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:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
μ
α
1
α
2
α
3
α
4
α
5
α
6
1
1
0
1
1
0
0
1
1
0
0
1
0
0
1
1
1
0
0
1
1
1
1
1
1
0
1
1
1
1
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
1
0
1
1
1
0
0
1
0
1
1
1
0
0
1
0
1
1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
The
X
matrix and data,
y
, are in the file ’messedup.txt’ on the class web site.
(a) Estimate a 90% confidence 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
Cβ
±
t
3
,
95%
σ
2
C
(
X X
)
−
C
= [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
= (0
,
1
,
0
,
0
,
0
,
0
,
0), thus the test statistic T is
T
=
Cβ
−
d
σ
2
C
(
X X
)
−
C
= 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
= 0
, α
2
= 0
, α
4
= 0
,
and
α
5
+
α
6
= 0. (N.B. This hypothesis is testable)
Report your test statistic and pvalue.
The corresponding C is
C
=
⎛
⎜
⎜
⎝
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
1
⎞
⎟
⎟
⎠
with full rank 4 and the test statistic F is
F
=
(
Cβ
−
d
) (
C
(
X X
)
−
C
)
−
1
(
Cβ
−
d
)
/
4
σ
2
= 32
.
7214
The pvalue is
P
(
F
4
,
3
> F
) = 0
.
008, then we reject the null hypothesis, namely, at least
one of
α
1
, α
2
, α
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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Stat 511
Homework 4 Solution
Spring 2011
of a particular group of monkeys is the fruit of the baobab tree. The researcher is measur
ing various nutrient characteristics of the baobab fruits.
One of the most important is the
sugar concentration. This is not easy to measure and even harder to measure precisely. The
researcher wanted to estimate the precision of her method of determining baobab sugar con
centration. There are no standard reference materials (samples with known chararacteristics)
for baobab fruits, so the measurement error has to be estimated from multiple measurements
on the same fruits. The researcher collected 8 fruits and measured each twice. (Ask if you
don’t remember how to estimate measurement variance from this sort of data). The data are
in baobab.txt on the class web site.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Staff
 Normal Distribution, Statistical hypothesis testing, researcher, class web site, measurement variance

Click to edit the document details