= Solution
Use the residual variance estimator
$$
s^2=\frac1{n-2}\sum_{i=1}^n
\{Y_i-\hat\alpha'-\hat\beta'(x_i-\bar x)\}^2.
$$
Under the Gaussian linear model, $\bar Y$ is independent of $s^2$ and
$$
\frac{(n-2)s^2}{\sigma^2}\sim\chi^2_{n-2}.
$$
Consequently the <Student t confidence interval for a centered regression intercept> follows from
$$
\frac{\bar Y-\alpha'}{s/\sqrt n}\sim t_{n-2}.
$$
It is
$$
\boxed{\bar Y\pm t_{n-2,0.975}\frac{s}{\sqrt n}}.
$$
The displayed pivotal quantity lies between its 2.5% and 97.5% quantiles with probability $0.95$, which proves the stated coverage.
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