Solution to Homework Assignment 3
1
Question 1
a. Since
g
0
(
p
) =
{
p
(1

p
)
}

1
/
3
, using delta method, we have that
V ar
{
g
(
p
)
}
=
{
g
0
(
π
)
}
2
V ar
(
p
) =
{
pi
(1

π
)
}

2
/
3
{
π
(1

π
)
}
n
=
{
p
(1

p
)
}
1
/
3
n
.
b. From part a., we have that
g
(
p
1
)
a
∼
N
g
(
π
1
)
,
{
π
1
(1

π
1
)
}
1
/
3
n
1
¶
, g
(
p
2
)
a
∼
N
g
(
π
2
)
,
{
π
2
(1

π
2
)
}
1
/
3
n
2
¶
,
where
p
1
and
p
2
are sample proportions from treatment 1 and 2, respectively. Under the null
hypothesis
H
0
:
π
1
=
π
2
, we know that the pooled sample proportion ¯
p
= (
p
1
n
1
+
p
2
n
2
)
/
(
n
1
+
n
2
)
is a better estimator for both
π
1
and
π
2
, so a test statistic for testing the null hypothesis
H
0
would be
T
=
g
(
p
1
)

g
(
p
2
)
p
{
¯
p
(1

¯
p
)
}
1
/
3
/n
1
+
{
¯
p
(1

¯
p
)
}
1
/
3
/n
2
.
With equal allocation of patients to the two treatments, i.e.
n
1
=
n
2
=
n/
2, we have the test
statistic be written as
T
=
g
(
p
1
)

g
(
p
2
)
p
4
{
¯
p
(1

¯
p
)
}
1
/
3
/n
,
where ¯
p
= (
p
1
+
p
2
)
/
2.
Clearly, the test statistic described above has asymptotic standard normal distribution
N
(0
,
1).
Therefore, we reject
H
0
at 2sided significance level
α
if

T
 ≥
z
α/
2
.
Note
: Another possible test statistic for testing
H
0
is
T
1
=
g
(
p
1
)

g
(
p
2
)
q
{
p
1
(1

p
1
)
}
1
/
3
n
1
+
{
p
2
(1

p
2
)
}
1
/
3
n
2
=
g
(
p
1
)

g
(
p
2
)
q
2
{
p
1
(1

p
1
)
}
1
/
3
n
+
2
{
p
2
(1

p
2
)
}
1
/
3
n
.
Under
H
0
:
π
1
=
π
2
,
T
1
also has the asymptotic standard normal distribution and will give more
evidence against (rejecting)
H
0
. However, the test statistic
T
has more stable denominator as it
uses larger sample to estimate the unknown parameters and hence will have a better asymptotic
normal distribution than
T
1
.
c. The asymptotic distribution of
T
under
H
A
:
π
1
=
π
1
A
, π
2
=
π
2
A
is
T
a
∼
N
ˆ
g
(
π
1
A
)

g
(
π
2
A
)
p
4
{
¯
π
(1

¯
π
)
}
1
/
3
/n
,
{
π
1
A
(1

π
1
A
)
}
1
/
3
+
{
π
2
A
(1

π
2
A
)
}
1
/
3
2
{
¯
π
(1

¯
π
)
}
1
/
3
!
,
1
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where ¯
π
= (
π
1
A
+
π
2
A
)
/
2.
Similarly, we may derive the asymptotic distribution for
T
1
as
T
1
a
∼
N
g
(
p
1
)

g
(
p
2
)
q
2
{
π
1
(1

π
1
)
}
1
/
3
n
+
2
{
π
2
(1

π
2
)
}
1
/
3
n
,
1
.
d. The sample size calculated based on the test statistic
T
has to satisfy that
φ
=
z
α/
2
+
z
β
σ
*
,
where
φ
=
g
(
π
1
A
)

g
(
π
2
A
)
p
4
{
¯
π
(1

¯
π
)
}
1
/
3
/n
σ
2
*
=
{
π
1
A
(1

π
1
A
)
}
1
/
3
+
{
π
2
A
(1

π
2
A
)
}
1
/
3
2
{
¯
π
(1

¯
π
)
}
1
/
3
.
That is,
n
{
g
(
π
1
A
)

g
(
π
2
A
)
}
2
4
{
¯
π
(1

¯
π
)
}
1
/
3
= (
z
α/
2
+
z
β
σ
*
)
2
.
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 Fall '11
 Staff
 Normal Distribution, Null hypothesis, Statistical hypothesis testing, NJ

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