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Concordia University
Department of Economics
ECON 221/4 – Sections AA, B, E
Instructors: M. DaPalma, V. Terekhov, G. Tsoublekas
Winter 2007  ASSIGNMENT 4  Answers
Due at noon on Thursday, April 5, 2007
1.
A cellular phone company advertises that the average age of its customers is fewer than 30.
It is known
that the customer ages follow a normal distribution.
A random sample of 16 of this company’s customers
responded as follows, when asked about their age:
35, 19, 25, 33, 32, 51, 65, 43, 44, 29, 33, 21, 34, 30,
25, 57.
(30 points)
a.
Calculate the sample mean and sample variance.
(5 points)
36
16
576
=
=
=
∑
n
X
X
i
91
.
12
67
.
166
1
16
2500
1
)
(
2
2
=
⇒
=

=


=
∑
s
n
X
X
s
i
b.
Develop an appropriate hypothesis to test the company’s claim at the 5% level of significance. (5 pts)
H
0
:
μ <
30
H
1
:
μ > 30
Reject H
0
if estimated
t > 1.753
859
.
1
2275
.
3
6
4
91
.
12
6
16
91
.
12
30
36
=
=
=

=

=
n
s
X
t
μ
and
1.859 > 1.753
Therefore, we reject
0
H
and we conclude that the average age of the company’s clients is more
than 30.
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Estimate and interpret the pvalue of the calculated test statistic in (b) above.
(4 points)
Pvalue = P(t
15
> 1.859) = about 4.3%
The pvalue here indicates that the significance level at which we could reject the null hypothesis is
4.3%, instead of the 5% used in (b) above.
This makes our conclusions somewhat weak.
d.
If the same sample results had been obtained from a random sample of 25 customers, could the
company’s claim be rejected at a lower level of significance than in part (b)?
(4 points)
H
0
:
μ <
30
H
1
:
μ > 30
Reject H
0
if estimated
t > 1.711
323
.
2
582
.
2
6
5
91
.
12
6
25
91
.
12
30
36
=
=
=

=

=
n
s
X
t
μ
Yes.
If the same result had come from a sample of 25, the estimated tratio would be 2.323, which
corresponds to a Pvalue of about 1.9%.
This means that we could reject the null hypothesis at the
1.9%, level of significance instead of the 5% used in (b) or the one we found in (c) above.
In such a
case, the conclusion of the test would be stronger.
e.
Suppose that the alternative hypothesis has been onesided and that it was set as H
a
:
μ
< 30.
Make a
graph to visualize the problem and state, without doing the calculations, whether the pvalue of the test
(the level of significance needed to reject the null hypothesis) would be higher than, lower than, or the
same as found in (c). (4 points)
H
0
: μ = 30
Pvalue
H
1
:
μ < 30
Reject H
0
if estimated
t <
–
1.753
30
36
The alternative hypothesis is on the leftward (negative) side of the tDistribution (which means that the
rejection area is also on that side).
On the other hand, our present sample mean is found on the rightward
(positive) side of the distribution of sample means.
Therefore, the pvalue of our present mean must be
greater than 50%.
2
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