Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 12H / Solution

Model the two samples as independent multinomial counts with fixed row totals $N_A=N_B=500$ and three category probabilities. The null hypothesis is $p_{Aj}=p_{Bj}=p_j$ for every category $j$. Up to combinatorial factors the likelihood is ${\prod }_{ij}{p}_{ij}^{O_{ij}}$. Without the null restriction its maximizers are $O_{ij}/N_i$; under the null, maximizing ${\sum }_j(O_{Aj}+O_{Bj})\log p_j$ subject to ${\sum }_j{p}_j=1$ givesThe observed column totals are $243,281,476$, giving fitted expected counts $(121.5,140.5,238)$ in each row.

To derive the null distribution, put $D=O_A-O_B$. Under the null its mean is zero and covariance is $1000(\mathrm{diag}p-p{p}^T)$. The multivariate central limit theorem gives, asymptotically, the standardized contrast $Z_j=D_j/\sqrt{1000p_j}$ a centered normal distribution with covarianceThis is the orthogonal projection onto a two-dimensional plane. In an orthonormal basis of that plane, the two nonzero coordinates are independent standard normal variables, so ${\sum }_j{Z}_j^2$ tends to ${\chi }_2^2$. The pooled estimates are consistent; substituting them for $p_j$ leaves this limit unchanged by Slutsky's theorem. Since $O_{Aj}-E_{Aj}=D_j/2$ and $O_{Bj}-E_{Bj}=-D_j/2$, the resulting statistic is precisely the Pearson chi-squared test of homogeneity statisticThis supplies the two degrees of freedom rather than treating all six cells as independent. Every fitted expected count is large, supporting the approximation.

For these data,It exceeds the two-degree-of-freedom 95th percentile $5.99$, but not the 99th percentile $9.21$. Hence reject equal score distributions at the 5% level, but not at the 1% level. The asymptotic p-value is $\mathrm{Pr}({\chi }_2^2\ge 7.56906)=e^{-7.56906/2}\simeq 0.02272$.