This preview shows pages 1–3. Sign up to view the full content.
STAT 410
Homework #14
Fall 2009
(due Wednesday, December 9, by 4:30 p.m.)
1 – 2.
Bert and Ernie noticed that the
following are satisfied when
Cookie Monster eats cookies:
(a)
the number of cookies eaten during
nonoverlapping time intervals are
independent;
(b)
the probability of exactly one cookie
eaten in a sufficiently short interval
of length
h
is approximately
λ
h
;
(c)
the probability of two or more cookies eaten in a sufficiently short interval is
essentially zero.
Therefore, X
t
, the number of cookies eaten by Cookie Monster by time
t
, is a Poisson
process, and for any
t
> 0, the distribution of X
t
is Poisson (
λ
t
).
However, Bert and Ernie could not agree on the value of
λ
, the average number of cookies
that Cookie Monster eats per minute.
Bert claimed that it equals 2, but Ernie insisted that it
is greater than 2.
Thus, the two friends decided to test
H
0
:
λ
= 2
vs.
H
1
:
λ
> 2.
Bert decided to count the number of cookies Cookie Monster would eat in 5 minutes, X, and
then Reject
H
0
if X is too large.
Ernie, who was the less patient of the two, decided to note
how much time Cookie Monster needs to eats the first 10 cookies, T, and then Reject
H
0
if T
is too small.
1.
a)
Help Bert to find the best (uniformly most powerful) Rejection Region with the
significance level
α
of the test closest to 0.05.
What is the actual value of the
significance level
α
associated with this Rejection Region?
X
has a Poisson
(
5
λ
) distribution.
0.05
=
α
=
P
(
Reject
H
0

H
0
is true
)
=
P
(
X
≥
c

λ
= 2
)
=
P
(
Poisson
(
10
)
≥
c
).
P
(
Poisson
(
10
)
≤
15
)
=
0.951.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
P
(
Poisson
(
10
)
≥
16
)
=
0.049.
Reject
H
0
if
X
≥
16
.
α
=
0.049
.
b)
Find the power of the test from part (a) if
λ
= 3.
Power
(
λ
= 3
)
=
P
(
X
≥
16

λ
= 3
)
=
P
(
Poisson
(
15
)
≥
16
)
=
1 – P
(
Poisson
(
15
)
≤
15
)
=
1 – 0.568
=
0.432
.
c)
Suppose Cookie Monster ate 17 cookies in 5 minutes.
Find the pvalue of the test.
Pvalue
=
P
(
X
≥
17

λ
= 2
)
=
P
(
Poisson
(
10
)
≥
17
)
=
1 – P
(
Poisson
(
10
)
≤
16
)
=
1 – 0.973
=
0.027
.
d)
Find the best
randomized
Rejection Region with the significance level
α
= 0.10.
P
(
Poisson
(
10
)
≤
14
)
=
0.917.
⇒
P
(
Poisson
(
10
)
≥
15
)
=
0.083.
P
(
Poisson
(
10
) = 14
)
=
P
(
Poisson
(
10
)
≤
14
) – P
(
Poisson
(
10
)
≤
13
)
=
0.917 – 0.864
=
0.053.
0.083
+
p
⋅
0.053
=
0.10.
⇒
p
≈
0.321.
Reject
H
0
if
X
≥
15
,
Reject
H
0
with probability
0.321
if
X = 14
.
2.
a)
Help Ernie to find the best (uniformly most powerful) Rejection Region
with the significance level
α
= 0.10.
Hint:
If
T
has a Gamma
(
α
,
θ
=
1
/
λ
) distribution, where
α
is an integer, then
2
T
/
θ
=
2
λ
T
has a
χ
2
(
2
α
)
distribution
(
a chisquare distribution with
2
α
degrees of freedom
).
T
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 04/15/2010 for the course STAT 410 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Monrad

Click to edit the document details