= Solution
Put $\widehat\beta=(X^TX)^{-1}X^TY$, $\widehat\sigma^2=\|Y-X\widehat\beta\|^2/(n-p)$. Orthogonal projections of the Gaussian error give independent variables
$$
\frac{(\widehat\beta-\beta)^TX^TX(\widehat\beta-\beta)}{\sigma^2}\sim\chi_p^2,\qquad\frac{(n-p)\widehat\sigma^2}{\sigma^2}\sim\chi_{n-p}^2.
$$
Hence the exact $(1-\alpha)$ <confidence set> is the ellipsoid
$$
\boxed{\left\{b:\frac{(b-\widehat\beta)^TX^TX(b-\widehat\beta)}{p\widehat\sigma^2}\leq F_{p,n-p}(1-\alpha)\right\}.}
$$
Here $F_{p,n-p}(q)$ denotes the $q$ quantile of the <F-distribution>.
Let $\widehat\beta_{(-i)}$ be the fit with observation $i$ removed, when that reduced design retains full rank. <Cook's distance> is
$$
D_i=\frac{(\widehat\beta_{(-i)}-\widehat\beta)^TX^TX(\widehat\beta_{(-i)}-\widehat\beta)}{p\widehat\sigma^2}
=\frac{e_i^2}{p\widehat\sigma^2}\frac{h_{ii}}{(1-h_{ii})^2},
$$
where $e_i$ is its residual and $h_{ii}=x_i^T(X^TX)^{-1}x_i$ 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. \b[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.
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