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
This preview has intentionally blurred sections. Sign up to view the full version.
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.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 KAILASHKAPUR
 Statistics, Normal Distribution, Interval finite element, Student's tdistribution

Click to edit the document details