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Unformatted text preview: Chapter 8 Sampling Distributions and Estimation 8.1 a. n = 32 4 = 1 6 b. n = 32 16 = 8 c. n = 32 64 = 4 8.2 a. 1.96 n = 12 200 1.960 36 or (196.08, 203.92). b. 1.96 n = 15 1000 1.960 9 or (990.2, 1009.80). c. 1.96 n = 1 50 1.960 25 or (49.608, 50.392). 8.3 a. 1.96 = 4.035 1.96*0.005 or (4.0252, 4.0448). b. 1.96 n = 0.005 4.035 1.96 25 or (4.03304, 4.03696). c. In either case, we would conclude that our sample came from a population that did not have a population mean equal to 4.035. 8.4 a. 1. No, for n = 1 the 100 samples dont represent a normal distribution 2.The distribution of the sample means becomes more normally distributed as n increases. 3. The standard error becomes closer to that predicted by the CLT the larger the sample becomes. 4. This demonstration reveals that if numerous samples were taken and analyzed we can confirm the CLT. In the real word, based on our notion of the true mean, we can assess this. We can generate the 95% range and determine if our values are within this range or not. Also, recognize that there is a low probability of this single range not being representative. b. 1. No, for n = 1 the 100 samples dont represent a normal distribution 2.The distribution of the sample means becomes more normally distributed as n increases. The standard error becomes closer to that predicted by the CLT the larger the sample becomes. 4. This demonstration reveals that if numerous samples taken and analyzed we can confirm the CLT. In the real word, based on our notion of the true mean, we can assess this. We can generate the 95% range and determine if our values are within this range or not. Also, recognize that there is a low probability of this single range not being representative. 8.5 a. 1.645 x n = 4 14 1.645 5 or (11.057, 16.943). b. 2.576 x n = 5 37 2.576 15 or (33.675, 40.325). 68 c. 1.96 x n = 15 121 1.96 25 or (115.12, 126.88). 8.6 Exam 1: 1.96 x n = 7 75 1.96 10 or (70.661, 79.339). Exam 2: 1.96 x n = 7 79 1.96 10 or (74.661, 83.339). Exam 3: 1.96 x n = 7 65 1.96 10 or (60.661, 69.339) The first two confidence intervals overlap. This suggests that the first two exams had the same population mean. 8.7 1.96 x n = 0.005 2.475 1.96 15 or (2.4725, 2.4775). 8.8 a. s x t n = 3 24 1.9432 7 or (21.797, 26.203). b. s x t n = 6 42 2.8982 18 or (37.901, 46.099). c. s x t n = 14 119 2.0518 28 or (113.571, 124.429). Note: t values are found using the Excel formula =tinv( (1 cc),n 1) where cc is the confidence coefficient. For a. this, this would be = tinv((1.90), 6) 8.9 a. Appendix D = 2.262, Excel = tinv(.05, 9) = 2.2622 b. Appendix D = 2.602, Excel = tinv(.02, 15) = 2.6025 c. Appendix D = 1.678 ,Excel = tinv(.10, 47) =1.6779 8.10 a. Appendix D = 2.021, Excel = tinv(.05, 40) = 2.0211 b. Appendix D = 1.990, Excel = tinv(.05, 80) = 1.9901 c. Appendix D = 1.984, Excel = tinv(.05, 100) = 1.984c....
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This note was uploaded on 12/13/2010 for the course LEEDS BCOR 1020 taught by Professor Heatheradams during the Spring '08 term at Colorado.
 Spring '08
 HEATHERADAMS

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