[R] deviance in polr method

[Prof Brian Ripley]

Withut looking up your reference, are you not comparing grouped and 
ungrouped deviances?  And polr() does not say anything about accepting 
a model (or not), only about the comparison between two models.

'Deviances' are in comparison with some 'saturated' model, and I would 
say that M&N are comparing with a separate fit to each group, polr() 
with correctly predicting each observation.  Both are valid, but refer 
to different questions.

On Tue, 13 Jan 2009, Gerard M. Keogh wrote:

>
> Dear all,
>
>
> I've replicated the cheese tasting example on p175 of GLM's by McCullagh
> and Nelder. This is a 4 treatment (rows) by 9 ordinal response (cols)
> table.
>
> Here's my simple code:
>
>      #### cheese
>      library(MASS)
>      options(contrasts = c("contr.treatment", "contr.poly"))
>      y = c(0,0, 1, 7, 8,8,19, 8,1, 6,9,12,11, 7,6, 1, 0,0, 1,1, 6, 8,23,7,
>      5, 1,0, 0,0, 0, 1, 3,7,14,16,11)
>      p =
>      c(1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9)
>      x1 =
>      c(1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
>      x2 =
>      c(0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
>      x3 =
>      c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0)
>      # catgeory 4 is used as baseline
>      ord.rp = ordered(p)
>      fit0.polr = polr(ord.rp ~ 1, weights=y)
>      fit1.polr = polr(ord.rp ~ x1 + x2+x3, weights=y)
>      summary(fit1.polr)
>      anova(fit0.polr,fit1.polr)
>
> This works and "summary" gives the correct parameter estimates but I have a
> problem with the deviance.
> Anova gives
>      Likelihood ratio tests of ordinal regression models
>
>      Response: ord.rp
>               Model Resid. df Resid. Dev   Test    Df LR stat. Pr(Chi)
>      1            1       200   859.8018
>      2 x1 + x2 + x3       197   711.3479 1 vs 2     3 148.4539       0
>
>
> In McCullagh the deviance for the null model is given as 168.8 on 24 d.f.
> and the fitted model is 20.3 on 21 d.f.
> This means the change in POLR and in the book is 148.5 on 3 d.f. so the
> model gives a significant improvement.
>
> But the model deviance from POLR is 711 vs. 20.3 from McCullagh.
> Thus POLR says we should reject the model fit with chi-sq = 35 while
> McCullagh say we should accept it
>
> Could someone explain what I should do when it comes to accepting or
> rejecting the model fit itself in R?
>
> I'm using R 2.8.0.
>
> Thanks.
>
> Gerard
>
>
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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