This preview shows page 1. Sign up to view the full content.
Unformatted text preview: k. For 50
randomly selected adult subjects, the sample mean is 16.40 minutes and the sample standard
deviation is 4.00 minutes. Construct the 95% confidence interval for the mean time required
by all adults to learn the task.
A. [8.56 24.24]
B. [15.47 17.33]
C. [16.24 16.56]
D. [15.29 17.51]
E. [17.12 48.48] AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Topic: Confidence interval for population mean σ unknown 49. Research has been conducted that studies the role that the age of workers has in
determining the hours per month spent on personal tasks. A sample of 1,686 adults were observed for one month. The data are:
Construct a 98% confidence interval for the mean hours spent on personal tasks for 2544
year olds.
A. [3.96 4.12]
B. [3.97 4.11]
C. [3.98 4.10]
D. [2.16 5.92]
E. [3.95 4.13] AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Topic: Confidence interval for population mean σ unknown 1667 Chapter 01  An Introduction to Business Statistics 50. 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. 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] AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Topic: Confidence interval for population mean σ unknown 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 AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Topic: Sample size determination for population mean 1668 Chapter 01  An Introduction to Business Statistics 52. In a random sample of 651 computer scientists who subscribed to a webbased 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] AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Topic: Confidence interval for population mean σ unknown 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] AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Topic: Confidence interval for population mean σ unknown 1669 Chapter 01  An Introduction to Business Statistics 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
nonheart 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 nonheart 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] AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Topic: Confidence interval for population mean σ unknown 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
nonheart 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 AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Topic: Sample size determination for population mean 1670 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...
View
Full
Document
This document was uploaded on 01/20/2014.
 Winter '14

Click to edit the document details