= Solution
{c}
Differentiating the least-squares residual norm gives the <normal equations>
$$
X^TX\hat\theta=X^TY.
$$
If $X$ has full column rank, $\hat\theta=(X^TX)^{-1}X^TY$, and therefore
$$
\operatorname{Var}(\hat\theta)
=(X^TX)^{-1}X^T(\sigma^2I)X(X^TX)^{-1}
=\sigma^2(X^TX)^{-1}.
$$
Here $\bar x=0$ and $\bar z=2$, so the <design matrix> and its Gram matrix are
$$
X=\begin{pmatrix}
1&-3&1\\1&-2&0\\1&-1&-1\\
1&1&-1\\1&2&0\\1&3&1
\end{pmatrix},\qquad
X^TX=\operatorname{diag}(6,28,4).
$$
The centered predictor columns are perpendicular, giving
$$
\boxed{\operatorname{Cov}(\hat\beta,\hat\gamma)=0,\qquad
\operatorname{Var}(\hat\beta)=\sigma^2/28,\quad
\operatorname{Var}(\hat\gamma)=\sigma^2/4.}
$$
Both estimates are linear combinations of the jointly normal errors, hence jointly normal. Zero covariance therefore also gives \b[independence]. The normality assumption is what permits this last step; orthogonality alone would only prove uncorrelatedness.
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