I remember have the same consternation using GLIM with binomial models
on grouped and ungrouped data, but I was counseled by my betters only
to consider differences in models. The differences in deviance are the
same up to rounding error.
> 859.8018 - 711.3479
[1] 148.4539
> 168.8-20.3
[1] 148.5
So I would ask if these really are different answers?
--
David Winsemius
Heritage Labs
On Jan 13, 2009, at 10:06 AM, Prof Brian Ripley wrote:
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, rip...@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______________________________________________
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.
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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