# Ch 2 Solutions to Slr

Topics: Regression analysis, Simple linear regression, Linear regression Pages: 16 (4899 words) Published: October 18, 2011
Solutions, Chapter 2/HL

ANSWERS TO CHAPTER 2 The Simple Regression Model

Econometrics Economics of Innovation and Growth

A = Problems B = Examples (from chapter 2) C = Cumputer Exercises

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Solutions, Chapter 2/HL

A: Problems
2.1 Let kids denote the number of children born to a woman, and let educ denote years of education for the woman. A simple model relating fertility to years of education is kids = β 0 + β1educ + u where u is the unobserved error. (i) (ii) What kind of factors are contained in u? Are these likely to be correlated with level of education? Will a simple regression analysis uncover ceteris paribus effects of education on fertility? Explain.

(i) Income, age, and family background (such as number of siblings) are just a few possibilities. It seems that each of these could be correlated with years of education. (Income and education are probably positively correlated; age and education may be negatively correlated because women in more recent cohorts have, on average, more education; and number of siblings and education are probably negatively correlated.) (ii) Not if the factors we listed in part (i) are correlated with educ. Because we would like to hold these factors fixed, they are part of the error term. But if u is correlated with educ then E(u|educ) ≠ 0, and so SLR.3 fails. --------------------------------------------------------------------------------------------------------------2.2 In the simple linear regression model y=β0+β1x + u, suppose that E(u) ≠ 0. Letting α0=e(u), show that the model can always be rewritten with the same slope, but new intercept and error, where the new error has a zero expected value. Answers In the equation y = β0 + β1x + u, add and subtract α0 from the right hand side to get y = (α0 + β0) + β1x + (u − α0). Call the new error e = u − α0, so that E(e) = 0. The new intercept is α0 + β0, but the slope is still β1.

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Solutions, Chapter 2/HL 2.3 The following table contains the ATC scores and the GPA (grade point average) for 8 college students. Grade point average is based on a four-point scale and has been rounded to the one digit after the decimal. Student 1 2 3 4 5 6 7 8 GPA 2.8 3.4 3.0 3.5 3.6 3.0 2.7 3.7 ACT 21 24 26 27 39 25 25 30

(i) Estimate the relationship between GPA and ACT using ols; that is, obtain the intercept and slope in the equation ˆ ˆ ˆ GPA = β 0 + β1 ACT

Comment on the direction of the relationship. Does the intercept have a useful interpretation here? Explain. How much higher is GPA predicted to be if the ACT score is increased by 5 points? (ii) Compute the fitted valued and the residuals for each observation, and verify that the residuals (approximately) sum to zero. (iii) What is the predicted value of GPA when ACT =20? (iv) How much of the variation in GPA for the 8 students is explained by ACT. Explain.

2.3 (i) Let yi = GPAi, xi = ACTi, and n = 8. Then x = 25.875, y = 3.2125, ∑ (xi – x )(yi – i=1

n

ˆ y ) = 5.8125, and ∑ (xi – x )2 = 56.875. From equation (2.9), we obtain the slope as β1 = i=1

n

ˆ 5.8125/56.875 ≈ .1022, rounded to four places after the decimal. From (2.17), β 0 = y – ˆ β x ≈ 3.2125 – (.1022)25.875 ≈ .5681. So we can write 1

ˆ GPA = .5681 + .1022 ACT

n = 8.
The intercept does not have a useful interpretation because ACT is not close to zero for the ˆ population of interest. If ACT is 5 points higher, GPA increases by .1022(5) = .511.

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Solutions, Chapter 2/HL
(ii) The fitted values and residuals — rounded to four decimal places — are given along with the observation number i and GPA in the following table:

i GPA 1 2.8
2 3.4 3 3.0 4 3.5 5 3.6 6 3.0 7 2.7 8 3.7

ˆ GPA

ˆ u

2.7143 .0857 3.0209 .3791 3.2253 –.2253 3.3275 .1725 3.5319 .0681 3.1231 –.1231 3.1231 –.4231 3.6341 .0659

You can verify that the residuals, as reported in the table, sum to −.0002, which is pretty close to zero given the inherent rounding error. ˆ (iii) When ACT = 20, GPA = .5681 +...