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Unformatted text preview: STAT 420 Fall 2008 Version B Name ANSWERS . Exam 2 Page Possible Earned 1 9 2 10 3 13 4 10 5 11 6 14 7 4 8 9 Total 80 Be sure to show all your work; your partial credit might depend on it. No credit will be given without supporting work. The exam is closed book and closed notes. You are allowed to use a calculator and one 8½" x 11" sheet with notes on it. ___________________________________________________________________________ Academic Integrity The University statement on your obligation to maintain academic integrity is: If you engage in an act of academic dishonesty, you become liable to severe disciplinary action. Such acts include cheating; falsification or invention of information or citation in an academic endeavor; helping or attempting to help others commit academic infractions; plagiarism; offering bribes, favors, or threats; academic interference; computer related infractions; and failure to comply with research regulations. Rule 33 of the Code of Policies and Regulations Applying to All Students gives complete details of rules governing academic integrity for all students. You are responsible for knowing and abiding by these rules. 1. A students wonders if people of similar heights tend to date each other. He measures himself, his dormitory roommate, and the men in the adjoining rooms; then he measures the next woman each man dates. Here are the data(heights in inches): Men, x 72 67 70 68 72 65 Women, y 68 64 68 67 70 65 Σ x = 414, Σ y = 402, Σ x 2 = 28,606, Σ y 2 = 26,958, Σ x y = 27,765, Σ ( x – x ) 2 = 40, Σ ( y – y ) 2 = 24, Σ ( x – x ) ( y – y ) = Σ ( x – x ) y = 27. Assume that ( X , Y ) have a bivariate normal distribution. a) (3) Find the sample correlation coefficient r between the heights of the men and women. ( ) ( ) ( ) ( ) 24 40 27 2 2 = = & & & y y x x y y x x r ≈ 0.87142 . b) (6) Test H : ρ = 0 vs. H 1 : ρ ≠ 0. Use α = 0.05. Test Statistic: T = 2 2 87142 . 1 2 6 87142 . 1 2 = r n r ≈ 3.553 . Rejection Region: Reject H if T < – t 0.025 or T > t 0.025 n – 2 = 4 degrees of freedom t 0.025 ( 4 ) = 2.776 . Reject H : ρ = 0 at α = 0.05. t 0.025 ( 4 ) = 2.776 < 3.553 < 3.747 = t 0.01 ( 4 ) pvalue = 2 tails. 0.02 < pvalue < 0.05 ( pvalue ≈ 0.02556 ) 1. (continued) c) (6) Test H : ρ = 0.3 vs. H 1 : ρ > 0.3. Use α = 0.05. r r + = 1 1 2 1 W ln = & ¡ ¢ £ ¤ ¥ + ⋅ 87142 . 1 87142 . 1 2 1 ln = 1.33895....
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This note was uploaded on 05/09/2010 for the course MATH 420 taught by Professor Stepanov during the Spring '10 term at University of Illinois, Urbana Champaign.
 Spring '10
 Stepanov

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