Stat 511
Homework 4 Solution
Spring 2011
1. Last HW had the
β
vector and
X
matrix for a very messed up experiment. They were:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
μα
1
α
2
α
3
α
4
α
5
α
6
1
101100
1
100100
1
110011
1
111011
1
100000
1
000000
1
001011
1
1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
The
X
matrix and data,
y
, are in the Fle ’messedup.txt’ on the class web site.
(a) Estimate a 90% conFdence interval for
α
4
−
(
α
2
+
α
5
+
α
6
).
The corresponding C is
C
=(0
,
0
,
−
1
,
0
,
1
,
−
1
,
−
1), thus the 90% C.I is
c
Cβ
±
t
3
,
95%
p
±
σ
2
C
(
X
0
X
)
−
C
0
=[0
.
7896
,
2
.
9104]
(b) Test Ho:
α
1
=0. (N
.B
.
α
1
is estimable). Report your test statistic and pvalue.
The corresponding C is
C
,
1
,
0
,
0
,
0
,
0
,
0), thus the test statistic T is
T
=
c
−
d
p
±
σ
2
C
(
X
0
X
)
−
C
0
=1
.
7259
,
thus the pvalue is
P
(

t
3

>T
)=0
.
1828, fail to reject the null hypothesis
α
1
=0
.
(c) Test Ho:
α
1
,α
2
4
,
and
α
5
+
α
6
= 0. (N.B. This hypothesis is testable)
Report your test statistic and pvalue.
The corresponding C is
C
=
⎛
⎜
⎜
⎝
0100000
0010000
0000100
0000011
⎞
⎟
⎟
⎠
with full rank 4 and the test statistic ± is
F
=
(
c
−
d
)
0
(
C
(
X
0
X
)
−
C
0
)
−
1
(
c
−
d
)
/
4
±
σ
2
=32
.
7214
The pvalue is
P
(
F
4
,
3
>F
.
008, then we reject the null hypothesis, namely, at least
one of
α
1
2
4
,
and
α
5
+
α
6
is not zero.
2. This problem is based on a research proposal I read last week. A researcher on campus is
studying whether monkeys are selective in what they eat. One of the major items in the diet
1