Chapter 5
Inference and Prediction
Exercises
1.
The estimated covariance matrix for the least squares estimator is
s
2
(
X
′
X
)
1
=
20
3900
3900 29
0
0
08
0
1
01
0
8
/
−
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
0
0
=
where
s
2
=
520/(293) = 20.
Then,
the test may be based on
t
= (.4 + .9  1)/[.410 + .256  2(.051)]
1/2
=
.399.
This is smaller than the critical
value of 2.056, so we would not reject the hypothesis.
.
..
69
0
0
0
40
051
00
5
1
2
5
6
−
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
2.
In order to compute the regression, we must recover the original sums of squares and cross products for y.
These are
X
′
y
=
X
′
Xb
=
[116, 29, 76]
′
.
The total sum of squares is found using
R
2
= 1 
e
′
e
/
y
′
M
0
y
, so
y
′
M
0
y
=
520 / (52/60)
=
600. The means are
x
1
=
0,
x
2
=
0,
y
=
4, so,
y
′
y
=
600 + 29(4
2
)
=
1064.
The
slope in the regression of
y
on
x
2
alone is
b
2
=
76/80, so the regression sum of squares is
b
2
2
⎡
⎣
⎢
⎤
⎦
⎥
+
−
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
1
2
0
0
β
β
(
(80)
=
72.2, and
the residual sum of squares is 600

72.2
=
527.8.
The test based on the residual sum of squares is
F
[(527.8  520)/1]/[520/26]
=
.390.
In the regression of the previous problem, the
t
ratio for testing the same
hypothesis would be
t
= .4/(.410)
1/2
= .624 which is the square root of .39.
3.
For the current problem,
R
= [
0
,
I
] where
I
is the last
K
2
columns.
Therefore,
R
(
X
′
X
)
1
R
N is the lower
right
K
2
×
K
2
block of (
X
′
X
)
1
.
As we have seen before, this is (
X
2
′
M
1
X
2)
1
.
Also, (
X
′
X
)
1
R
′
is the last
K
2
columns of (
X
′
X
)
1
.
These are (
X
′
X
)
1
R
′
=
Finally, since
q
=
0
,
Rb

q
= (
0b
1
+
Ib
2
) 
0
=
b
2
.
Therefore, the constrained estimator is
X X
X X X MX
XMX
11
12 2 12
21
2
(') ' ('
)
('
)
−−
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
1
b
*
=
(
X
2
′
M
1
X
2
)
b
2
,
where
b
1
and
b
2
are the multiple regression
coefficients in the regression of
y
on both
X
1
and
X
2
. Collecting terms, this produces
b
*
=
.
But, we have from Section 6.3.4 that
b
1
(
X
1
′
X
1
)
1
X
1
′
y
 (
X
1
′
X
1
)

1
X
1
′
X
2
b
2
so the preceding reduces to
b
*
=
which was to be shown.
b
b
1
2
⎡
⎣
⎢
⎤
⎦
⎥

2
)
)
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
1
b
b
1
2
⎡
⎣
⎢
⎤
⎦
⎥
XX
XXb
b
12
2
2
(') '
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
1
XX Xy
0
1
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
1
If, instead, the restriction is
β
2
=
β
2
0
then the preceding is changed by replacing
R
β

q
=
0
with
R
β

β
2
0
=
0
.
Thus,
Rb

q
=
b
2

β
2
0
.
Then, the constrained estimator is
b
*
=
(
X
2
′
M
1
X
2
)(
b
2

β
2
0
)
b
b
1
2
⎡
⎣
⎢
⎤
⎦
⎥
2
)
)
−
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
1
or
b
*
=
b
b
1
2
2
2
Using the result of the previous paragraph, we can rewrite the first part as
(') ' (
)
)
XX XXb
 b
122
b
=
(
X
1
′
X
1
)
1
X
1
′
y
 (
X