Stat 104 Fall 2010 Midterm Solutions

Stat 104 Fall 2010 Midterm Solutions - Stat 104:...

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Stat 104: Quantitative Methods for Economists Fall 2010 Midterm Examination Solutions 1. (e) 11 ) 2 . 0 ( 55 = = = np µ , 97 . 2 ) 8 . 0 )( 2 . 0 ( 55 ) 1 ( = = = p np σ 2. (a) 90 . 0 10 . 0 1 ) ( 1 ) ( = = = tennis no and golf no P tennis or golf P 24 . 0 90 . 0 45 . 0 69 . 0 ) ( ) ( ) ( ) ( = + = + = tennis or golf P tennis P gofl P tennis and golf P 533 . 0 45 . 0 24 . 0 ) ( ) ( ) | ( = = = tennis P tennis and golf P tennis golf P 3. (e) There are no guarantees in this situation since they are randomly generated tickets. 4. (b) The magnitude of correlation cannot be interpreted directly into anything specific (yet), but can be interpreted to talk about general strength and direction of the relationship. 5. (a) 1.6 = 4(0.2) + 2(0.4) + 0(0.4) ) ( = = = x X xP 2.24 = (0.2) 1.6) - (4 + (0.4) 1.6) - (2 + (0.4) 1.6) - 0 ( ) ( ) ( 2 2 2 2 2 = = = x X P x 6. (a) P (all six believe a good job) = [ P (one believes a good job)] 6 = (0.72) 6 = 0.139 7. (b) ) 11 . 2 , 59 . 1 ( 32 75 . 0 96 . 1 85 . 1 96 . 1 = ± = ± n s x 8. (a) Entire length of the confidence interval goes from n s 96 . 1 to + n s 96 . 1 . So the length is twice the margin of error, or = n s Length ) 96 . 1 ( 2 . Solving for n , we get: 385 ) 002 . 0 ( ) 01 . 0 )( 96 . 1 ( 2 ) 96 . 1 ( 2 2 2 = = Length s n *Note, if you used z = 2 instead of 1.96, you would have gotten
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This note was uploaded on 03/27/2012 for the course STATS 104 taught by Professor Michaelparzen during the Fall '11 term at Harvard.

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Stat 104 Fall 2010 Midterm Solutions - Stat 104:...

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