Message-ID: <alpine.OSX.1.00.0901131500300.74128@tystie.local>
Date: 2009-01-13T15:06:04Z
From: Brian Ripley
Subject: deviance in polr method
In-Reply-To: <200901131404.n0DE4tvG000791@hypatia.math.ethz.ch>
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