Step 1: Instead of raw sums, work with deviations from the means. With $n=9$, the mean temperature is $\bar{x} = 779/9 = 86.56$ and the mean absences is $\bar{y} = 77/9 = 8.56$.
Step 2: For each point compute $d_x = x - \bar{x}$ and $d_y = y - \bar{y}$, then form $d_x^2$, $d_y^2$ and $d_x d_y$. Summing all nine rows gives:
\[ \sum d_x^2 = 736.22, \qquad \sum d_y^2 = 154.22, \qquad \sum d_x d_y = 330.22 \]
These are exactly $S_{xx}$, $S_{yy}$ and $S_{xy}$, confirming this mean-deviation method agrees with the raw-sum method.
Step 3: The least-squares slope is the ratio of the cross-product deviation sum to the sum of squared $x$-deviations.
\[ b = \frac{\sum d_x d_y}{\sum d_x^2} = \frac{330.22}{736.22} = 0.449 \]
Step 4: The unexplained (residual) variation is what remains of $\sum d_y^2$ after removing the part explained by the regression.
\[ SSE = \sum d_y^2 - b\sum d_x d_y = 154.22 - (0.449)(330.22) = 6.07 \]
The residual standard deviation with $n - 2 = 7$ degrees of freedom is:
\[ s = \sqrt{6.07/7} = 0.931 \]
Step 5: The standard error of the slope is $s$ scaled by the spread of the $x$ values:
\[ SE(b) = \frac{0.931}{\sqrt{736.22}} = 0.0343 \]
Step 6: Using the $t$-distribution with 7 degrees of freedom, the critical value for a 95% two-sided interval is $t = 2.365$. The half-width of the interval is:
\[ 2.365 \times 0.0343 = 0.081 \]
Step 7: Adding and subtracting this half-width from the slope estimate gives the confidence interval.
\[ 0.449 - 0.081 = 0.368, \qquad 0.449 + 0.081 = 0.530 \]
Allowing for rounding, this matches option (B).
\[ \boxed{(0.367,\ 0.530)} \]