ass4 sol - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT 1302 PROBABILITY AND STATISTICS II Assignment 4 (Sketch Solution) Section A Hypotheses & likelihood ratios A1. Let X be the no. of marked fish among a sample of 2,000 caught from the lake and N be the total no. of fish in the lake. (a) (i) H 0 : N 21000 vs H 1 : N > 21000. (ii) Note that the likelihood is ( N ) = ( 1000 100 )( N - 1000 1900 ) / ( N 2000 ) . Consider ( N +1) /‘ ( N ) = 1 - 100( N - 19999)( N +1) - 1 ( N - 2899) - 1 > 1 , 2900 N 19998 , = 1 , N = 19999 , < 1 , N 20000 . Thus ( N ) increases as N increases up to 19,999, stays constant at N = 19999 and 20000, and then decreases as N increases beyond 20,000. Under H 1 , ( N ) is maximized at N = ˆ N 1 = 21000. Under H 0 , ( N ) is maximized at N = ˆ N 0 = 20000 (or 19999). The likelihood ratio is ( ˆ N 1 ) /‘ ( ˆ N 0 ) = 0 . 872511. (b) (i) H 0 : N > 21000 vs H 1 : N 21000. (ii) Under H 0 , ( N ) is maximized at N = ˆ N 0 = 21000. Under H 1 , ( N ) is maximized at N = ˆ N 1 = 20000 (or 19999). The likelihood ratio is ( ˆ N 1 ) /‘ ( ˆ N 0 ) = 1 . 146118. A2. Let X s , X m , X n be nos. substantially, mildly and not improved by the ordinary treatment re- spectively. Define Y s , Y m , Y n for the new treatment similarly. Then ( X s , X m , X n ) Multinomial(54; p s , p m , p n ) and ( Y s , Y m , Y n ) Multinomial(54; q s , q m , q n ) . (i) H 0 : p s = q s , p m = q m vs H 1 : p s < q s , p m < q m . (ii) The likelihood is ( p, q ) p 12 s p 6 m p 36 n q 18 s q 12 m q 24 n . Under H 1 , is max’d at p s = 12 / 54, p m = 6 / 54, p n = 36 / 54, q s = 18 / 54, q m = 12 / 54 and q n = 24 / 54. Under H 0 , is max’d at p s = q s = 30 / 108, p m = q m = 18 / 108 and p n = q n = 60 / 108. The likeihood ratio is 16 . 9724. A3. Let X be the no. of heads. Then X Binomial(100 , p ), p [0 , 1]. (i) H 0 : p = 1 / 2 vs H 1 : p 6 = 1 / 2. (ii) The likelihood is ( p ) p 60 (1 - p ) 40 , maximized at p = ˆ p = 0 . 6. The likelihood ratio is p ) /‘ (1 / 2) = 7 . 48987. 1
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A4. (i) H 0 : θ = ψ vs H 1 : θ < ψ . (ii) The likelihood is ( θ, ψ ) = θ 5 e - 30 θ ψ 11 e - 30 ψ . Under H 0 , ( θ, θ ) is max’d at θ = ˆ θ 0 = 4 / 15. Under H 1 , ( θ, ψ ) is max’d at θ = ˆ θ = 1 / 6 and ψ = ˆ ψ = 11 / 30. The likelihood ratio is ( ˆ θ, ˆ ψ ) /‘ ( ˆ θ 0 , ˆ θ 0 ) = 3 . 1676. A5. Let X 1 , X 2 be the nos. of breakdowns on two consecutive days. Assume X i Poisson( λ ), i = 1 , 2, independently. (i) The probability in question is P ( X 1 = 0) = e - λ . We want to test H 0 : e - λ > 3 / 4 (i.e. λ < ln(4 / 3)) vs H 1 : e - λ 3 / 4 (i.e. λ ln(4 / 3)). (ii) The likelihood is ( λ ) = λ 9 e - 2 λ / (3!6!). Under H 1 , is max’d at λ = 9 / 2. Under H 0 , is max’d at λ = ln(4 / 3). The likelihood ratio is 12300551 . 2. A6. Let X 1 , . . . , X 4 be heights of male students and Y 1 , . . . , Y 5 be heights of female students. (i) H 0 : μ 1 = μ 2 vs H 1 : μ 1 > μ 2 . (ii) The likelihood is ( μ 1 , μ 2 , σ 2 1 , σ 2 2 ) = (2 πσ 2 1 ) - 4 / 2 (2 πσ 2 2 ) - 5 / 2 exp {- i ( X i - μ 1 ) 2 / (2 σ 2 1 ) - j ( Y j - μ 2 ) 2 / (2 σ 2 2 ) } . Under H 0 , is max’d at μ 1 = μ 2 = ˆ μ = 1 . 56753 (a cubic equation needs to be solved for this answer); σ 2 1 = S 01 = i ( X i - ˆ μ ) 2 / 4, σ 2 2 = S 02 = j ( Y j - ˆ μ ) 2 / 5. Un- der H 1 , is max’d at μ 1 = ¯ X , μ 2 = ¯ Y , σ 2 1 = S 1 = i ( X i - ¯ X ) 2 / 4 and σ 2 2 = S 2 = j ( Y j - ¯ Y ) 2 / 5, provided ¯ X > ¯ Y . The likelihood ratio is ( ¯ X, ¯ Y , S 1 , S 2 ) /‘ μ, ˆ μ, S 01 , S 02 ) = 108 . 287.
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