(c) The value suggests that a 1% increase in the percentage of the population with a bachelor’s degree or more will lead to an increase of $1028.96 in the mean income per capita.
(d)
(e) Since , we have
(f) For Georgia
Exercise 2.8
(a) The sample means from the two data parts are
and
Using these values, we find and . The fitted line is shown in Figure xr2.8.
Figure xr2.8 Fitted regression line and mean
(b) The values of the residuals, computed from , are:
16611
244−3−6
3111126
499−2−8
5131300
61717212
The required sums are ,
Exercise 2.8 (continued)
(c) The least squares estimates are
For the least squares residuals , .
(d) The sum of squared residuals from the mean regression is . The sum of the least squares residuals is . The least squares estimator is designed to provide the smallest value.
Exercise 2.9
(a)
Similarly, . Then
Finally,
We have shown that conditional on x the estimator is unbiased.
(b) Use the law of iterated expectations.
Because the estimator is conditionally unbiased it is unconditionally unbiased also.
(c)
Similarly, . So that
Exercise 2.9(c) (continued)
We know that is larger than the variance of the least squares estimator because is a linear estimator. To show this note that
Where and
Furthermore is an unbiased estimator. From the Gauss-Markov theorem we know that the least squares estimator is the “best” linear unbiased estimator, the one with the smallest variance. Therefore, we know that is larger than the variance of the least squares estimator.
Exercise 2.10
(a) If the model reduces to
(b) Graphically, setting implies the regression model is a horizontal line when plotted against at the height .
(c)
Figure xr2.10 Sum of squares for
The minimum appears to be at
(d) To find the minimum, we find the value of such that the slope of the sum of squares function is zero.
Solving, we find
To ensure that this is a minimum the second derivative must be positive. as long as N > 0, so that we have at least one data point.
Exercise 2.10 (Continued)
(e) The least-squares estimate is