Past exam of the statistics course of the University of Oxford / 2014 / 1

Write $I_1(\theta )$ for the Fisher information in one observation:Under the stipulated regularity assumptions, the maximum-likelihood estimator is consistent and has asymptotic normality:Equivalently, its large-sample normal approximation has mean $\theta$ and variance $1/[n{I}_1(\theta )]$. The information for the whole random sample is $n{I}_1(\theta )$; keeping the one-observation convention explicit prevents an extra factor of $n$.

Let $z={\Phi }^{-1}(1-\alpha /2)$, where $\Phi$ is the standard normal distribution's cumulative distribution function. Statistical consistency and Slutsky's theorem allow replacement of $I_1(\theta )$ by $I_1(\hat{\theta })$:Thus the probability that this standardized estimation error lies between $-z$ and $z$ tends to $1-\alpha$. Rearranging gives the Wald confidence intervalThe coverage probability is approximate, with limiting value $1-\alpha$; it is a repeated-sampling statement about the random confidence interval.

Here the exponential distribution is parametrized by its mean, so $f(x;\theta )={\theta }^{-1}{e}^{-x/\theta }$ for $x\ge 0$ and $\theta >0$. Its log-likelihood and maximum-likelihood estimate areFor one observation, ${\partial }_{\theta }\log f=-1/\theta +X/{\theta }^2$. Since $\mathrm{Var}(X)={\theta }^2$, its Fisher information is $I_1(\theta )=1/{\theta }^2$. Consequently the estimated standard error is $\overline{X}/\sqrt{n}$, givingThis is a large-sample confidence interval, not an exact one. If its lower endpoint is negative, intersecting with the parameter space $(0,\infty )$ removes impossible values without excluding any possible true statistical parameter.

Define $R={\sum }_i(Y_i-\overline{Y}{)}^2$ and $s_Y^2=R/(n-1)$, with $n\ge 2$. The unrestricted maximum-likelihood estimates are $\hat{\mu }=\overline{Y}$ and ${\hat{\sigma }}^2=R/n$. Under the null hypothesis, maximizing the normal distribution's likelihood over the remaining variance givesbecause the cross term vanishes when ${\sum }_i(Y_i-\overline{Y})=0$. At either fitted variance, the exponential factor in the likelihood function is $e^{-n/2}$, so it cancels in the ratio:The standardized test statistic isThe sample variance $s_Y^2$ uses divisor $n-1$, whereas the maximum-likelihood estimate uses divisor $n$; this explains the $n-1$ in the final expression. Under a nondegenerate normal distribution, $R>0$ with probability one.

For fixed $n$, the likelihood ratio is strictly decreasing as $t^2$ increases. A likelihood-ratio test rejects for small $\lambda$, so its rejection rule is equivalently large $|t|$. Under the null hypothesis, the standard normal distribution and chi-squared distribution results give independent quantitiesTheir ratio therefore has Student's t-distribution with $n-1$ statistical degrees of freedom. The exact level-$\alpha$ rule isThis is precisely the two-sided Student's t-test; choosing the corresponding cutoff for $\lambda$ yields the same rejection region and the same p-value.

The equal-variance two-sample Student's t-test pools the within-group sums of squares. If $s_A^2$ and $s_B^2$ are the two unbiased sample variances, the required positive square root isFor ten observations in each group, $S^2=(s_A^2+s_B^2)/2$ and the usual unpaired reference Student's t-distribution has $18$ statistical degrees of freedom, provided the assumed independence between groups is justified.

At the conventional $5\%$ significance level, Student 1's p-value $0.72$ gives no rejection of equal mean wear, while Student 2's p-value $0.009$ gives rejection of equal mean wear. The latter also rejects at the $1\%$ significance level. More generally, compare each p-value with the prespecified level $\alpha$; no decision threshold was fixed in the data supplied.

Failure to reject does not establish equality, and a p-value is not the posterior probability that the null hypothesis is true. These two-sided p-values alone also do not identify which material has larger mean wear.

Prefer Student 2's paired analysis, because the observational units are matched pairs: two shoes exposed to the same child's activity. A paired t-test uses $Y_i=A_i-B_i$, so shared child-to-child variation is removed before estimating the standard error. The independent units are the ten pairs, giving $9$ statistical degrees of freedom under the model of independent normally distributed pair differences.

The role of covariance is explicit:With common marginal variance ${\sigma }^2$ and within-pair correlation coefficient $\rho$, the true variance of the average difference is $2{\sigma }^2(1-\rho )/10$. The independent-groups calculation ignores the covariance term and uses $2{\sigma }^2/10$. Positive within-pair association can therefore make the paired standard error much smaller, explaining how the reported p-values can differ without arithmetic mistakes. The preference follows from the paired design, not from selecting the smaller p-value. Marginal normality alone need not imply normally distributed differences; exact paired inference uses the normal-difference assumption, for example justified by a joint multivariate normal distribution.