[R] deviance in polr method

[Gerard M. Keogh]

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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