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Unformatted text preview: nes to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. 225 flight records are randomly selected and the number of unoccupied seats is noted with a sample mean of 11.6 seats and a standard deviation of 4.1 seats. Calculate a 90% confidence interval for , the mean number of unoccupied seats per flight during the past year. A. [4.86 18.34] B. [11.25 11.95] C. [11.57 11.63] D. [11.15 12.05] E. [11.30 12.20] 51. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. 225 flight records are randomly selected and the number of unoccupied seats is noted with a sample mean of 11.6 seats and a standard deviation of 4.1 seats. How many flights should we select if we wish to estimate µ to within 2 seats and be 95% confident? A. 130 B. 65 C. 33 D. 17 E. 12 52. In a random sample of 651 computer scientists who subscribed to a web-based daily news update, it was found that the average salary was \$46,816 with a standard deviation of \$12,557. Calculate a 91% confidence interval for the mean salary of computer scientists. A. [\$25,469 \$68,163] B. [\$46,592 \$47,040] C. [\$46,157 \$47,475] D. [\$46,783 \$46,849] E. [\$45,979 \$47,653] 1-624 Chapter 01 - An Introduction to Business Statistics 53. At the end of 1990, 1991, and 1992 the average prices of a share of stock in a money market portfolio were \$34.83, \$34.65 and \$31.26 respectively. To investigate the average share price at the end of 1993, a random sample of 30 stocks was drawn and their closing prices on the last trading day of 1993 were observed with a mean of 33.583 and a standard deviation of 19.149. Estimate the average price of a share of stock in the portfolio at the end of 1993 with a 90% confidence interval. A. [27.832 39.334] B. [26.732 40.434] C. [32.514 34.651] D. [32.533 34.633] E. [32.269 34.897] 54. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. In 1996, the average LOS for nonheart patient was 4.6 days. A random sample of 20 hospitals in one state had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days. Calculate a 95% confidence interval for the population mean LOS for non-heart patients in the state's hospitals in 2000. A. [3.24 4.36] B. [3.67 3.93] C. [3.34 4.26] D. [3.38 4.22] E. [3.27 4.33] 55. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. In 1996, the average LOS for nonheart patient was 4.6 days. A random sample of 20 hospitals in one state had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days. How large a sample of hospitals would we need to be 99% confident that the sample mean is within 0.5 days of the population mean? A. 3 B. 7 C. 32 D. 48 E. 96 1-625 Chapter 01 - An Introduction to Business Statistics 56. The coffee/soup machine at the local bus station is supposed to fill cups with 6 ounces of soup. Ten cups of soup are brought with results of a mean of 5.93 ounces and a standard deviation of 0.13 ounces. Construct a 99% confidence interval for the true machine-fill amount. A. [5.888 5.972] B. [5.814 6.046] C. [5.716 6.144] D. [5.824 6.036] E. [5.796 6.064] 57. The coffee/soup machine at the local bus station is supposed to fill cups with 6 ounces of soup. Ten cups of soup are brought with results of a mean of 5.93 ounces and a standard deviation of 0.13 ounces. How large a sample of soups would we need to be 95% confident that the sample mean is within 0.03 ounces of the population mean? A. 97 B. 96 C. 73 D. 62 E. 10 58. A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of 0.996 ounces with a standard deviation of 0.004 ounces. Construct a 98% confidence interval for the mean weight of all such candy. A. [0.9645 1.0275] B. [0.9956 0.9964] C. [0.9860 1.0060] D. [0.9940 0.9980] E. [0.9942 0.9978] 1-626 Chapter 01 - An Introduction to Business Statistics 59. A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of 0.996 ounces with a standard deviation of 0.004 ounces. How many pieces of candy must we sample if we want to be 99% confident that the sample mean is within 0.001 ounces of the true mean? A. 126 B. 124 C. 107 D. 12 E. 6 60. An environmental group at a local college is conducting independent tests to determine the distance a particular make of automobile will travel while consuming only 1 gallon of gas. A sample of five cars is tested and a mean of 28.2 miles is obtained. Assuming that the standard deviation is 2.7 miles, find the 95% confidence interval for the mean distance traveled by all such cars using 1 gallon of gas. A. [26.16 30.24] B. [20.70 35.70] C. [24.85 31.55] D. [26.70 29.70] E. [25.83 30.57] 61. An environmental group at a local co...
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## This document was uploaded on 01/20/2014.

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