# [R] Comparing output from linear regression to output from quasipoisson to determine the model that fits best.

John Sorkin jsorkin at grecc.umaryland.edu
Tue Dec 2 00:00:05 CET 2008

```R 2.7
Windows XP

I have two model that have been run using exactly the same data, both fit using glm(). One model is a linear regression (gaussian(link = "identity"))  the other a quasipoisson(link = "log"). I have log likelihoods from each model. Is there any way I can determine which model is a better fit to the data? anova() does not appear to work as the models have the same residual degrees of freedom:

fit1<-glm(PHYSFUNC~HIV,data=KA)
summary(fit1)

fitQP<-glm(PHYSFUNC~HIV,data=KA,family=quasipoisson)
summary(fitQP)

anova(fit1,fitOP)

Program OUTPUT:
> fit1<-glm(PHYSFUNC~HIV,data=KA)
> summary(fit1)

Call:
glm(formula = PHYSFUNC ~ HIV, data = KA)

Deviance Residuals:
Min      1Q  Median      3Q     Max
-4.197  -4.192  -2.192   2.808  19.808

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  4.19670    0.08508   49.33   <2e-16 ***
HIV         -0.00487    0.12071   -0.04    0.968
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for gaussian family taken to be 22.78134)

Null deviance: 142429  on 6253  degrees of freedom
Residual deviance: 142429  on 6252  degrees of freedom
(213 observations deleted due to missingness)
AIC: 37302

Number of Fisher Scoring iterations: 2

>
> fitQP<-glm(PHYSFUNC~HIV,data=KA,family=quasipoisson)
> summary(fitQP)

Call:
glm(formula = PHYSFUNC ~ HIV, family = quasipoisson, data = KA)

Deviance Residuals:
Min      1Q  Median      3Q     Max
-2.897  -2.895  -1.193   1.250   6.644

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  1.434297   0.020280   70.72   <2e-16 ***
HIV         -0.001161   0.028780   -0.04    0.968
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for quasipoisson family taken to be 5.432011)

Null deviance: 35439  on 6253  degrees of freedom
Residual deviance: 35439  on 6252  degrees of freedom
(213 observations deleted due to missingness)
AIC: NA

Number of Fisher Scoring iterations: 5

> anova(fit1,fitQP)
Analysis of Deviance Table

Model 1: PHYSFUNC ~ HIV
Model 2: PHYSFUNC ~ HIV
Resid. Df Resid. Dev   Df Deviance
1      6252     142429
2      6252      35439    0   106989
>

Thanks,
John

John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)

Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}

```