STAT 426 HW1 Solutions (provided by TA for your information)
1.3
Let
X
be the number of defects,
λ
= 1.
(a)
P
(
X
= 0) =
e

1
1
0
0!
=
e

1
= 0
.
3679
(b)
P
(
X
= 1) =
e

1
1
1
1!
=
e

1
= 0
.
3679
(c)
P
(
X
≥
2) = 1

P
(
X
= 0)

P
(
X
= 1) = 1

2
e

1
= 0
.
2642
1.8
a.
Let
π
be the proportion of women reported greater relief with the STAN
DARD analgesic
H
0
:
π
= 0
.
5 vs
H
1
:
π
= 0
.
5
Sample proportion for
π
:
p
=
40
40+60
= 0
.
4

z

=

p

π

π
(1

π
)
n
=

0
.
4

0
.
5

(0
.
5)(0
.
5)
100
= 2
pvalue=
P
(

z
 ≥
2) = 0
.
0456
≤
0
.
05
Therefore, we reject
H
0
at the level of significance 0.05, that is, the probabil
ity of greater relief with the standard analgesic is significantly different from
that with the new one.
b.
Let
π
*
be the proportion of women reported greater relief with the NEW
analgesic
Sample proportion for
π
*
:
p
*
=
60
40+60
= 0
.
6
95% C.I. for
π
*
=
p
*
±
Z
0
.
025
p
*
(1

p
*
)
n
= 0
.
6
±
1
.
96
(0
.
6)(1

0
.
6)
100
= (0
.
504
,
0
.
696)
Since this interval contains only positive values and larger than 0
.
5
,
we con
clude with 05% confidence that
π
*
≥
0
.
5.
2.2
Let
p
1
be the estimated annual probability that a woman over the age 35 dies
of lung cancer for current smokers, and
p
2
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 Spring '10
 Xe
 Epidemiology, TA

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