# Chapter 9 Test Handout

Pages: 2 (711 words) Published: February 12, 2012
Chapter 9 Estimation and Confidence Intervals
30. A random sample of 85 group leaders, supervisors, and similar personnel revealed that a person spent an average 6.5 years on the job before being promoted. The population standard deviation was 1.7 years. Using the 0.95 degree of confidence, what is the confidence interval for the population mean? A) 6.99 and 7.99 B) 4.15 and 7.15 C) 6.14 and 6.86 D) 6.49 and 7.49 Answer: C 31. The mean weight of trucks traveling on a particular section of I-475 is not known. A state highway inspector needs an estimate of the mean. He selects a random sample of 49 trucks passing the weighing station and finds the mean is 15.8 tons. The population standard deviation is 3.8 tons. What is the 95 percent interval for the population mean? A) 14.7 and 16.9 B) 13.2 and 17.6 C) 10.0 and 20.0 D) 16.1 and 18.1 Answer: A 41. Suppose 1,600 of 2,000 registered voters sampled said they planned to vote for the Republican candidate for president. Using the 0.95 degree of confidence, what is the interval estimate for the population proportion (to the nearest tenth of a percent)? A) 78.2% to 81.8% B) 69.2% to 86.4% C) 76.5% to 83.5% D) 77.7% to 82.3% Answer: A Use the following to answer questions 81-83: A student wanted to quickly construct a 95% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a standard deviation of 1.5 years. 81. What is the best point estimate for the population mean? A) 2.1 years B) 1.5 years C) 19.1 years D) 9 years Answer: C 82. What is the 95% confidence interval for the population mean? A) [0.97, 3.27] B) [15.64, 22.56] C) [17.97, 20.23] D) [17.95, 20.25] Answer: D

Statistical Techniques in Business & Economics, Lind/Marchal/Wathen, 13/e

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83. What is the 99% confidence interval for the population mean? A) [17.42, 20.78] B) [17.48, 20.72] C) [14.23, 23.98] D) [0.44, 3.80] Answer: A Use the following to answer...

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