CHAPTER 15
SOLUTIONS TO PROBLEMS
15.1
(i) It has been fairly well established that socioeconomic status affects student performance.
The error term
u
contains, among other things, family income, which has a positive effect on
GPA
and is also very likely to be correlated with PC ownership.
(ii) Families with higher incomes can afford to buy computers for their children.
Therefore,
family income certainly satisfies the second requirement for an instrumental variable:
it is
correlated with the endogenous explanatory variable [see (15.5) with
x
=
PC
and
z
=
faminc
].
But as we suggested in part (i),
faminc
has a positive affect on
GPA
, so the first requirement for a
good IV, (15.4), fails for
faminc
.
If we had
faminc
we would include it as an explanatory
variable in the equation; if it is the only important omitted variable correlated with
PC
, we could
then estimate the expanded equation by OLS.
(iii) This is a natural experiment that affects whether or not some students own computers.
Some students who buy computers when given the grant would not have without the grant.
(Students who did not receive the grants might still own computers.)
Define a dummy variable,
grant
, equal to one if the student received a grant, and zero otherwise.
Then, if
grant
was
randomly assigned, it is uncorrelated with
u
.
In particular, it is uncorrelated with family income
and other socioeconomic factors in
u
.
Further,
grant
should be correlated with
PC
:
the
probability of owning a PC should be significantly higher for student receiving grants.
Incidentally, if the university gave grant priority to low-income students,
grant
would be
negatively correlated with
u
, and IV would be inconsistent.
15.3
It is easiest to use (15.10) but where we drop
z
.
Remember, this is allowed because
1
()
n
i
i
zz
=
−
∑
i
x
x
−
=
1
(
n
ii
i
zx x
=
−
∑
)
and similarly when we replace
x
with
y
.
So the numerator in
the formula for
1
ˆ
β
is
11
1
1
nn
n
i
i
i
i
zy y
zy
ny ny
==
=
⎛⎞
−=
−
=
−
⎜⎟
⎝⎠
∑∑
∑
,
where
n
1
=
is the number of observations with
z
i
= 1, and we have used the fact that
/
n
1
=
1
n
i
i
z
=
∑
1
n
i
=
∑
1
y
, the average of the
y
i
over the
i
with
z
i
= 1.
So far, we have shown that the
numerator in
1
ˆ
is
n
1
(
1
y
–
y
).
Next, write
y
as a weighted average of the averages over the
two subgroups:
y
=
(
n
0
/
n
)
0
y
+ (
n
1
/
n
)
1
y
,
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