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Unformatted text preview: Kata Bognar email@example.com Economics 41 Statistics for Economists UCLA Winter 2010 Final Exam- suggested solutions - Part I - Multiple Choice Questions (3 points each) 1. Which of the following is an observational study? B (a) You investigate the effect of aspirin on preventing heart attacks by giving one group of people a daily dose of aspirin and another group a placebo. Then you compare the rates of heart attacks between the two groups. (b) You investigate the effect of college GPA on starting salary by questioning ran- dom recent graduates what their starting salary is and you find their GPA from school records. (c) You are the dictator of a country and you decide to investigate the effect of smoking on lung cancer. You force half of your population to smoke and the other half not to. Then you compare the rates of lung cancer. (d) Both (a) and (c). 2. Which of the following is an example of a discrete variable? C (a) The weight of a box of cookies (b) The length of a window frame (c) The number of horses owned by a farmer (d) The distance from home to work for a worker 3. Which of the following is true? C (a) 5 i =1 i = 5 (b) 5 i =1 2 = 2 (c) ( 5 4 ) = 5 (d) 5! = 100 4. Assume there are two urns. Urn A has one white and three black balls and Urn B has two black and two white balls. You randomly take out two balls (without replacement) from Urn A and (independently so) one ball from Urn B. What is the probability of the event among the three balls taken out of the urns there are exactly two black ones? C (a) 0.15 (b) 0.25 (c) 0.50 (d) none of the above Explanation : One can select 2 balls form Urn A and one ball from Urn B in ( 4 2 )( 4 1 ) different ways. One can select two balls form Urn A and one ball from Urn B such that there are exactly two black balls in ( 3 1 )( 2 1 ) + ( 3 2 )( 2 ) . Notice that one can either pick exactly one black ball from each urn or exactly two form Urn A and none form Urn B. 5. A random variable X that takes on four different values has the following distribution: P ( X =- 1) = 2 c, P ( X = 0) = 3 c, P ( X = 1) = 0 . 4 and P ( X = 2) = 0 . 1 . The value of the constant c is: B (a) 1 1 (b) 0.1 (c) 0.3 (d) 2.5 Explanation : Since X can only take the for values above, the total probability of those must sum up to 1. Thus, 2 c + 3 c + 0 . 4 + 0 . 1 = 1 and c = 0 . 1 . 6. Suppose that A and B are events such that P ( A ) = 1 / 2 , P ( B ) = 1 / 4 and P ( A | B )+ P ( B | A ) = 2 / 3 , then P ( A B ) is D (a) 3/4 (b) 1/9 (c) 8/9 (d) none of the above Explanation : Recall the addition rule: P ( A B ) = P ( A )+ P ( B )- P ( A B ) . P ( A ) and P ( B ) are given. How can you get P ( A B )? According to the conditional probability rule P ( A | B ) = P ( A B ) P ( B ) and P ( B | A ) = P ( A B ) P ( A ) . Therefore, P ( A | B ) + P ( B | A ) = P ( A B ) P ( A ) + P ( A B ) P ( B ) . Denote by x the probability of A B. Then, x 1 / 2 + x 1 / 4 = 2 3 and x = 1 9 . Finally, P ( A...
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This note was uploaded on 12/28/2010 for the course ECON 41 taught by Professor Guggenberger during the Fall '07 term at UCLA.
- Fall '07