Econometrics H.W. 2
Soumil Agarwal
February 2022
Conceptual Problems
Q1. Given that,
\ = 1.392 − 0.0135hsperc + 0.00148sat, n = 4137
colgpa
i) It makes sense for the coefficient on hsperc to be negative since a lower hsperc implies that the student is better at academics which translates to a high college GPA as well.
\ = 1.392 − 0.0135 × 20 + 0.00148 × 1050
ii) colgpa
\
colgpa = 2.676
\ to find out the difference in college GPA between
iii) Given: ∆sat = 140. We will calculate ∆colgpa
these students, keeping their hsperc constant.
\ = 0.00148∆sat
∆colgpa
\ = 0.00148 × 140
=⇒ ∆colgpa
\
=⇒ ∆colgpa = 0.2072
The predicted difference in college GPA between student A and B is 0.2072. This difference is not very
large considering that the difference in SAT scores between the two is 140 points owing to the small value of βˆ2
\ = 0.00148∆sat
iv) ∆colgpa
\ = 0.5
. Given: ∆colgpa
∴ 0.00148∆sat = 0.5
0.5
=⇒ ∆sat = 0.00148
=⇒ ∆sat = 337.83
This shows that for a change in predicted grade point by 0.5, the SAT score change should be 337.83, which
is quite significant. The reason for this is that grade point is measured between 0 and 4, whereas range for
SAT scores is large, i.e. 0–1600.
Q2. Given that,
d = 10.36 − 0.094sibs + 0.131meduc + 0.210f educ, n = 4137, R2 = 0.214
educ
i) sibs does have the necessary effect. We would expect that larger number of siblings would mean lesser
time and money invested by parents towards each child’s education and hence the coefficient on sibs, βˆ1
d using the expression
is negative. Holding meduc and f educ fixed, we can find the effect of sibs on educ
d
−0.094∆sibs = ∆educ
d = −1, we can use this expression to find ∆sibs
Given ∆educ
−1
∆sibs = −0.094
= 10.63
This implies that to reduce the years of education, the number of siblings must increase by 10.63.
ii) The coefficient on meduc, βˆ2 = 0.131 means that the predicted increase in educ as a result of 1 unit
increase in meduc, keeping other variables constant, is 0.131.
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iii) For Man A:
d = 10.36 − 0.094 × 0 + 0.131 × 12 + 0.210 × 12
educ
d = 14.452
educ
For Man B:
d = 10.36 − 0.094 × 0 + 0.131 × 16 + 0.210 × 16
educ
d = 15.816
educ
The predicted difference in years of education between man A and man B is 1.364, where man B’s years of
education are higher than A.
Q3. Given that,
sleep = β0 + β1 totwork + β2 educ + β3 age + u
i) The sign of β1 will be negative because a rise in totwork causes sleep to fall.
ii) Sign for β2 will be negative which means that a rise in years of education leads to a fall in the number of
minutes of sleep per week. The reason for this can be that as number of years of education rises, the more
rigorous the curriculum becomes and hence the numbers of hours of sleep.
Sign for β3 will also be negative as with a rise in age one sleeps lesser and owing to the bodily requirements
of an individual, coupled with the increase in responsibilities and work.
[ = −0.148∆totwork, keeping other variiii) From the given estimated equation, we can say that ∆sleep
[ = −0.148 × 300 = −44.4.
ables constant. Given that ∆totwork = 5 × 60 = 300, we can estimate ∆sleep
This means that minutes of sleep per week fall by 44.4 if an individual works for 5 more hours in a week.
iv) The sign on βˆ2 is negative because as the number of years of education increase, we can logically say
that the number of minutes an individual sleeps per week will fall, owing to a more rigorous course workload
as one attains a more advanced degree. The high magnitude of βˆ2 can be simply explained by the fact that
we measure educ by years of education, the range for which is quite limited, mostly from 0–30 (considering
extreme scenarios), whereas the range for sleep is much higher as the minutes of sleep can go from 0–6300
(taking 15 hours a day of sleep to be the extreme scenario). If we consider the range here, then we can see
that the effect of educ on sleep is not as significant as one might expect.
v) totwork, educ and age explain significant variation in sleep but not entirely. There are other factors
that affect the time sleeping which are ambient noise levels around an individual’s neighbourhood, their
weight, their height, the number of hours they spend doing physical exercises. These factors are likely to be
correlated with totwork because they affect sleep.
Q6. Given that,
y = β0 + β1 x1 + β2 + x2 + β3 x3 + u
i) Given assumptions MLR.1 through MLR.4, can write that:
E(βˆ1 ) = β1
E(βˆ2 ) = β2
Given: θ1 = β1 + β2
E(θˆ1 ) = E(βˆ1 + βˆ2 )
=⇒ E(θˆ1 ) = E(βˆ1 ) + E(βˆ2 )
∴ θ 1 = β1 + β2
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Therefore, we can say that θˆ1 is an unbiased estimator of β1 + β2
ii)
θˆ1 = βˆ1 + βˆ2
=⇒ V ar(θˆ1 ) = V ar(βˆ1 + βˆ2 )
=⇒ V ar(θˆ1 ) = V ar(βˆ1 ) + V ar(βˆ2 ) + 2Cov(βˆ1 βˆ2 )
∴ V ar(θˆ1 ) = V ar(βˆ1 ) + V ar(βˆ2 ) + 2Corr(βˆ1 βˆ2 )σβˆ1 σβˆ2
Conceptual Problems
Q1. i) The sign for β2 should be negative because better nutrition for the mother means that the child will
be healthy and will have more birth weight.
ii) cigs and f aminc are correlated. There is expected to be a negative correlation because we can think that
high income families will be more educated and would avoid smoking cigarettes as it is harmful for health.
Similarly, there could also be a positive correlation because if we consider cigarettes to be a normal good
then an increase in income would raise the demand for them.
iii) The equation without f aminc is:
\ = 119.7719 − 0.514cigs, R2 = 0.0227, n = 1388
bwght
The equation with f aminc is:
\ = 116.9741 − 0.463cigs + 0.927f aminc, R2 = 0.0298, n = 1388
bwght
Adding f aminc reduced the effects of cigs on bwght by some amount but the effect is not very large because
of the fact that cigs and f aminc are not heavily correlated and the coefficient on f aminc is quite small.
[ = −19.315 + 0.128sqrf t + 15.198bdrms, R2 = 0.6319, n = 88
Q2. i) price
[ i.e. ∆price
[ with a 1-unit change in bdrms,
ii) Keeping sqrf t constant, we need to find the change in price,
i.e. ∆bdrms = 1.
[ = 15.198∆bdrms = 15.198 × 1 = 15.198. Since house price is measured in USD 1000s, we can say
∆price
that price increases by USD 15,198 with the addition of one more bedroom, keeping sqrf t constant.
[ i.e. ∆price
[ with a 1-unit increase in bdrms, i.e. ∆bdrms = 1
iii) We need to find the change in price,
and an increase in sqrf t by 140, i.e. ∆sqrf t = 140.
[ = 0.128∆sqrf t + 15.198∆bdrms = 0.128(140) + 15.198(1) = 33.118. Since house price is measured
∆price
in USD 1000s, we can say that price increases by USD 33,118 with the addition of one more bedroom and
an increase in sqrf t by 140 units.
iv) Since R2 = 0.6319, this means that 63.19% of the variation in price is explained by square footage
and number of bedrooms.
[ = −19.315 + 0.128sqrf t + 15.198bdrms. We now know that sqrf t = 2438
v) The OLS regression line is price
[ = −19.315 + 0.128(2438) + 15.198(4) = 353.541. This means that the price of
and bdrms = 4. Hence, price
this house is USD 353541.
vi) Residual value here is 300 − 353.541 = 53.541. The buyer underpaid for this house because the actual price at which it was sold was 300,000 whereas the predicted price is 353,541.
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Q5. Regressing educ on exper and tenure, we get:
d = 13.574 − 0.0737exper + 0.0477tenure, n = 526, R2 = 0.1013
educ
Regressing lwage on u educ, we get:
\ = 1.623 + 0.920u educ, n = 526, R2 = 0.2066
lwage
Regressing lwage on educ, exper and tenure, we get:
\ = 0.284 + 0.092educ + 0.004exper + 0.022tenure, n = 526, R2 = 0.3160
lwage
When we compare the coefficients of r1 and educ, we see that the coefficients are exactly equal to 0.092 for
both. The value of R2 for regression on rˆ1 or u educ is lower than the value in regression on educ, tenure
and exper. This points to the fact that rˆ1 or u educ explains log(wage) for those part of educ that are
independent of exper and tenure.
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