Discussion10%20Solutions

# Discussion10%20Solutions - IE300/GE331 Discussion Problem...

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Unformatted text preview: IE300/GE331 Discussion Problem Set 10 7-14 (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. 7-20 (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 = 7-28 ( - )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 - 7-30 () = (1 + ) = (1 + ) (1 2 ... ) ln () = ln(1 + ) + ln =1 =1 =1 ln () - =0 + ln = 0 = -1 =1 ln (1 + ) 7-48 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 8-6 (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 8-12 (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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