Unformatted text preview: AD Number________ Name_________________________ Score________ IE 300 / GE 331 HW8 718 (1 point) 726 (4 points) 1 , 2 are unbiased, 3 is biased 1 = 1 = 12, 2 = 2 = 10, 3 = 6 We select 3 since it has the smallest MSE (a) () = (1 ) + (1  )(2 ) = + (1  ) = =
2 2 2 2 +(1 ) 1 1 1 2 (b) () = 2 (1 ) + (1  )2 (2 ) = 2 1 + (1  )2 2 = 2 1 + (1  )2 1 2
1 2
2 2
1 2
2 (c) SE( ) = 1 ) ( 8 9 (d) = SE( ) = 1 1 = 0 = 2 2 +(1)2 1 1 2 1 2 + 1 2 2 +(1)2 1 1 2 734 (3 points) (a). (1 + ) = 1 = 0.5 1 ln () = ln 0.5 + =1 = 0.5 results in a much larger standard error: SE( ) = = 0.6671 2 1.0607 1 2 1 1 (c). = xf(x)dx = 0.5 x(1 + )dx = , E()= (3) = 3 = 1 1 3 (d). () = 0.5 (1 + )(1 + 2 ) ... (1 + ) ln ( ) =0 =0 =1 (1 + ) The maximum likelihood estimate solves the above equation ln(1 + ) 82 (1 point) (b) = 0.2 /2 = 0.1 = 1.29 88 (1 point) = 0.05 /2 = 0.025 = 1.96, = 9, = 2, = 98 95% confidence interval on the mean breaking strength is  0.025 + 0.025 96.7 99.3 ...
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This note was uploaded on 05/04/2010 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

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