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

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Unformatted text preview: 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 } is equivalent to { S > k } = C k . (ii) Type I error probability of the test is P ( S > k | θ = 1) = R ∞ k se- s ds = (1 + k ) e- k . (iii) Type II error probability of the test is P ( S ≤ k | θ = 2) = R k se- 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 at 5% level. P-value = P ( S > 5 . 3 | θ = 1) = R ∞ 5 . 3 se- s ds = 0 . 031447. P-value < . 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 ) /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 ( s ) =- se- 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 ) =      , θ < , θ/ 2 , ≤ θ ≤ 2 , 1 , θ > 2 . (e) The likelihood is ‘ ( θ ) = 2- 1 1 { θ- 1 ≤ U ≤ θ + 1 } . The likelihood ratio is 1 { ≤ U ≤ 2 } / 1 {- 1 ≤ U ≤ 1 } =      ,- 1 < U < , 1 , ≤ 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 (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- . 5 | , k = 0 . 4. (iii) T is observed to be t = 0, so that pv (0) = sup H ' 1- P ( ¯ X = 0 . 5 | θ ) “ = 1- inf H P (Binomial(10 ,θ ) = 5) = 1- inf H 10 5 ¶ θ 5 (1- θ ) 5 = 1- 10 5 ¶ 1 3 ¶ 5 2 3 ¶ 5 = 5665 6561 ≈ . 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 : N ≤ 21000 vs H 1 : N > 21000....
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This note was uploaded on 11/17/2011 for the course MUSIC 1001 taught by Professor Dr.yun during the Spring '11 term at Princess Sumaya University for Technology.

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

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