PR1A_S_11_SOL

PR1A_S_11_SOL - ESE 302 SOLUTIONS TO THE ADDITIONAL...

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ESE 302 SOLUTIONS TO THE ADDITIONAL PRACTICE QUESTIONS FOR EXAM I Spring, 2011 1. Given the no-intercept model, , 1,. ., , ii i Yx i n    a. the least-squares estimation problem for this case reduces to: Minimize:  2 1 () n i Sy x  So the first-order condition becomes 1 02 ( ) n iii i dS y xx d  2 11 0 nn i y   1 2 1 ˆ n i n i i x y x  b. To establish unbiasedness of ˆ observe that 2 22 2 ˆ i i jj j j x EE x      ˆ E  c. To determine the variance of ˆ , observe from the assumed independence of 1 ( ,. ., ) n YY that
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 2 22 11 2 2 2 2 2 1 2 ˆ var var var var nn ii jj n i i i j j j j xx YY x x x x        2 2 1 ˆ var n i i x  2. Given that 1 1 B if a the bottle is broken during labeling and that 2 1 B if the bottle is broken during crating, it follows that: a. The overall probability, p , that a bottle is broken is given by  12 1 2 1 1 1 Pr( 0, 0) 1 Pr( 0) Pr( 0 | 0) 11P r ( 1 )1P r | 0 ) pB B B B B BB B           But since Pr( 1) and 1 Pr( 1| 0) B by definition, it then follows that 1 2 1 2 1( 1 ) ( 1 ) p pp p p p p    b. Using this result, observe next that 1 2 1 2 121 2 1 2 1 2 1 1 ˆ () [by independence] XXX X X X X X Ep E E E E NNN N N N N N X EEE E N      But since the relative frequency, / X N , is the natural unbiased estimator of , 1,2 i pi , it then follows that 2 ˆ p p pp p and hence that ˆ p is an unbiased estimator of p .
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c. Observe that by definition 21 2 12 1 1 2 Pr( 1) Pr( 0, Pr( 0)Pr( 1| 0) (1 ) BB B B p p  Hence in this case: 1 2 11 1 1 1 1 1 2 ˆ () 1 XX X X Ep E NN X N N X X E X N N X E X N X X EE E N N         But since / i X N is again the natural unbiased estimator of Pr( 1), 1,2, i Bi it follows that 2 121 2 ˆ () P r ( 1 ) P r ( 1 ) ) B B pp p ppp p p   So even though 1 X and 2 X are no longer independent, ˆ p is still an unbiased estimator of p . 3. Given the unit transformations, ii y ay  and ( ) tb t c  : a. observe first that
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11 () nn ii y y ay ay tt b t c b t c    and hence that the new estimate for 1 is given by    1 2 1 22 ( ) ( ) us i i i eng i i i bt c bt c ay ay bt c c ttyy ab a t b     and similarly, the new estimate for 0 is given by  01 1 1 1 us us eng eng eng aa ay c ay c a y t c bb       But eng eng y t  , so that we must have 00 1 us eng eng ac a b  b.
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This note was uploaded on 03/03/2011 for the course ESE 302 taught by Professor Ese302 during the Spring '08 term at UPenn.

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PR1A_S_11_SOL - ESE 302 SOLUTIONS TO THE ADDITIONAL...

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