 Solution

Put $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$, ${\hat{\sigma }}^2=\Vert Y-X\hat{\beta }{\Vert }^2/(n-p)$. Orthogonal projections of the Gaussian error give independent variablesHence the exact $(1-\alpha )$ confidence set is the ellipsoidHere $F_{p,n-p}(q)$ denotes the $q$ quantile of the F-distribution.

Let ${\hat{\beta }}_{(-i)}$ be the fit with observation $i$ removed, when that reduced design retains full rank. Cook's distance iswhere $e_i$ is its residual and $h_{ii}=x_i^T(X^{T}X{)}^{-1}{x}_i$ is its leverage. Thus $D_i$ measures the deleted estimate's distance from the full estimate in exactly the metric of the confidence ellipsoids. The deleted estimate lies on the confidence contour whose $F$ quantile equals $D_i$; this is a geometric influence diagnostic, not an assertion that $D_i$ itself has an $F$ law.

For $p=2,n=50$, $0.70<1.3<2.42$, so deletion moves the estimate outside the 50% confidence ellipsoid but inside the 90% one. The observation warrants investigation for influence: the shift is substantial, though the supplied figures do not justify calling it outside the 90% region. The reciprocal $F_{48,2}$ and the approximate chi-square figures are unnecessary.