EF 507 PS-Chapter 8 FALL 2008 1. The amount of material used in making a custom sail for a sailboat is normally distributed. For a random sample of 15 sails, you find that the mean amount material is 912 square feet, with a standard deviation of 64 square feet. Which of the following represents a 99% confidence interval for the population mean amount of material used in a custom sail? A) B) C) D) 912 49.2 912 42.6 912 44.3 912 46.8 ANSWER: A 2. Let X 1 , X 2 , X 3 , and X 4 be a random sample of observations from a population with mean and variance 2 . Consider the following estimator of : ˆ1 = 0.15 X 1 + 0.35 X 2 + 0.20 X 3 + 0.30 X 4 . What is the variance of ˆ1 ? A) 0.275 B) 0.275 2 C) 0.55 2 D) 0.55 ANSWER: C 3. Which of the following statements is false? A) An estimator ˆ that is a function of the sample data X1 , X 2 ,....., X n is called an unbiased estimator of the population parameter ˆ ; that is , E ˆ The statistic X is an unbiased estimator of the population mean The sample statistic s is an unbiased estimator of the population parameter when (n – 1) is used as a divisor in the calculation of the sample standard deviation D) None of the above ANSWER: C B) C) 4. In developing an interval estimate for a population mean, the population standard deviation was assumed to be 8. The interval estimate was 45.82 2.36. Had equaled 16, the interval estimate would be A) 45.82 4.72 B) 55.82 2.36 C) 45.85 9.44 D) 91.64 4.72 ANSWER: A 5. The lower limit of a 95% confidence interval for the population proportion P given a sample size n = 100 and sample proportion P̂ = 0.62 is equal to A) B) C) D) 0.715 0.699 0.525 0.540 1 ANSWER: C 6. In order to estimate the average daily down time, a manufacturer randomly samples 40 days of production records and found a mean of 51.576 minutes and standard deviation of 75.9 minutes. An 80% confidence interval is given by: A) 51.675 1.28(12.0) B) 51.675 1.28(75.9) C) 51.675 1.304(12.0) D) 51.675 1.304(8.7) ANSWER: C 7. The Daytona Beach Tourism Commission is interested in the average amount of money a typical college student spends per day during spring break. They survey 35 students and find that the mean spending is $63.57 with a standard deviation of $17.32. a) Develop a 95% confidence interval for the population mean daily spending. ANSWER: X tn1 , / 2 s / n = 63.57 2.032(17.32) / 5.92 = 63.57 35 = 63.57 5.95. Then, UCL = 69.52 and LCL = 57.62. b) Interpret the confidence level in 7 (a). ANSWER: If independent random samples of size 35 are repeatedly selected from the population and 95% confidence intervals for each of these samples are determined, then over a very large number of repeated trials, 955 of these intervals will contain the value of the true average amount of money a typical college student spends per day during spring break. 8. What level of confidence is associated with an interval of $58.62 to $68.52 for the population mean daily spending. ANSWER: t34, / 2 17.32 35 tn1 , / 2 s / n = – (68.52 58.62) / 2 = 4.95 4.95 t34, / 2 1.69 . Hence, / 2 = 0.05, or = 0.10. Then the confidence level is 90%. 9. In a recent survey of 600 adults, 16.4% indicated that they had fallen asleep in front of the television in the past month. Develop a 95% confidence interval for the population proportion. ANSWER: pˆ z / 2 pˆ 1 pˆ / n = 0.164 1.96(0.0151) = 0.164 0.0296. Hence, UCL = 0.1936 and LCL = 0.1344. 2 QUESTIONS 10 AND 11 ARE BASED ON THE FOLLOWING INFORMATION: A researcher is interested in determining the percentage of all households in the U.S. that have more than one home computer. In a survey of 492 households, 27% indicated that they own more than one home computer. 10. Develop a 90% confidence interval for the proportion of all households in the U.S. with more than one computer. ANSWER: n = 492, p̂ =0.27, z / 2 z0.05 =1.645 pˆ z / 2 pˆ 1 pˆ / n 0.27 1.645 0.27 0.73 / 492 0.27 0.033 . Hence, UCL= 0.303 and LCL = 0.237 11. The researcher reports a confidence interval of 0.2312 to 0.3096 but neglects to tell you the confidence level. What is the confidence level associated with this interval? ANSWER: Width= z / 2 pˆ 1 pˆ / n =0.3096-0.2312 = 0.0784 Then, 2 z / 2 0.27 0.76 / 492 = 0.0784 0.04 z / 2 = 0.0784 z / 2 =1.96. confidence level is 95%. Hence, / 2 = 0.025, or = 0.05. Then, the QUESTIONS 12 AND 13 ARE BASED ON THE FOLLOWING INFORMATION: Suppose that the amount of time teenagers spend on the internet is normally distributed with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean computed as 6.5 hours. 12. Determine the 95% confidence interval estimate of the population mean. ANSWER: x z / 2 / n 6.5 (1.96)(1.5/ 100) 6.5 0.294 6.206 < < 6.794 13. Interpret what the interval estimate in Question 102 tells you. ANSWER: If we repeatedly draw independent random samples of size 100 from the population of teenagers, and confidence interval for each of these samples are determined, then over a very large number of repeated trials, 95% of these confidence intervals will contain the value of the true population mean amount of time teenagers spend on the internet. QUESTIONS 14 THROUGH 17 ARE BASED ON THE FOLLOWING INFORMATION: A furniture mover calculates the actual weight as a proportion of estimated weight for a sample of 31 recent jobs. The sample mean is 1.13 and the sample standard deviation is 0.16. 14. Calculate a 95% confidence interval for the population mean using t tables. 3 ANSWER: Given: n = 31, x 1.13, and s = 0.16. To calculate a 95% confidence interval, we need the t table value for df = 30. With .05, this value is t.025,30 2.042. The confidence interval is x t / 2 s / n 1.13 (2.042)(0.16 / 31) 1.13 .059 or 1.071 1.189. 15. Assume that the population standard deviation is known to be 0.16. Calculate a 95% confidence interval for the population mean using the z- table. ANSWER: If is known to be 0.16, we may use the z table value z.025 1.96 (rather than the t table value) in the 95% confidence interval. x z / 2 / n 1.13 (1.96)(0.16 / 31) 1.13 .056 or 1.074 1.186. 4