Unformatted text preview: IE300/GE331 Discussion Problem Set 10 714 (1 ) =
1 1 2 = (2) = =1 2 2 1 1 ( ) = () = =1 (2 ) = Both 1 and 2 are unbiased estimators of 1 1 (1 ) = (2)2 2 = (2)2 (2 2 ) = , (1 ) = . =1 1 1 (2 ) = ()2 ( ) = ()2 ( 2 ) = , (2 ) = . =1 2 2 2 1 is a better estimator of since it has a smaller variance and hence MSE. It is more precise. 720 (a). (  )2 = 2  2 (see discussion 8) =1 =1 = 2 + 2  =1 2 2 2 (  )2 1 =1 2 2 = 2  ( 2 ) = (1 )  () + () 2 Note that (1 ) = (1 ) + ((1 ))2 (b). Bias = 728 (  )2 =1 (c). Bias decreases to zero as n increases () = (1  )1 () = (1  )=1  = (1  )  ln () = ln + (  ) ln(1  )  2 =  2 1 + 2 = 1  2 2 ln ()  1 =0  = 0 = 1  730 () = (1 + ) = (1 + ) (1 2 ... ) ln () = ln(1 + ) + ln =1 =1 =1 ln ()  =0 + ln = 0 = 1 =1 ln (1 + ) 748
25
2 1.5 2 1.5 2 1 ~ 100, , 2 ~ 105, 1  2 ~ 100  105, + = (5,0.2233) Standard error is 0.2233 = 30 2 25 2 30 2 86 (a) Sample mean is the midpoint of the confidence interval (b) The first interval since it has a smaller length 3124.9 + 3251.7 3110.5 + 3230.1 = 3170.3 = 2 2 812 (a) = 0.05 /2 = 0.025 = 1.96 = 1014, = 20, = 25 The 95% confidence interval on the mean life is  0.025 + 0.025 1003 1025 ...
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 Spring '09
 NegarKayavash
 Normal Distribution

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