= Solution
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,
$$
\overline D\sim N(\mu,\sigma^2/n),\qquad
S_D^2=\frac1{n-1}\sum_{i=1}^n(D_i-\overline D)^2.
$$
The normal-<sample variance> identity gives
$$
\boxed{c=\frac{n-1}{\sigma^2},\qquad cS_D^2\sim\chi^2_{n-1}.}
$$
Moreover $\overline D$ 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 $U=(n-1)S_D^2/\sigma^2\sim\chi^2_{n-1}$. Consequently
$$
\boxed{\frac{\overline D-\mu}{S_D/\sqrt n}
=\frac N{\sqrt{U/(n-1)}}\sim t_{n-1}.}
$$
This distribution is free of both unknown parameters.
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