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CHAPTER 8
Section 82
81
a) The confidence level for
n
x
n
x
/
14
.
2
/
14
.
2
σ
μσ
+
≤
≤
−
is determined by the
value of z
0
which is 2.14.
From Table III,
Φ
(2.14) = P(Z<2.14) = 0.9838 and the
confidence level is 2(0.98380.5) = 96.76%.
b) The confidence level for
n
x
n
x
/
49
.
2
/
49
.
2
+
≤
≤
−
is determined by the
by the value of z
0
which is 2.14.
From Table III,
Φ
(2.49) = P(Z<2.49) = 0.9936 and the
confidence level is is 2(0.99360.5) = 98.72%.
c) The confidence level for
n
x
n
x
/
85
.
1
/
85
.
1
+
≤
≤
−
is determined by the
by the value of z
0
which is 2.14.
From Table III,
Φ
(1.85) = P(Z<1.85) = 0.9678 and the
confidence level is 93.56%.
82
a) A z
α
= 2.33 would give result in a 98% twosided confidence interval.
b) A z
α
= 1.29 would give result in a 80% twosided confidence interval.
c) A z
α
= 1.15 would give result in a 75% twosided confidence interval.
83
a) A z
α
= 1.29 would give result in a 90% onesided confidence interval.
b) A z
α
= 1.65 would give result in a 95% onesided confidence interval.
c) A z
α
= 2.33 would give result in a 99% onesided confidence interval.
84
a) 95% CI for
96
.
1
,
1000
20
,
10
,
=
=
=
=
z
x
n
μ
4
.
1012
6
.
987
)
10
/
20
(
96
.
1
1000
)
10
/
20
(
96
.
1
1000
/
/
≤
≤
+
≤
≤
−
+
≤
≤
−
σμ
n
z
x
n
z
x
b) .95% CI for
96
.
1
,
1000
20
,
25
,
=
=
=
=
z
x
n
8
.
1007
2
.
992
)
25
/
20
(
96
.
1
1000
)
25
/
20
(
96
.
1
1000
/
/
≤
≤
+
≤
≤
−
+
≤
≤
−
n
z
x
n
z
x
c) 99% CI for
58
.
2
,
1000
20
,
10
,
=
=
=
=
z
x
n
3
.
1016
7
.
983
)
10
/
20
(
58
.
2
1000
)
10
/
20
(
58
.
2
1000
/
/
≤
≤
+
≤
≤
−
+
≤
≤
−
n
z
x
n
z
x
d) 99% CI for
58
.
2
,
1000
20
,
25
,
=
=
=
=
z
x
n
3
.
1010
7
.
989
)
25
/
20
(
58
.
2
1000
)
25
/
20
(
58
.
2
1000
/
/
≤
≤
+
≤
≤
−
+
≤
≤
−
n
z
x
n
z
x
e) When n is larger, the CI will get narrower. The higher the confidence level, the wider the CI.
81
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View Full Document85.
a) Find n for the length of the 95% CI to be 40.
Z
a/2
= 1.96
84
.
3
20
2
.
39
20
2
.
39
20
/
)
20
)(
96
.
1
(
length
1/2
2
=
⎟
⎠
⎞
⎜
⎝
⎛
=
=
=
=
n
n
n
Therefore,
n
= 4.
b) Find n for the length of the 99% CI to be 40.
Z
a/2
= 2.58
66
.
6
20
6
.
51
20
6
.
51
20
/
)
20
)(
58
.
2
(
length
1/2
2
=
⎟
⎠
⎞
⎜
⎝
⎛
=
=
=
=
n
n
n
Therefore,
n
= 7.
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 Spring '09
 KAILASHKAPUR

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