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CHEN 3010
Applied Data Analysis
Fall 2009
Name:________________________________
Midterm Examination 2 Solutions
1. (10 pts) Minivans represented 29 percent of the 9080 vehicles torn into by bears in
Yosemite National Park between 2001 and 2007, even though they made up just 7
percent of the cars that visited Yosemite during that time.
Is there evidence that bears
preferentially break into minivans over other vehicles?
Solution: This is just a simple hypothesis test on a binomial proportion.
Since 7% of all
cars entering Yosemite are minivans, then the proportion of cars broken into by bears
should also be 7% if bears do not preferentially break into minivans over other vehicles.
Since 29% of all bear breakins occur to minivans, it appears that bears preferentially
break into minivans, so we want to prove that the proportion is significantly greater than
7%.
We can set up the following hypothesis test on the binomial proportion:
ܪ
: ൌ 0.07; ܪ
ଵ
:0.07
We can use the normal approximation to the binomial and compare our test statistic to
the standard normal distribution:
ݖ
ൌ
ܺെ݊
ඥ݊
ሺ1 െ
ሻ
ൌ
0.29 · 9080 െ 9080 · 0.07
ඥ9080 · 0.07ሺ1 െ 0.07ሻ
ൌ82
.2
We compare this to the critical zvalue:
ݖ
.ହ
ൌ 1.645
.
Since our test statistic far exceeds
the critical zvalue of 1.645, we accept the alternate hypothesis and claim that bears
preferentially break into minivans.
2. (20 pts) The breaking strength of hockey stick shafts made of two different graphite
Kevlar composites yield the following results (in newtons).
Assume that breaking
strength is normally distributed.
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View Full Document CHEN 3010
Midterm Exam 1
Fall 2009
Name:________________________________
a.
(10 pts) Construct a 95% confidence interval on the ratio
2
2
2
1
/
σ
to determine if
the variances are unequal.
Solution: First, we need to determine the sample variances, which are:
ݏ
ଵ
ଶ
ൌ 130.4, ݏ
ଶ
ଶ
ൌ 319.4
.
It appears that
ݏ
ଵ
ଶ
൏ݏ
ଶ
ଶ
.
We will calculate a 95%
confidence interval on
2
2
2
1
/
to determine whether or not this interval contains
1.
If the interval contains 1, then we conclude that the ratio of the variances
could be equal.
Our 95% confidence interval is:
ݏ
ଵ
ଶ
ݏ
ଶ
ଶ
݂
ଵିఈ ଶ
⁄ ,
మ
ିଵ,
భ
ିଵ
ߪ
ଵ
ଶ
ߪ
ଶ
ଶ
ݏ
ଵ
ଶ
ݏ
ଶ
ଶ
݂
ఈଶ
⁄ ,
మ
ିଵ,
భ
ିଵ
130.4
319.4
·
1
3.85
ߪ
ଵ
ଶ
ߪ
ଶ
ଶ
130.4
319.4
·4
.30
0.106
ߪ
ଵ
ଶ
ߪ
ଶ
ଶ
1
.75
Conclusion: Since the confidence interval contains 1, we can assume that the
variances are equal.
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This note was uploaded on 04/25/2010 for the course CHEN 3010 at Colorado.
 '08
 KOMPALA,DH

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