Exercise:
The department of Transportation reported Amtrack trains had a
78% ontime arrival record over the previous 12 months. Assume
that in 2004, a study found 330 of 400 Amtrack trains arrived on
time. Does the sample indicate the Amtrack arrival rate has change?
Test
0
:
0.78
:
0.78
a
Hp
at
0.05
(a)
What is the sample proportion of Amtrack trains arriving
on time?
(b)
Compute the value of the test statistic (
z
)
(c)
What is the
p
value?
(d)
What is your conclusion?
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View Full DocumentSolution:
(a)
330/ 400
0.825
p
0
00
330 400(0.78)
(1
)
400(0.78)(0.22)
18
2.17
8.28
x np
z
np
p
p
value = 2(0.50.4850)=0.03
(b)
z
value of 2.17 > 1.96 critical value,
(c)
Reject
H
0
Power calculation
Power=1β=1P(Type II error)=1P(not reject  H
α
is true)
Steps. The following are the steps for computing the β for large
sample test with σ unknown. Power of other types of tests can be
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 Spring '11
 FaridAlizadeh
 Null hypothesis, Hypothesis testing, Statistical hypothesis testing, Statistical significance, Amtrack

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