Fall_2005_Exam_C_solutions

# Fall_2005_Exam_C_solutions - FALL 2005 EXAM C SOLUTIONS...

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Unformatted text preview: FALL 2005 EXAM C SOLUTIONS Question #1 Key: D ˆ (300) 3/10 S = (there are three observations greater than 300) ˆ ˆ (300) ln[ (300)] ln(0.3) 1.204 H S = − = − = . Question #2 Key: A 2 ( | ) ( | ) ( ) ; ( ) ; / 1/ 1/ 1 1 0.15 (1) 1 1 1 2 1 4 0.20 (2) 2 1 2 1 2 1 From the first equation, 0.15 0.15 and so 0.15 0.85 Then the second equat E X Var X v E a Var k v a n n Z n n λ λ λ µ λ α θ λ α θ θ θ θ θ θ θ µ µ θ θ θ θ θ µ µ θ θ θ θ θ µ µ θ = = = = = = = = = = = + + + = + = + + + + = + = + + + + = + = − ion becomes 0.4 0.2 4 0.15 0.85 0.05 2.75 ; 0.01818 θ θ θ θ θ + = + − = = Question #3 Key: E 1 1 0.75 ; 0.25 1 (100/ ) 1 (500/ ) (100/ ) 1/3; (500/ ) 3 γ γ γ γ θ θ θ θ = = + + = = Taking the ratio of these two equalities produces 5 9 γ = . From the second equality, 2 2 9 [(500/ ) ] 5 ; (500/ ) 5; 223.61 γ γ θ θ θ = = = = Question #4 Key: B 2 3 ( ) ( ) 2 ( 2 ) 2 4 8 1 ' ( ) 24 ''(2) 2 12 ; 12 6 f x a bx cx dx f a f a b c d f b f c d c d = + + + = = = = + + + = = − = = + = − − Insert the values for a , b , and c into the second equation to obtain 2 2 4( 12 6 ) 8 ; 48 16 ; 3 d d d d = + − − + = − = − Then 6 c = and 2 3 ( ) 6 3 ; (1) 4 f x x x x f = + − = Question #5 Key: E Begin with y 350 500 1000 1200 1500 s 2 2 1 1 1 r 10 8 5 2 1 Then 1 8 6 4 1 ˆ (1250) 0.24 10 8 5 2 S = = The likelihood function is 2 2 2 1 350/ 1 500/ 500/ 1 1000/ 1000/ 1 1200/ 1 1500/ 7 7900/ 2 1250(7)/7900 2 ( ) 7900 7 7900 ˆ ( ) 7ln ; '( ) 0; 7900/ 7 ˆ (1250) 0.33 L e e e e e e e e l l S e θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ − − − − − − − − − − − − − − − ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ = ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ = = − − = − + = = = = The absolute difference is 0.09. Question #6 Key: E 4 2 2 3 4 4 4 1 2 2 2 3 2 2 3 2 2 2 2 3 2 5 2 2 4 2 4 2 4 2 2 4 ( ) '( ) ( ) 4(2) 4(4) 128 ( ) (2) (4) (4) ( 2 ) ( 4 ) ( 4 ) ( 4) ( 16) ( ) ln128 12ln 3ln( 4) 5ln( 16) 12 6 10 '( ) 0;12( 20 64) 6( 16 ) 10( 4 4 1 6 x f x S x x L f f S l l θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ = − = + = = = + + + + + = + − + − + = − − = + + − + − + + + 2 4 2 4 2 2 2 ) 4 104 768 26 192 26 26 4(192) 32; 5.657 2 θ θ θ θ θ θ θ = = − + + = − − ± + = = = Question #7 Key: A 2 2 2 2 2 2 2 1 4 1 2 5 2 2 2 2 2 4 4 ( | ) ; ( | ) 3 9 2 9 1 8 (2/3) ( ) (2/3) 4 8/15 (1/18) ( ) (1/18) 4 1/ 27 (2/3) ( ) (4/9) 4/ 6 (4/5) 8/ 675 1/ 27 1 25/8; 8/ 675 1 25/8 x x E X x dx Var X x dx E d EVPV v E d VHM a Var k Z θ θ θ θ θ θ θ θ θ θ θ µ θ θ θ θ θ θ θ = = = − = − = = = = = = = = ⎡ ⎤ = = = − = ⎣ ⎦ = = = = + ∫ ∫ ∫ ∫ 8/33 Estimate is (8/33)(0.1) (25/33)(8/15) 0.428. + = Question #8 Key: D From the Poisson(4) distribution the probabilities at 0, 1, and 2 are 0.0183, 0.0733, and 0.1463....
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Fall_2005_Exam_C_solutions - FALL 2005 EXAM C SOLUTIONS...

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