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Y
is a Bernoulli random variable with probability Pr
(
1)
Y
p
=
=
and Pr
(
0)
1
.
Y
p
=
=
−
From the solution to Exercise 3.2,
Y
has mean
p
and variance
(1
).
p
p
−
(a)
.
(b)
ˆ
ˆ
(1
)
0.5375
(1
0.5375)
4
400
ˆ
var(
)
6 2148
10 .
p
p
n
p
−
×
−
−
=
=
=
.
×
The standard error is
.
(c) The computed
t
-statistic is
Because of the large sample size
(
400),
n
=
we can use Equation (3.14) in the text to get the
p
-value for the test
0
0 5
H
p
:
= .
vs.
1
0 5:
H
p
:
≠
.
-value
2
( |
|)
2
( 1 506)
2
0 066
0 132
act
p
t
=
Φ −
=
Φ − .
=
× .
=
.
(d) Using Equation (3.17) in the text, the
p
-value for the test
0
0 5
H
p
:
= .
vs.
1
0 5
H
p
:
>
.
is

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