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### tut_7_Chap_7_sol

Course: STATISTICS 101, Spring 2010
School: Qatar University
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Word Count: 759

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- CollegeofEngineering,QatarUniversity ProbabilityandStatistics,GENG200 Tutorial 7 2B Question Sections 7-3 Page 242: 1) 7.14 SOLUTION: 7-14 2n Xi X 1 = E i =1 2n E( ) E (X 2 ) n Xi = E i =1 n 2n = 1 E X = 1 (2n ) = 2n i 2n i =1 n = 1 E X = 1 (n ) = , i n i =1 n X 1 and X 2 are unbiased estimators of . The variances are V (X1 ) = and V (X 2 ) = ; compare...

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- CollegeofEngineering,QatarUniversity ProbabilityandStatistics,GENG200 Tutorial 7 2B Question Sections 7-3 Page 242: 1) 7.14 SOLUTION: 7-14 2n Xi X 1 = E i =1 2n E( ) E (X 2 ) n Xi = E i =1 n 2n = 1 E X = 1 (2n ) = 2n i 2n i =1 n = 1 E X = 1 (n ) = , i n i =1 n X 1 and X 2 are unbiased estimators of . The variances are V (X1 ) = and V (X 2 ) = ; compare the MSE (variance in this case), 2 2 2n n MSE (1 ) 2 / 2n n 1 = = = ) 2 /n 2n 2 MSE ( 2 Since both estimators are unbiased, examination of the variances would conclude that the better estimator with the smaller variance. X1 is Exercises Sections 7-3 Page 242: 2) 7-15 Answer: 7-15 E (1 ) = 1 [E ( X 1 ) + E ( X 2 ) + L + E ( X 7 )] = 1 (7 E ( X )) = 1 (7 ) = 7 7 7 E ( 2 ) = 1 [E (2 X 1 ) + E ( X 6 ) + E ( X 7 )] = 1 [2 + ] = 2 2 a) Both 1 and 2 are unbiased estimates of since the expected values of these statistics are equivalent to the true mean, . b) V (1 ) = V X 1 + X 2 + ... + X 7 1 1 12 2 = 2 (V ( X 1 ) + V ( X 2 ) + L + V ( X 7 ) ) = 49 (7 ) = 7 7 7 V (1 ) = 7 2 V ( 2 ) = V = = 2 X1 X 6 + X 4 1 1 = 2 (V (2 X 1 ) + V ( X 6 ) + V ( X 4 ) ) = 4 (4V ( X 1 ) + V ( X 6 ) + V ( X 4 )) 2 2 1 4 2 + 2 + 2 4 1 (6 2 ) 4 ( ) 3 2 V ( 2 ) = 2 Since both estimators are unbiased, the variances can be compared to decide to select the better estimator. The variance of 1 is smaller than that of 2 , 1 is the better estimator. Page 1 of 3 CollegeofEngineering,QatarUniversity Exercises Chapter 7: 1) 7-21 2) 7-34, 7-49, 7-50, 7-51, 7-54 Answer: 7-21 a) Show that X 2 is a biased estimator of 2. Using () E X 2 = V( X ) [ + E ( X) ] 2 2 2 n 1 n 1 n V X + E X E X = 2 E X i = 2 i i n i =1 n i =1 i =1 2 n 1 1 2 = 2 n 2 + = 2 n 2 + (n ) n n i =1 2 1 = 2 n 2 + n 2 2 E X 2 = + 2 n n () 2 ( ( ) )( ) Therefore, X 2 is a biased estimator of .2 b) Bias = () E X 2 2 = 2 n + 2 2 = 2 n c) Bias decreases as n increases. 1 x2 a) c(1 + x)dx = 1 = (cx + c ) = 2c 2 1 1 1 7-34 so that the constant c should equal 0.5 b) E( X ) = 1n Xi = 3 n i =1 = 3 1n Xi n i =1 c) 1n E () = E 3 X i = E (3 X ) = 3E ( X ) = 3 = n 3 i =1 d) n1 L( ) = (1 + X i ) i =1 2 n 1 ln L( ) = n ln( ) + ln(1 + X i ) 2 i =1 n ln L( ) Xi = i =1(1 + X i ) By inspection, the value of that maximizes the likelihood is max (Xi) Page 2 of 3 CollegeofEngineering,QatarUniversity 7-49 X ~ N (50,144) P (47 X 53) = P ( 47 50 12 / 36 53 Z 12 /50 36 ) = P (1.5 Z 1.5) = P ( Z 1.5) P ( Z 1.5) = 0.9332 0.0668 = 0.8664 No, because Central Limit Theorem states that with large samples (n 30), approximately normally distributed. 7-50 Assume X X is is approximately normally distributed. 4985 5500 ) 100 / 9 P ( X > 4985) = 1 P ( X 4985) = 1 P ( Z = 1 P ( Z 15.45) = 1 0 = 1 z= X = 52 50 7-51 = 5.6569 s/ n 2 / 16 P(Z > z) ~0. The results are very unusual. 7-54 E (aX 1 + (1 a) X 2 ) = a + (1 a ) = V ( X ) = V [aX 1 + (1 a) X 2 ] = a 2V ( X 1 ) + (1 a) 2 V ( X 2 ) = a 2 ( 1 ) + (1 2a + a 2 )( 2 ) n n 2 = 2 2 a 2 2 2 2a 2 a 2 2 + + = (n2 a 2 + n1 2n1a + n1a 2 )( ) n2 n2 n2 n1n2 n1 2 V ( X ) = ( )(2n2 a 2n1 + 2n1a) 0 n1n2 a 0 = 2n2 a 2n1 + 2n1a 2a(n2 + n1 ) = 2n1 a(n2 + n1 ) = n1 a= n1 n2 + n1 Page 3 of 3
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