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ECON 2300 – CUMULATIVE FINAL EXAM SOLUTIONS
FALL SEMESTER, 2009
1. A consumer products manufacturer wants to estimate the mean water absorbency of its new
brand of paper towels.
For a random sample of 64 paper towels, the mean water absorbency
was 450 grams. Assume the population standard deviation is
σ
= 21 grams.
(Assume the
water absorbency rate is normally distributed.)
a.
Compute
and interpret
a 99% confidence interval estimate for the mean water absorbency
of its new brand of paper towels.
x
σ
=
n
= 21/√64 = 2.625 grams
Z
.01/2
= 2.576
99% confidence interval estimate of the mean water absorbency
= [450 +/ 2.576 * 2.625] =[4506.762; 450+6.762] = [443.2 grams; 456.8 grams]
It is estimated, with 99% confidence, that the mean water absorbency rate for the
new paper towels is 443.2 to 456.8 grams.
b.
Explain how you determined whether to use a zstatistic or a tstatistic to develop your
confidence interval estimate.
Because we know the population standard deviation,
σ
= 21 grams, we use a z
statistic.
c.
Suppose you want to estimate the mean water absorbency of the new paper towels to
within 5 grams (maximum margin of error). Use a 99% confidence level and compute the
necessary sample size.
n = (z
α
/2
*
σ
) / E )
2
or (2.576 * 21) / 5 )
2
= 117.1 or 118 paper towels
2.
A car dealer currently bases its labor charge for a tire rotation and alignment job on its belief
that the mean time spent on such a job is 40 minutes.
For a random sample of 36 tire rotation
and alignment jobs performed at the dealership, the mean time spent on the jobs was 46
minutes with a standard deviation of 15 minutes.
Does that sample data provide evidence
that the mean time needed to rotate and align a set of tires is greater
(> 40)
than 40 minutes?
Use a significance level of α = .10 in addressing that question.
a.
Define the null and alternative hypotheses below.
Ho:
μ
<
40 minutes
Ha:
μ
> 40 minutes
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b.
State the decision / rejection rule for this test.
Because we do not know the population standard deviation, we must use a critical t
value.
Critical t
.10,35
= 1.69
Decision rule: Do not reject Ho if computed t <
t
.10,35
= 1.69
μ=40
=46
0
t
.10,35
= 1.69
2.4
t
c.
Calculate the computed z or t value for the sample.
x
σ
= 15/√36 = 2.5 minutes
t = (
x
–
μ29/
x
= (4640)/2.5 = 2.4
d.
Explain whether or not the sample data provide evidence that the mean time needed to
rotate and align a set of tires is greater than 40 minutes?
Explain completely.
Because the computed t = 2.4 is > t
.10,35
= 1.69, we must reject the null hypothesis and
conclude, with 90% confidence, that the mean time to rotate and align a set of tires
is significantly greater than 40 minutes
3.
The marketing director for an automobile manufacturer wants to determine the proportion of
car customers that will purchase a special extended warranty for their new car. A random
sample of 169 potential new car buyers was surveyed and, for the sample, only 20 expressed
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 Spring '08
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