From pdalgd at gmail.com  Tue Mar 15 12:17:46 2011
From: pdalgd at gmail.com (peter dalgaard)
Date: Tue, 15 Mar 2011 12:17:46 +0100
Subject: [Rd] Standardized Pearson residuals
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Message-ID: <58393ED6-2E85-4C80-843B-51DA023274FF@gmail.com>


On Mar 15, 2011, at 04:40 , Brett Presnell wrote:

> 
>>> Background: I'm currently teaching an undergrad/grad-service course from
>>> Agresti's "Introduction to Categorical Data Analysis (2nd edn)" and
>>> deviance residuals are not used in the text.  For now I'll just provide
>>> the students with a simple function to use, but I prefer to use R's
>>> native capabilities whenever possible.
>> 
>> Incidentally, chisq.test will have a stdres component in 2.13.0 for
>> much the same reason.
> 
> Thank you.  That's one more thing I won't have to provide code for
> anymore.  Coincidentally, Agresti mentioned this to me a week or two ago
> as something that he felt was missing, so that's at least two people who
> will be happy to see this added.
> 

And of course, I was teaching a course based on Agresti & Franklin: "Statistics, The Art and Science of Learning from Data", when I realized that R was missing standardized residuals. 


> It would also be nice for teaching purposes if glm or summary.glm had a
> "pearsonchisq" component and a corresponding extractor function, but I
> can imagine that there might be arguments against it that haven't
> occured to me.  Plus, I doubt that anyone wants to touch glm unless it's
> to repair a bug. If I'm wrong about all that though, ...
> 

Hmm, how would that work? If there was one, I'd worry that people would start subtracting them which is usually not the right thing to do. I do miss having a test on the residual deviance occasionally (even though it is only sometimes meaningful), having to fit a saturated model explicitly can be a bit silly. E.g. in this case (homogeneity of birth rates):

> anova(glm(births~month,poisson,data=bb), test="Chisq")
...
      Df Deviance Resid. Df Resid. Dev P(>|Chi|)    
NULL                     11     225.98              
month 11   225.98         0       0.00 < 2.2e-16 ***
> anova(glm(births~1,poisson,data=bb), test="Chisq")
...
     Df Deviance Resid. Df Resid. Dev P(>|Chi|)
NULL                    11     225.98          

Notice that the latter version gives me the correct deviance but no p-value.


A better support for generic score tests could be desirable too. I suspect that this would actually be the Pearson Chi-square in the interesting cases.  

-- 
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com



[Rd] Standardized Pearson residuals
peter dalgaard