luis <- expand.grid(mar=factor(c("Yes","No")),cig=factor(c("Yes","No")),alc=factor(c("Yes","No")))
count <- c(911,538,44,456,3,43,2,279)
seniors <- data.frame(luis,count)
# Q1: Just using the sample, estimate the conditional odds ratio btw. alcohol use and marijuana use for
#a.) people that use cigarettes
#b.) people that do not use cigaretts

> seniors
  mar cig alc count
1 Yes Yes Yes   911
2  No Yes Yes   538
3 Yes  No Yes    44
4  No  No Yes   456
5 Yes Yes  No     3
6  No Yes  No    43
7 Yes  No  No     2
8  No  No  No   279

fit <- glm(count~alc*cig*mar,family=poisson,data=seniors)
anova(fit, test="Chisq") #shows model (AC,AM,CM) fit reasonable, any simpler one not

Analysis of Deviance Table
            Df Deviance Resid. Df Resid. Dev  Pr(>Chi)
NULL                            7    2851.46
alc          1  1281.71         6    1569.75 < 2.2e-16 ***
cig          1   227.81         5    1341.93 < 2.2e-16 ***
mar          1    55.91         4    1286.02 7.575e-14 ***
alc:cig      1   442.19         3     843.83 < 2.2e-16 ***
alc:mar      1   346.46         2     497.37 < 2.2e-16 ***
cig:mar      1   497.00         1       0.37 < 2.2e-16 ***
alc:cig:mar  1     0.37         0       0.00    0.5408
---

fit.ac.am.cm <- glm(count~alc*cig+alc*mar+cig*mar,family=poisson,data=seniors)
summary(fit.ac.am.cm)

Coefficients:
              Estimate Std. Error z value Pr(>|z|)
(Intercept)    5.63342    0.05970  94.361  < 2e-16 ***
alcYes         0.48772    0.07577   6.437 1.22e-10 ***
cigYes        -1.88667    0.16270 -11.596  < 2e-16 ***
marYes        -5.30904    0.47520 -11.172  < 2e-16 ***
alcYes:cigYes  2.05453    0.17406  11.803  < 2e-16 ***
alcYes:marYes  2.98601    0.46468   6.426 1.31e-10 ***
cigYes:marYes  2.84789    0.16384  17.382  < 2e-16 ***
---

    Null deviance: 2851.46098  on 7  degrees of freedom
Residual deviance:    0.37399  on 1  degrees of freedom
AIC: 63.417

Number of Fisher Scoring iterations: 4

data.frame(luis,obs=count,fitted=fitted(fit.ac.am.cm))

  mar cig alc obs    fitted
1 Yes Yes Yes 911 910.38317
2  No Yes Yes 538 538.61683
3 Yes  No Yes  44  44.61683
4  No  No Yes 456 455.38317
5 Yes Yes  No   3   3.61683
6  No Yes  No  43  42.38317
7 Yes  No  No   2   1.38317
8  No  No  No 279 279.61683

# Q2: Based on model (AC,AM,CM), give an estimate of the conditonal odds ratios above.
# Interpret this common value.
exp(2.98)

# Q3: Test if it is plausible that alcohol and marijuana use are conditionally independent,
# controlling for smoking status
fit.ac.cm <- glm(count~alc*cig+cig*mar,family=poisson,data=seniors)
deviance(fit.ac.cm)-deviance(fit.ac.am.cm)