 Solution

Let $\mathrm{RSS}=\Vert Y-X\hat{\theta }{\Vert }^2$. The usual unbiased error-variance estimator isIndeed the residual is the orthogonal projection of $ϵ$ onto the $(n-p)$-dimensional complement of the design column space, giving $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$ and expected RSS equal to $(n-p){\sigma }^2$. The maximum-likelihood estimator instead divides RSS by $n$; the unbiased divisor above is the one used for the exact t-interval in part (d).