Module 8: Linear least squares regression practice problems
Ordinary least squares estimators from intermediate values
(The attached PDF file has better formatting.)
** Exercise 1.1: Ordinary Least Squares Estimates
We have 4 pairs of points
i | Xi | Yi |
1 | 0.5 | 1.5 |
2 | 1.0 | 2.5 |
3 | 1.5 | 3.5 |
4 | 2.0 | 5.5 |
Note that = 5.0, = 13.0, = 19.5, and = 7.5
What is (xi – )2?
What is (xi – )(yi – )?
What is B, the ordinary least squares estimate of â?
What is A, the ordinary least squares estimate of á?
Part A: Derive from and as
= – ()2 / N = 7.5 – 5.02 / 4 = 7.5 – 25/4 = 1.250
Part B: Derive from , , and .as
= – × / N = 19.5 – 5 × 13 / 4 = 3.250
Part C: B = = 3.250 / 1.250 = 2.600
Alternatively, we use the formula involving the expressions above:
B = =
= [ 19.5 – 4 × 5/4 × 13/4 ] / (7.5 – 4 × (5/4)2 ] = 2.600
Part D: B = 2.600, so A = – B× = ¼ × 13 – 2.6 × ¼ × 5 = 0.000.
Jacob: Could we also solve this problem by regressing the four y values on the four x values?
Rachel: This exercise is deliberately simple, so you can verity the computation using the intermediate values with a regression analysis (using Excel or R or any statistical package) of the four observations. Final exam problems may give just the intermediate values (not the raw observations), with an N or 100 or 200.