[R] Anova function and glm.nb

[Prof Brian Ripley]

The re-estimated version is the principled one -- the other is a 
computational shortcut.

However, what you model might choose depends on why you are doing model 
selection, and also if there is a subject-specific reason to consider the 
terms in this order.

And if you look in MASS (the book) you will see a similar example and 
discussion of how to treat it.  It uses dropterm(), which is likely to be 
preferable to either of these approaches.

On Mon, 7 Apr 2008, Nelson, Gary (FWE) wrote:

> 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.
>
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-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595