= Solution
Let $H=X(X^TX)^{-1}X^T$ be the orthogonal projection onto the design column space. <Linear transformations> of the Gaussian errors give
$$
\boxed{\hat\theta\sim N_p(\theta,\sigma^2(X^TX)^{-1}),\qquad
\hat\epsilon\sim N_n(0,\sigma^2(I-H)).}
$$
The residual distribution is a singular multivariate normal supported on the orthogonal complement of that column space. Their cross-<covariance> is $\sigma^2(X^TX)^{-1}X^T(I-H)=0$. Joint normality even makes the two vectors independent.
Order the observations as $11,12,21,22$. The two <normal linear models> have design <matrices>
$$
X_I=\begin{pmatrix}1&0&0\\1&0&1\\0&1&2\\0&1&3\end{pmatrix},\qquad
X_{II}=\begin{pmatrix}1&0&0\\1&0&2\\0&1&3\\0&1&1\end{pmatrix}.
$$
Removing each row mean leaves the information for the common slope equal to the sum of within-row squared fertilizer deviations. These sums are $1$ and $4$, respectively. Therefore
$$
\boxed{\operatorname{Var}_I\hat\beta=\sigma^2,\qquad
\operatorname{Var}_{II}\hat\beta=\sigma^2/4.}
$$
The second design estimates the slope four times as precisely in <variance> terms.
The observed data in the second design are fitted exactly by $\hat\theta=(100,300,100)^T$. With $a=(2,2,0)^T$, the no-fertilizer total forecast is
$$
\boxed{\widehat T_{\rm next}=a^T\hat\theta=800.}
$$
The future four independent errors contribute <variance> $4\sigma^2$. Parameter-estimation uncertainty contributes $\sigma^2a^T(X_{II}^TX_{II})^{-1}a=13\sigma^2$, so the prediction-error <variance> is $17\sigma^2$, not merely the <variance> of the estimated mean. Normally, with $s^2=\|\hat\epsilon\|^2/(4-3)$, the <prediction interval in a normal linear model> is
$$
800\pm t_{1,0.975}\,s\sqrt{17},\qquad t_{1,0.975}=12.7062\ldots.
$$
Here the printed data have zero residual and hence $s=0$. \b[The formal plug-in interval collapses to $[800,800]$], exhibiting <zero-residual degeneracy of regression prediction>. Under the stated model with unknown positive <variance>, this exact fit is a probability-zero sample and the Studentized statistic is undefined at it; the collapsed expression is not evidence that future yield is certain. No positive-width numerical 95% interval can be obtained from these exact data by the usual residual-<variance> method without extra <variance> information or an explicit observation-rounding model.
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