course4solutions_1102 - November 2002 Course 4 solutions...

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November 2002 Course 4 solutions Question # 1 Answer: B ρ φ ρφ 1 1 2 22 1 2 2 1 05 1 02 = = =+ =− . . Solving simultaneously gives: 1 2 08 06 = . . Question # 2 Answer: C [12(.45)] [5.4] 5; 5.4 5 0.4. gh == = = = .45 (5) (6) ˆ .6 .4 .6(360) .4(420) 384. xx π =+= + = Question # 3 Answer: D N is distributed Poisson ( λ ) ( ) 1(1.2) 1.2. E µ λα θ = = ( ) 1.2; ( ) 1.44. vE aV a r λλ α = = = = 1.2 5 2 12 ;. 1.44 6 2 5/ 6 17 kZ = = + Thus, the estimate for Year 3 is 12 5 (1.5) (1.2) 1.41. 17 17 += Note that a Bayesian approach produces the same answer.
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Question # 4 Answer: C At the time of the second failure, 11 23 ˆ ( ) 12. 1 132 H tn nn = + = ⇒= - At the time of the fourth failure, 11 11 ˆ ( ) .3854. 1 2 1 1 1 09 Ht =++ += Question # 5 Answer: E 2 2 22 2 42 ˆ 2.06 5 .9841. 182 i i x R y b = == Question # 6 Answer: B The likelihood is: ( 1 ) ( 1) ( 1). !( j jj j x x rx j j r r L x b bb b -- + + +- = ∝+ + ∏∏ L The loglikelihood is: 1 1 l n ( )ln( 0 1 0 ( 1 ) () ˆ 0 ; /. n j n j j l x x rx l x r x x rn n x r n xr b = =  = - ++  + =-=  + = + - + =- = -= ∑∑
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Question # 7 Answer: C The Bühlmann credibility estimate is (1 ) Z xZ µ + where x is the first observation. The Bühlmann estimate is the least squares approximation to the Bayesian estimate. Therefore, Z and must be selected to minimize 22 2 11 1 [( 1)1 . 5 ] [ 2( . 5 ] [ 3( 1)3 ] . 33 3 ZZ µµµ +− + + Setting partial derivatives equal to zero will give the values. However, it should be clear that is the average of the Bayesian estimates, that is, 1 (1.5 1.5 3) 2. 3 =+ + = The derivative with respect to Z is (deleting the coefficients of 1/3): 2( .5)( 1) 2(.5)(0) 2( 1)(1) 0 .75. Z −+ −+ + = = The answer is .75(1) .25(2) 1.25. += Question # 8 Answer: E The confidence interval is 1/ 00 ˆˆ (() ,()) . St θθ Taking logarithms of both endpoints gives the two equations 0 0 1 ˆ ln.695 .36384 ln ( ) ˆ ln.843 .17079 ln ( ). θ =− = = Multiplying the two equations gives 2 0 0 0 ˆ .06214 [ln ( )] ˆ ln ( ) .24928 ˆ ( ) .77936. = = The negative square root is required in order to make the answer fall in the interval (0,1).
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Question # 9 Answer: E Because k k ρ φ = there are a number of ways to get the value of . 1/2 1/3 .1 .215 .46368; ( .1) .46416; .46512. .215 φφ == = = = = Also, because the mean is zero, δ must be zero. Then (using the first choice for ), 1 ˆ .46368( .431) 0 .1998. T y + =− + = Question # 10 Answer: B The likelihood is: 111 33 1 150 150 150 (150 225) (150 525) (150 950) 150 . (375 675 1100) L ααα α +++ + = = ii The loglikelihood is: 3ln 3 ln150 ( 1)ln(375 675 1100) 3ln150 ln(375 675 1100) 4.4128 ˆ 3/ 4.4128 .6798. l l αα =+ + Question # 11 Answer: D For this problem, r = 4 and n = 7. Then, 33.60 ˆ 1.4 4(7 1) v and 3.3 1.4 ˆ .9.
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course4solutions_1102 - November 2002 Course 4 solutions...

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