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Graded Assignment 3
In your outline there are 6 methods to compare means or medians, methods D1, D2, D3, D4, D5a and D5b.
Method D6 compares proportions and method D7 compares variances or standard deviations. In all the
following cases, identify H 0 and H 1 and identify which method to use. If the hypotheses involve a mean,
state the hypotheses in terms of both  and D  1   2 . If the hypotheses involve a proportion, state
them in terms of both p and p  p1  p 2 . If the hypotheses involve standard deviations or variances,
state them in terms of both  2 and
 12
 22
or
 22
 12
. All the questions involve means, medians, proportions or
variances. (Most problems are highly edited versions of problems in McClave, et. al.)
Note: Look at 252thngs ( 252thngs) in the syllabus supplement before you start (and before you take
exams).
1. Of 200 MBA students, 110 are males. Of 500 managers, 300 are males. Is there a significant difference
between the fraction of males in the population of MBA students and the population of managers? (What
are H 0 and H 1 and what is the identifier of the method you would use?)
Solution: You are comparing two proportions. If p1 refers to MBA students and p 2 refers to managers,
 p  x1  110
 1
n1
200  H 0 : p1  p 2
H : p  p 2  0
or  0 1
. If p  p1  p 2 , then


 H 1 : p1  p 2
H 1 : p1  p 2  0
 p2  x2
 300
n2
500

we are comparing proportions, use Method D6.
 H 0 : p  0
. Since

 H 1 : p  0
2. You have two machines that plop fruit into bottles, a new one and an old one. A sample of weights of 10
bottles from the old machine is taken, the average weight is 971.375 grams with a standard deviation of
15.250 grams. A sample of weights is taken from the new machine and the average weight turns out to be
971.374 grams with a standard deviation of 11.001 grams. Is the new machine more reliable than the old
one?
Solution: The variance (or standard deviation)is a measure of variability, which is the opposite of
consistency or reliability especially in this case where the difference in means hardly exists. You need to
test the equality of the variances. ‘Less consistent or reliable’ means a larger variance, so the new machine
 H 0 :  12   22
(machine 2) being more reliable than the old one translates as " 12   22 ?" so we have 
or
 H 1 :  12   22
H 0 :   
2
2
1
2
. In terms of the variance ratio 12 or 22 , you must test only the first to see if it is above

2
1
H 1 :  1   2
1 as the alternate hypothesis implies. This is Method D7
3. The Wallaby Shock Absorber company takes 6 of its own shock absorbers and tests them for durability
by driving different cars 20000 miles with them. The mean and variance of the strength of the shock was
recorded giving a mean of 10.716 and a variance of 3.069. 6 of a competitor’s shocks were tested the same
way, and a mean of 10.3 and a variance of 3.304 were found. The manufacturer wants to compare the
means, and assumes an underlying Normal distribution, but needs to find out first whether to use method D2
or D3. What should the manufacturer do to decide?
252solngr3 10/27/03
Solution: To make the decision as to which method to use, you need to test the equality of the variances.
H 0 :   
 H 0 :  12   22
2
2
1
2
or 
. In terms of the variance ratio 12 or 22 , you must test both, but

2
1
 H 1 :  12   22
H 1 :  1   2
only the ratio of sample variances that is actually above 1 really needs to be tested, since the other ratio
cannot possibly lead to a rejection. This is Method D7.
4. The manufacturer in the previous example never did decide what to do. Instead Wallaby continued the
experiment by testing 120 of its own shocks and 90 of the competitors. What method can they now use to
compare the average strength of the shocks?
Solution: The word ‘average’ makes you think of the mean. You are comparing two means, with a total
 H 0 : 1   2
sample size of 200. There is no reason to assume that one mean is larger than the other. So 
 H 1 : 1   2
 H 0 : 1   2  0
H 0 : D  0
or 
. If D  1   2 , then 
. You can certainly get away with method D1
 H 1 : 1   2  0
H 1 : D  0
which works for large sample sizes. D3 would also work, but it’s more effort.
5. Assume that the situation is identical to problem 3 above, but that an analysis of the data indicates that
the distribution of strengths is highly skewed to the right. What method should be used now to compare the
strength of the shocks?
Solution: The skewness of the data should alert you to the fact that you should compare medians rather than
means, since it is unlikely for this small a sample that the sample means are normally distributed. If  1 is
 H 0 : 1   2
the median for urban restaurants, we have 
. Since we are comparing medians and the data are
 H 1 : 1   2
not paired, use Method D5a.
6. A group of supervisors are given the exams on management skills before and after taking a course in
management. Scores are as follows.
Supervisor
Before
After
1
63
78
2
93
92
3
84
91
4
72
80
5
65
69
6
72
85
7
91
99
8
84
82
9
71
81
10
80
87
11
68
93
If we assume that the distribution of results is Normal, what method should we use to answer the question
“Has the course improved the scores of the managers?”
Solution: You are comparing means before and after the course. You can get away with using means
because the parent distributions are Normal. If  2 is the mean of the second sample, you are hoping that
 2  1 , which, because it contains no equality is an alternate hypothesis. So your hypotheses are
2
 H 0 : 1   2
H :    2  0
H : D  0
or  0 1
. If D  1   2 , then  0
. The important thing to notice

 H 1 : 1   2
 H 1 : 1   2  0
H 1 : D  0
here is that the data are in before and after pairs, so you use Method D4.
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252solngr3 10/27/03
7. What method would we use in problem 6 above if we assumed that the underlying distribution was not
Normal?
 H :   2
Solution: If  2 is the median after and  1 is the median after, we have  0 1
. Since we are
 H 1 : 1   2
comparing medians and the data are paired, use Method D5b.
8. We have data for 15 recent oil spills caused by fire and 15 oil spills caused by collision. We want to
show that the spills caused by collision are worse than those caused by fire, and we have evidence that the
variability of the spills caused by collision is far larger than the variability of spills caused by fire. What
method should we use?
Solution: It is hard to think of anything but average size of spills that would be compared here, though per
cent of environment damages is a possibility. You can get away with using means if the parent distributions
are Normal. If  2 is the mean of the second sample, you are expecting that  2  1 , which, because it
 H 0 : 1   2
contains no equality is an alternate hypothesis. So your hypotheses are 
or
 H 1 : 1   2
 H 0 : 1   2  0
H 0 : D  0
. If D  1   2 , then 
An important thing to notice here is that there is no

 H 1 : 1   2  0
H 1 : D  0
obvious way to pair the data and that the variances seem different, so you use Md3ethod D3 unless you are
really suspicious that the data are not Normal, in which case you use method D5a.
9. Students are asked to rate the reality shows produced by ABC and Fox on a scale of 1 to 10. Analysis
of the data set, which can be considered ordinal data, indicates that the distribution is not symmetrical. A
random sample of 30 ratings of Fox shows and 25 ratings of ABC show is given and the researcher wants
to prove that Fox produces better shows.
Solution: The things that should alert you to the fact that you should compare medians rather than means
are the fact that this is ordinal data and the fact that the distribution is not symmetrical. If  1 is the median
 H 0 : 1   2
for Fox and  2 is the median for ABC, we have 
. Since we are comparing medians and the
 H 1 : 1   2
data are not paired, use Method D5a.
10. Unemployment rates are found for 20 urban communities and 10 university communities in
Pennsylvania. The researcher wants to show that workers in the second group of communities are better off
than workers in the first group. Average unemployment rates are computed and variances of employment
rates between communities of each type seem similar.
Solution: The word ‘average’ makes you think of the mean. You are comparing two means, with a total
sample size of 30, which is relatively small. Workers in the second group of communities are better off if
the unemployment rate is lower – this translates as " 1   2 ?" . Since it does not contain an equality, it
 H 0 : 1   2
 H 0 : 1   2  0
must be an alternate hypothesis. So 
or 
. If D  1   2 , then
 H 1 : 1   2
 H 1 : 1   2  0
H 0 : D  0
. You can certainly get away with method D3, but it’s more work than D2, which the similar

H 1 : D  0
variance tells us ought to work.
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