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ECON513-无代写
时间：2023-03-06
ECON 513: Practice of Econometrics
(Marco Angrisani)
Midterm Exam Example
This exam consists of
• multiple choice questions
• true/false questions
• short answer questions
To answer multiple choice and true/false questions, simply cross or color the box corresponding
to your answer.
For short answer questions, write your answers in the box provided below each question.
Only what is inside the boxes will be evaluated for grading. Try to provide concise
and precise answers.
You have 2 hours to complete the exam.
Good luck!
1
QUESTION1 [5 points]
Consider the following linear model in matrix notation:
y = Xβ + u with E[u|X] = 0 and E[uu′|X] = σ2I
where y is a N × 1 vector, β is a k × 1 vector, X is a N × k matrix and u is a N × 1 vector.
Under the assumptions of this model, the OLS estimator βˆ = (X′X)−1 X′y is
Unbiased
Consistent
Asymptotically normally distributed
All of the above
None of the above
QUESTION2 [5 points]
Consider the following linear model in matrix notation:
y = Xβ + u with E[u|X] = 0 and E[uu′|X] = σ2I
where y is a N × 1 vector, β is a k × 1 vector, X is a N × k matrix and u is a N × 1 vector.
Under the assumptions of this model, the OLS estimator βˆ = (X′X)−1 X′y is the best unbi-
ased estimator.
True
False
QUESTION3 [5 points]
Consider the following linear model in matrix notation:
y = Xβ + u with E[u|X] = 0 and E[uu′|X] = σ2I
where y is a N × 1 vector, β is a k × 1 vector, X is a N × k matrix and u is a N × 1 vector.
The variance of the OLS estimator is V ar
(
βˆ|X
)
= σ2 (X′X)−1. In order to compute V ar
(
βˆ|X
)
,
you need to estimate σ2. Explain how an unbiased estimator of σ2 can be computed.
2
QUESTION4 [5 points]
You want to use the following linear model:
y = Xβ + u with E[u|X] = 0 and E[uu′|X] = σ2I
where y is a N × 1 vector, β is a k× 1 vector, X is a N × k matrix and u is a N × 1 vector, to
study the effect of education on earnings.
You have data on 481 individuals for whom you observe yearly earnings in thousand dollars
and years of education. You run a regression of earnings on education in Stata and obtain the
following output:
. reg earnings education
Source | SS df MS Number of obs = 481
-------------+---------------------------------- F(1, 479) = 51.55
Model | 4400.07607 1 4400.07607 Prob > F = 0.0000
Residual | 40885.9606 479 85.3569114 R-squared = 0.0972
-------------+---------------------------------- Adj R-squared = 0.0953
Total | 45286.0366 480 94.3459096 Root MSE = 9.2389
------------------------------------------------------------------------------
earnings | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
education | 1.491731 .2077685 7.18 0.000 1.08348 1.899981
_cons | -1.535019 2.564343 -0.60 0.550 -6.573771 3.503734
------------------------------------------------------------------------------
Interpret the estimated coefficient for education. Under the assumptions of the model above, can
you give a causal interpretation to this estimated coefficient? Do you think these assumptions
are plausible for this specific application? Explain briefly.
3
QUESTION5 [5 points]
Consider the Stata output
. reg earnings education
Source | SS df MS Number of obs = 481
-------------+---------------------------------- F(1, 479) = 51.55
Model | 4400.07607 1 4400.07607 Prob > F = 0.0000
Residual | 40885.9606 479 85.3569114 R-squared = 0.0972
-------------+---------------------------------- Adj R-squared = 0.0953
Total | 45286.0366 480 94.3459096 Root MSE = 9.2389
------------------------------------------------------------------------------
earnings | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
education | 1.491731 .2077685 7.18 0.000 1.08348 1.899981
_cons | -1.535019 2.564343 -0.60 0.550 -6.573771 3.503734
------------------------------------------------------------------------------
Using the information in this output, compute the unbiased estimator σˆ2 for the variance of the
error term.
The matrix (X′X)−1 for this application is:
education _cons
education .00050573
_cons -.00615711 .07703954
Using this matrix and the value of σˆ2 you found, compute the standard error of the estimated
coefficients for education and the constant term. Verify that they are the same as the those
provided in the Stata output.
4
QUESTION6 [5 points]
Consider the Stata output
. reg earnings education
Source | SS df MS Number of obs = 481
-------------+---------------------------------- F(1, 479) = 51.55
Model | 4400.07607 1 4400.07607 Prob > F = 0.0000
Residual | 40885.9606 479 85.3569114 R-squared = 0.0972
-------------+---------------------------------- Adj R-squared = 0.0953
Total | 45286.0366 480 94.3459096 Root MSE = 9.2389
------------------------------------------------------------------------------
earnings | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
education | 1.491731 .2077685 7.18 0.000 1.08348 1.899981
_cons | -1.535019 2.564343 -0.60 0.550 -6.573771 3.503734
------------------------------------------------------------------------------
Explain how the value in the top-right corner F (1, 479) is computed and use it to test the
hypothesis that education has no explanatory power for earnings. Do you reject the null? How
is the value of F (1, 479) related to the value of t-statistic for the coefficient of education provided
in the Stata output? Explain.
QUESTION7 [5 points]
You have data from an extensive medical survey on health and economic status. A cross section
of 500 men provides information on wage in $ per week, age in years, and an index of fitness,
fit, which ranges from 0 (least fit) to 10 (most fit).
You consider the following linear model to study the relationship between wages and fitness:
ln(wagei) = α+ βfitfiti + ui where ui|fiti ∼ N(0, σ2), ∀i = 1, ..., N
Assume random sampling.
5
The expected value of ln(wagei) conditional on fiti is α + βfitfiti and the distribution of
ln(wagei) conditional on fiti is N(0, 1).
True
False
QUESTION8 [5 points]
You estimate the model
ln(wagei) = α+ βfitfiti + ui where ui|fiti ∼ N(0, σ2), ∀i = 1, ..., N
by OLS in Stata (using the command regress) and obtain the following results:
ln(wagei) = 6.1− 0.31
(0.084)
fiti + ei,
where ei represents a residual and the standard error is reported in brackets under the estimated
coefficient of fit, βˆfit.
Knowing that the 99.5% quantile of a t498 is 2.58, construct the exact 99% confidence interval
for βfit and use it to test the null hypothesis H0 : βfit = 0 against the alternative H1 : βfit 6= 0.
Do you reject the null?
QUESTION9 [5 points]
Consider again the regression output
ln(wagei) = 6.1− 0.31
(0.084)
fiti + ei,
where ei represents a residual and the standard error is reported in brackets under the estimated
coefficient of fit.
6
Interpret the estimated coefficient of fit. Knowing that 1500
∑500
i=1 fiti = 5.55, what is the value
of 1500
∑500
i=1 ln(wagei)? Explain how you derive this result.
QUESTION10 [5 points]
Consider the following two models:
ln(wagei) = α+ βfitfiti + ui where ui|fiti ∼ N(0, σ2), ∀i = 1, ..., N (1)
ln(wage)i = α+βfitfiti+βageagei+ui with E [u|fit, age] = 0 and V ar [u|fit, age] = σ2 (2)
Besides adding the variable age, what is the main difference between model (1) and model (2)?
What does this difference imply for how you estimate the unknown parameters of the model
and conduct inference? Explain briefly.
7
QUESTION11 [5 points]
You estimate the two models described in the previous question by OLS and obtain the following
results:
ln(wagei) = 6.1− 0.31
(0.084)
fiti + ei, (3)
ln(wage)i = 5.3 + 0.12
(0.132)
fiti + 0.07
(0.015)
agei + ei, (4)
where standard errors re reported in brackets under the estimated coefficients.
Referring to the results in equation (4) and knowing that the 95% quantile of a standard normal
is qΦ(0.95) = 1.64, test the null hypothesis H0 : βfit = 0 against the alternative H1 : βfit > 0.
Do you reject the null?
Comparing the results in (3) with those in equation (4), would you say that in (3) the estimated
coefficient of fit is biased due to the omission of age? Would this be a downward or upward
bias? Motivate your answer.
8
QUESTION12 [5 points]
Consider again the regression output
ln(wage)i = 5.3 + 0.12
(0.132)
fiti + 0.07
(0.015)
agei + ei,
where standard errors are reported in brackets under the estimated coefficients.
The R2 for this regression is 0.19. Recall that sample size is 500. The F statistic to test whether
the explanatory variables fit and age are jointly significant is
116.58 and distributed as F1,497
116.34 and distributed as F1,496
58.29 and distributed as F2,497
58.17 and distributed as F2,496
Question 13 [5 points]
You are using a data set including 5,587 individuals age 18-64. For each individual in the sample
you observe:
• age in years;
• education categories:
1.edu cat individuals with high school or less;
2.edu cat individuals with some college;
3.edu cat individuals with a bachelor degree or more.
• health:
1.bad health individuals with poor or fair health;
0.bad health individuals with good, very good, or excellent health.
• log of annual family income: lfaminc;
• log of out-of-pocket medical expenses in a year: lmedexp oop.
You obtain the following regression output from Stata:
. reg lmedexp_oop ib1.edu_cat
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(2, 5584) = 64.45
Model | 347.039085 2 173.519543 Prob > F = 0.0000
Residual | 15034.6106 5,584 2.6924446 R-squared = 0.0226
-------------+---------------------------------- Adj R-squared = 0.0222
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.6409
--------------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
---------------------+----------------------------------------------------------------
edu_cat |
2. Some College | .2305112 .0546213 4.22 0.000 .1234322 .3375903
3. Bachelor or More | .5888344 .0518661 11.35 0.000 .4871566 .6905121
|
_cons | 5.211075 .0326674 159.52 0.000 5.147034 5.275116
--------------------------------------------------------------------------------------
9
The average log of out-of-pocket medical expenses in a year for individuals with some college
is:
0.230511
5.799909
5.441586
4.852751
Question 14 [10 points]
Using the output
. reg lmedexp_oop ib1.edu_cat
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(2, 5584) = 64.45
Model | 347.039085 2 173.519543 Prob > F = 0.0000
Residual | 15034.6106 5,584 2.6924446 R-squared = 0.0226
-------------+---------------------------------- Adj R-squared = 0.0222
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.6409
--------------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
---------------------+----------------------------------------------------------------
edu_cat |
2. Some College | .2305112 .0546213 4.22 0.000 .1234322 .3375903
3. Bachelor or More | .5888344 .0518661 11.35 0.000 .4871566 .6905121
|
_cons | 5.211075 .0326674 159.52 0.000 5.147034 5.275116
--------------------------------------------------------------------------------------
test the null hypothesis that the coefficient of 3.edu cat is equal to the coefficient of 2.edu cat
(Hint: education categories are mutually exclusive, which implies that 1.edu cat, 2.edu cat and
3.edu cat can be considered independent). In order to test the null, construct an approximate
95% confidence interval for the difference between 3.edu cat and 2.edu cat (the 97.5% quantile
of a standard normal is qΦ(0.975) = 1.959964).
10
Question 15 [10 points]
Consider the regression output below:
. reg lmedexp_oop c.age##i.bad_health
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(3, 5583) = 146.97
Model | 1125.85779 3 375.285931 Prob > F = 0.0000
Residual | 14255.7919 5,583 2.55342861 R-squared = 0.0732
-------------+---------------------------------- Adj R-squared = 0.0727
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.5979
----------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-----------------+----------------------------------------------------------------
age | .031237 .0017656 17.69 0.000 .0277758 .0346982
1.bad_health | -.2616268 .2509612 -1.04 0.297 -.7536084 .2303548
|
bad_health#c.age |
1 | .0106244 .0051547 2.06 0.039 .0005191 .0207297
|
_cons | 4.081756 .0769414 53.05 0.000 3.930921 4.232591
----------------------------------------------------------------------------------
Knowing that average age in the sample is 42.48 and that the proportion of individuals in bad
health is 14.27%, compute the marginal effects of age and bad health and interpret the values
you obtain.
11
Question 16 [5 points]
Consider now the following two regression outputs:
. reg lmedexp_oop c.age##i.bad_health
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(3, 5583) = 146.97
Model | 1125.85779 3 375.285931 Prob > F = 0.0000
Residual | 14255.7919 5,583 2.55342861 R-squared = 0.0732
-------------+---------------------------------- Adj R-squared = 0.0727
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.5979
----------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-----------------+----------------------------------------------------------------
age | .031237 .0017656 17.69 0.000 .0277758 .0346982
1.bad_health | -.2616268 .2509612 -1.04 0.297 -.7536084 .2303548
|
bad_health#c.age |
1 | .0106244 .0051547 2.06 0.039 .0005191 .0207297
|
_cons | 4.081756 .0769414 53.05 0.000 3.930921 4.232591
----------------------------------------------------------------------------------
. reg lmedexp_oop c.age i.bad_health
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(2, 5584) = 218.21
Model | 1115.01062 2 557.505308 Prob > F = 0.0000
Residual | 14266.6391 5,584 2.55491388 R-squared = 0.0725
-------------+---------------------------------- Adj R-squared = 0.0722
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.5984
------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
age | .0324834 .0016593 19.58 0.000 .0292306 .0357361
1.bad_health | .2395768 .0620486 3.86 0.000 .1179374 .3612162
_cons | 4.029943 .0727399 55.40 0.000 3.887344 4.172541
------------------------------------------------------------------------------
Based on these outputs, the F -statistic to test the null hypothesis that the interaction between
age and bad health has a zero coefficient is (2.06)2 = 4.25 and you reject the null.
True
False
Question 17 [5 points]
Consider the following model: y = Xβ + u with E [u|X] = 0 and V ar [u|X] = Ω 6= σ2I. Under
these assumptions, the OLS is asymptotically normally distributed
True
False
12
Question 18 [5 points]
Which of the following is true about bootstrap standard errors
They are robust to heteroskedasticity
They are computed as the standard deviation of estimates obtained from a number of
bootstrap samples drawn from the original sample
They approach heteroskedasticity-robust standard errors as the number of bootstrap sam-
ples increases
All of the above
(Marco Angrisani)
Midterm Exam Example
This exam consists of
• multiple choice questions
• true/false questions
• short answer questions
To answer multiple choice and true/false questions, simply cross or color the box corresponding
to your answer.
For short answer questions, write your answers in the box provided below each question.
Only what is inside the boxes will be evaluated for grading. Try to provide concise
and precise answers.
You have 2 hours to complete the exam.
Good luck!
1
QUESTION1 [5 points]
Consider the following linear model in matrix notation:
y = Xβ + u with E[u|X] = 0 and E[uu′|X] = σ2I
where y is a N × 1 vector, β is a k × 1 vector, X is a N × k matrix and u is a N × 1 vector.
Under the assumptions of this model, the OLS estimator βˆ = (X′X)−1 X′y is
Unbiased
Consistent
Asymptotically normally distributed
All of the above
None of the above
QUESTION2 [5 points]
Consider the following linear model in matrix notation:
y = Xβ + u with E[u|X] = 0 and E[uu′|X] = σ2I
where y is a N × 1 vector, β is a k × 1 vector, X is a N × k matrix and u is a N × 1 vector.
Under the assumptions of this model, the OLS estimator βˆ = (X′X)−1 X′y is the best unbi-
ased estimator.
True
False
QUESTION3 [5 points]
Consider the following linear model in matrix notation:
y = Xβ + u with E[u|X] = 0 and E[uu′|X] = σ2I
where y is a N × 1 vector, β is a k × 1 vector, X is a N × k matrix and u is a N × 1 vector.
The variance of the OLS estimator is V ar
(
βˆ|X
)
= σ2 (X′X)−1. In order to compute V ar
(
βˆ|X
)
,
you need to estimate σ2. Explain how an unbiased estimator of σ2 can be computed.
2
QUESTION4 [5 points]
You want to use the following linear model:
y = Xβ + u with E[u|X] = 0 and E[uu′|X] = σ2I
where y is a N × 1 vector, β is a k× 1 vector, X is a N × k matrix and u is a N × 1 vector, to
study the effect of education on earnings.
You have data on 481 individuals for whom you observe yearly earnings in thousand dollars
and years of education. You run a regression of earnings on education in Stata and obtain the
following output:
. reg earnings education
Source | SS df MS Number of obs = 481
-------------+---------------------------------- F(1, 479) = 51.55
Model | 4400.07607 1 4400.07607 Prob > F = 0.0000
Residual | 40885.9606 479 85.3569114 R-squared = 0.0972
-------------+---------------------------------- Adj R-squared = 0.0953
Total | 45286.0366 480 94.3459096 Root MSE = 9.2389
------------------------------------------------------------------------------
earnings | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
education | 1.491731 .2077685 7.18 0.000 1.08348 1.899981
_cons | -1.535019 2.564343 -0.60 0.550 -6.573771 3.503734
------------------------------------------------------------------------------
Interpret the estimated coefficient for education. Under the assumptions of the model above, can
you give a causal interpretation to this estimated coefficient? Do you think these assumptions
are plausible for this specific application? Explain briefly.
3
QUESTION5 [5 points]
Consider the Stata output
. reg earnings education
Source | SS df MS Number of obs = 481
-------------+---------------------------------- F(1, 479) = 51.55
Model | 4400.07607 1 4400.07607 Prob > F = 0.0000
Residual | 40885.9606 479 85.3569114 R-squared = 0.0972
-------------+---------------------------------- Adj R-squared = 0.0953
Total | 45286.0366 480 94.3459096 Root MSE = 9.2389
------------------------------------------------------------------------------
earnings | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
education | 1.491731 .2077685 7.18 0.000 1.08348 1.899981
_cons | -1.535019 2.564343 -0.60 0.550 -6.573771 3.503734
------------------------------------------------------------------------------
Using the information in this output, compute the unbiased estimator σˆ2 for the variance of the
error term.
The matrix (X′X)−1 for this application is:
education _cons
education .00050573
_cons -.00615711 .07703954
Using this matrix and the value of σˆ2 you found, compute the standard error of the estimated
coefficients for education and the constant term. Verify that they are the same as the those
provided in the Stata output.
4
QUESTION6 [5 points]
Consider the Stata output
. reg earnings education
Source | SS df MS Number of obs = 481
-------------+---------------------------------- F(1, 479) = 51.55
Model | 4400.07607 1 4400.07607 Prob > F = 0.0000
Residual | 40885.9606 479 85.3569114 R-squared = 0.0972
-------------+---------------------------------- Adj R-squared = 0.0953
Total | 45286.0366 480 94.3459096 Root MSE = 9.2389
------------------------------------------------------------------------------
earnings | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
education | 1.491731 .2077685 7.18 0.000 1.08348 1.899981
_cons | -1.535019 2.564343 -0.60 0.550 -6.573771 3.503734
------------------------------------------------------------------------------
Explain how the value in the top-right corner F (1, 479) is computed and use it to test the
hypothesis that education has no explanatory power for earnings. Do you reject the null? How
is the value of F (1, 479) related to the value of t-statistic for the coefficient of education provided
in the Stata output? Explain.
QUESTION7 [5 points]
You have data from an extensive medical survey on health and economic status. A cross section
of 500 men provides information on wage in $ per week, age in years, and an index of fitness,
fit, which ranges from 0 (least fit) to 10 (most fit).
You consider the following linear model to study the relationship between wages and fitness:
ln(wagei) = α+ βfitfiti + ui where ui|fiti ∼ N(0, σ2), ∀i = 1, ..., N
Assume random sampling.
5
The expected value of ln(wagei) conditional on fiti is α + βfitfiti and the distribution of
ln(wagei) conditional on fiti is N(0, 1).
True
False
QUESTION8 [5 points]
You estimate the model
ln(wagei) = α+ βfitfiti + ui where ui|fiti ∼ N(0, σ2), ∀i = 1, ..., N
by OLS in Stata (using the command regress) and obtain the following results:
ln(wagei) = 6.1− 0.31
(0.084)
fiti + ei,
where ei represents a residual and the standard error is reported in brackets under the estimated
coefficient of fit, βˆfit.
Knowing that the 99.5% quantile of a t498 is 2.58, construct the exact 99% confidence interval
for βfit and use it to test the null hypothesis H0 : βfit = 0 against the alternative H1 : βfit 6= 0.
Do you reject the null?
QUESTION9 [5 points]
Consider again the regression output
ln(wagei) = 6.1− 0.31
(0.084)
fiti + ei,
where ei represents a residual and the standard error is reported in brackets under the estimated
coefficient of fit.
6
Interpret the estimated coefficient of fit. Knowing that 1500
∑500
i=1 fiti = 5.55, what is the value
of 1500
∑500
i=1 ln(wagei)? Explain how you derive this result.
QUESTION10 [5 points]
Consider the following two models:
ln(wagei) = α+ βfitfiti + ui where ui|fiti ∼ N(0, σ2), ∀i = 1, ..., N (1)
ln(wage)i = α+βfitfiti+βageagei+ui with E [u|fit, age] = 0 and V ar [u|fit, age] = σ2 (2)
Besides adding the variable age, what is the main difference between model (1) and model (2)?
What does this difference imply for how you estimate the unknown parameters of the model
and conduct inference? Explain briefly.
7
QUESTION11 [5 points]
You estimate the two models described in the previous question by OLS and obtain the following
results:
ln(wagei) = 6.1− 0.31
(0.084)
fiti + ei, (3)
ln(wage)i = 5.3 + 0.12
(0.132)
fiti + 0.07
(0.015)
agei + ei, (4)
where standard errors re reported in brackets under the estimated coefficients.
Referring to the results in equation (4) and knowing that the 95% quantile of a standard normal
is qΦ(0.95) = 1.64, test the null hypothesis H0 : βfit = 0 against the alternative H1 : βfit > 0.
Do you reject the null?
Comparing the results in (3) with those in equation (4), would you say that in (3) the estimated
coefficient of fit is biased due to the omission of age? Would this be a downward or upward
bias? Motivate your answer.
8
QUESTION12 [5 points]
Consider again the regression output
ln(wage)i = 5.3 + 0.12
(0.132)
fiti + 0.07
(0.015)
agei + ei,
where standard errors are reported in brackets under the estimated coefficients.
The R2 for this regression is 0.19. Recall that sample size is 500. The F statistic to test whether
the explanatory variables fit and age are jointly significant is
116.58 and distributed as F1,497
116.34 and distributed as F1,496
58.29 and distributed as F2,497
58.17 and distributed as F2,496
Question 13 [5 points]
You are using a data set including 5,587 individuals age 18-64. For each individual in the sample
you observe:
• age in years;
• education categories:
1.edu cat individuals with high school or less;
2.edu cat individuals with some college;
3.edu cat individuals with a bachelor degree or more.
• health:
1.bad health individuals with poor or fair health;
0.bad health individuals with good, very good, or excellent health.
• log of annual family income: lfaminc;
• log of out-of-pocket medical expenses in a year: lmedexp oop.
You obtain the following regression output from Stata:
. reg lmedexp_oop ib1.edu_cat
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(2, 5584) = 64.45
Model | 347.039085 2 173.519543 Prob > F = 0.0000
Residual | 15034.6106 5,584 2.6924446 R-squared = 0.0226
-------------+---------------------------------- Adj R-squared = 0.0222
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.6409
--------------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
---------------------+----------------------------------------------------------------
edu_cat |
2. Some College | .2305112 .0546213 4.22 0.000 .1234322 .3375903
3. Bachelor or More | .5888344 .0518661 11.35 0.000 .4871566 .6905121
|
_cons | 5.211075 .0326674 159.52 0.000 5.147034 5.275116
--------------------------------------------------------------------------------------
9
The average log of out-of-pocket medical expenses in a year for individuals with some college
is:
0.230511
5.799909
5.441586
4.852751
Question 14 [10 points]
Using the output
. reg lmedexp_oop ib1.edu_cat
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(2, 5584) = 64.45
Model | 347.039085 2 173.519543 Prob > F = 0.0000
Residual | 15034.6106 5,584 2.6924446 R-squared = 0.0226
-------------+---------------------------------- Adj R-squared = 0.0222
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.6409
--------------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
---------------------+----------------------------------------------------------------
edu_cat |
2. Some College | .2305112 .0546213 4.22 0.000 .1234322 .3375903
3. Bachelor or More | .5888344 .0518661 11.35 0.000 .4871566 .6905121
|
_cons | 5.211075 .0326674 159.52 0.000 5.147034 5.275116
--------------------------------------------------------------------------------------
test the null hypothesis that the coefficient of 3.edu cat is equal to the coefficient of 2.edu cat
(Hint: education categories are mutually exclusive, which implies that 1.edu cat, 2.edu cat and
3.edu cat can be considered independent). In order to test the null, construct an approximate
95% confidence interval for the difference between 3.edu cat and 2.edu cat (the 97.5% quantile
of a standard normal is qΦ(0.975) = 1.959964).
10
Question 15 [10 points]
Consider the regression output below:
. reg lmedexp_oop c.age##i.bad_health
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(3, 5583) = 146.97
Model | 1125.85779 3 375.285931 Prob > F = 0.0000
Residual | 14255.7919 5,583 2.55342861 R-squared = 0.0732
-------------+---------------------------------- Adj R-squared = 0.0727
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.5979
----------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-----------------+----------------------------------------------------------------
age | .031237 .0017656 17.69 0.000 .0277758 .0346982
1.bad_health | -.2616268 .2509612 -1.04 0.297 -.7536084 .2303548
|
bad_health#c.age |
1 | .0106244 .0051547 2.06 0.039 .0005191 .0207297
|
_cons | 4.081756 .0769414 53.05 0.000 3.930921 4.232591
----------------------------------------------------------------------------------
Knowing that average age in the sample is 42.48 and that the proportion of individuals in bad
health is 14.27%, compute the marginal effects of age and bad health and interpret the values
you obtain.
11
Question 16 [5 points]
Consider now the following two regression outputs:
. reg lmedexp_oop c.age##i.bad_health
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(3, 5583) = 146.97
Model | 1125.85779 3 375.285931 Prob > F = 0.0000
Residual | 14255.7919 5,583 2.55342861 R-squared = 0.0732
-------------+---------------------------------- Adj R-squared = 0.0727
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.5979
----------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-----------------+----------------------------------------------------------------
age | .031237 .0017656 17.69 0.000 .0277758 .0346982
1.bad_health | -.2616268 .2509612 -1.04 0.297 -.7536084 .2303548
|
bad_health#c.age |
1 | .0106244 .0051547 2.06 0.039 .0005191 .0207297
|
_cons | 4.081756 .0769414 53.05 0.000 3.930921 4.232591
----------------------------------------------------------------------------------
. reg lmedexp_oop c.age i.bad_health
Source | SS df MS Number of obs = 5,587
-------------+---------------------------------- F(2, 5584) = 218.21
Model | 1115.01062 2 557.505308 Prob > F = 0.0000
Residual | 14266.6391 5,584 2.55491388 R-squared = 0.0725
-------------+---------------------------------- Adj R-squared = 0.0722
Total | 15381.6497 5,586 2.75360718 Root MSE = 1.5984
------------------------------------------------------------------------------
lmedexp_oop | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
age | .0324834 .0016593 19.58 0.000 .0292306 .0357361
1.bad_health | .2395768 .0620486 3.86 0.000 .1179374 .3612162
_cons | 4.029943 .0727399 55.40 0.000 3.887344 4.172541
------------------------------------------------------------------------------
Based on these outputs, the F -statistic to test the null hypothesis that the interaction between
age and bad health has a zero coefficient is (2.06)2 = 4.25 and you reject the null.
True
False
Question 17 [5 points]
Consider the following model: y = Xβ + u with E [u|X] = 0 and V ar [u|X] = Ω 6= σ2I. Under
these assumptions, the OLS is asymptotically normally distributed
True
False
12
Question 18 [5 points]
Which of the following is true about bootstrap standard errors
They are robust to heteroskedasticity
They are computed as the standard deviation of estimates obtained from a number of
bootstrap samples drawn from the original sample
They approach heteroskedasticity-robust standard errors as the number of bootstrap sam-
ples increases
All of the above
