= Solution
Using the paper's named weights $w_i$, take
$$
W=\operatorname{diag}(1,1,1,\tfrac12,\tfrac12),\qquad
A=\begin{pmatrix}1&1\\1&2\\1&3\\1&4\\1&5\end{pmatrix},\qquad
b=\begin{pmatrix}3\\5\\5\\4\\3\end{pmatrix}.
$$
The <least-squares normal equations> for <weighted least squares> become
$$
\begin{pmatrix}7/2&33/4\\33/4&97/4\end{pmatrix}
\begin{pmatrix}\widehat\beta_0\\\widehat\beta_1\end{pmatrix}
=\begin{pmatrix}59/4\\143/4\end{pmatrix}.
$$
Multiplying by four gives $14\widehat\beta_0+33\widehat\beta_1=59$ and $33\widehat\beta_0+97\widehat\beta_1=143$. Their <determinant> is $269$, giving
$$
\widehat\beta_0=\frac{1004}{269},\qquad \widehat\beta_1=\frac{55}{269}.
$$
Thus, to two significant digits, \b[$\boxed{\widehat\beta_0=3.7,\quad\widehat\beta_1=0.20}$], with the slope measured in expression units per hour.
There is a wording convention to make explicit. Because the stated objective squares $w_i$, the choice above gives the final two \i[squared residuals] one-quarter of the influence of the others. If “half as much” instead means half the coefficient of each squared residual, use $W=\operatorname{diag}(1,1,1,1/\sqrt2,1/\sqrt2)$. Then the <least-squares normal equations> are
$$
\begin{pmatrix}4&21/2\\21/2&69/2\end{pmatrix}
\begin{pmatrix}\widehat\beta_0\\\widehat\beta_1\end{pmatrix}
=\begin{pmatrix}33/2\\87/2\end{pmatrix},
$$
giving $\widehat\beta_0=150/37\approx\boxed{4.1}$ and $\widehat\beta_1=1/37\approx\boxed{0.027}$. Stating which quantity is halved resolves the ambiguity.
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