HW6-solutions1 - IE 300 GE 331 Spring 2009 Homework#6...

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IE 300 / GE 331 Spring 2009 Homework #6 Solutions Question 1 (a) First we calculate the values of c 1 and c 2 . We have c 1 = 1 R 60 5 e - 0 . 04 x dx 0 . 0549 c 2 = 1 R 60 5 e - 0 . 16 x dx 0 . 3561 Next we derive the posterior probability of each hypothesis, p Θ | T (1 | 20) = 0 . 3 f T | Θ ( x | Θ = 1) 0 . 3 f T | Θ ( x | Θ = 1) + 0 . 7 f T | Θ ( x | Θ = 2) = 0 . 3 · 0 . 0549 e - 0 . 04 · 20 0 . 3 · 0 . 0549 e - 0 . 04 · 20 + 0 . 7 · 0 . 3561 e - 0 . 16 · 20 = 0 . 4214 and p Θ | T (2 | 20) = 0 . 7 f T | Θ ( x | Θ = 2) 0 . 3 f T | Θ ( x | Θ = 1) + 0 . 7 f T | Θ ( x | Θ = 2) = 0 . 3 · 0 . 3561 e - 0 . 16 · 20 0 . 3 · 0 . 0549 e - 0 . 04 · 20 + 0 . 7 · 0 . 3561 e - 0 . 16 · 20 = 0 . 5786 Therefore she would accept the hypothesis that the problem is not difficult, and the probability of error is p e = p Θ | T (1 | 20) = 0 . 4214 . 1
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(b) We write the posterior probability of each hypothesis, p Θ | T 1 ,T 2 ,T 3 ,T 4 ,T 5 (1 | 20 , 10 , 25 , 15 , 35) = 0 . 3 f T 1 ,T 2 ,T 3 ,T 4 ,T 5 | Θ (20 , 10 , 25 , 15 , 35 | Θ = 1) 0 . 3 f T 1 ,T 2 ,T 3 ,T 4 ,T 5 | Θ (20 , 10 , 25 , 15 , 35 | Θ = 1) + 0 . 7 f T 1 ,T 2 ,T 3 ,T 4 ,T 5 | Θ (20 , 10 , 25 , 15 , 35 | Θ = 2) = 0 . 3 · 0 . 0549 5 e - 0 . 04 · (20+10+25+15+35) 0 . 3 · 0 . 0549 5 e - 0 . 04 · (20+10+25+15+35) + 0 . 7 · 0 . 3561 5 e - 0 . 16 · (20+10+25+15+35) = 0 . 9171 , and similarly p Θ | T 1 ,T 2 ,T 3 ,T 4 ,T 5 (2 | 20 , 10 , 25 , 15 , 35) = 0 . 3 f T 1 ,T 2 ,T 3 ,T 4 ,T 5 | Θ (20 , 10 , 25 , 15 , 35 | Θ = 2) 0 . 3 f T 1 ,T 2 ,T 3 ,T 4 ,T 5 | Θ (20 , 10 , 25 , 15 , 35 | Θ = 1) + 0 . 7 f T 1 ,T 2 ,T 3 ,T 4 ,T 5 | Θ (20 , 10 , 25 , 15 , 35 | Θ = 2) = 0 . 0829 . So this time the professor would accept the hypothesis that the problem is difficult. The probability of error is 0 . 0829 , much lower than the case of a single observation.
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