6 solution.pdf

6 solution.pdf - HW 6 Solution[100 points total Haoyang...

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HW 6 Solution [100 points total] Haoyang Liu 2/17/2017 [please email me if there is an error.] problem 1 [ grading scheme :5 points each for (a),(b),(c), 1 points free, total 16] (a) step 1: realize we should use Neyman Pearson lemma. step 2: derive log likelihood ratio as log( LR ) = c +log( σ 0 1 )+ ( x - μ 0 ) 2 2 σ 2 0 - ( x - μ 1 ) 2 2 σ 2 1 . step 3: realize that log( LR ) η for some η is equivalent to x η 0 for some η 0 step 4: look at normal table to find η 0 9 . 29, thus reject if x 9 . 29 (b) step 1: Express out the power as Power = Pr N (9 , 4) ( x η 0 ) = 1 - Φ( 9 . 29 - 9 2 ) step 2: Look up the table to find its value 1 - 0 . 56 = 0 . 44. (c) step 1: Express out the power as Power = 1 - Φ( 9 . 29 - μ 1 2 ), where μ 1 is the mean for H 1 step 2: Roughly draw a graph 1

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2 problem 2 [ grading scheme :10 points for finding test, 6 for calculating power, total 16] step 1: realize we should use Neyman Pearson lemma. step 2: actually calculate likelihood ratio Pr ( x | 6) /Pr ( x | 0) for each x , whose value are ( NA, , 0 . 5 , 4 . 33 , 4 , 4 , 0 , 0 , 0 . 25 , 1 . 67 , 0 . 5 , 0 . 77 , 2 , 0 . 48 , 0 , 0 . 53) step 3: choose the threshold η such that LR η has 0 . 1 level. And get η (1 . 67 , 2].
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