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Assignment 4 solution

# Assignment 4 solution - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT 1302 PROBABILITY AND STATISTICS II (2010-11) Assignment 4 (Sketch Solution) Section A Test statistics, p-values & critical regions A1. (a) (i) The likelihood is ( θ ) = θ - 2 e - S/θ . Likelihood ratio is LR= (2) /‘ (1) = 2 - 2 e S/ 2 . Thus { LR > k 0 } is equivalent to { S > k } = C k . (ii) Type I error probability of the test is P ( S > k | θ = 1) = R k s e - s ds = (1 + k ) e - k . (iii) Type II error probability of the test is P ( S k | θ = 2) = R k 0 s e - s/ 2 / 4 ds = 1 - e - k/ 2 ( k + 2) / 2. (iv) Substitute k = 4 . 744 into the type I error probability to get approximately 0 . 05. (v) From the data we have S = 5 . 3 > 4 . 744. We reject H 0 at 5% level. P-value = P ( S > 5 . 3 | θ = 1) = R 5 . 3 s e - s ds = 0 . 031447. P-value < 0 . 05 above conclusion. (b) (i) The power function of the test is w ( θ ) = P ( S > k | θ ) = R k f ( s ) ds = e - k/θ ( k + θ ) . To check it is increasing, we may examine the derivative of its logarithm. (ii) The size of the test is sup θ 1 w ( θ ) = w (1) = e - k ( k + 1). (iii) pv ( s ) = sup θ 1 P ( S > s | θ ) = e - s ( s + 1) (from (i) and (ii)). Note that pv 0 ( s ) = - s e - s < 0, so that pv ( s ) is decreasing in s . A2. (a) Type I error probability = P ( U > 1 | θ = 0) = 0. (b) Type II error probability = P ( U 1 | θ = 1) = 1 / 2. (c) The size of the test is type I error probability = 0. (d) The power function of the test is w ( θ ) = P ( U > 1 | θ ) = 0 , θ < 0 , θ/ 2 , 0 θ 2 , 1 , θ > 2 . (e) The likelihood is ( θ ) = 2 - 1 1 { θ - 1 U θ + 1 } . The likelihood ratio is 1 { 0 U 2 } / 1 {- 1 U 1 } = 0 , - 1 < U < 0 , 1 , 0 U 1 , , 1 < U < 2 . A3. (a) The power function is w 1 ( θ ) = P (Binomial(10 , θ ) > 8) = 10 θ 9 (1 - θ ) + θ 10 . 1

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(b) The power function w 1 is increasing on (0 , 1). Thus the size of the test is w (1 / 2) = 11 / 2 10 . (c) The power function w 1 is increasing on (0 , 1). Thus the size is w (2 / 3). (d) (i) The power function is w 2 ( θ ) = θ 10 + (1 - θ ) 10 . The size is sup { w 2 ( θ ) : 1 / 3 θ 2 / 3 } = w 2 (1 / 3). (ii) T = | ¯ X - 0 . 5 | , k = 0 . 4. (iii) T is observed to be t = 0, so that pv (0) = sup H 0 ' 1 - P ( ¯ X = 0 . 5 | θ ) = 1 - inf H 0 P (Binomial(10 , θ ) = 5) = 1 - inf H 0 10 5 θ 5 (1 - θ ) 5 = 1 - 10 5 1 3 5 2 3 5 = 5665 6561 0 . 863435 . Section B Hypotheses & likelihood ratios B1. 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. B2. Let X s , X m , X n be nos. substantially, mildly and not improved by the ordinary treatment respectively.
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Assignment 4 solution - THE UNIVERSITY OF HONG KONG...

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