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Stat 5102 (Geyer) 2003 Final Exam
Problem 1
(a)
We need to fnd the probability under
H
0
, that is, assuming
θ
= 1 that
X
≤
c
. That is
Z
c
0
e

x
dx
= 1

e

c
and set this equal to
α
and solve For
c
.
The solution is
c
=

log(1

α
) =

log(0
.
95) = 0
.
05129329.
(b)
The
P
value is
P
θ
(
X
≤
x
), where
θ
= 1 is the value hypothesized under
H
0
and
x
= 0
.
45 is the observed value oF the test statistic. By the same argument
as above, that is 1

e

x
= 1

exp(

0
.
45) = 0
.
3623718.
Problem 2
The distribution oF one data point
X
i
is Normal(0
,
1
/θ
) which has density
f
(
x
) =
1
√
2
πσ
e

x
2
/
2
σ
2
where
σ
2
= 1
/θ
. Plugging in
σ
2
= 1
/θ
and
σ
= 1
/
√
θ
gives
f
(
x
) =
r
θ
2
π
e

θx
2
/
2
For the density oF each
x
i
.
The likelihood is then
n
Y
i
=1
f
(
x
i
) =
n
Y
i
=1
r
θ
2
π
e

θx
2
i
/
2
=
µ
θ
2
π
¶
n/
2
n
Y
i
=1
e

θx
2
i
/
2
=
µ
θ
2
π
¶
n/
2
exp
ˆ

θ
2
n
X
i
=1
x
2
i
!
As always we can throw away multiplicative Factors that don’t contain the pa
rameter, in this case the 2
π
bits, leaving
θ
n/
2
exp
ˆ

θ
2
n
X
i
=1
x
2
i
!
The prior distribution is
g
(
θ
) =
e

θ
1
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View Full Document(the same as the prior for problem 2 on the Frst midterm).
So likelihood times prior is
θ
n/
2
exp
ˆ

θ
2
n
X
i
=1
x
2
i
!
e

θ
=
θ
n/
2
exp
ˆ

θ
"
1 +
1
2
n
X
i
=1
x
2
i
#!
This is recognizable as having the form
θ
α

1
exp(

βθ
)
where
α
=
n
2
+ 1
β
= 1 +
1
2
n
X
i
=1
x
2
i
Thus the posterior distribution is
Gamma
ˆ
n
2
+ 1
,
1 +
1
2
n
X
i
=1
x
2
i
!
Problem 3
(a)
The sample regression coe±cient is 2.2370 with standard error 0.2254. There
are 18 degrees of freedom for error. Thus we look up the
t
critical value for 95%
conFdence and 18 degrees of freedom, which is found in the column headed
.
975
in the table on p. 776 in DeGroot and Schervish, and is 2.101. The conFdence
interval is then 2
.
2370
±
2
.
101
×
0
.
2254 which is (1
.
763435
,
2
.
710565).
(b)
The
P
value is just read out of the printout,
P
= 0
.
4218, no calculations
necessary.
Since the
P
value is much larger than any interesting signiFcance
level, we conclude that the data give no evidence that the quadratic model Fts
better than the linear model. Thus we should use the linear model.
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 Spring '02
 Staff
 Probability

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