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# Ch 08 Answers - Chapter 8 Sampling Distributions and...

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Chapter 8 Sampling Distributions and Estimation 8.1 a. n σ = 32 4 = 16 b. n σ = 32 16 = 8 c. n σ = 32 64 = 4 8.2 a. 1.96 n σ μ ± = 12 200 1.960 36 ± or (196.08, 203.92). b. 1.96 n σ μ ± = 15 1000 1.960 9 ± or (990.2, 1009.80). c. 1.96 n σ μ ± = 1 50 1.960 25 ± or (49.608, 50.392). 8.3 a. 1.96 μ σ ± = 4.035 1.96*0.005 ± or (4.0252, 4.0448). b. 1.96 n σ μ ± = 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 don’t 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 don’t 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 σ ± = 4 14 1.645 5 ± or (11.057, 16.943). b. 2.576 x n σ ± = 5 37 2.576 15 ± or (33.675, 40.325). 68

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c. 1.96 x n σ ± = 15 121 1.96 25 ± or (115.12, 126.88). 8.6 Exam 1: 1.96 x n σ ± = 7 75 1.96 10 ± or (70.661, 79.339). Exam 2: 1.96 x n σ ± = 7 79 1.96 10 ± or (74.661, 83.339). Exam 3: 1.96 x n σ ± = 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 σ ± = 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) = 1.984 All are fairly close to 1.96. 8.11 a. s x t n ± = 27.79 45.66 2.0860 21 ± or (33.01, 58.31). b. The confidence interval could be narrower increasing the size of the sample or decreasing the confidence level.
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Ch 08 Answers - Chapter 8 Sampling Distributions and...

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