Ch08 - Chapter8 IntervalEstimation LearningObjectives 1...

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Chapter 8 Interval Estimation Learning Objectives 1. Know how to construct and interpret an interval estimate of a population mean and / or a population proportion. 2. Understand and be able to compute the margin of error. 3. Learn about the t distribution and its use in constructing an interval estimate when σ is unknown for a population mean. 4. Be able to determine the size of a simple random sample necessary to estimate a population mean and/or a population proportion with a specified margin of error. 5. Know the definition of the following terms: confidence interval margin of error confidence coefficient degrees of freedom confidence level Solutions: 8 - 1
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Chapter 8 1. a. / 5/ 40 .79 x n σ σ = = = b. At 95%, z n σ / . ( / ) . = = 196 5 40 155 2. a. 32 ± 1.645 ( / ) 6 50 32 ± 1.4 or 30.6 to 33.4 b. 32 ± 1.96 ( / ) 6 50 32 ± 1.66 or 30.34 to 33.66 c. 32 ± 2.576 ( / ) 6 50 32 ± 2.19 or 29.81 to 34.19 3. a. 80 ± 1.96 ( / ) 15 60 80 ± 3.8 or 76.2 to 83.8 b. 80 ± 1.96 ( / ) 15 120 80 ± 2.68 or 77.32 to 82.68 c. Larger sample provides a smaller margin of error. 4. Sample mean 160 152 156 2 x + = = Margin of Error = 160 – 156 = 4 1.96( / ) 4 n σ = 1.96 / 4 1.96(15) / 4 7.35 n σ = = = n = (7.35) 2 = 54 5. a. 1.96 / 1.96(5/ 49) 1.40 n σ = = b. 24.80 ± 1.40 or 23.40 to 26.20 6. .025 ( / ) x z n σ ± 8.5 ± 1.96(3.5/ 300 ) 8.5 ± .4 or 8.1 to 8.9 7. .025 ( / ) 1.96(4000 / 60) 1012 z n σ = = A larger sample size would be needed to reduce the margin of error. Section 8.3 can be used to 8 - 2
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Interval Estimation show that the sample size would need to be increased to n = 246. 1.96(4000/ ) 500 n = Solving for n , shows n = 246 8. a. Since n is small, an assumption that the population is at least approximately normal is required. b. .025 ( / ) 1.96(5/ 10) 3.1 z n σ = = c. .005 ( / ) 2.576(5/ 10) 4.1 z n σ = = 9. x ± .025 z ( / ) n σ 3.37 ± 1.96 (.28/ 120) 3.37 ± .05 or 3.32 to 3.42 10. a. x z n ± α σ /2 119,155 ± 1.645 (30,000/ 80) 119,155 ± 5517 or $113,638 to $124,672 b. 119,155 ± 1.96 (30,000/ 80) 119,155 ± 6574 or $112,581 to $125,729 c. 119,155 ± 2.576 (30,000/ 80) 119,155 ± 8640 or $110,515 to $127,795 d. The confidence interval gets wider as we increase our confidence level. We need a wider interval to be more confident that it will contain the population mean. 11. a. .025 b. 1 - .10 = .90 c. .05 d. .01 e. 1 – 2(.025) = .95 f. 1 – 2(.05) = .90 12. a. 2.179 b. -1.676 c. 2.457 8 - 3
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Chapter 8 d. Use .05 column, -1.708 and 1.708 e. Use .025 column, -2.014 and 2.014 13. a. 80 10 8 i x x n Σ = = = b. i x ( ) i x x - 2 ( ) i x x - 10 0 0 8 -2 4 12 2 4 15 5 25 13 3 9 11 1 1 6 -4 16 5 -5 25 84 2 ( ) 84 3.464 1 7 i x x s n Σ - = = = - c. .025 ( / ) 2.365(3.464/ 8) 2.9 t s n = = d. .025 ( / ) x t s n ± 10 ± 2.9 or 7.1 to 12.9 14. / 2 ( / ) x t s n α ± df = 53 a. 22.5 ± 1.674 (4.4/ 54) 22.5 ± 1 or 21.5 to 23.5 b. 22.5 ± 2.006 (4.4/ 54) 22.5 ± 1.2 or 21.3 to 23.7 c. 22.5 ± 2.672 (4.4/ 54) 22.5 ± 1.6 or 20.9 to 24.1 d. As the confidence level increases, there is a larger margin of error and a wider confidence interval. 15. / 2 ( / ) x t s n α ± 90% confidence df = 64 t .05 = 1.669 19.5 ± 1.669 (5.2/ 65) 8 - 4
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Interval Estimation 19.5 ± 1.08 or 18.42 to 20.58 95% confidence df = 64 t .025 = 1.998 19.5 ± 1.998 (5.2/ 65) 19.5 ± 1.29 or 18.21 to 20.79 16. a.
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