Chapter7

# Chapter7 - CHAPTER 7 Section 7.1 1 a z 2 = 2.81 implies...

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219 CHAPTER 7 Section 7.1 1. a. 81 . 2 2 = a z implies that ( 29 0025 . 81 . 2 1 2 = Φ - = a , so 005 . = a and the confidence level is ( 29 % 5 . 99 % 1 100 = - a . b. 44 . 1 2 = a z for ( 29 [ ] 15 . 44 . 1 1 2 = Φ - = a , and ( 29 % 85 % 1 100 = - a . c. 99.7% implies that 003 . = a , 0015 . 2 = a , and 96 . 2 0015 . = z . (Look for cumulative area .9985 in the main body of table A.3, the Z table.) d. 75% implies 25 . = a , 125 . 2 = a , and 15 . 1 125 . = z . 2. a. The sample mean is the center of the interval, so 115 2 6 . 115 4 . 114 = + = x . b. The interval (114.4, 115.6) has the 90% confidence level. The higher confidence level will produce a wider interval. 3. a. A 90% confidence interval will be narrower (See 2b, above) Also, the z critical value for a 90% confidence level is 1.645, smaller than the z of 1.96 for the 95% confidence level, thus producing a narrower interval. b. Not a correct statement. Once and interval has been created from a sample, the mean m is either enclosed by it, or not. The 95% confidence is in the general procedure, for repeated sampling. c. Not a correct statement. The interval is an estimate for the population mean, not a boundary for population values. d. Not a correct statement. In theory, if the process were repeated an infinite number of times, 95% of the intervals would contain the population mean m .

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Chapter 7: Statistical Intervals Based on a Single Sample 220 4. a. ( 29 ( 29 5 . 59 , 1 . 57 18 . 1 3 . 58 25 3 96 . 1 3 . 58 = ± = ± b. ( 29 ( 29 9 . 58 , 7 . 57 59 . 3 . 58 100 3 96 . 1 3 . 58 = ± = ± c. ( 29 ( 29 1 . 59 , 5 . 57 77 . 3 . 58 100 3 58 . 2 3 . 58 = ± = ± d. 82% confidence 09 . 18 . 82 . 1 2 = = = - a a a , so 34 . 1 09 . 2 = = z z a and the interval is ( 29 ( 29 7 . 58 , 9 . 57 100 3 34 . 1 3 . 58 = ± . e. ( 29 62 . 239 1 3 58 . 2 2 2 = = n so n = 240. 5. a. ( 29( 29 = ± = ± 33 . 85 . 4 20 75 . 96 . 1 85 . 4 (4.52, 5.18). b. 33 . 2 01 . 2 02 . 2 = = = z z z a , so the interval is ( 29( 29 = ± 16 75 . 33 . 2 56 . 4 (4.12, 5.00). c. ( 29( 29 02 . 54 40 . 75 . 96 . 1 2 2 = = n , so n = 55. d. ( 29( 29 61 . 93 2 . 75 . 58 . 2 2 2 = = n , so n = 94. 6. a. ( 29( 29 = ± = ± 9 . 32 8439 25 100 645 . 1 8439 (8406.1, 8471.9). b. 04 . 08 . 92 . 1 2 = = = - a a a so 75 . 1 04 . 2 = = z z a
Chapter 7: Statistical Intervals Based on a Single Sample 221 7. If n z L s a 2 2 = and we increase the sample size by a factor of 4, the new length is 2 2 1 2 4 2 2 2 L n z n z L = = = s s a a . Thus halving the length requires n to be increased fourfold. If n n 25 = , then 5 L L = , so the length is decreased by a factor of 5.

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Chapter7 - CHAPTER 7 Section 7.1 1 a z 2 = 2.81 implies...

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