Economics 30331: Econometrics

Professor Kasey Buckles

Spring 2007

Midterm Solutions

1. A. Since schoolQ and black are negatively correlated, and schoolQ is positively correlated with score, we expect the coefficient on black to be negatively biased. Since schoolQ and faminc are positively correlated, and schoolQ is positively correlated with score, we expect the coefficient on faminc to be positively biased.

2. B. The model violates the no perfect multicollinearity assumption (Assumption 4). Therefore, the parameter estimates will be biased, and B is false.

3. B. The R-squared is the ESS/TSS = 52.3/148.3 = .3526

4. E. All of these would be consequences of including an irrelevant variable.

5. A. Normality is required for valid inference—that is, for doing t-tests and F-tests. Note that if the sample size is large enough, we can relax this assumption, but the question asked you if inference is valid for any sample size.

6. C. OLS is not BLUE when you have heteroskedasticity. You do not need assumption 5 for unbiasedness or consistency of OLS, and inference will be invalid if this assumption is broken.

7. C. We have to be careful about the ceteris paribus interpretation here. If two players play the same number of minutes, but one player does it over the course of many more games, that player must not be as talented and therefore is likely to receive a lower salary.

II. Short Answer

8. (4 points) This is problem C7.4, part (iv) from your homework. To be able to directly test this null hypothesis, you could create four dummy categories (femaath, femnoath, maleath, malenoath) and leave out either femath or femnoath. The coefficient on the other female category variable is the one you want to test, so if this coefficient were b4, the null hypothesis would be b4 = 0. You could also create an interaction of male and athlete (maleath), and include the variables male, athlete, and maleath. The coefficient on athlete will be the one you want to test. Again, if this coefficient were b4, the null hypothesis would be b4 = 0.

9. (8 points)

  1. The two ways were Method of Moments and Minimizing the Sum of Squared Residuals. For the MoM, we impose population moments on the sample. This gives us two equations and two unknowns, which we can solve to derive the OLS estimators. For minimizing the SSR, we take the derivative of SSR with respect to b0-hat and b1-hat and set them equal to zero. Again, this gives us two equations and two unknowns, which we can solve to derive the OLS estimators.
  2. We needed assumptions 1-4 to derive the OLS estimators.

10. (9 points).

  1. A 1% increase in nox causes a .72% decrease in housing prices, ceteris paribus. Increasing average rooms by one increases housing prices by 30.6%, ceteris paribus.
  2. In the first equation, the coefficient on log(nox) is suffering from omitted variable bias due to the omission of rooms. If rooms and log(nox) are negatively correlated, and rooms and log(price) are positively correlated, this would result in a negative bias in the coefficient on log(nox) in equation [1].
  3. False (this is homework problem 3.9, part (iii)). As these estimates are only for one sample, we can never know which is closer to b1. But if this is a “typical” sample, b1 is closer to -.718.

11. (15 points)

  1. lprice = β0 + β1 lassess + β2 bdrms + β3 lsqrft + β4 llotsize + β5 colonial + u.
  2. A 1% increase in assessed values leads to a 1.04% increase in prices, ceteris paribus. At the 5% level, we fail to reject Ho: β1 = 1. (Because t = (1.036-1)/.151 = .238, t < 1.987.)
  3. Adding a bedroom to the house, ceteris paribus, increases housing prices by 2.5%. Since t = .025/.023 = 1.09, t<1.987, fail to reject Ho.
  4. Ho: b2=b3=b4=b5=0. H1: Ho is false. Restricted model: lprice = β0 + β1 lassess +u. You need to create an F-statistic, which would have q = 4 numerator d.o.f., and 88-5-1 = 82 denominator d.o.f.

12. (18 points)

  1. The coefficient on agemale has a p-value of .091, which tells us that it is not significant at the 5% level for a two-sided test.
  2. Taking the derivative of divorced with respect to male, we get -.027 + .0002(age). For males, the effect of an extra year of age is .0010+.0002 = .0012.
  3. Restricted: divorced = β0 + β1 age + β2educ + u.

Unrestricted: divorced = β0 + β1 age + β2educ +β3male + β4educmale +β5 agemale + u.

Ho: b3=b4=b5=0.

d. She should estimate the restricted model for everyone together (pooled version) and then for men and women separately (group 1 and group2). Taking the SSRs from these regressions, she creates the Chow statistic. The statistic will have k + 1 = 3 numerator d.o.f. and n-2(k+1) = 101687 denominator d.o.f. If the Chow statistic is greater than 2.60, she rejects the null hypothesis, where 2.60 is the critical value for the 5% test.