= Solution
`read.table(...,header=T)` reads four grouped rows with column names. The displayed outcome counts give the observed risks $0.1519,0.2145,0.2611,0.5111$ in the indicated factor order. `rbind` in the first test pools over predisposition: the two rows represent no cannabis use and some use, with event/non-event counts $(341,1775)$ and $(82,238)$. Thus the pooled risks are $341/2116=0.1612$ and $82/320=0.2563$. This test concerns the marginal <statistical association> between use and outcome, not an association adjusted for predisposition.
Under independence, expected counts in the two-by-two <contingency table> are $E_{ij}=O_{i+}O_{+j}/N$. The ordinary <Pearson chi-squared statistic> is $17.5183$. The printed $16.8618$ is reproduced by the default two-by-two <Yates continuity correction>:
$$
X_Y^2=\sum_{i,j}\frac{(|O_{ij}-E_{ij}|-1/2)^2}{E_{ij}}=16.8618.
$$
All absolute differences here exceed one half. Comparison with a <chi-squared distribution> on one <statistical degree of freedom> gives $p=4.02\times10^{-5}$. \b[There is strong evidence of marginal association.] The simplified output omits the usual continuity-correction wording, but its numerical value shows which statistic was used.
`attach` makes the data columns available by name, and `tot=with+without` obtains group sizes $1936,275,180,45$. The grouped <binomial regression> fits $Y_g\sim\operatorname{Bin}(n_g,\pi_g)$, with response $Y_g/n_g$ and prior weights $n_g$. These weights tell R the binomial denominator; treating the four proportions as four equally precise individual observations would be wrong. The default <logit link> is
$$
\eta_g=\log\frac{\pi_g}{1-\pi_g},\qquad
\pi_g=\frac{e^{\eta_g}}{1+e^{\eta_g}}.
$$
Let $A$ indicate some use and $B$ indicate predisposition. The additive <logistic regression> specifies $\eta=a+bA+cB$. Its <log-likelihood>, including the binomial constants, is
$$
\ell=\sum_g\left[\log\binom{n_g}{Y_g}+Y_g\eta_g-n_g\log(1+e^{\eta_g})\right].
$$
Differentiation gives <score equations> $X^T(Y-n\pi)=0$ and <Fisher information matrix> $X^TWX$, where $W_{gg}=n_g\pi_g(1-\pi_g)$. <Fisher scoring> solves these equations, and the inverse fitted <Fisher information matrix> gives the reported approximate coefficient <variances>. The number of scoring iterations records numerical convergence, not degrees of freedom or statistical evidence.
The intercept corresponds to no use and no predisposition and gives baseline fitted risk $\operatorname{logit}^{-1}(-1.73881)=0.1495$. The additive coefficients imply common adjusted <odds ratios>:
$$
\boxed{\operatorname{OR}_{A\mid B}=e^{0.53847}=1.713,\qquad
\operatorname{OR}_{B\mid A}=e^{0.82824}=2.289.}
$$
They multiply the odds, not the risks. The <Wald tests> divide the estimates by their <standard errors> and use a <standard normal distribution>; both predictors have small two-sided <p-values>. Approximate $95\%$ <confidence intervals> for these <odds ratios> are $(1.296,2.266)$ and $(1.685,3.110)$, obtained by exponentiating the log-scale intervals. The intercept test against zero asks whether the reference risk equals one half, and is not a test of either exposure association. The binomial dispersion is assumed to be one because $\operatorname{Var}(Y_g)=n_g\pi_g(1-\pi_g)$; the output does not demonstrate independence or rule out extra-binomial variation.
There are four grouped probabilities and three fitted coefficients, leaving one <residual degree of freedom>. The fitted risks are approximately $(0.1495,0.2314,0.2869,0.4080)$. The <binomial deviance> compares this fit with the four unrestricted observed proportions:
$$
D=2\sum_g\left[Y_g\log\frac{Y_g}{n_g\widehat\pi_g}
+(n_g-Y_g)\log\frac{n_g-Y_g}{n_g(1-\widehat\pi_g)}\right]=3.0733.
$$
The <deviance goodness-of-fit test> against $\chi^2_1$ has <p-value> $0.0796$. The common-odds-ratio fit is not rejected at $5\%$, although the doubly exposed group's observed risk exceeds its fitted risk appreciably and suggests possible <interaction>.
The second formula adds $\delta AB$. Its four coefficients fit all four probabilities, making a <saturated statistical model> with zero grouped <residual degrees of freedom>. The near-zero deviance is roundoff from exact interpolation; it supplies no remaining grouped lack-of-fit test. Unlike Q1's unreplicated Gaussian fit, however, these grouped responses represent many independent Bernoulli trials, and their binomial <variances> are specified by their sizes and probabilities. Coefficient <standard errors> remain available from binomial information without estimating dispersion from the zero residual degrees of freedom.
Indeed, let $\eta_{ab}=\log[Y_{ab}/(n_{ab}-Y_{ab})]$ be the fitted cell <logit link> value. The saturated coefficients are
$$
a=\eta_{00},\qquad b=\eta_{10}-\eta_{00},\qquad
c=\eta_{01}-\eta_{00},\qquad
\delta=\eta_{11}-\eta_{10}-\eta_{01}+\eta_{00}.
$$
These calculations yield the printed estimates. The <delta method> gives $\operatorname{Var}(\widehat\eta_{ab})\approx1/Y_{ab}+1/(n_{ab}-Y_{ab})$, and independent groups make the <variance> of $\widehat\delta$ the sum of all eight reciprocal counts. Its square root is $0.37857$, reproducing the printed <standard error>.
The saturated <logistic regression> describes the conditional <odds ratios> as follows:
$$
\operatorname{OR}_{A\mid B=0}=e^{0.42235}=1.526,\qquad
\operatorname{OR}_{A\mid B=1}=e^{0.42235+0.66230}=2.958,
$$
$$
\operatorname{OR}_{B\mid A=0}=e^{0.67989}=1.974,\qquad
\operatorname{OR}_{B\mid A=1}=e^{0.67989+0.66230}=3.827.
$$
Thus the main coefficients now concern effects at the other factor's reference level. <Logistic interaction as a ratio of odds ratios> gives $e^\delta=1.939$, with approximate $95\%$ <confidence interval> $(0.923,4.073)$. The interaction <Wald test> has $p=0.0802$, while the nested <likelihood-ratio test> uses deviance difference $3.0733$ on one <statistical degree of freedom> and has $p=0.0796$. \b[The data suggest a larger odds association in predisposed subjects, but do not establish this interaction at the conventional $5\%$ level.] This conclusion is specific to the odds scale; absolute-risk differences can vary even with no logit-scale interaction.
The <Akaike information criterion> is $-2\widehat\ell+2k$. The additive fit has $k=3$ and AIC $31.766$; the interaction fit has $k=4$ and AIC $30.692$. The extra coefficient improves twice the log-likelihood by $3.0733$ but incurs penalty $2$, so AIC falls by only $1.0733$. It weakly favors the larger fit rather than decisively establishing an interaction, and AIC is not a <p-value>.
The marginal <odds ratio> $1.793$ and the fitted conditional <odds ratios> are different quantities. Predisposition is associated with both use frequency and outcome, so <confounding> is a substantive possibility; <noncollapsibility of the odds ratio> also means marginal and conditional odds ratios need not agree. This is an observational association analysis with only one adjustment factor. It does not by itself establish a causal effect, exclude other <confounding>, or validate the assumed independent binomial sampling.
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