R help - May 1999 - surrogate poisson models

Dear R-help,

I'm applying the surrogate Poisson glm, by following Venables & Ripley
(7.3
pp238-42). 
>overall_cbind(expand.grid(treatment=c("Pema","control"),age=c("young","adult","old"),repair=c("excellent","good","poor")),Fr=c(8,0,7,1,2,0,2,7,1,4,7,1,
0,3,2,5,1,9))
>overall$age_ordered(overall$age,levels=c("young","adult","old"))
>overall$repair_ordered(overall$repair,levels=c("poor","good","excellent"))
>overall.lm1_glm(terms(Fr~treatment*age+(treatment+age)*repair,
keep.order=T),family=poisson,data=overall)
> summary(overall.lm1)
Call:
glm(formula = terms(Fr ~ treatment * age + (treatment + age) * 
    repair, keep.order = T), family = poisson, data = overall)

.
.
.  

Coefficients:
                   Estimate Std. Error z value Pr(>|z|)    
(Intercept)         0.68441    0.28979   2.362  0.01819 *  
treatment          -0.28096    0.43716  -0.643  0.52043    
age.L               0.55595    0.42669   1.303  0.19259    
age.Q               0.03404    0.42817   0.079  0.93664    
treatment.age.L    -1.31805    0.66813  -1.973  0.04853 *  
treatment.age.Q    -0.37452    0.67332  -0.556  0.57805    
repair.L            1.53962    0.54255   2.838  0.00454 ** 
repair.Q           -0.49447    0.40128  -1.232  0.21787    
treatment.repair.L -3.93138    0.96310  -4.082 4.46e-05 ***
treatment.repair.Q -0.58937    0.62739  -0.939  0.34753    
age.L.repair.L     -2.08339    0.67093  -3.105  0.00190 ** 
age.Q.repair.L     -0.47257    0.59116  -0.799  0.42406    
age.L.repair.Q     -0.04208    0.42881  -0.098  0.92183    
age.Q.repair.Q     -0.64314    0.42800  -1.503  0.13293    
- ---
.
.
.

How do you interpret the suffixes .L  and .Q  in the  summary? I tried
redefining overall$age_ordered(.... labels=c(...)), to no avail. When the
factors are unordered, the suffixes are the labels, but the fitted model is
different.

Any help is much appreciated.


Simon Bond.
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On Tue, 4 May 1999, Simon  Bond wrote:
> 
> Call:
> glm(formula = terms(Fr ~ treatment * age + (treatment + age) * 
>     repair, keep.order = T), family = poisson, data = overall)
> 
> 
> Coefficients:
>                    Estimate Std. Error z value Pr(>|z|)    
> (Intercept)         0.68441    0.28979   2.362  0.01819 *  
> treatment          -0.28096    0.43716  -0.643  0.52043    
> age.L               0.55595    0.42669   1.303  0.19259    
> age.Q               0.03404    0.42817   0.079  0.93664    
> treatment.age.L    -1.31805    0.66813  -1.973  0.04853 *  
> treatment.age.Q    -0.37452    0.67332  -0.556  0.57805    
> repair.L            1.53962    0.54255   2.838  0.00454 ** 
> repair.Q           -0.49447    0.40128  -1.232  0.21787    
> treatment.repair.L -3.93138    0.96310  -4.082 4.46e-05 ***
> treatment.repair.Q -0.58937    0.62739  -0.939  0.34753    
> age.L.repair.L     -2.08339    0.67093  -3.105  0.00190 ** 
> age.Q.repair.L     -0.47257    0.59116  -0.799  0.42406    
> age.L.repair.Q     -0.04208    0.42881  -0.098  0.92183    
> age.Q.repair.Q     -0.64314    0.42800  -1.503  0.13293    
> - ---
> 
> How do you interpret the suffixes .L  and .Q  in the  summary? I tried
> redefining overall$age_ordered(.... labels=c(...)), to no avail. When the
> factors are unordered, the suffixes are the labels, but the fitted model is
> different.
For an ordered factor polynomial contrasts are used by default (this can
be set with options(contrasts=)).  Polynomial contrasts decompose the
effect of a variable into orthogonal linear, quadratic, cubic,... terms.
Based on the analysis above you might argue that all the quadratic terms
are relatively small and not statistically significant and that the linear
terms capture most of the variation.

You can see what the contrast matrix looks like by typing eg
R> contr.poly(3)
                .L         .Q
[1,] -7.071068e-01  0.4082483
[2,] -7.850462e-17 -0.8164966
[3,]  7.071068e-01  0.4082483

to get the polynomial contrasts for a three-level ordered factor.

For an unordered factor we use treatment contrasts by default (that is,
indicator variables for all but one level of the factor).  Any full rank
set of contrasts gives the same model, but the coefficients are different,
and allow you to answer different questions.  The choice of contrasts
depends on which questions you want to answer.

Thomas Lumley
Assistant Professor, Biostatistics
University of Washington, Seattle

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