Question

The annual mean daily high and low temperatures by selected cities are given in the following table. What is the estimate of the intercept coefficient a for the regression of the mean daily high temperature as a function of the mean daily low temperature? 8 8 City Low (degrees) High (degrees) 78 82 68 Mobile Jacksonville Portland Omaha Cincinnati 49 74 75 Dallas 49 50.0 45.0 47.5 -47.50 -50.0.

Answer 1

The correct option is

• - 47.50

The calculations are as below.

The mean of high and low temperature would be $\dpi{80} \bar h = \frac{\sum h}{n} = \frac{450}{6} = 75$ and $\dpi{80} \bar l = \frac{\sum l}{n} = \frac{294}{6} = 49$ . Also, $\dpi{80} \sum (h - \bar h)(l - \bar l) = 30$ , and $\dpi{80} \sum (l - \bar l)^2 = 12$ . For the regression be $\dpi{80} \hat h = \hat b_0 + \hat b_1 l$ , the slope coefficient would be $\dpi{80} \hat b_1 = \frac{\sum (h - \bar h)(l - \bar l)}{\sum (l - \bar l)^2} = \frac{30}{12} = 2.5$ . The intercept would be $\dpi{80} \hat b_0 = \bar h - \hat b_1 \bar l$ or $\dpi{80} \hat b_0 = 75 - 2.5 *49 = -47.5$ .