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

Each before-and-after observation belongs to the same individual, so pairing removes individual baseline variation and allows dependence within a person's two readings. Under the stated independent normal model for the differences,The independence of the normal sample mean and sample variance theorem gives independence of $\overline{D}$ and $S_D^2$, andAlso $(\overline{D}-\mu )/(\sigma /\sqrt{n})$ is standard normal. Dividing it by the square root of the independent ${\chi }_{n-1}^2/(n-1)$ variable givesIts law contains neither unknown parameter, providing the pivotal quantity for a paired t-test.

Write $s_e=S_D/\sqrt{17}$. By symmetry of the $t_{16}$ distribution, $\mathbb{P}(Z>-1.746)=0.95$. Rearranging $Z=(\overline{D}-\mu )/s_e>-1.746$ gives $\mu <\overline{D}+1.746s_e$. Thus the interval printed in the question has endpointThere is a direction mismatch in the question. Since $D_i=X_i-Y_i$ is before minus after, a reduction corresponds to $H_1:\mu >0$. The upper interval $(-\infty ,b)$ excludes zero when $b<0$, which instead tests against a negative mean difference.

For the stated reduction alternative, use the lower 95% interval $(\overline{D}-1.746s_e,\infty )$. It excludes zero exactly whenThis is the level-$0.05$ one-sided paired t-test of $H_0:\mu =0$ against $H_1:\mu >0$. Both the requested upper endpoint and the correctly directed test are thus determined without silently reversing the difference convention.

The full column rank makes $X^{\mathsf{T}}X$ invertible. The least-squares normal equations from minimizing squared residuals giveThe errors have covariance matrix ${\sigma }^2{I}_n$ by independence and equal variance. Applying the covariance transformation rule to this linear estimator yieldsThe parameter estimator is unbiased because $\mathbb{E}Y=X\theta$.

Put $z_i=e^{-t_i}-c$. Assume the measurement errors are independent and centered, so $Y_i=\alpha +\beta z_i+{ϵ}_i$ with independent ${ϵ}_i\sim N(0,{\sigma }^2)$. Treat the observation times as fixed. The design matrix has rows $(1,z_i)$ and parameter vector $(\alpha ,\beta {)}^{\mathsf{T}}$. Require at least two distinct times so its columns are independent, equivalently ${\sum }_i{z}_i^2>0$. This gives the required Gaussian linear regression model.

Centering gives ${\sum }_i{z}_i=0$, and henceThe least-squares covariance formula from part (a) therefore givesExplicitly $\hat{\alpha }=\overline{Y}$ and $\hat{\beta }={\sum }_i{z}_i{Y}_i/{\sum }_i{z}_i^2$. Their zero covariance follows directly from the centered predictor, and their joint normality additionally makes them independent.

With independent Gaussian errors, maximizing the likelihood over $\alpha ,\beta$ minimizes the residual sum of squares, so the maximum-likelihood estimators are the same as least squares. The invariance property of maximum likelihood estimation givesThe two estimates have zero covariance by part (ii). ThereforeThis is the uncertainty in the estimated expected response. Estimating an additional noisy observation would add its own error variance, but the quantity requested is the mean response $E_t$.

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.