final2006sol - PROFESSOR HONG FALL 2007 TA Jae-ho Yun...

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PROFESSOR HONG FALL 2007 TA: Jae-ho Yun Economics 319: Selected Suggested Solutions for 2006 Final 4. [10 pts] (a) The sample mean is de°ned as follows: X n = P n i =1 X i n : The expectation of sample mean can be derived as follows: E ( X n ) = E °P n i =1 X i n ± = P n i =1 E [ X i ] n = P n i =1 ° n = °: Therefore, X n is unbiased for °: (b) For this, we have to °gure out whether E ( X 2 n ) = ° 2 : We know that that V ar ( X n ) = E [( X n ° ° ) 2 ] = ° 2 n , since X i is iid. Let±s use this result. E [( X n ° ° ) 2 ] = ± 2 n ; E [( X n ° ° ) 2 ] = E [ X 2 n ° 2 ° X n + ° 2 ] = E [ X 2 n ] ° E [2 ° X n ] + E [ ° 2 ] = E [ X 2 n ] ° 2 °E [ X n ] + ° 2 = E [ X 2 n ] ° 2 ° 2 + ° 2 = E [ X 2 n ] ° ° 2 : Therefore, we have E [ X 2 n ] ° ° 2 = ° 2 n , which implies E [ X 2 n ] = ° 2 + ° 2 n 6 = ° 2 : So, X 2 n is biased for ° 2 : 5. [10 pts] 1
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(a) (1 ° ² )100% con°dence interval when ² = 5% is as follows: X n ° ± p n z ±= 2 < ° < X n + ± p n z ±= 2 ; for every n ± 1 , where z ±= 2 is the upper-tail critical value of N (0 ; 1) at level ²= 2 ( z 2 : 5% = 1 : 96 ). Note that we use standard normal distribuion since ± is known. (Otherwise, we should have used student-t distribution) Now, we have that n is 4, ± is 3 (Note that standard deviation should be used rather than variance), and X n = 5+10+6+7 4 = 7 : Hence, con°dence interval can be computed as follows: 7 ° 3 p 4 ² 1 : 96 < ° < 7 + 3 p 4 ² 1 : 96 ; 4 : 06 < ° < 9 : 94 : (b) The con°dence level is the probability that ° will lie in the interval between X n ° °
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