Least squares picks the line with the smallest \(\sum_{i=1}^{n} (y_i - \hat y_i)^2\):
\[\hat\beta_1 = \frac{\sum_{i=1}^{n}(x_i - \bar x)(y_i - \bar y)}{\sum_{i=1}^{n}(x_i - \bar x)^2},
\qquad
\hat\beta_0 = \bar y - \hat\beta_1 \bar x\]
95% confidence interval for the slope, and the t-test of \(H_0: \beta_1 = 0\) (temperature has no linear effect):
\[\hat\beta_1 \pm t_{0.975,\,n-2}\;\mathrm{SE}(\hat\beta_1),
\qquad
t = \frac{\hat\beta_1 - 0}{\mathrm{SE}(\hat\beta_1)} \sim t_{n-2} \text{ under } H_0\]
How much it explains: \(R^2 = 1 - \dfrac{\sum (y_i - \hat y_i)^2}{\sum (y_i - \bar y)^2}\). With one predictor, \(R^2 = r^2\).