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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?

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  • 1
    $\begingroup$ Aren't your models perfectly nested? Why not use a standard test? $\endgroup$ – gung - Reinstate Monica Oct 26 '14 at 22:45

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