Question

You have the following data T2 T3 and the true DGP is: Your friend calculated two estimates of ^: ^1- 1 and B2 2 (a) Find the sum of squared residuals for A1. (hint: you should get an actual number) ind the sum of squared residuals for ß2. (hint: you should get an actual number) (c) Which one of these two estimates is more likely to be the OLS estimate of 3? Why? (d) Find the estimate of β that has the smallest sum of squared residuals

Answer 1

(a)

For $\hat{\beta_1} = 1$ , the estimated equation is, y = x

The estimated y from x are 1, 6, 4

Residuals are (2 - 1) , (6 - 6) , (8 -4)

Sum of squared residuals for $\hat{\beta_1}$ is,

(2 - 1)² + (6 - 6)² + (8 -4)² = 17

(b)

For $\hat{\beta_2} = 2$ , the estimated equation is, y = 2x

The estimated y from x are 2, 12, 8

Residuals are (2 - 2) , (6 - 12) , (8 -8)

Sum of squared residuals for $\hat{\beta_2}$ is,

(2 - 2)² + (6 - 12)² + (8 -8)² = 36

(c)

Since the sum of squared residuals for $\hat{\beta_1}$ is less than that of $\hat{\beta_2}$,  $\hat{\beta_1}$ is more likely to be the OLS estimate of $\beta$.

(d)

For the smallest sum of squared residuals, $\beta$ is, given as,

$\hat{\beta} = \frac{\sum_{i=1}^{n}x_i y_i}{\sum_{i=1}^{n}x_i^2} = \frac{1*2 + 6 * 6 + 4 * 8}{1^2 + 6^2 + 4^2} = \frac{70}{53} = 1.320755$