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Unformatted text preview: 98 Instructors Solutions Manual for Statistics for Business & Economics, 5 th Edition Chapter 7: Sampling and Sampling Distributions 71 a. Probability distribution for one die: Die outcome Probability 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 b. Sampling distribution of the sample means from rolling a pair of dice: Total Sample x Prob. of x 2 1 1 1 1/36 3 1 2, 2 1 1.5 2/36 4 1 3, 3 1, 2 2 2 3/36 5 1 4, 4 1, 2 3, 3 2 2.5 4/36 6 1 5, 5 1, 2 4, 4 2, 3 3 3 5/36 7 1 6, 6 1, 2 5, 5 2, 3 4, 4 3 3.5 6/36 8 2 6, 6 2, 3 5, 5 3, 4 4 4 5/36 9 3 6, 6 3, 4 5, 5 4 4.5 4/36 10 4 6, 6 4, 5 5 5 3/36 11 5 6, 6 5 5.5 2/36 12 6 6 6 1/36 7.2 The sampling distribution of the sample mean can be generated by listing out all possible samples of size n, calculate each possible x , determine the probability of each possible x and generate the sampling distribution. Alternatively, the probabilities of each x can be generated by use of the binomial formula. a. When n = 5: Use the binomial formula for x = 0, x = 1, etc.: X P(X) x .07776 1 .25920 .2 2 .34560 .4 3 .23040 .6 4 .07680 .8 5 .01024 1.0 ( ) x E p = np = (5)(.4) = 2.0, 2 (1 ) p p p n  = = .4(.6) 5 = = .048, .2191 p = b. Using the result from part a: ( ) x E p = np = (100)(.4) = 40, 2 (1 ) p p p n  = = .4(.6) 100 = = .0024, .04899 p = 73 The response should note that there will be errors in taking a census of the entire population as well as errors in taking a sample. Improved accuracy can be achieved via sampling methods versus taking a complete census (see reference to Hogan, 90). By 99 Instructors Solutions Manual for Statistics for Business & Economics, 5 th Edition using sample information, we can make valid inferences about the entire population without the time and expense involved in taking a census. 74 a. E( X ) = x = 92. c. x = x n = 3.6 2 = 1.8 b. 2 x = 2 x n = ( 29 2 3.6 4 = 3.24 d. P(Z > 93 92 1.8 ) = P(Z > .56) = .2877 75 a. E( X ) = x = 1,200 c. x = x n = 400 3 = 133.33 b. 2 x = 2 x n = ( 29 2 400 9 = 17,778 d. P(Z> 1,050 1,200 133.33 ) = P(Z<1.13) =.1292 7.6 a. i) P(Z > 24 25 2 ) = P(Z < .5) = .3085. ii) P(Z < 24 25 2 4 ) = P(Z < 1) =.1587 iii) P(Z < 24 25 2 16 ) = P(Z < 2) = .0228 b. As the sample size increases, the standard error of the sampling distribution will decrease. That is, as the sample size increases, the sampling distribution of the sample means will clump up tighter around the true population mean. 7.7 a. P(Z > 110 115 25 100 ) = P(Z < 2) = .9772 b. P( 113 115 25 100 < Z < 117 115 25 100 ) = P(.8 < Z < .8) = .5762 c. P( 114 115 25 100 < Z < 116 115 25 100 ) = P(.4 < Z < .4) = .3108 d. $114,000  $116,000 e. Even with nonnormal populations, the sampling distribution of the sample means will be normal for sufficient sample n. Since n is 30, the sampling distribution of the sample means can assumed to be a normal distribution....
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This note was uploaded on 03/25/2009 for the course FIN FIN534 taught by Professor Stevenjordan during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 StevenJordan

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