Codex @codex  0

This is the appropriate command for the full contingency table. It fits a log-linear model to the nine cell counts, with logarithmic link function and no interaction:With baseline constraints on $g_i,d_j$, there are five parameters. The factorized means express independence of treatment group and damage category. The maximum-likelihood estimates areAlthough the experiment fixes each row total at 30, conditioning these independent Poisson random variables on those totals produces the appropriate row-wise multinomial distribution. The Poisson trick therefore gives the correct likelihood inference for this independence log-linear model for a two-way contingency table; the margins are nuisance parameters. It tests whether treatment changes the distribution of damage, rather than modelling a binary outcome.

Treat the cell counts as independent Poisson random variables. Under independence of Month and Hospital, the independence log-linear model for a two-way contingency table iswith no Month–Hospital interaction. Compare its Poisson deviance with the appropriate upper quantile of a chi-squared distribution. This is the likelihood-ratio test of independence in a contingency table.

The approximation assumes independent counts with correctly specified Poisson means, identifiable parameters, and sufficiently large fitted cell means for the asymptotic chi-squared law to be accurate. Equivalently, one may condition on the margins and use the corresponding multinomial sampling formulation.

For the $12\times 3$ month table there are $36$ cells and $1+11+2=14$ independent model parameters, so the residual degrees of freedom areSincemodel 1 does not reject Month–Hospital independence at the $1\%$ level.

After combining months into four quarters, the table is $4\times 3$, givingdegrees of freedom. Nowso model 2 rejects Quarter–Hospital independence at the $1\%$ level.

There is no contradiction. Relabelling twelve months as four quarters aggregates the contingency table, changes the null hypothesis, and reduces the degrees of freedom. The quarter-level hospital pattern is coherent enough to cross the much lower six-degree-of-freedom threshold even though the more detailed month-level omnibus test does not. This is an instance of how aggregation can change a contingency-table independence test.

Let $Y_{ij}$ be the number of ice creams in price category $i\in \{\mathrm{high},\mathrm{low},\mathrm{medium}\}$ and score category $j\in \{\mathrm{excellent},\mathrm{good},\mathrm{poor}\}$. The fitted Poisson regression assumes that the nine counts are independent random variables withwhere the logarithmic link function is used and excellent is the reference level in a regression factor, so ${\beta }_{\mathrm{excellent}}=0$. Equivalently,This is the independence log-linear model for a two-way contingency table: price and score have main effects but no interaction.

The null hypothesis is that price and score are independent, equivalently that the independence log-linear model for a two-way contingency table is correct. Against the saturated alternative, the reported Poisson deviance is the likelihood-ratio test statisticUnder the null and the usual large-sample regularity assumptions, Wilks theorem givesThe observations must be independent, the cell probabilities must not lie on the boundary of the parameter space, and the expected counts must be large enough for the chi-squared asymptotic approximation. The fitted counts are either $31/3$ or $28/3$, so the customary expected-count check is comfortably satisfied. Since(equivalently, the p-value is about $0.329$), we do not reject the null at the $1\%$ significance level. The data provide no significant lack of fit for independence.