382 0.001734351 3 p + sal 0.2214626 115 -515.7942 2 vs 3 1 2.140706 0.143435894 4 p + sal + tem 0.2261900 114 -513.8846 3 vs 4 1 1.909643 0.167002884 5 p + sal + tem + ph 0.2344827 113 -511.3633 4 vs 5 1 2.521237 0.112322429 The conclusions are the same, but I'd like to know if one method is favored over the other. Thanks, Gary Nelson. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.