c The condition for using normal distribution is np 5 and n1 p 5 Therefore the

C the condition for using normal distribution is np 5

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c. The condition for using normal distribution is: np > 5 and n(1 – p) > 5 Therefore, the condition is satisfied.
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200 * 0.25 = 50 200 0.75 = 150 4. a. Construct a 95% confidence interval: 23.491 to 27.509 25.5 ± 1.96(10.25 / √100) = 25.5 ± 2.009 = 23.491 to 27.509 b. Margin of error for 90% confidence interval: 1.686 = 1.645 (10.25 / √100) = 1.686 c. Construct a 95% confidence interval for a random sample of 150: 23.86 to 27.14 25.5 ± 1.96(10.25 / √150) = 25.5 ± 1.640 = 23.86 to 27.14 d. The length of the confidence interval decreases as the sample size increases. 5. a. Construct a 99% confidence interval: 19.307 to 27.6927 23.5 ± 2.662 (12.2 / √60) = 23.5 ± 4.1927 b. Margin for error on a 95% confidence interval: 3.15 = 2.001(12.2 / √60) = 3.15 6. a. With a 95% confidence interval what is the margin for error? 0.5118 2.064(1.24 / √25) = 0.5118 b. What is the 95% confidence interval estimate of the population mean score? 3.238 to 4.261 3.75 ± 2.064 (1.24 / √25) = 3.75 ± 0.5118 = 95% confidence interval. 7. a. Approximate mean = range / 4 = (55,000 – 35,000) / 4 = 5000 The planning value for the population standard deviation is 5000. b. n = (z α /2) 2 ( σ) 2 / E 2 = (1.96) 2 (5000) 2 / 1000 2 = 96.04 The sample size should contain 96.04 or 97 university graduates if the margin for error is 100 c. n = (1.96) 2 (5000) 2 / (500) 2 = 384.16
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The sample size should contain 384.16 or 385 university graduates if the margin for effort is 500 8.a. Construct a 95% confidence interval: (0.4912, 0.5888) p ± 1.96 √p(1-p) / √n = 0.54 ± 1.96 √0.54(0.46) / √400 = 0.048842948 0.54 + 0.048842948 = 0.5888 0.54 – 0.048842948 = 0.4912 (0.4912, 0.5888) b. What is the margin of error for the 90% confidence interval? 0.0409932 Margin of Error = / 2 p(1-p) / √n = 1.645 √(0.54)(0.46) / √400 = 0.0409932 c. Required sample size if desired margin of error is 3%, at the 95% confidence interval: 1061 (1.96 2 * 0.54(1 – 0.54) / 0.03 2 = 3.8416 * (0.54)(0.46) / 0.0009 = 1060.2816 9. a. Construct a 95% confidence interval to estimate the population proportion: (0.1253, 0.1747) p = x / n = 120 / 800 = 0.15 p ± 1.96 √p(1-p) / √n 0.15 ± 1.96 √(0.15*0.85) / √800 = 0.024743787 0.15 + 0.024743787 = 0.1747 0.15 - 0.024743787 = 0.1253 (0.1253, 0.1747) b. What is the margin of error of the estimate at the 99% confidence interval? 0.03252 E = 2.576* √(0.15*0.85) / √800 = 0.03252406 10. How large should the sample size be?
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