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)) () 11 0.6 54 43 0.048 j j jj j s SS rr s = = ⎛⎞ =+ ⎜⎟ ×× ⎝⎠ = (c) 111 (4) 0.65 545 H =++= ± 222 ˆ Var( (4)) 0.1425 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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A 95% log-transformed confidence interval for S (2) is given by (0.6 1/ U , 0.6 U ) = (0.1257, 0.8818) (f) Using the results from part (c), a 95% symmetric linear confidence interval for H (4) is given by (0, 1.3899). The interval of ( 0.0899, 1.3899) is also acceptable. Let 1.96 0.1425 exp 0.65 U ⎛⎞ = ⎝⎠ . A 95% log-transformed confidence interval for H (4) is given by (0.65/ U , 0.65 U ) = (0.2082, 2.0289). Question 2 (a) 22 2 5 ... 50 ˆ 22.9 10 5 ... 50 ˆ 777.9 10 μ ++ == (b) For the 25 th percentile, we have ( n +1) p = 11 × 0.25 = 2.75, which gives g = 2. Interpolating between x (2) and x (3) , the smoothed estimate of the 25 th percentile is given by (3 – 2.75) × 8 + (2.75 – 2) × 10 = 9.5. For the 75 h percentile, we have ( n +1) p = 11 × 0.7 = 8.25, which gives g = 8. Interpolating between x (8) and x (9) , the smoothed estimate of the 75 th percentile is given by (9 – 8.25) × 35 + (8.25 – 8) × 50 = 38.75.
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This note was uploaded on 10/30/2009 for the course ACTSC 433 taught by Professor Johnny during the Winter '09 term at Waterloo.

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

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