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Solutions to Homework #8
(W4150 ‘Intro to Probability and Statistics’, S04)
Sec. 9.7. (4)
n = 30;
X = 780;
σ
= 40; z
α
/2
=
z
0.02
= 2.054
Hence by applying formula for CI from p.235 we obtain:
)
30
40
)(
05
.
2
(
780
)
30
40
)(
05
.
2
(
780
+
<
<
−
µ
or
795
765
<
<
Sec.9.7. (6)
n = 50;
X = 174.5;
σ
= 6.9; z
α
/2
=
z
0.01
= 2.33
(a) The confidence interval is given by the same formula as above, which in this case yields:
)
50
9
.
6
)(
33
.
2
(
5
.
174
)
50
9
.
6
)(
33
.
2
(
5
.
174
+
≤
≤
−
or
77
.
176
23
.
172
<
<
(b) By Theorem 9.1 we can be 98% sure that the error
27
.
2
50
/
)
9
.
6
)(
33
.
2
(
=
<
e
Sec. 9.7. (7)
Applying the same methods for n = 100;
X = 23,500;
σ
= 3,900 and z
α
/2
=
z
0.005
= 2.575
we get:
(a) 99% CI for mean:
22,496 <
µ
< 24,504; and
(b) estimation for error
1004
<
e
Sec. 9.7 (8)
In this as well as in two next problems we use formula given in Theorem 9.1
z
α
/2
=
z
0.02
= 2.054; e = 10;
σ
= 40 and so
n = [(2.05)(40)/10]
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 Spring '04
 GALLEGO

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