= Solution
{c}
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 endpoint
$$
\boxed{b=\overline D+1.746S_D/\sqrt{17}.}
$$
There 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 when
$$
\boxed{\frac{\overline D}{S_D/\sqrt{17}}>1.746.}
$$
This 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.
Back to article page
