Economics 30331, Fall 2007
Professor Kasey Buckles
Midterm Solutions
I. Multiple Choice (3 points each)
1. D An increase of 100 runs increases salary by 16%. The dependent variable is in logs,
which is why the coefficient on “runs” is interpreted as a percent increase (rather than
dollars). The t-statistic on runs is statistically significant at all conventional levels.
2. B Including an irrelevant variable that is not correlated with the dependent variable will
not bias the coefficients, but will likely increase the standard errors on those coefficients.
3. B. This is like homework problem 3.7. Omitting an important variable can cause bias
when the omitted variable is correlated with the included explanatory variables. This
produces a violation of the zero conditional mean assumption. The homoskedasticity
assumption played no role in showing that the OLS estimators are unbiased. Further, the
degree of collinearity between the explanatory variables in the sample, even if it is reflected
in a correlation as high as .95, does not violate the Gauss-Markov assumptions. Only if there
is a perfect linear relationship among two or more explanatory variables is Assumption 4
violated. Likewise, interaction terms do not produce a perfect linear combination of the
variables, and are not a violation of Assumption 4.
4. B. Larger variation in the independent variable x1. See the formula for variance.
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.
II. Short Answers
6. (8 points) This is like homework problem 4.6.
a. price – assess = u
b. We use the SSR form of the F statistic. We are testing q = 2 restrictions and the
df in the unrestricted model is 86. We are given SSRr = 209,448.99 and SSRur =
165,644.51. Therefore,
(209, 448.99  165, 644.51)  86 
F 
    11.37,
165, 644.51
 2 
which is a strong rejection of H0: the 1% critical value with 2 and 90 df is 4.85.
7. (8 points)
a. 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.
b. We needed assumptions 1- 4 to derive the OLS estimators.
8. (10 points)
a. deans = β0 + β1ACT + β2hsGPA + u
b. A one-point increase in hsGPA increases the probability of making the dean’s list
by 46.3%. The t-statistic is .4628784/.133973 = 3.46. The critical value for a 1% test with
141–2 –1 d.o.f. is 2.576 (or 2.617). Since t>c, we can reject the null at the 1% level.
c. The coefficient on ACT is neither practically nor statistically significant (t = .011).
It seems surprising that students with higher ACT scores are no more likely to make
the dean’s list. However, remember that we are controlling for hsGPA. For students
with the same hsGPA, doing better on the ACT does not affect their probability of
making the dean’s list. We have to keep in mind the ceteris paribus interpretation.
d. The R-squared is .089 (ESS/TSS). This means that our model explains 8.9% of
the variation in making the dean’s list.
9. (10 points)
a. The model exhibits the “dummy variable trap,” and therefore violates the no perfect
collinearity assumption. She needs to omit one of the dummy categories to use as a
base case.
b. β3 tells us whether the effect of being married on salary is different for men and
women (or, whether the effect of being female is different for married and singles). The
effect of being a married female vs. a single female is β2 + β3. The effect of being a married
female is β1 + β2 + β3, and the effect of being a single female is β1. The difference is β2 +
β3. Ho: β3 = 0.
c. In part A, she could have dropped femsing, and the coefficient on femmarr would give
her the same result. The t-statistic on femsing would tell us whether the difference is
statistically significant. Likewise, she could have dropped femmarr.
10. (14 points)
a. This would generate negative omitted variable bias in the coefficient on yrsmarr.
Thus, when adding kids, we would expect the coefficient on yrsmarr to increase.
b. This is like homework problem 3.9, part iii. No, because the results are estimates for
only one sample, and we can never know which is closer to 1 . But if this is a
“typical” sample, 1 is closer to its true value in the second regression.
c. unrestricted model: affair = β0 + β1 yrsmarr + β2 religious + β3 kids + β4 kidsyrs +
β5 kidsrelig + u
restricted model: affair = β0 + β1 yrsmarr + β2 religious + u
H0: β3 = β4 = β5 = 0.
d. The economist would perform an F-test (or Chow test). For an F-test, there are three
exclusion restrictions (q = 3), so there are 3 numerater d.o.f. and 601-5-1 = 595
denominator d.o.f. The critical value for a 5% test is 2.60. For a Chow test, you
would run the restricted model for people with kids, people without kids, and for all
people together. The formula is: [(SSRp – (SSRk + SSRnk))/(k+1)]/[(SSRk +
SSRnk)/(n-2(k+1))]. k + 1 = 3, n-2(k+1) = 595, same numerator and denominator
d.o.f as above, same critical value.