Past exam of the options II course of the University of Oxford / 2008 / O (Statistics)

For $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$, and its log derivative is $n/\lambda -S$. Thus, almost surely,The large-sample distribution is that of the estimator, rather than of the fixed parameter: $\sqrt{n}(\hat{\lambda }-\lambda )\stackrel{d}{\to }N(0,{\lambda }^2)$. A Wald confidence interval is consequentlyintersected with the positive parameter space if necessary. The unadorned parameter in the source's distribution request is interpreted as this estimator.

A single aphid survives to $T$ with probability $r=e^{-\lambda T}$, independently of the others. Hence $N_T\sim \mathrm{Bin}(n,r)$, andfor $0<n_T<n$. If $n_T=n$, the optimum is at the boundary $\lambda =0$; if $n_T=0$, the likelihood increases to its supremum as $\lambda \to \infty$, so there is no finite maximum likelihood estimate. Fisher information transforms from the binomial probability parameter asThe given strict inequality showsThresholding discards information about how long each survival time is. The second experiment therefore has a larger information bound on variance and larger asymptotic uncertainty for an efficient estimator. This is an expected-information comparison, rather than a claim that every realised finite-sample confidence interval must be wider.

Independence of the two samples makes their sample means independent normal random variables. A difference of independent normal random variables is normal, with the means subtracted and the variances added. Thus

For each normal sample, its sample mean is independent of its sample variance, and the scaled residual sum of squares is chi-squared. Independence of the samples also makes their two residual sums independent. Adding their chi-squared distributions givesConsequentlyThe pooled variance is also independent of the difference of sample means.

For a normal sample of size $m$,Then $\bar{X}\sim N({\mu }_X,{\sigma }^2/m)$ and $c{S}_X^2\sim {\chi }_{m-1}^2$ with $c=(m-1)/{\sigma }^2$. These two statistics are independent.

Combining the difference distribution and pooled variance calculation gives, with $\nu =m+n-2$,The printed statistic has the sign of the centering term reversed. Its numerator subtracts ${\mu }_Y-{\mu }_X$, so it is not generally a central Student statistic; the corrected centering above is necessary. Under the null hypothesis ${\mu }_X={\mu }_Y$, the two expressions coincide, so the requested test is unaffected. Reject the two-sided null at significance level $\alpha$ whenHere $\nu =8$ and $|T|=2.19<2.31$, so do not reject at the 5% level. This does not establish equality of the population means.

Age matching means that the two observations in a pair share an age effect, while ages vary across pairs. The independent, identically distributed two-sample model no longer represents this design: random pair effects induce within-pair dependence, and fixed differing ages can instead produce different individual means. Pairing by itself need not cause stochastic dependence when ages are treated as fixed. The appropriate alternative is a paired t-test on the within-pair differences. Removing shared age variation can reduce the noise and improve power when age strongly affects volume and the members of a pair are positively associated. Improvement is not automatic, since the paired test also uses fewer degrees of freedom.