Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / O / 3 (O3)

Under the independence log-linear model for a two-way contingency table, the six cell probabilities factor asThe row distribution has one free parameter and the column distribution two, giving three free parameters under $H_0$. Under the unrestricted alternative, six cell probabilities sum to one, giving five free parameters under $H_1$. The difference is two, which is the degrees of freedom of the resulting chi-squared test.

Let $N_{ij}$ be the observed cell counts, $N_{i+}$ the row totals, $N_{+j}$ the column totals and $N=33009$. The fitted expected frequencies under independence are $E_{ij}=N_{i+}{N}_{+j}/N$. For example the three-child, one-boy cell has expected count $7541\cdot 16472/33009=3763.1$.

The Pearson chi-squared statistic isUnder $H_0$ it has asymptotically a chi-squared distribution with $(2-1)(3-1)=2$ degrees of freedom, assuming the multinomial sampling model and sufficiently large expected counts. All expected counts here are large. The distribution is an asymptotic approximation, rather than an exact finite-sample identity.

The observed statistic $94.34$ exceeds the supplied $0.999$ quantile $13.82$ for two degrees of freedom. Thus reject independence, with $p<0.001$.

For one child of each sex among the first two, the three-child count is $3393$, below its independence expectation $3763.1$. Equivalently, the two-child count is $13079$, above its expectation $16472-3763.1=12708.9$. These departures have exactly the direction predicted by the suggestion. Thus the data support the proposed association: mixed-sex first pairs are associated with a greater tendency to stop at two children.