# compare quasi poisson models

I have two models:

frm.mE <- glm(frm ~ age + education + socialrole + countedmembers +
offset(log(words)), family=quasipoisson, data=daten.alle.kom)
frm.oE <- glm(frm ~ age + socialrole + countedmembers +
offset(log(words)), family=quasipoisson, data=daten.alle.kom)


now I want to know which model is the better one, but because of quasipoisson, AIC don't work

summary(frm.mE)
Call:
glm(formula = frm ~ age + education + socialrole + countedmembers +
offset(log(words)), family = quasipoisson, data = daten.alle.kom)

Deviance Residuals:
Min       1Q   Median       3Q      Max
-6.7040  -1.6727  -0.2329   1.0003   7.4897

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)    -3.95362    0.21432 -18.448  < 2e-16 ***
age             0.01293    0.07041   0.184  0.85454
education1      0.11532    0.11647   0.990  0.32367
socialrole1    -0.28367    0.23685  -1.198  0.23287
socialrole2    -0.80474    0.29054  -2.770  0.00629 **
countedmembers -0.03716    0.06120  -0.607  0.54461
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for quasipoisson family taken to be 5.792638)

Null deviance: 909.51  on 160  degrees of freedom
Residual deviance: 841.35  on 155  degrees of freedom
AIC: NA

Number of Fisher Scoring iterations: 5


and the second model:

Call:
glm(formula = frm ~ age + socialrole + countedmembers + offset(log(words)),
family = quasipoisson, data = daten.alle.kom)

Deviance Residuals:
Min       1Q   Median       3Q      Max
-6.4844  -1.6613  -0.3583   1.1036   7.1557

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)    -3.89079    0.20350 -19.119  < 2e-16 ***
age             0.00540    0.06966   0.078  0.93832
socialrole1    -0.33991    0.22947  -1.481  0.14054
socialrole2    -0.75470    0.28553  -2.643  0.00905 **
countedmembers -0.02634    0.05996  -0.439  0.66104
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for quasipoisson family taken to be 5.761264)

Null deviance: 909.51  on 160  degrees of freedom
Residual deviance: 847.08  on 156  degrees of freedom
AIC: NA

Number of Fisher Scoring iterations: 5


is there another way to compare them? or to know if I should keep the variable "education"? thanks for any help!

I tried a F test, but not sure if it makes sense:

 anova(frm.mE, frm.oE, test="F")
Analysis of Deviance Table

Model 1: frm ~ age + education + socialrole + countedmembers + offset(log(words))
Model 2: frm ~ age + socialrole + countedmembers + offset(log(words))
Resid. Df Resid. Dev Df Deviance      F Pr(>F)
1       155     841.35
2       156     847.08 -1  -5.7368 0.9904 0.3212


but I'm not sure how to understand it, does it mean that I should keep "education" because model 2 has a too big p-value?

• Aren't your models perfectly nested? Why not use a standard test? – gung - Reinstate Monica Oct 26 '14 at 22:45