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ADM 2304
APPLICATIONS OF STATISTICAL METHODS
TO BUSINESS
October 21, 2006
NAME (printed): _______________________________________
Student Number: __________________ SECTION:
H
___
I
____
Time allowed: Two (2) hours.
Length of Midterm: Total 6 pages, plus 2page Minitab supplement (please return).
You are encouraged to use the Minitab output as much as possible.
However, it should
be noted that not all Minitab procedures given are appropriate for the situations
described.
Please show all work on the midterm itself.
You are permitted to have calculators, rulers, and 1 sheet (8.5 x 11
@
paper) of notes.
Statistical tables (normal, t and Chisquare) are provided separately (please keep).
If the
value you want from a table cannot be found, use the nearest value.
Page
Value
Mark
2
11
3
6
4
4
5
7
6
7
Total
35
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This work conforms to the rules on academic integrity of the University of Ottawa:
Signature: __________________________________
Please note
that any submission in a course (homework, assignment, report, etc) that does not include that
signed statement will not be corrected and will get a grade of zero.
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1.
[11 marks]
The data below show the 1 year returns (as of September
30, 2006) for randomly selected Canadian and U.S. equity mutual funds.
It
has been suggested that Canadian equity funds have outperformed U.S. equity
funds in the past year.
The Minitab output gives some appropriate and
inappropriate analyses of the data.
CdnEq1
3.46
5.84
0.01
16.97
14.6
4.39
10.26
13.06
4.85
11.33
5.19
3.87
USEq2
13.04
0.45
4.96
6.09
4.68
0.32
0.29
3.52
9.87
4.23
2.81
1.44
(a)
[5 marks] Test whether Canadian funds have outperformed U.S. funds.
Use
the .05 level of significance.
Show how the test statistic is calculated.
H0: mu1=mu2, H1: mu1 > mu2;
pooled stdev is sqrt [(5.25^2 + 4.25^2)/2] = 4.77
tstat is (7.824.02)/sqrt[4.77^2 * (1/12 + 1/12)] = 1.95
pvalue is .032 (half of the twosided pvalue) < .05, or critical value is 1.717 (22df)
decide to reject the null H and conclude Cdn funds have outperformed US funds
accept a test not assuming equal variance for full marks.
Here the critical value is
1.721 (21 df) and the pvalue is the same.
deduct 2 marks for paired ttest here if everything else is correct.
(b)
[3 marks] Explain how you selected the specific test above and explain
whether the assumptions of the test are warranted.
data are from 2 random samples as there is no evidence of matched data;
since these are small samples, we must assume the data come from normally
distributed populations
(.5 mark)
from the first two boxplots, there is a little bit of
skewness but no outliers; therefore, it is reasonable to assume these data come from
normal distributions (.5 mark)
the sample variances are close in value; it is reasonable to assume equal variances, or
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 Winter '05
 Phansalker

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