[R] p values of plor

Prof Brian Ripley ripley at stats.ox.ac.uk
Wed May 29 16:09:41 CEST 2013

```AIC is a different story.  To do hypothesis tests on terms, use anova()
or dropterm() (as done in the book):

library(MASS)
example(polr)
dropterm(house.plr, test = "Chisq")

Single term deletions

Model:
Sat ~ Infl + Type + Cont
Df    AIC     LRT   Pr(Chi)
<none>    3495.1
Infl    2 3599.4 108.239 < 2.2e-16
Type    3 3545.1  55.910 4.391e-12
Cont    1 3507.5  14.306 0.0001554

That is an object of class "anova", so can be saved and subsetted.

On 28/05/2013 22:30, David Winsemius wrote:
>
> On May 27, 2013, at 11:05 PM, Prof Brian Ripley wrote:
>
>> On 28/05/2013 06:54, David Winsemius wrote:
>>>
>>> On May 27, 2013, at 7:59 PM, meng wrote:
>>>
>>>> Hi all:
>>>> As to the polr {MASS} function, how to find out p values of every
>>>> parameter?
>>>>
>>>>
>>>>>  From the example of R help:
>>>> house.plr <- polr(Sat ~ Infl + Type + Cont, weights = Freq, data =
>>>> housing)
>>>> summary(house.plr)
>>>>
>>>>
>>>> How to find out the p values of house.plr?
>>>
>>> Getting  p-values from t-statistics should be fairly straight-forward:
>>>
>>> summary(house.plr)\$coefficients
>>
>> And what distribution are you going to use to compute the p-values?
>
> I should have responded with my first impulse: "If the authors didn't provide p-values, then perhaps they don't think they are credible."
>
>>
>> Hint: there is no exact distribution theory for POLR fits and the asymptotic theory can be far enough off to be seriously misleading (just as for the two-class case, logistic regression: see MASS the book). That is why likelihood-ratio tests are recommended in MASS, not Wald tests.
>
> And so the more correct answer would be to use stepAIC? I would have thought sequential removal of terms with comparisons of deviance estimates might be informative. This is what I get with that data:
>
>> house.AIC.1 <- stepAIC(house.plr, list(upper=~., lower=~1) )
> Start:  AIC=3495.15
> Sat ~ Infl + Type + Cont
>
>         Df    AIC
> <none>    3495.1
> - Cont  1 3507.5
> - Type  3 3545.1
> - Infl  2 3599.4
>>
>
> So something along  those lines seems to be happening, but I am not able to extract those values programmatically, nor am I able to see how they even get displayed.
>
>> class(house.AIC.1)
> [1] "polr"
>> str(house.AIC.1\$anova)
> Classes ‘Anova’ and 'data.frame':	1 obs. of  6 variables:
>   \$ Step      : Factor w/ 1 level "": 1
>   \$ Df        : num NA
>   \$ Deviance  : num NA
>   \$ Resid. Df : num 1673
>   \$ Resid. Dev: num 3479
>   \$ AIC       : num 3495
>
>
> Which lead me to look at:
>
> getAnywhere(print.polr)
>
> But that was uninformative to my level of reading R code. The AIC trials seem to get printed by stepAIC() but are not saved in the returned object.
>
> ---
>
> David Winsemius
> Alameda, CA, USA
>

--
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

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