STAT 409
Examples for 10/21/2009
Fall 2009
1.
An advertisement for a particular brand of automobile states that it accelerates from
0 to 60 mph in an average of 5.0 seconds.
Makers of a competing automobile feel
that the true average number of seconds it takes to reach 60 mph from zero is above
5.0.
Suppose the population standard deviation is believed to be 0.43 seconds.
We
wish to test
H
0
:
μ
≤
5.0
vs.
H
1
:
μ
> 5.0.
a)
Find the power of this test if the true average time is 5.15 seconds, a 5% level of
significance will be used, and 50 automobiles will be tested (each automobile to
be tested a single time).
σ
= 0.43.
n
= 50.
α
= 0.05.
σ
is known.
Test Statistic:
50
43
.
0
0
.
5
X
X
Z
σ
μ
0

=

=
n
.
Rejection Region:
Right  tailed.
Reject
H
0
if
Z
z
α
= 1.645.
50
43
.
0
645
.
1
0
.
5
X
⋅
+
=
5.1
.
Power = P
(
Reject
H
0

H
0
is NOT true
)
= P
(
X > 5.1

μ
= 5.15
)
=

50
43
.
0
15
.
5
1
.
5
Z
P
= P
(
Z > –
0.82
)
=
0.7939
.
b)
Fifty automobiles were tested (each automobile was tested a single time).
The sample
mean time to reach 60 mph from zero was 5.09 seconds.
Find the pvalue of this test.
Test Statistic:
σ
is known.
50
43
.
0
0
.
5
09
.
5
X
Z
σ
μ
0

=

=
n
=
1.48
.
Right – tailed test.
pvalue =
(
area of the right tail
)
= P
(
Z > 1.48
)
=
0.0694
.
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2.
Suppose
n
= 49
observations are taken from a normal distribution where
σ
= 8.0
for the purpose of testing
H
0
:
μ
= 60
versus
H
1
:
μ
≠
60
at the
α
= 0.05
level
of significance.
What is the power of the appropriate test when
μ
= 59.2?
H
0
:
μ
= 60
vs.
H
1
:
μ
≠
60
2 – tailed.
α
= 0.05.
Rejection Region
:
Reject
H
0
if
Z =
σ
μ
0
X
n

<
2
α
z

or
Z =
σ
μ
0
X
n

>
2
α
z
⇒
σ
μ
2
0
α
X
n
z

<
or
σ
μ
2
0
α
X
n
z
+
⇒
49
8
96
.
1
60
X

<
or
49
8
96
.
1
60
X
+
⇒
57.76
<
X
or
62.24
X
Power
= P
(
Reject
H
0

H
0
is false
)
= P
(
76
.
57
X
<

μ
= 59.2
)
+ P
(
24
.
62
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 Spring '11
 Stepanov
 Statistics, Normal Distribution, Statistical hypothesis testing, Statistical significance, Type I and type II errors, rejection region

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