statistics for business and economics 11th edition chapter 7 - exercise solution

Statistics for business and economics 11th edition chapter 7 - exercise solution

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Unformatted text preview: Chapter 7 Sampling and Sampling Distributions Solutions: 15. a. $45,500 / $4,550 10 i x x n = = = b. 2 ( ) 9,068,620 $1003.80 1 10 1 i x x s n - = = =-- 16. a. We would use the sample proportion for the estimate. 5 .10 50 p = = (Authors' note: The actual proportion from New York is 52 .104. 500 p = = ) b. The sample proportion from Minnesota is 2 .04 50 p = = Our estimate of the number of Fortune 500 companies from New York is (.04)500 = 20. (Authors' note: The actual number from Minnesota is 18.) c. Fourteen of the 50 in the sample come from these 4 states. So 36 do not. 36 .72 50 p = = (Authors' note: The actual proportion from the other states is 366 .732. 500 p = = ) 18. a. E x ( ) = = 200 b. x n = = = / / 50 100 5 c. Normal with E ( x ) = 200 and x = 5 d. It shows the probability distribution of all possible sample means that can be observed with random samples of size 100. This distribution can be used to compute the probability that x is within a specified from . 19. a. The sampling distribution is normal with E ( ) x = = 200 / 50 / 100 5 x n = = = For 5, 195 205 x Using Standard Normal Probability Table: At x = 205, 5 1 5 x x z - = = = ( 1) P z = .8413 At x = 195, 5 1 5 x x z -- = = = - ( 1) P z < - = .1587 (195 205) P x = .8413 - .1587 = .6826 b. For 10, 190 210 x Using Standard Normal Probability Table: At x = 210, z x x =- = = 10 5 2 ( 2) P z = .9772 At x = 190, 10 2 5 x x z -- = = = - ( 2) P z < - = .0228 (190 210) P x = .9772 - .0228 = .9544 Chapter 7 26. a. 939 / x z n - = Within 25 means x- 939 must be between -25 and +25. The z value for x- 939 = -25 is just the negative of the z value for x- 939 = 25. So we just show the computation of z for x- 939 = 25. n = 30 25 .56 245/ 30 z = = P (-.56 z .56) = .7123 - .2877 = .4246 n = 50 25 .72 245/ 50 z = = P (-.72 z .72) = .7642 - .2358 = .5284 n = 100 25 1.02 245/ 100 z = = P (-1.02 z 1.02) = .8461 - .1539 = .6922 n = 400 25 2.04 245/ 400 z = = P (-2.04 z 2.04) = .9793 - .0207 = .9586 b. A larger sample increases the probability that the sample mean will be within a specified distance of the population...
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This note was uploaded on 02/01/2011 for the course STAT 101 taught by Professor Oba during the Fall '10 term at Athens University of Econ and Bus.

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Statistics for business and economics 11th edition chapter 7 - exercise solution

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