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OR312_Final_2008_solutions

OR312_Final_2008_solutions - ORIE 312 Final Exam Solutions...

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ORIE 312 — Final Exam — Solutions Spring 2008 Professors Peter Jackson and David Ruppert 1. (a) P (defective | X = 10 . 5) = h 1 + exp { 25 - 1 . 5 * 10 . 5 - 0 . 6 / 10 . 5 } i - 1 = { 1 + exp(9 . 19) } - 1 (b) Need to find x to minimize H ( x ) := β 0 + β 1 x + β 2 x - 1 , so solve 0 = H 0 ( x ) = β 1 - β 2 /x 2 so that x = s β 2 β 1 = q 0 . 6 / 1 . 5 = 0 . 4 2. (a) 10.028 + 0.081 (b) reference cell constraint. The model is y ij = μ + α i + ² ij and the constraint is α 1 = 0 (c) b μ = 5 * 10 . 02 + 0 . 081 + 3 . 03 + 0 . 198 + 0 . 576 5 b α 1 = 10 . 02 - b μ b α 5 = (0 . 577 + 10 . 02) - b μ 3. (a) 1 . 279 2 + 0 . 165 2 (b) 1 . 279 2 1 . 279 2 + 0 . 165 2 4. (a) d ( αβ ) 2 , 1 = 0 (b) b μ 11 = 25 b μ 12 = 25 - 4 (minimum) b μ 21 = 25 + 2 b μ 22 = 25 - 4 + 2 + 1 . 5 Use A=1 and B=2 (c) b α 1 = - (25 + 2) - (25 - 4 + 2 + 1 . 5) + (25) + (25 - 4) 4 = - 5 . 5 4 5. (A) (5) (B) (3) (C) (4) (D) (2) (E) (6) (F) (1) 1
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6. (a) A (b) AB and BC (c) The first boxplot tells us to use A=1 and B=2. The second boxplot tells us to use either B=2 and C=1 or B=1 and C=2. Therefore, combining the information in the two boxplots tells us to use: A=1 B=2 C=1 7. 1 month ahead: x 120 (1) = { b μ 120 + b β 120 } b S 120 1 - 12 = (100 + 0 . 4)(1 . 2) 2 months ahead: x 120 (2) = { b μ 120 + b β 120 (2) } b S 120 2 - 12 = (100 + (0 . 4)(2))(1 . 0) 3 months ahead: x 120 (3) = { b μ 120 + b β 120 (3) } b
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