Unformatted text preview: percent
confidence interval for the true proportion favoring new laws.
A. [.570 .630]
B. [.576 .624]
C. [.552 .648]
D. [.566 .634] 113. We want to estimate with 99 percent confidence the percentage of buyers of cars who are
under 30 years of age. A margin of error of 5 percentage points is desired. What sample size
is needed? In an earlier sample we found a 99 percent confidence interval of buyers under 30
years of age to be [.18 .27].
A. 104
B. 664
C. 392
D. 523 114. A cable TV company wants to estimate the percentage of cable boxes in use during an
evening hour. An approximation is 20 percent. They want the estimate to be at the 90 percent
confidence level and within 2 percent of the actual proportion. What sample size is needed?
A. 22
B. 1692
C. 1537
D. 1083 115. An insurance company estimates 45 percent of its claims have errors. The insurance
company wants to estimate with 99 percent confidence the proportion of claims with errors.
What sample size is needed if they wish to be within 5 percent of the actual?
A. 664
B. 657
C. 163
D. 1084 1641 Chapter 01  An Introduction to Business Statistics 116. In a randomly selected group of 650 automobile deaths, 180 were alcohol related.
Construct a 95 percent confidence interval for the true proportion of all automobile accidents
caused by alcohol.
A. [.243 .311]
B. [.262 .292]
C. [.259 .294]
D. [.263 .291] 117. You want to estimate the proportion of customers who are satisfied with their
supermarket at α = .10 and within .025 of the true value. It has been estimated that p = .85.
How large of a sample is needed?
A. 1083
B. 553
C. 71
D. 336 118. Find a 99 percent confidence interval for µ if
= 98.6, s = 2, and n = 5. Assume that
the sample is randomly selected from a normally distributed population.
A. [95.69 101.51]
B. [95.25 101.95]
C. [94.48 102.72]
D. [89.39 107.81] 119. The customer service manager for the XYZ Fastener Manufacturing Company examined
60 vouchers and found nine vouchers containing errors. Find a 98 percent confidence interval
for the proportion of vouchers with errors.
A. [.147 .153]
B. [.080 .220]
C. [.043 .257]
D. [.112 .188] 1642 Chapter 01  An Introduction to Business Statistics 120. The weight of a product is measured in pounds. A sample of 50 units is taken from a
batch. The sample yielded the following results:
= 75 lbs., but we know that σ2 = 100 lbs.
Calculate a 90 percent confidence interval for
.
A. [73.19 76.81]
B. [51.74 98.26]
C. [72.67 77.33]
D. [67.50 82.50] 121. The weight of a product is measured in pounds. A sample of 50 units is taken from a
batch. The sample yielded the following results:
= 75 lbs., but we know that σ2 = 100 lbs.
Calculate a 95 percent confidence interval for
.
A. [71.25 78.75]
B. [72.23 77.77]
C. [47.28 102.72]
D. [72.67 77.33] 122. The weight of a product is measured in pounds. A sample of 50 units is taken from a
batch. The sample yielded the following results:
= 75 lbs., but we know that σ2 = 100 lbs.
Calculate a 99 percent confidence interval for
.
A. [71.36 78.64]
B. [74.25 75.75]
C. [38.58 111.42]
D. [71.71 78.29] 123. The 95% confidence interval for the average weight of a product is from 72.23 lbs. to
77.77 lbs. Can we conclude that
= 77 using a 95 percent confidence interval?
A. Yes
B. No 1643 Chapter 01  An Introduction to Business Statistics 124. The 99% confidence interval for the average weight of a product is from 71.36 lbs. to
78.64 lbs. Can we conclude that
is equal to 71 using a 99 percent confidence interval?
Briefly explain.
A. Yes
B. No 125. A sample of 2,000 people yielded
proportion?
A. .0056
B. .0112
C. .00012
D. .0161 = .52. What is the variance of the population 126. A sample of 2,000 people yielded
for p.
A. [.506 .534]
B. [.468 .572]
C. [.511 .529]
D. [.502 .538] = .52. Calculate a 90 percent confidence interval 127. A sample of 2,000 people yielded
for p.
A. [.509 .531]
B. [.494 .546]
C. [.498 .542]
D. [.502 .538] = .52. Calculate a 95 percent confidence interval 128. A sample of 2,000 people yielded
for p.
A. [.515 .525]
B. [.494 .546]
C. [.506 .534]
D. [.491 .549] = .52. Calculate a 99 percent confidence interval 1644 Chapter 01  An Introduction to Business Statistics 129. A random sample of size 15 is taken from a population assumed to be normal, and
1.2 and s = 0.6. Calculate a 90 percent confidence interval for
.
A. [.927 1.473]
B. [.848 1.552]
C. [.992 1.408]
D. [1.045 1.355] = 130. A random sample of size 15 is taken from a population assumed to be normal, and
1.2 and s = .6. Calculate a 98 percent confidence interval for
.
A. [.675 1.725]
B. [.794 1.606]
C. [.876 1.524]
D. [.882 1.518] = 131. A random sample of size 15 is taken from a population assumed to be normal, and
1.2 and the sample variance is .36. Calculate a 99 percent confidence interval for
.
A. [.923 1.477]
B. [.794 1.607]
C. [.801 1.600]
D. [.739 1.661] = 132. If the 95% confidence interval for a mean is from .771 to 1.629, can we conclude that
= .5, using a 95 percent confidence interval?
A. Y...
View
Full
Document
This document was uploaded on 01/20/2014.
 Winter '14

Click to edit the document details