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a2.w09.sol - ACTSC 433/833 Winter 09 Solutions to...

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ACTSC 433/833 – Winter 09 Solutions to Assignment 2 Question 1 (a) j y j r j s j 1 1 5 1 2 2 4 1 3 4 5 1 4 5 3 1 5 8 2 1 (b) S 10 (2) = (1 – 1/5)(1 – 1/4) = 0.6. 2 2 10 10 1 2 ˆ Var( (2)) ( (2)) ( ) 1 1 0.6 5 4 4 3 0.048 j j j j j s S S r r s = = = + × × = (c) 1 1 1 (4) 0.65 5 4 5 H = + + = ± 2 2 2 1 1 1 ˆ Var( (4)) 0.1425 5 4 5 H = + + = ± (d) Using the results from part (b), a 95% symmetric linear confidence interval for S (2) is given by (0.1706, 1). The interval of (0.1706, 1.0294) is also acceptable. (e) 1.96 0.048 exp 0.2463 0.6ln(0.6) U = =
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