HW3 - 8.39 yn unif (0, ), a fy = 1 , y (0, ) Let Y(n) =...

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8.39 y n unif (0 ) , f y = 1 θ , y (0 ) Let Y ( n ) = max( Y 1 , ··· ,Y n ) , U = 1 θ Y ( n ) a F U ( u ) = P ( 1 θ Y ( n ) u ) = P ( Y ( n ) θu ) = P ( Y 1 θu, ··· Y n θu ) ( * ) = P ( Y θu ) n ( ** ) = ± θu θ ² n = u n Because (*) : Y ( n ) is a maximum in Y 1 , ··· ,Y n ( ** ) : Y is random sample(independent and identically distributed) Therefore F U ( u ) = 0 u < 0 u n 0 u < 1 1 u 1 b P ( U a ) = 0 . 95 a n = 0 . 95 a = n 0 . 95 8.42 Y Bin ( n,p ) , E ( ¯ Y ) = p,V ar ( ¯ Y ) = pq n a By large sample theory, Z = ¯ Y - p pq/n N (0 , 1) ˆ p ± Z 0 . 1 q ˆ p (1 - ˆ p ) n = 0 . 536 ± 0 . 052 b C.I includes 0.51, so there is no evidence of being a difference in graduation rates before and after proposition 48. 8.46 ¯ y ± Z 0 . 025 s/ n = 4 . 2 ± 0 . 34 8.47 ¯ y 1 N ( μ 1 2 / n 1 ) , ¯ y 2 N ( μ 2 2 / n 2 ) ¯ y 1 - ¯ y 2 N ( μ 1 - μ 2 2 / n 1 + σ 2 / n 2 ) By large sample theory, we can use S 2 instead of σ 2 . (167 . 1 - 140 . 9) ± 1 . 96 q 24 . 3 2 30 + 17 . 6 2 30 = 26 . 2 ± 10 . 74 8.71 Assume X,Y N with same variance. a ¯ x
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HW3 - 8.39 yn unif (0, ), a fy = 1 , y (0, ) Let Y(n) =...

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