= Solution
Set $s=S_D/\sqrt{17}$. Symmetry of $t_{16}$ gives $P(T>-1.746)=0.95$, so
$$
P_\mu(\mu<\overline D+1.746s)=0.95.
$$
The requested upper confidence bound is therefore
$$
\boxed{b=\overline D+1.746\frac{S_D}{\sqrt{17}}.}
$$
There 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% <interval>
$$
\left(\overline D-1.746\frac{S_D}{\sqrt{17}},\infty\right).
$$
It excludes zero in the appropriate direction exactly when
$$
\boxed{\frac{\overline D}{S_D/\sqrt{17}}>1.746.}
$$
This 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>.
Back to article page
