= Solution
The preferred <cumulative-response logistic model> estimates
$$
\operatorname{logit}(p_{it})=-1.417630-0.364092S_i+0.001224A_i
+0.327118T_i+0.399296C_{it},\qquad C_{it}=\sum_{s<t}Y_{is}.
$$
Its <logit link> describes conditional smoking-free <probability> given baseline predictors and prior successful weeks. Holding the other predictors fixed, males have $e^{-0.364092}=0.695$ times the female success <odds>; the reported <p-value> is $0.0152$, and the approximate 95% <odds ratio> <confidence interval> is $(0.518,0.932)$. Age has estimated <odds ratio> $e^{0.001224}=1.0012$ per additional year, with <p-value> $0.918$, giving little evidence for an age association in this fit.
The combined treatment has conditional success <odds ratio> $e^{0.327118}=1.387$ versus the reference treatment, with approximate 95% <confidence interval> $(1.035,1.859)$ and <p-value> $0.0285$. \b[The fitted conditional odds are about 39% higher for the combined treatment.] The trial randomization supports treatment comparisons, but conditioning on accumulated post-treatment outcomes means this coefficient is not directly the marginal total treatment effect.
Each previous successful week multiplies current success <odds> by $e^{0.399296}=1.491$, with approximate 95% <confidence interval> $(1.329,1.673)$ and very small <p-value> $1.06\times10^{-11}$. This is strong fitted persistence. It can reflect <true state dependence>, <unobserved heterogeneity>, or an omitted calendar-time trend; this fit alone cannot distinguish them. The baseline intercept implies success <probability> $\operatorname{logit}^{-1}(-1.417630)\simeq0.195$ for a reference-treatment female aged zero with no previous success. That age is outside the study's useful interpretation range, so the intercept chiefly anchors the regression. The <Bernoulli distribution> <dispersion parameter> is fixed at one, and the residual <binomial deviance> is 1122 on 995 <statistical degrees of freedom>. With individual binary outcomes, comparing that <binomial deviance> mechanically to a <chi-squared distribution> is not a reliable general goodness-of-fit test.
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