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For cell counts $Y_{ij}$ in a two-way contingency table, the independence model isThe absence of an interaction makes ${\mu }_{ij}$ factor as a row effect times a column effect. Its maximum-likelihood fitted values are

The prose names a second year that differs from the table label. The calculations use the four supplied counts, interpreting the first and second rows as the two displayed periods; no numerical conclusion depends on resolving that label discrepancy.

The first generalized linear model treats the four counts as independent Poisson random variables, with a log link and both row and column regression factors. The row-by-column interaction makes the log-linear model saturated. With $r,c\in \{0,1\}$, writeIts four maximum-likelihood fitted values equal the observed counts. Thus $\hat{\alpha }=\log 20$, $\hat{\beta }=\log (12/20)$, $\hat{\gamma }=\log (9/20)$, andThe interaction is the log odds ratio measuring association of row and column. Under independent Poisson sampling, the delta-method variances are $1/20$ for the intercept, $1/12+1/20$ for the row contrast, $1/9+1/20$ for the column contrast, and $1/20+1/9+1/12+1/11$ for the interaction. Their square roots reproduce the printed standard errors. Although the software labels the standardized ratios as t values, these are asymptotic normal Wald statistics, with known Poisson dispersion one, not exact finite-df Student tests.

The statistical saturated model has zero residual deviance and zero residual degrees of freedom; this is a perfect interpolation, not evidence of predictive adequacy. The null deviance $5.016056$ compares the four observed counts with an intercept-only model, whose four fitted means are $52/4=13$. Fisher scoring is the numerical score/information iteration used to fit the model; the iteration count is a convergence report, not a test statistic.

Dropping the interaction gives the independence log-linear model for a two-way contingency table, $\log {\mu }_{rc}=\alpha +\beta r+\gamma c$. Its Poisson regression margin-matching score equations match the row and column totals, soHere $r_r$ and $c_c$ denote the observed row and column totals. In particular the fitted row multiplier is $23/29$, the column multiplier is $20/32$, and the intercept is $\log (29\cdot 32/52)=2.881788$. These produce the second coefficient table. Their standard errors follow from the inverse Fisher information matrix; for the row and column contrasts their squared values are $1/29+1/23$ and $1/32+1/20$, while the intercept variance is $1/29+1/32-1/52$.

The Poisson deviance relative to the saturated fit isThe linear terms cancel because fitted and observed totals agree. The additive model has three coefficients, leaving one degree of freedom. Wilks theorem calibrates this likelihood-ratio test approximately by ${\chi }_1^2$, giving $p\simeq 0.2164$. The interaction Wald test instead gives $p\simeq 0.2192$; the two are only asymptotically equivalent.

Fisher's exact test conditions on both margins under independence. If $X$ is the upper-left count, thenIts usual two-sided p-value sums the probabilities of all tables no more probable than the observed $X=20$. That sum is $0.259724$, reproducing the output without an asymptotic approximation. It tests the same absence of row-column association, but conditioning and discreteness explain the numerical difference from the likelihood-ratio test.

In ordinary language, the male proportion among the cases is $20/29\simeq 69.0\%$ in the first row and $12/23\simeq 52.2\%$ in the second. The estimated case-sex odds ratio is $2.037$, but its approximate 95-percent confidence interval is wide, $(0.655,6.338)$. These small samples do not give convincing evidence that the sex composition of cases changed between the two periods. Failure to reject is not evidence that the compositions are exactly equal. Nor do these counts establish a male-to-female disease-incidence ratio or an incidence trend: population denominators and comparable ascertainment would be needed for those interpretations.

Write $y_{i+}={\sum }_j{y}_{ij}$ and ${\mu }_{i+}={\sum }_j{\mu }_{ij}$. Under the model in part (a),and $p_j$ is independent of the price category $i$. The Poisson-multinomial conditioning identity givesMore explicitly, the likelihood function for row $i$ factors asThe first factor is the likelihood of the row total and the second is the conditional multinomial likelihood. The free price effect ${\alpha }_i$ makes the first factor maximal at ${\hat{\mu }}_{i+}=y_{i+}$, while maximizing the remaining factors givesConsequently both fits have the same maximum-likelihood fitted valueHere every row total is $30$, the column totals are $31,28,31$, and $y_{++}=90$, so every fitted row isThis factorization is the Poisson trick.