2006solution - ADM 2304 APPLICATIONS OF STATISTICAL METHODS...

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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 2-page 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 Chi-square) 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 Statement of Academic Integrity: 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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p.2 of 6 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 -t-stat is (7.82-4.02)/sqrt[4.77^2 * (1/12 + 1/12)] = 1.95 -p-value is .032 (half of the two-sided p-value) < .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 p-value is the same. -deduct 2 marks for paired t-test 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)
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