Past exam of the options I course of the University of Oxford / 2010 / O (Statistics)

Each before/after pair concerns the same individual, so subject-specific baseline blood pressure can affect both measurements. Analyzing the within-person difference retains this pairing and removes a common baseline contribution. Under the stated independent normal-difference model,The normal-sample variance identity givesMoreover $\overline{D}$ and $S_D^2$ are independent, because the sample-mean projection and centered residual projection of a Gaussian vector are orthogonal and hence independent. Thus the standardized normal variable $N=\sqrt{n}(\overline{D}-\mu )/\sigma$ is independent of $U=(n-1)S_D^2/{\sigma }^2\sim {\chi }_{n-1}^2$. ConsequentlyThis distribution is free of both unknown parameters.

Set $s=S_D/\sqrt{17}$. Symmetry of $t_{16}$ gives $P(T>-1.746)=0.95$, soThe requested upper confidence bound is thereforeThere is a direction inconsistency in connecting this particular interval to the stated treatment alternative. Since $D$ is before minus after, a blood-pressure reduction means $\mu >0$. An upper interval $(-\infty ,b)$ excludes the null value zero only when $b<0$, which instead supports $\mu <0$.

For the requested reduction alternative, use the lower 95% intervalIt excludes zero in the appropriate direction exactly whenThis is the correct level-5% paired one-sided $t$ test of no mean change against a reduction. The requested upper bound is valid as a confidence statement, but its ordinary exclusion-of-zero test has the opposite direction.

The sign follows the duality between confidence intervals and hypothesis tests for the paired t-test.

Full column rank gives the least-squares estimatorWrite $A=(X^{T}X{)}^{-1}{X}^T$. Since $AX=I$ and $\mathrm{Cov}(ϵ)={\sigma }^{2}I$, we have $\hat{\theta }=\theta +Aϵ$ andThis calculation needs the stated uncorrelated equal-variance errors, but not normality.

Put $z_i=e^{-t_i}-c$ and take independent measurement errors ${ϵ}_i\sim N(0,{\sigma }^2)$, independent across the prescribed measurement times. ThenAssume the errors have zero mean so the stated activity curve is the observation mean. Also assume at least two distinct times, making the two design columns linearly independent. The times and the centering constant $c$ are treated as known.

By the definition of $c$, ${\sum }_i{z}_i=0$. Put $S={\sum }_i{z}_i^2>0$. ThenPart (a) therefore givesIn particular the two estimates are uncorrelated, and under the assumed normal-error model they are independent.

The expected activity is $E_t=\alpha +\beta (e^{-t}-c)$. Under normal errors, least squares is maximum likelihood, and invariance of maximum-likelihood estimation givesThe covariance term vanishes by part (ii), soThis is uncertainty in the estimated mean activity; the variance of a future individual observation would include its additional measurement-error variance.