= Solution
{c}
Each before-and-after observation belongs to the same individual, so pairing removes individual baseline variation and allows dependence within a person's two readings. Under the stated independent normal model for the differences,
$$
\overline D\sim N(\mu,\sigma^2/n),\qquad
S_D^2=\frac1{n-1}\sum_{i=1}^n(D_i-\overline D)^2.
$$
The <independence of the normal sample mean and sample variance> theorem gives <independence> of $\overline D$ and $S_D^2$, and
$$
\boxed{c=\frac{n-1}{\sigma^2},\qquad cS_D^2\sim\chi^2_{n-1}.}
$$
Also $(\overline D-\mu)/(\sigma/\sqrt n)$ is standard normal. Dividing it by the square <root> of the independent $\chi^2_{n-1}/(n-1)$ variable gives
$$
\boxed{Z=\frac{\overline D-\mu}{S_D/\sqrt n}\sim t_{n-1}.}
$$
Its law contains neither unknown parameter, providing the <pivotal quantity> for a <paired t-test>.
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