Loglinear and Logit models for COntingency tables

Exercise 9.1 Consider the data set DaytonSurvey (described in Example 2.6), giving results of a survey of use of alcohol (A), cigarettes (C), and marijuana (M) among high school seniors. For this exercise, ignore the variables sex and race, by working with the marginal table Dayton.ACM, a 2 × 2 × 2 table in frequency data frame form.

data("DaytonSurvey",package="vcdExtra")
Dayton.ACM = aggregate(Freq~cigarette+alcohol+marijuana,data=DaytonSurvey, FUN=sum)
Dayton.ACM 
##   cigarette alcohol marijuana Freq
## 1       Yes     Yes       Yes  911
## 2        No     Yes       Yes   44
## 3       Yes      No       Yes    3
## 4        No      No       Yes    2
## 5       Yes     Yes        No  538
## 6        No     Yes        No  456
## 7       Yes      No        No   43
## 8        No      No        No  279

Fit the following models:

Joint independence [AC][M]

COnditional independence [AM][CM]

Homogeneous model [AC][AM][CM]

Mutual independence [A][C][M]

Saturated model [ACM]

library(MASS)
Dayton.joint=loglm(Freq~alcohol*cigarette+marijuana, data = Dayton.ACM)
Dayton.cond=loglm(Freq~marijuana*(alcohol+cigarette),data=Dayton.ACM)
Dayton.hom=loglm(Freq~(alcohol+cigarette+marijuana)^2,data=Dayton.ACM)
Dayton.mutual=loglm(Freq~alcohol+cigarette+marijuana,data=Dayton.ACM)
Dayton.sat=loglm(Freq~(alcohol*cigarette*marijuana),data=Dayton.ACM)

Prepare a table comparing the GOF of these models:

anova(Dayton.joint,Dayton.cond,Dayton.hom,Dayton.mutual,Dayton.sat, test="chisq")
## LR tests for hierarchical log-linear models
## 
## Model 1:
##  Freq ~ alcohol + cigarette + marijuana 
## Model 2:
##  Freq ~ alcohol * cigarette + marijuana 
## Model 3:
##  Freq ~ marijuana * (alcohol + cigarette) 
## Model 4:
##  Freq ~ (alcohol + cigarette + marijuana)^2 
## Model 5:
##  Freq ~ (alcohol * cigarette * marijuana) 
## 
##               Deviance df  Delta(Dev) Delta(df) P(> Delta(Dev)
## Model 1   1286.0199544  4                                     
## Model 2    843.8266437  3 442.1933108         1        0.00000
## Model 3    187.7543029  2 656.0723408         1        0.00000
## Model 4      0.3739859  1 187.3803170         1        0.00000
## Model 5      0.0000000  0   0.3739859         1        0.54084
## Saturated    0.0000000  0   0.0000000         0        1.00000

The Mutual independence model shows to be the most acceptable model in terms of fit. This model has the lowest pearson chi square and highestr X^2 than the rest of the models.