Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / O / 1 / b / Solution

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.