= Solution
For a new independent tumour grown for ten days in B, use $x_0=(1,10,10,0)^T$. Its fitted mean is
$$
\widehat\mu_0=x_0^T\widehat\beta=0.6464+10(15.2174)=152.8204.
$$
Write $h_0=x_0^T(X^TX)^{-1}x_0$. The <prediction interval in a normal linear model> is
$$
\boxed{152.8204\ \pm\ t_{26,\,1-\alpha/2}\,s\sqrt{1+h_0},}
$$
where $s^2=\mathrm{RSS}/26$ and $t_{26,1-\alpha/2}$ is the indicated <Student's t-distribution> <quantile>. The design information needed to evaluate $h_0$ is not supplied, so a fully numerical interval cannot be recovered from the coefficient <standard errors> alone: the coefficient covariances are also needed.
To prove coverage, write $Y_0=x_0^T\beta+\varepsilon_0$, with $\varepsilon_0\sim N(0,\sigma^2)$ independent of the original sample. The error $Y_0-x_0^T\widehat\beta$ has a <normal distribution> with mean zero and <variance> $\sigma^2(1+h_0)$. By the <joint distribution of least-squares and variance estimators>, $26s^2/\sigma^2\sim\chi^2_{26}$ and is independent of $\widehat\beta$; it is also independent of $\varepsilon_0$. Hence the <pivotal quantity>
$$
T=\frac{Y_0-x_0^T\widehat\beta}{s\sqrt{1+h_0}}\sim t_{26}.
$$
Consequently $\Pr(-t_{26,1-\alpha/2}\le T\le t_{26,1-\alpha/2})=1-\alpha$, which is exactly the claimed <prediction interval> coverage. This is repeated-sampling coverage conditional on the design, assuming the model; it need not hold conditional on the particular observed training responses. The extra $1$ represents new-observation noise and must not be omitted as in a mean <confidence interval>.
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