ESBE4eAISEsm08 - Chapter 8 Interval Estimation Learning...

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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 level of precision. 5. Know the definition of the following terms: confidence interval margin of error confidence coefficient degrees of freedom confidence level 8 - 1 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher.
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Chapter 8 Solutions: 1. a. /5 / 4 0 . x n σσ == = 7 9 b. At 95%, zn σ /. ( / ) . == 196 5 40 155 2. a. 32 ± 1.645 (/ ) 65 0 32 ± 1.4 or 30.6 to 33.4 b. 32 ± 1.96 ) 0 32 ± 1.66 or 30.34 to 33.66 c. 32 ± 2.576 ) 0 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 = 1.96 / 4 1.96(15)/ 4 7.35 n = n = (7.35) 2 = 54 5. a. 1.96 / 1.96(5/ 49) 1.40 n b. 24.80 ± 1.40 or 23.40 to 26.20 6. .025 (/ ) x ± 8.5 ± 1.96(3.5/ 300 ) 8.5 ± .4 or 8.1 to 8.9 8 - 2 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher.
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Interval Estimation 7. .025 ( / ) 1.96(4000/ 60) 1012 zn σ == A larger sample size would be needed to reduce the margin of error. Section 8.3 can be used to 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 c. .005 (/ )2 . 5 7 6 ( 5 /1 0 )4 . 1 9. x ± .025 z (/ ) n 3.37 ± 1.96 (.28/ 120) 3.37 ± .05 or 3.32 to 3.42 10. a. xz n ± α /2 119,155 ± 1.645 (30,000/ 80) 119,155 ± 5517 or $113,638 to $124,672 b. 119,155 ± 1.96 119,155 ± 6574 or $112,581 to $125,729 c. 119,155 ± 2.576 119,155 ± 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 8 - 3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher.
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Chapter 8 12. a. 2.179 b. -1.676 c. 2.457 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 17 i xx s n Σ− == = c. .025 ( / ) 2.365(3.464/ 8) 2.9 tsn d. .025 (/ ) x ± 10 ± 2.9 or 7.1 to 12.9 14. /2 ) x 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 22.5 ± 1.2 or 21.3 to 23.7 c. 22.5 ± 2.672 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.
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ESBE4eAISEsm08 - Chapter 8 Interval Estimation Learning...

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