= Solution
{c}
Minimize $\|Y-X\theta\|^2$. Differentiation gives the <normal equations> $X^TX\widehat\theta=X^TY$. Full column rank makes $X^TX$ positive definite, so
$$
\boxed{\widehat\theta=(X^TX)^{-1}X^TY,\quad
\mathbb E\widehat\theta=\theta,\quad
\operatorname{Cov}(\widehat\theta)=\sigma^2(X^TX)^{-1}.}
$$
The expectation uses the zero error mean; the <covariance> uses $\operatorname{Cov}(\epsilon)=\sigma^2I$. With independent <Gaussian> errors of common <variance>, maximizing likelihood minimizes exactly the same residual sum of squares, so this is also the maximum likelihood estimator, whether the common <variance> is known or estimated.
For the quadratic design, decompose each group's residual sum of squares into its within-group constant plus the group size times the squared residual of its <sample mean>. The three parameter combinations $\alpha$, $\alpha+\beta+\gamma$ and $\alpha-\beta+\gamma$ can fit all three group means exactly. Therefore
$$
\widehat\alpha=\overline Y_0,\qquad
\boxed{\widehat\beta=\frac{\overline Y_1-\overline Y_{-1}}2,\qquad
\widehat\gamma=\frac{\overline Y_1+\overline Y_{-1}-2\overline Y_0}2.}
$$
The group means are independent normal variables, with variances $\sigma^2/m,\sigma^2/n,\sigma^2/n$. Hence their two stated linear combinations are jointly normal, and
$$
\operatorname{Cov}(\widehat\beta,\widehat\gamma)
=\tfrac14(\sigma^2/n-\sigma^2/n)=0.
$$
Joint normality therefore proves \b[the two estimators are independent]. Their variances are $\sigma^2/(2n)$ and $\sigma^2/(2n)+\sigma^2/m$, respectively. Equal replication at the two off-center temperatures is what cancels the <covariance>.
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