= Solution
{c}
Since $\sum_{i=1}^nD_i=0$, we have $D_1=-\sum_{i=2}^nD_i$. Squaring and adding the other residual squares gives
$$
\boxed{\sum_{i=1}^nD_i^2=\left(\sum_{i=2}^nD_i\right)^2+\sum_{i=2}^nD_i^2.}
$$
Therefore $S^2$ is a measurable <function> of $(D_2,\ldots,D_n)$. By the block <independence> proved in part <8/b/i>, it is independent of $\overline X$. This is the <independence of the normal sample mean and sample variance>; pairwise independence alone would not justify passing to an arbitrary function of several residuals.
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