STAT 409
Homework #8
Fall 2011
(due Friday, October 28, by 4:00 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?
(
Hint:
X
≥
c
.
)
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.
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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
.
2.
a)
Help Ernie to find the best (uniformly most powerful) Rejection Region
with the significance level
α
= 0.10.
(
Hint:
T
≤
c
.
)
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
has a Gamma distribution with parameters
α
= 10
and
θ
=
1
/
λ
.
2
λ
T
has a chisquare distribution with
2
α
= 20
degrees of freedom
0.10
=
α
=
P
(
Reject
H
0

H
0
is true
)
=
P
(
T
≤
c

λ
= 2
)
=
P
(
4 T
≤
4
c

λ
= 2
)
=
P
(
χ
2
(
20
)
≤
4
c
).
⇒
4
c
=
( )
20
2
90
.
0
χ
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 Fall '11
 STEPHANOV
 Normal Distribution, Exponential distribution, Statistical hypothesis testing, Chisquare distribution, rejection region

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