Sample Final ECON 120C, Summer 2022
代做assignment – , 该题目是值得借鉴的assignment代写的题目
- The actual exam will be a Canvas quiz
- You have 80 minutes (incl. self-proctoring routines).
- For the multiple choice questions, please check all answers that are correct. There can be multiple correct answers. At least one answer is correct.
- (25 points) Stata output Consider the following Stata output. . reg wage tenure collgrad, r Linear regression Number of obs = 2, F((a), 2228) = 122. Prob > F = 0. R-squared = 0. Root MSE = 5.
| Robust wage | Coef. Std. Err. t P>|t| [95% Conf. Interval] ————-+—————————————————————- tenure | .1629178 .019697 (b) 0.000 .1242914. collgrad | 3.43025 .3056852 11.22 0.000 (c) (d) _cons | (e) .1782825 33.67 0.000 5.654036 6.
Compute the missing numbers. Round to two digits.
- (24 points) Misc questions(There could also be questions on other topics such as panel data. See, for example, the WI22 final exam.)
(a) (8 points) Consider the following model: Yi= 0 + 1 Xi+ui, whereXiN(0,1),ui
N(0,1), andCov(ui,Xi) = 0.3. In addition, you have a candidate instrumentZi. The
instrument satisfies relevance,Cov(Xi,Zi) = 0.5, but not exogeneity,Cov(ui,Zi) = 0.3.
We assume that the data are iid and that large outliers are unlikely.
i. 1 T SLSp 1
ii. 1 T SLSp 1 + 0. 3
iii. 1 T SLSp 1 + 0. 6
iv. 1 p 1
v. 1 p 1 + 0. 3
(b) (8 points) Consider the following model: Yi = 1 + 3Xi+ui, whereXi N(0,1) and
ui N(0,1) are independent. Suppose that we have an instrumentZi N(0,1) and
thatCov(Zi,Xi) = 0.5. Instead ofXi, we observeX i=Xi+mi, wheremiN(0,1) is
independent of (ui,Xi,Zi). Suppose you run TSLS ofYionX iincluding a constant and
usingZias the instrument. Denote the resulting TSLS estimator as 1 T SLS. Choose the
correct answers.
i. 1 T SLSp 3
ii. 1 T SLSp CovCov((XYii,Z,Zii))
iii. 1 T SLSp 1
iv. 1 T SLSp CovCov((XY ii,Z,Zii))
v. 1 T SLSp 2
(c) (8 points) Some questions
i. Mean independence implies independence
ii. Suppose thatXN(0,1), thenE(X) = 0.
iii. LetX 1 andX 2 be independent normal random variables with mean 0 and variance 2.
ThenV ar(X 1 +X 2 ) = 2.
iv. The CLT states that, in large samples, sample averages are approximately normally
distributed.
v. Random assignment ofDiimpliesE(Yi^1 |Di= 1) =E(Yi^1 )
- (26 points) Instrumental variables
Here we analyze the impact of economic conditions on the likelihood of a civil conflict using
data from African countries. Our analysis is based on Miguel et al. (2004, JPE). For simplicity,
we ignore the panel data dimension and pool all country-year observations together. We further
restrict our attention to a simplified version of their model.
Specifically, we consider the following model:
anyprioi= 1 + 2 gdpgi+ 3 Oili+ui,
whereanyprioiis an indicator for the incidence of a civil war (anyprioi= 1 if there was a civil
war and 0 otherwise),gdpgiis GDP growth, andOiliis a indicator for whether the country is
a major oil producer. In addition, we observe annual rainfall growthGPCPgiwhich will serve
as an instrument for GDP growth.
(a) (6 points) The following output presents the results of the first stage regression. Are you
worried about weak instruments? Conduct a formal test.
. regress gdp_g GPCP_g Oil, robust Linear regression Number of obs = 743 F(2, 740) = 6. Prob > F = 0. R-squared = 0. Root MSE =.
| Robust
gdp_g | Coef. Std. Err. t P>|t| [95% Conf. Interval] ————-+—————————————————————- GPCP_g | .0416126 .0130415 3.19 0.001 .0160097. Oil | -.0091104 .0076873 -1.19 0.236 -.0242018. _cons | -.0044324 .0027591 -1.61 0.109 -.009849.
. test GPCP_g ( 1) GPCP_g = 0 F( 1, 740) = 10. Prob > F = 0. (b) (7 points) Next, we investigate an instrumental variables strategy to estimate 1 which usesGPCPgas an instrument forgdpg. Do you thinkGPCPgsatisfies exogeneity? Why, or why not? (max. 3 sentences) (c) (6 points) The next output presents the TSLS estimates of the model of interest. Interpret the coefficient ongdpg. . ivregress 2sls any_prio (gdp_g = GPCP_g) Oil, robust Instrumental variables (2SLS) regression Number of obs = 743 Wald chi2(2) = 0. Prob > chi2 = 0. R-squared =. Root MSE =.
| Robust any_prio | Coef. Std. Err. z P>|z| [95% Conf. Interval] ————-+—————————————————————- gdp_g | .4719936 1.950132 0.24 0.809 -3.350195 4. Oil | -.0029555 .0534177 -0.06 0.956 -.1076522. _cons | .2704271 .0188512 14.35 0.000 .2334795.
Instrumented: gdp_g Instruments: Oil GPCP_g
(d) (7 points) Another potential instrument for GDP growth is lagged rainfall growthGPCPgl, that is rainfall growth in the previous period. Thus, we have two instruments. Consider the following Stata outputs:
. ivregress 2sls any_prio (gdp_g = GPCP_g GPCP_g_l) Oil, robust Instrumental variables (2SLS) regression Number of obs = 743 Wald chi2(2) = 0. Prob > chi2 = 0. R-squared =. Root MSE =.
| Robust
any_prio | Coef. Std. Err. z P>|z| [95% Conf. Interval] ————-+—————————————————————- gdp_g | -.9376921 1.584072 -0.59 0.554 -4.042415 2. Oil | -.0160465 .0524952 -0.31 0.760 -.1189353. _cons | .2652756 .0182278 14.55 0.000 .2295498.
Instrumented: gdp_g Instruments: Oil GPCP_g GPCP_g_l . predict uhat, resid . reg uhat GPCP_g GPCP_g_l Oil Source | SS df MS Number of obs = 743 ————-+———————————- F(3, 739) = 0. Model | .320959862 3 .106986621 Prob > F = 0. Residual | 146.45314 739 .198177457 R-squared = 0. ————-+———————————- Adj R-squared = -0. Total | 146.7741 742 .197808761 Root MSE =.
uhat | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————- GPCP_g | .0227963 .0855004 0.27 0.790 -.1450563. GPCP_g_l | -.0889741 .0866411 -1.03 0.305 -.259066. Oil | -.0005938 .0505506 -0.01 0.991 -.0998336. _cons | .0006584 .0175388 0.04 0.970 -.0337735.
. test GPCP_g GPCP_g_l ( 1) GPCP_g = 0 ( 2) GPCP_g_l = 0 F( 2, 739) = 0. Prob > F = 0.
Based on the information given, conduct a formal test for instrument exogeneity at the
1%-level. What do you conclude?
- (25 points) Difference-in-differences In this question we study the effect of the minimum legal drinking age (MLDA) on death rates of 18-20-year-olds. Our analysis exploits differences in the MLDA across U.S. states and a difference-in-differences setting. Our control group is Arkansas (AR), which has had a 21 MLDA since 1933. Our treatment group is Alabama (AL) which lowered its MLDA from 21 to 19 in 1975. The following table presents mortality rates per 100,000 for both states before and after 1975.^1
Year AL AR
1974 143 119
1976 132 137
(a) (6 points) Mortality rates in AL dropped from 143 to 132 between 1974 and 1976. Does this
suggest that decreasing the MLDA decreased mortality rates? Why, or why not? (max. 3
sentences)
(b) (6 points) Compute the difference-in-difference estimate and interpret the result.
(c) (6 points) Based on the following figure, which displays mortality rates for both states from
1972 to 1978, assess the plausibility of the common trends assumption.
(d) (7 points) Now suppose that you have data on two additional states: California (CA) and
Pennsylvania (PA) both of which have had a 21 MLDA since 1933 (like AR). Based on
the information in the following table, which presents mortality rates from 1972 to 1978,
which control group (state) would you choose and why?
Year AL AR CA PA
1972 159 194 149 107
1974 143 119 132 112
1976 132 137 126 102
1978 138 141 135 100
(^1) This question is inspired by Angrist and Pischke (2015).