sam.mid1.06.sol

# sam.mid1.06.sol - . (c). Let Y be the number of applicants...

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Solutions to STAT 200 Midterm Exam I Problem 1. Answers: T F F T F Problem 2. Answers: c, d, a, b, b, a, b, b, a, c Problem 3. (b). Based on the given information, we have b = - 0 . 03 , 0 . 42 = a - 0 . 03 × 5 . So a = 0 . 42 + 0 . 15 = 0 . 57 , and the regression line is given by y = 0 . 57 - 0 . 03 x. (c). The slope changes sign (from negative to positive), and the intercept changes substantially, so the 5th observation is inﬂuential. The dotted line is more reliable for prediction. (d). r 2 = ( - 0 . 75) 2 = 56 . 25% . If we remove the last observation, the number in (i) will increase. (e). Answers: T T F T Problem 4. (a) Let X be the GRE score. We have 500 < X < 700 500 - 560 100 < Z < 700 - 560 100 - 0 . 6 < Z < 1 . 4 . From Table A, we have 0 . 9192 - 0 . 2743 = 0 . 6449 = 64 . 49% . (b). X - 560 100 1 . 28 X 560 + 128 = 688

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Unformatted text preview: . (c). Let Y be the number of applicants with GRE scores higher than 750. Then p = P ( X > 750) = p ( Z > 750-560 100 ) = P ( Z > 1 . 9) = 0 . 0287 . P ( Y 1) = 0 . 90 P ( Y = 0) = 0 . 10 1 . 0287 (1-. 0287) n = 0 . 10 . 9713 n = 0 . 10 n = ln (0 . 10) ln (0 . 9713) = 79 . 07 80 . Problem 5. (a) P ( X = 0) = 3 2 ! 5 2 ! = 0 . 3 , 1 P ( X = 1) = 2 1 ! 3 1 ! 5 2 ! = 0 . 6 , P ( X = 2) = 2 2 ! 5 2 ! = 0 . 1 . (b) E ( X ) = 0 . 3 + 1 . 6 + 2 . 1 = 0 . 8 = q V ( X ) = q (0-. 8) 2 . 3 + (1-. 8) 2 . 6 + (2-. 8) 2 . 1 = 0 . 6 2...
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## This note was uploaded on 05/09/2011 for the course STAT 200 taught by Professor Karim during the Spring '08 term at The University of British Columbia.

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sam.mid1.06.sol - . (c). Let Y be the number of applicants...

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