Homework 5
Stat 3022
February 25, 2010
Problem 6.57
1. P(Z
>
1.73)=pnorm(1.73,lower.tail=F)=.958
2. P(Z
<
1.73)=pnorm(1.73)=.0418
3. P(Z
<
1.73)+P(Z
>
1.73)=2*P(Z
<
1.73)=2*.0418=.0836
Problem 6.59
1. Regardless of the alternative we would reject the null hypothesis because
24 falls well inside the CI.
2. If the alternative is twosided, the answer is yes. If the alternative is
μ >
30
the answer is no. The answer is yes if the alternative is
μ <
30
.
Problem 6.68 For testing these hypotheses, we ﬁnd
Z
=
10
.
2

8
.
9
2
.
5
/sqrt
6
= 1
.
27
. This is not
signiﬁcant because the pvalue is .1020. Thus there is not enough evidence
to conclude that these sonets were not written by our poet.
Problem 6.95
1.
Z
=
508

505
100
/sqrt
100
=
.
3
, so P=P(Z
>
.03)=.3821
2.
Z
=
508

505
100
/sqrt
1000
=
.
95
, so P=P(Z
>
.95)=.1711
3.
Z
=
508

505
100
/sqrt
10000
= 3
, so P=P(Z¿3)=.0013
Problem 6.112
1
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α
=P(Type 1 Error)=P(
¯
X >
26
when
μ
= 25
)= P(
Z >
26

25
50
/sqrt
900
)=P(Z¿.6)=.2743
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 Spring '08
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 Economics, Standard Deviation, SCI Upper SCI, Lower SCI Upper, Size Lower SCI

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