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Sampling Distributions 187
CHAPTER 7: SAMPLING AND SAMPLING
DISTRIBUTIONS
1. Sampling distributions describe the distribution of
a) parameters.
b) statistics.
c) both parameters and statistics.
d) neither parameters nor statistics.
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: statistics, sampling distribution
2. The standard error of the mean
a) is never larger than the standard deviation of the population.
b) decreases as the sample size increases.
c) measures the variability of the mean from sample to sample.
d) All of the above.
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: standard error, mean
3. The Central Limit Theorem is important in statistics because
a) for a large n, it says the population is approximately normal.
b) for any population, it says the sampling distribution of the sample mean is approximately
normal, regardless of the sample size.
c) for a large n, it says the sampling distribution of the sample mean is approximately
normal, regardless of the shape of the population.
d) for any sized sample, it says the sampling distribution of the sample mean is
approximately normal.
ANSWER:
c
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: central limit theorem
4. If the expected value of a sample statistic is equal to the parameter it is estimating, then we call
that sample statistic
a) unbiased.
b) minimum variance.
c) biased.
d) random.
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
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Sampling Distributions
KEYWORDS: unbiased
5. For air travelers, one of the biggest complaints is of the waiting time between when the airplane
taxis away from the terminal until the flight takes off. This waiting time is known to have a
skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes.
Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the
mean waiting time between when the airplane taxis away from the terminal until the flight takes
off for these 100 flights.
a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes.
b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes.
c) Distribution is approximately normal with mean = 10 minutes and standard error = 0.8
minutes.
d) Distribution is approximately normal with mean = 10 minutes and standard error = 8
minutes.
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: central limit theorem
6. Which of the following statements about the sampling distribution of the sample mean is
incorrect?
a) The sampling distribution of the sample mean is approximately normal whenever the
sample size is sufficiently large ( n  30 ).
b) The sampling distribution of the sample mean is generated by repeatedly taking samples
of size n and computing the sample means.
c) The mean of the sampling distribution of the sample mean is equal to  .
d) The standard deviation of the sampling distribution of the sample mean is equal to  .
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: sampling distribution, properties
7. Which of the following is true about the sampling distribution of the sample mean?
a) The mean of the sampling distribution is always  .
b) The standard deviation of the sampling distribution is always  .
c) The shape of the sampling distribution is always approximately normal.
d) All of the above are true.
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, properties
8. True or False: The amount of time it takes to complete an examination has a skewed-left
distribution with a mean of 65 minutes and a standard deviation of 8 minutes. If 64 students were
randomly sampled, the probability that the sample mean of the sampled students exceeds 71
minutes is approximately 0.
ANSWER:
Sampling Distributions 189
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, central limit theorem
9. Suppose the ages of students in Statistics 101 follow a skewed-right distribution with a mean of
23 years and a standard deviation of 3 years. If we randomly sampled 100 students, which of the
following statements about the sampling distribution of the sample mean age is incorrect?
a) The mean of the sampling distribution is equal to 23 years.
b) The standard deviation of the sampling distribution is equal to 3 years.
c) The shape of the sampling distribution is approximately normal.
d) The standard error of the sampling distribution is equal to 0.3 years.
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: sampling distribution, central limit theorem
10. Why is the Central Limit Theorem so important to the study of sampling distributions?
a) It allows us to disregard the size of the sample selected when the population is not
normal.
b) It allows us to disregard the shape of the sampling distribution when the size of the
population is large.
c) It allows us to disregard the size of the population we are sampling from.
d) It allows us to disregard the shape of the population when n is large.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: central limit theorem
11. A sample that does not provide a good representation of the population from which it was
collected is referred to as a(n)
sample.
ANSWER:
biased
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: unbiased
12. True or False: The Central Limit Theorem is considered powerful in statistics because it works
for any population distribution provided the sample size is sufficiently large and the population
mean and standard deviation are known.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: central limit theorem
13. Suppose a sample of n = 50 items is drawn from a population of manufactured products and the
weight, X, of each item is recorded. Prior experience has shown that the weight has a probability
distribution with  = 6 ounces and  = 2.5 ounces. Which of the following is true about the
sampling distribution of the sample mean if a sample of size 15 is selected?
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Sampling Distributions
a)
b)
c)
d)
The mean of the sampling distribution is 6 ounces.
The standard deviation of the sampling distribution is 2.5 ounces.
The shape of the sample distribution is approximately normal.
All of the above are correct.
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, unbiased
14. The average score of all pro golfers for a particular course has a mean of 70 and a standard
deviation of 3.0. Suppose 36 golfers played the course today. Find the probability that the
average score of the 36 golfers exceeded 71.
ANSWER:
0.0228
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
15. The distribution of the number of loaves of bread sold per day by a large bakery over the past 5
years has a mean of 7,750 and a standard deviation of 145 loaves. Suppose a random sample of n
= 40 days has been selected. What is the approximate probability that the average number of
loaves sold in the sampled days exceeds 7,895 loaves?
ANSWER:
Approximately 0
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
16. Sales prices of baseball cards from the 1960s are known to possess a skewed-right distribution
with a mean sale price of $5.25 and a standard deviation of $2.80. Suppose a random sample of
100 cards from the 1960s is selected. Describe the sampling distribution for the sample mean sale
price of the selected cards.
a) skewed-right with a mean of $5.25 and a standard error of $2.80
b) normal with a mean of $5.25 and a standard error of $0.28
c) skewed-right with a mean of $5.25 and a standard error of $0.28
d) normal with a mean of $5.25 and a standard error of $2.80
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: sampling distribution, central limit theorem
17. Major league baseball salaries averaged $1.5 million with a standard deviation of $0.8 million in
1994. Suppose a sample of 100 major league players was taken. Find the approximate probability
that the average salary of the 100 players exceeded $1 million.
a) approximately 0
b) 0.2357
c) 0.7357
d) approximately 1
Sampling Distributions 191
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
18. At a computer manufacturing company, the actual size of computer chips is normally distributed
with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12
computer chips is taken. What is the standard error for the sample mean?
a) 0.029
b) 0.050
c) 0.091
d) 0.120
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: standard error, mean
19. At a computer manufacturing company, the actual size of computer chips is normally distributed
with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12
computer chips is taken. What is the probability that the sample mean will be between 0.99 and
1.01 centimeters?
ANSWER:
0.2710 using Excel or 0.2736 using Table E.2
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
20. At a computer manufacturing company, the actual size of computer chips is normally distributed
with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12
computer chips is taken. What is the probability that the sample mean will be below 0.95
centimeters?
ANSWER:
0.0416 using Excel or 0.0418 using Table E.2
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
21. At a computer manufacturing company, the actual size of computer chips is normally distributed
with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12
computer chips is taken. Above what value do 2.5% of the sample means fall?
ANSWER:
1.057
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, value
22. The owner of a fish market has an assistant who has determined that the weights of catfish are
normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample
of 16 fish is taken, what would the standard error of the mean weight equal?
a) 0.003
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Sampling Distributions
b) 0.050
c) 0.200
d) 0.800
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: standard error, mean
23. The owner of a fish market has an assistant who has determined that the weights of catfish are
normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample
of 25 fish yields a mean of 3.6 pounds, what is the Z-score for this observation?
a) 18.750
b) 2.500
c) 1.875
d) 0.750
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean
24. The owner of a fish market has an assistant who has determined that the weights of catfish are
normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample
of 64 fish yields a mean of 3.4 pounds, what is probability of obtaining a sample mean this large
or larger?
a) 0.0001
b) 0.0013
c) 0.0228
d) 0.4987
ANSWER:
c
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
25. The owner of a fish market has an assistant who has determined that the weights of catfish are
normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. What
percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds?
a) 84%
b) 67%
c) 29%
d) 16%
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
Sampling Distributions 193
26. The use of the finite population correction factor when sampling without replacement from finite
populations will
a) increase the standard error of the mean.
b) not affect the standard error of the mean.
c) reduce the standard error of the mean.
d) only affect the proportion, not the mean.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: finite population correction
27. For sample size 16, the sampling distribution of the mean will be approximately normally
distributed
a) regardless of the shape of the population.
b) if the shape of the population is symmetrical.
c) if the sample standard deviation is known.
d) if the sample is normally distributed.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, central limit theorem
28. The standard error of the mean for a sample of 100 is 30. In order to cut the standard error of the
mean to 15, we would
a) increase the sample size to 200.
b) increase the sample size to 400.
c) decrease the sample size to 50.
d) decrease the sample to 25.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: standard error, mean
29. Which of the following is true regarding the sampling distribution of the mean for a large sample
size?
a) It has the same shape, mean, and standard deviation as the population.
b) It has a normal distribution with the same mean and standard deviation as the population.
c) It has the same shape and mean as the population, but has a smaller standard deviation.
d) It has a normal distribution with the same mean as the population but with a smaller
standard deviation.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, central limit theorem
30. For sample sizes greater than 30, the sampling distribution of the mean will be approximately
normally distributed
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Sampling Distributions
a)
b)
c)
d)
regardless of the shape of the population.
only if the shape of the population is symmetrical.
only if the standard deviation of the samples are known.
only if the population is normally distributed.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, central limit theorem
31. For sample size 1, the sampling distribution of the mean will be normally distributed
a) regardless of the shape of the population.
b) only if the shape of the population is symmetrical.
c) only if the population values are positive.
d) only if the population is normally distributed.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, central limit theorem
32. The standard error of the population proportion will become larger
a) as population proportion approaches 0.
b) as population proportion approaches 0.50.
c) as population proportion approaches 1.00.
d) as the sample size increases.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: standard error, proportion
33. True or False: As the sample size increases, the standard error of the mean increases.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: standard error, mean
34. True or False: If the population distribution is symmetric, the sampling distribution of the mean
can be approximated by the normal distribution if the samples contain 15 observations.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, central limit theorem
35. True or False: If the population distribution is unknown, in most cases the sampling distribution
of the mean can be approximated by the normal distribution if the samples contain at least 30
observations.
Sampling Distributions 195
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, central limit theorem
36. True or False: If the amount of gasoline purchased per car at a large service station has a
population mean of $15 and a population standard deviation of $4, then 99.73% of all cars will
purchase between $3 and $27.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
37. True or False: If the amount of gasoline purchased per car at a large service station has a
population mean of $15 and a population standard deviation of $4 and a random sample of 4 cars
is selected, there is approximately a 68.26% chance that the sample mean will be between $13
and $17.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
EXPLANATION: The sample is too small for the normal approximation.
KEYWORDS: sampling distribution, mean, probability
38. True or False: If the amount of gasoline purchased per car at a large service station has a
population mean of $15 and a population standard deviation of $4 and it is assumed that the
amount of gasoline purchased per car is symmetric, there is approximately a 68.26% chance that
a random sample of 16 cars will have a sample mean between $14 and $16.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
39. True or False: If the amount of gasoline purchased per car at a large service station has a
population mean of $15 and a population standard deviation of $4 and a random sample of 64
cars is selected, there is approximately a 95.44% chance that the sample mean will be between
$14 and $16.
ANSWER:
True
TYPE: TF DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, probability
40. True or False: As the sample size increases, the effect of an extreme value on the sample mean
becomes smaller.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
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Sampling Distributions
KEYWORDS: sampling distribution, law of large numbers
41. True or False: If the population distribution is skewed, in most cases the sampling distribution of
the mean can be approximated by the normal distribution if the samples contain at least 30
observations.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, central limit theorem
42. True or False: A sampling distribution is a distribution for a statistic.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution
43. True or False: Suppose  = 50 and  = 100 for a population. In a sample where n = 100 is
randomly taken, 95% of all possible sample means will fall between 48.04 and 51.96.
2
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
44. True or False: Suppose  = 80 and  = 400 for a population. In a sample where n = 100 is
randomly taken, 95% of all possible sample means will fall above 76.71.
2
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
45. True or False: Suppose  = 50 and  = 100 for a population. In a sample where n = 100 is
randomly taken, 90% of all possible sample means will fall between 49 and 51.
2
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
46. True or False: The Central Limit Theorem ensures that the sampling distribution of the sample
mean approaches normal as the sample size increases.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: central limit theorem
Sampling Distributions 197
47. True or False: The standard error of the mean is also known as the standard deviation of the
sampling distribution of the sample mean.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: standard error, mean
48. True or False: A sampling distribution is defined as the probability distribution of possible
sample sizes that can be observed from a given population.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution
49. True or False: As the size of the sample is increased, the standard deviation of the sampling
distribution of the sample mean for a normally distributed population will stay the same.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: standard error, properties
50. True or False: For distributions such as the normal distribution, the arithmetic mean is
considered more stable from sample to sample than other measures of central tendency.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean
51. True or False: The fact that the sample means are less variable than the population data can be
observed from the standard error of the mean.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, standard error
52. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110
grams and  = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that
the sample mean will be between 100 and 120 grams?
ANSWER:
0.9545 using Excel or 0.9544 using Table E.2
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
198
Sampling Distributions
53. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110
grams and  = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that
the sample mean will be less than 100 grams?
ANSWER:
0.0228
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
54. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110
grams and  = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that
the sample mean will be greater than 100 grams?
ANSWER:
0.9772
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
55. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110
grams and  = 25 grams. A sample of 25 vitamins is to be selected. So, 95% of all sample
means will be greater than how many grams?
ANSWER:
101.7757
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, value
56. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110
grams and  = 25 grams. A sample of 25 vitamins is to be selected. So, the middle 70% of all
sample means will fall between what two values?
ANSWER:
104.8 and 115.2
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, value
57. The amount of time required for an oil and filter change on an automobile is normally distributed
with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is
selected. What would you expect the standard error of the mean to be?
ANSWER:
2.5 minutes
TYPE: PR DIFFICULTY: Easy
KEYWORDS: standard error, mean
58. The amount of time required for an oil and filter change on an automobile is normally distributed
with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is
selected. What is the probability that the sample mean is between 45 and 52 minutes?
ANSWER:
0.4974
Sampling Distributions 199
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
59. The amount of time required for an oil and filter change on an automobile is normally distributed
with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is
selected. What is the probability that the sample mean will be between 39 and 48 minutes?
ANSWER:
0.8767
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
60. The amount of time required for an oil and filter change on an automobile is normally distributed
with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is
selected. So, 95% of all sample means will fall between what two values?
ANSWER:
40.1 and 49.9 minutes
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, probability
61. The amount of time required for an oil and filter change on an automobile is normally distributed
with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is
selected. So, 90% of the sample means will be greater than what value?
ANSWER:
41.8 minutes
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, probability
62. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a
standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this
machine. The sampling distribution of the sample mean has a mean of 36 oz.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, unbiased
63. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a
standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this
machine. The sampling distribution of the sample mean has a standard error of 0.15.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, standard error
64. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a
standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this
200
Sampling Distributions
machine. The sampling distribution of the sample mean will be approximately normal only if the
population sampled is normal.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, central limit theorem
65. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation
of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The
probability that the mean of the sample exceeds 36.01 oz. is __________.
ANSWER:
0.3446
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
66. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation
of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The
probability that the mean of the sample is less than 36.03 is __________.
ANSWER:
0.8849
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
67. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation
of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The
probability that the mean of the sample is between 35.94 and 36.06 oz. is __________.
ANSWER:
0.9836
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
68. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation
of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The
probability that the mean of the sample is between 35.95 and 35.98 oz. is __________.
ANSWER:
0.1891
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
69. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation
of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. So, 95% of the
sample means based on samples of size 36 will be between __________ and __________.
ANSWER:
35.951 and 36.049 ounces
TYPE: FI DIFFICULTY: Difficult
Sampling Distributions 201
KEYWORDS: sampling distribution, mean, value, central limit theorem
70. A manufacturer of power tools claims that the average amount of time required to assemble their
top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a
random sample of 64 purchasers of this table saw is taken. The mean of the sampling distribution
of the sample mean is __________ minutes.
ANSWER:
80
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, unbiased
71. A manufacturer of power tools claims that the average amount of time required to assemble their
top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a
random sample of 64 purchasers of this table saw is taken. The standard deviation of the
sampling distribution of the sample mean is __________ minutes.
ANSWER:
5
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, standard error
72. A manufacturer of power tools claims that the average amount of time required to assemble their
top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a
random sample of 64 purchasers of this table saw is taken. The probability that the sample mean
will be less than 82 minutes is __________.
ANSWER:
0.6554
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
73. A manufacturer of power tools claims that the average amount of time required to assemble their
top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a
random sample of 64 purchasers of this table saw is taken. The probability that the sample mean
will be between 77 and 89 minutes is __________.
ANSWER:
0.6898
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
74. A manufacturer of power tools claims that the average amount of time required to assemble their
top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a
random sample of 64 purchasers of this table saw is taken. The probability that the sample mean
will be greater than 88 minutes is __________.
ANSWER:
0.0548
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
202
Sampling Distributions
75. A manufacturer of power tools claims that the average amount of time required to assemble their
top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a
random sample of 64 purchasers of this table saw is taken. So, 95% of the sample means based
on samples of size 64 will be between __________ and __________.
ANSWER:
70.2 and 89.8 minutes
TYPE: FI DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, value, central limit theorem
76. To use the normal distribution to approximate the binomial distribution, we need ______ and
______ to be at least 5.
ANSWER:
np and n(1-p)
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
77. True or False: The sample mean is an unbiased estimate of the population mean.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: mean, unbiased
78. True or False: The sample proportion is an unbiased estimate of the population proportion.
ANSWER:
True
TYPE: TF DIFFICULTY: Difficult
KEYWORDS: proportion, unbiased
79. True or False: The mean of the sampling distribution of a sample proportion is the population
proportion,  .
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion
80. True or False: The standard error of the sampling distribution of a sample proportion is
p 1  p 
where p is the sample proportion.
n
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion
Sampling Distributions 203
81. True or False: The standard deviation of the sampling distribution of a sample proportion is
 1   
n
where  is the population proportion.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion
82. True or False: A sample of size 25 provides a sample variance of 400. The standard error, in this
case equal to 4, is best described as the estimate of the standard deviation of means calculated
from samples of size 25.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: standard error
83. True or False: An unbiased estimator will have a value, on average across samples, equal to the
population parameter value.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: unbiased
84. True or False: In inferential statistics, the standard error of the sample mean assesses the
uncertainty or error of estimation.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: mean, standard error
85. Assume that house prices in a neighborhood are normally distributed with standard deviation
$20,000. A random sample of 16 observations is taken. What is the probability that the sample
mean differs from the population mean by more than $5,000?
ANSWER:
0.3173 using Excel or 0.3174 using Table E.2
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
TABLE 7-1
Times spent studying by students in the week before final exams follow a normal distribution with
standard deviation 8 hours. A random sample of 4 students was taken in order to estimate the mean
study time for the population of all students.
204
Sampling Distributions
86. Referring to Table 7-1, what is the probability that the sample mean exceeds the population mean
by more than 2 hours?
ANSWER:
0.3085
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
87. Referring to Table 7-1, what is the probability that the sample mean is more than 3 hours below
the population mean?
ANSWER:
0.2266
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
88. Referring to Table 7-1, what is the probability that the sample mean differs from the population
mean by less than 2 hours?
ANSWER:
0.3829 using Excel or 0.3830 using Table E.2
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
89. Referring to Table 7-1, what is the probability that the sample mean differs from the population
mean by more than 3 hours?
ANSWER:
0.4533 using Excel or 0.4532 using Table E.2
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
TABLE 7-2
The mean selling price of new homes in a city over a year was $115,000. The population standard
deviation was $25,000. A random sample of 100 new home sales from this city was taken.
90. Referring to Table 7-2, what is the probability that the sample mean selling price was more than
$110,000?
ANSWER:
0.9772
TYPE: PR DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, probability, central limit theorem
91. Referring to Table 7-2, what is the probability that the sample mean selling price was between
$113,000 and $117,000?
ANSWER:
0.5763 using Excel or 0.5762 using Table E.2
TYPE: PR DIFFICULTY: Easy
Sampling Distributions 205
KEYWORDS: sampling distribution, mean, probability, central limit theorem
92. Referring to Table 7-2, what is the probability that the sample mean selling price was between
$114,000 and $116,000?
ANSWER:
0.3108
TYPE: PR DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, probability, central limit theorem
93. Referring to Table 7-2, without doing the calculations, state in which of the following ranges the
sample mean selling price is most likely to lie?
a) $113,000 -- $115,000
b) $114,000 -- $116,000
c) $115,000 -- $117,000
d) $116,000 -- $118,000
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, probability, central limit theorem
TABLE 7-3
The lifetimes of a certain brand of light bulbs are known to be normally distributed with a mean of
1,600 hours and a standard deviation of 400 hours. A random sample of 64 of these light bulbs is
taken.
94. Referring to Table 7-3, what is the probability that the sample mean lifetime is more than 1,550
hours?
ANSWER:
0.8413
TYPE: PR DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, probability
95. Referring to Table 7-3, the probability is 0.15 that the sample mean lifetime is more than how
many hours?
ANSWER:
1,651.82 hours using Excel or 1,652 hours using Table E.2
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
96. Referring to Table 7-3, the probability is 0.20 that the sample mean lifetime differs from the
population mean lifetime by at least how many hours?
ANSWER:
64.08 hours using Excel or 64 hours using Table E.2
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, value
206
Sampling Distributions
TABLE 7-4
According to a survey, only 15% of customers who visited the web site of a major retail store made a
purchase. Random samples of size 50 are selected.
97. Referring to Table 7-4, the average of all the sample proportions of customers who will make a
purchase after visiting the web site is _______.
ANSWER:
0.15 or 15%
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, proportion, mean
98. Referring to Table 7-4, the standard deviation of all the sample proportions of customers who
will make a purchase after visiting the web site is ________.
ANSWER:
0.05050
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, proportion, standard error
99. True of False: Referring to Table 7-4, the requirements for using a normal distribution to
approximate a binomial distribution is fulfilled.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
100. Referring to Table 7-4, what proportion of the samples will have between 20% and 30% of
customers who will make a purchase after visiting the web site?
ANSWER:
0.1596
TYPE: PR DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, probability
101. Referring to Table 7-4, what proportion of the samples will have less than 15% of customers
who will make a purchase after visiting the web site?
ANSWER:
0.5
TYPE: PR DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, probability
102. Referring to Table 7-4, what is the probability that a random sample of 50 will have at least
30% of customers who will make a purchase after visiting the web site?
ANSWER:
0.0015
Sampling Distributions 207
TYPE: PR DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, probability
103. Referring to Table 7-4, 90% of the samples will have less than what percentage of customers
who will make a purchase after visiting the web site?
ANSWER:
21.47%
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, proportion, value
104. Referring to Table 7-4, 90% of the samples will have more than what percentage of customers
who will make a purchase after visiting the web site?
ANSWER:
8.528% using Excel or 8.536 using Table E.2
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, proportion, value
105. A study at a college in the west coast reveals that, historically, 45% of their students are
minority students. The expected percentage of minority students in their next batch of freshmen
is _______.
ANSWER:
45%
TYPE: FI DIFFCULTY: Moderate
KEYWORDS: sampling distribution, proportion, mean
106. A study at a college in the west coast reveals that, historically, 45% of their students are
minority students. If random samples of size 75 are selected, the standard error of the
proportions of students in the samples who are minority students is _________.
ANSWER:
0.05745
TYPE: FI DIFFCULTY: Moderate
KEYWORDS: sampling distribution, proportion, standard error
107. A study at a college in the west coast reveals that, historically, 45% of their students are
minority students. If a random sample of size 75 is selected, the probability is _______ that
between 30% and 50% of the students in the sample will be minority students.
ANSWER:
0.8034 using Excel or 0.8033 using Table E.2
TYPE: FI DIFFCULTY: Easy
KEYWORDS: sampling distribution, proportion, probability
108. A study at a college in the west coast reveals that, historically, 45% of their students are
minority students. If a random sample of size 75 is selected, the probability is _______ that more
than half of the students in the sample will be minority students.
ANSWER:
208
Sampling Distributions
0.1920 using Excel or 0.1922 using Table E.2
TYPE: FI DIFFCULTY: Easy
KEYWORDS: sampling distribution, proportion, probability
109. A study at a college in the west coast reveals that, historically, 45% of their students are
minority students. If random samples of size 75 are selected, 80% of the samples will have less
than ______% of minority students.
ANSWER:
49.83
TYPE: FI DIFFCULTY: Difficult
KEYWORDS: sampling distribution, proportion, value
110. A study at a college in the west coast reveals that, historically, 45% of their students are
minority students. If random samples of size 75 are selected, 95% of the samples will have more
than ______% of minority students.
ANSWER:
35.55
TYPE: FI DIFFCULTY: Difficult
KEYWORDS: sampling distribution, proportion, value
111. Which of the following is NOT a reason for the need for sampling?
a) It is usually too costly to study the whole population.
b) It is usually too time consuming to look at the whole population.
c) It is sometimes destructive to observe the entire population.
d) It is always more informative by investigating a sample than the entire population.
ANSWER:
d
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: reasons for sampling
112. Which of the following is NOT a reason for drawing a sample?
a) A sample is less time consuming than a census.
b) A sample is less costly to administer than a census.
c) A sample is usually not a good representation of the target population.
d) A sample is less cumbersome and more practical to administer.
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: reasons for sampling
113. Which of the following sampling methods is a probability sample?
a) Chunk
b) Quota sample
c) Stratified sample
d) Judgment sample
ANSWER:
Sampling Distributions 209
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: probability sample
114. A sample of 300 subscribers to a particular magazine is selected from a population frame of
9,000 subscribers. If, upon examining the data, it is determined that no subscriber had been
selected in the sample more than once,
a) the sample could not have been random.
b) the sample may have been selected without replacement or with replacement.
c) the sample had to have been selected with replacement.
d) the sample had to have been selected without replacement.
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling method
115. For a population frame containing N = 1,007 individuals, what code number should you assign
to the first person on the list in order to use a table of random numbers?
a) 0
b) 1
c) 01
d) 0001
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: random number
116. Which of the following types of samples can you use if you want to make valid statistical
inferences from a sample to a population?
a) A judgment sample
b) A quota sample
c) A chunk
d) A probability sample
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: probability sample, sampling method
117. The evening host of a dinner dance reached into a bowl, mixed all the tickets around, and
selected the ticket to award the grand door prize. What sampling method was used?
a) Simple random sample
b) Systematic sample
c) Stratified sample
d) Cluster sample
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
210
Sampling Distributions
KEYWORDS: simple random sample, probability sample, sampling method
118. The Dean of Students mailed a survey to a total of 400 students. The sample included 100
students randomly selected from each of the freshman, sophomore, junior, and senior classes on
campus last term. What sampling method was used?
a) Simple random sample
b) Systematic sample
c) Stratified sample
d) Cluster sample
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: stratified sample, probability sample, sampling method
119. A telemarketer set the company’s computerized dialing system to contact every 25th person
listed in the local telephone directory. What sampling method was used?
a) Simple random sample
b) Systematic sample
c) Stratified sample
d) Cluster sample
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: systematic sample, probability sample, sampling method
120. Since a _______ is not a randomly selected probability sample, there is no way to know how
well it represents the overall population.
a) Simple random sample
b) Quota sample
c) Stratified sample
d) Cluster sample
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: quota sample, nonprobability sample, sampling method
121. A population frame for a survey contains a listing of 72,345 names. Using a table of random
numbers, how many digits will the code numbers for each member of your population contain?
a) 3
b) 4
c) 5
d) 6
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: random number
Sampling Distributions 211
122. A population frame for a survey contains a listing of 6,179 names. Using a table of random
numbers, which of the following code numbers will appear on your list?
a) 06
b) 0694
c) 6946
d) 61790
ANSWER:
b
TYPE: MC DIFFICULTY: Easy
KEYWORDS: random number
123. Which of the following can be reduced by proper interviewer training?
a) Sampling error
b) Measurement error
c) Both of the above
d) None of the above
ANSWER:
b
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: measurement error, survey worthiness
TABLE 7-5
The manager of the customer service division of a major consumer electronics company is interested
in determining whether the customers who have purchased a videocassette recorder made by the
company over the past 12 months are satisfied with their products.
124. Referring to Table 7-5, the manager decides to ask a sample of customers, who have bought a
videocassette recorder made by the company and filed a complaint over the past year, to fill in a
survey about whether they are satisfied with the product. This method will most likely suffer
from
a) non-response error.
b) measurement error.
c) coverage error.
d) non-probability sampling.
ANSWER:
c
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: coverage error, survey worthiness
125. Referring to Table 7-5, if there are 4 different brands of videocassette recorders made by the
company, the best sampling strategy would be to use a
a) a simple random sample.
b) a stratified sample.
c) a cluster sample.
d) a systematic sample.
ANSWER:
212
Sampling Distributions
b
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: stratified sample, probability sample, sampling method
126. Referring to Table 7-5, which of the following questions in the survey will NOT likely induce
a measurement error?
a) How many times have you illegally copied copyrighted sporting events?
b) What is your exact annual income?
c) How many times have you brought the videocassette recorder back for service?
d) How many times have you failed to set the time on the videocassette recorder?
ANSWER:
c
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: measurement error, survey worthiness
127. Referring to Table 7-5, if a customer survey questionnaire is included in all the videocassette
recorders made and sold by the company over the past 12 months, this method of collecting data
will most like suffer from
a) nonresponse error.
b) measurement error.
c) coverage error.
d) nonprobability sampling.
ANSWER:
a
TYPE: MC DIFFICULTY: Difficult
KEYWORDS: nonresponse error, survey worthiness
128. True or False: As a population becomes large, it is usually better to obtain statistical
information from the entire population.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: population, sample, reasons for samplings
129. True or False: If a simple random sample is chosen with replacement, each individual has the
same chance of selection on every draw.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: simple random sample, probability sample, sampling method
130. True or False: When dealing with human surveys, we are usually interested in sampling with
replacement.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
Sampling Distributions 213
KEYWORDS: sampling with replacement, sampling method, survey worthiness
131. True or False: The only reliable way a researcher can make statistical inferences from a sample
to a population is to use nonprobability sampling methods.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: nonprobability, probability sample, sampling method
132. True or False: A sample is always a good representation of the target population.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sample, population, sampling method
133. True or False: There can be only one sample drawn from a population.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sample, sampling method
134. True or False: Using different frames to generate data can lead to totally different conclusions.
ANSWER:
True
TYPE:TF DIFFICULTY: Easy
KEYWORDS: frame, sampling method
135. True or False: Sampling error can be completely eliminated by taking larger sample sizes.
ANSWER:
False
TYPE: TF DIFFICULTY: Difficult
KEYWORDS: sampling error
136. True or False: Sampling error can be reduced by taking larger sample sizes.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling error
137. True or False: Chunk sample is a type of probability sample.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: chunk sample
214
Sampling Distributions
138. True or False: Items or individuals in a judgment sample are chosen with regard to their
probability of occurrence.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: judgment sample, nonprobability sample
139. True or False: When participants are allowed to self-select into the sample, you have a
nonprobability sample.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: nonprobability sample
140. True or False: Systematic samples are less efficient than stratified sample.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: systematic sample, stratified sample
141. True or False: The professor of a business statistics class wanted to find out the average
amount of time per week her students spent studying for the class. Among the 50 students in her
class, 20% were freshmen, 50% were sophomores and 30% were juniors. She decided to draw 2
students randomly from the freshmen, 5 randomly from the sophomores and 3 randomly from the
juniors. This is an example of a systematic sample.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: stratified sample
142. ________ results from the exclusion of certain groups of subjects from a population frame.
ANSWER:
Coverage error
TYPE: FI DIFFICULTY: Difficult
KEYWORDS: coverage error, survey worthiness
143. Coverage error results in a ________.
ANSWER:
selection bias
TYPE: FI DIFFICULTY: Difficult
KEYWORDS: selection bias, survey worthiness
144. ________ results from the failure to collect data on all subjects in the sample.
Sampling Distributions 215
ANSWER:
Nonresponse error or bias
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: nonresponse error, survey worthiness
145. The sampling process begins by locating appropriate data sources called ___________.
ANSWER:
frames
TYPE: FI DIFFICULTY: Easy
KEYWORDS: frames, sampling method
146. True or False: If you randomly select a student from the first row of a business statistics class
and then every other 5 students thereafter until you get a sample of 20 students, this is an
example of a chunk sample.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: chunk sample
147. True or False: You stand at the main entrance to a departmental store and pick the first 20
customers that enter the store after it has opened its door for business on a single day. This is an
example of a systematic sample.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: systematic sample
148. True or False: An electronic appliance chain gathered customer opinions on their services
using the customer feedback forms that are attached to the product registration forms. This is an
example of a convenience sample.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: convenience sample
149. True or False: To gather opinions on the efficacy of U.S. foreign policies, a sample of 50
faculty members is selected from the pool of university professors who have taught political
science at the graduate level. This is an example of a judgment sample.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: judgment sample
150. True or False: A sample is selected by including everybody who sits in the first row of a
business statistics class. This is an example of a cluster sample.
216
Sampling Distributions
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: cluster sample
151. True or False: The question “How many times have you abused your spouse in the last 6
months?” will most likely result in nonresponse error.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: nonresponse error, survey worthiness
152. True or False: The question “Is your household income last year somewhere in between
$25,000 and $35,000?” will most likely result in coverage error.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: coverage error, survey worthiness
153. True or False: The only way one can eliminate sampling error is to take the whole population
as the sample.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderte
KEYWORDS: sampling error, survey worthiness
154. True or False: Coverage error can become an ethical issue if a particular group is intentionally
excluded from the frame.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: ethical issue, coverage error, survey worthiness
155. True or False: Measurement error will become an ethical issue when the findings are
presented without reference to sample size and margin of error.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: ethical issue, measurement error, sampling error, survey worthiness
156. Which of the following sampling methods will more likely be susceptible to ethical violation?
a) Simple random sample
b) Cluster sample
c) Convenience sample
d) Stratified sample
Sampling Distributions 217
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: ethical issues, sampling method
157. True or False: The professor of a business statistics class wanted to find out the average
amount of time per week her students spent studying for the class. She divided the students into
the left, right and center groups according to the location they sat in the class that day. One of
these 3 groups was randomly selected and everyone in the group was asked the average amount
of time per week he/she spent studying for the class. This is an example of a cluster sample.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: cluster sample
158. True or False: The professor of a business statistics class wanted to find out the average
amount of time per week her students spent studying for the class. She divided the fifty students
on her roster into ten groups starting from the first student on the roster. The first student was
randomly selected from the first group. Then every tenth student was selected from the
remaining students. This is an example of a cluster sample.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: systematic sample
159. True or False: Selection bias occurs more frequently in systematic samples than in simple
random samples.
ANSWER:
True
TYPE: TF DIFFICULTY: easy
KEYWORDS: simple random sample, systematic sample
160. True or False: The question: “Have you used any form of illicit drugs over the past 2 months?”
will most likely result in measurement error if the question is answered.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: measurement error, survey worthiness
161. True or False: The question: “How much did you make last year rounded to the nearest
hundreds of dollars?” will most likely result in measurement error.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: measurement error, survey worthiness
218
Sampling Distributions
TABLE 7-6
According to an article, 19% of the entire U.S. population have high-speed access to the Internet.
Random samples of size 200 are selected from the U.S. population.
162. Referring to Table 7-6, the population mean of all the sample proportions is ______.
ANSWER:
19% or 0.19
TYPE: FI DIFFICULTY: Easy
KEYWORDS: mean, sampling distribution, proportion, central limit theorem
163. Referring to Table 7-6, the standard error of all the sample proportions is ______.
ANSWER:
0.0277
TYPE: FI DIFFICULTY: Easy
KEYWORDS: standard error, sampling distribution, proportion, central limit theorem
164. Referring to Table 7-6, among all the random samples of size 200, ______ % will have
between 14% and 24% who have high-speed access to the Internet.
ANSWER:
92.85 using Excel or 92.82 using Table E.2
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
165. Referring to Table 7-6, among all the random samples of size 200, ______ % will have
between 9% and 29% who have high-speed access to the Internet.
ANSWER:
99.97
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
166. Referring to Table 7-6, among all the random samples of size 200, ______ % will have more
than 30% who have high-speed access to the Internet.
ANSWER:
0.0000 or virtually zero
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
167. Referring to Table 7-6, among all the random samples of size 200, ______ % will have less
than 20% who have high-speed access to the Internet.
ANSWER:
64.08 using Excel or 64.06 using Table E.2
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
Sampling Distributions 219
168. Referring to Table 7-6, among all the random samples of size 200, 90 % will have less than
_____% who have high-speed access to the Internet.
ANSWER:
22.56 using Excel or 22.55 using Table E.2
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, proportion, central limit theorem
169. Referring to Table 7-6, among all the random samples of size 200, 90 % will have more than
_____% who have high-speed access to the Internet.
ANSWER:
15.45
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, proportion, central limit theorem
TABLE 7-7
Online customer service is a key element to successful online retailing. According to a marketing
survey, 37.5% of online customers take advantage of the online customer service. Random samples
of 200 customers are selected.
170. Referring to Table 7-7, the population mean of all possible sample proportions is ______.
ANSWER:
0.375 or 37.5%
TYPE: FI DIFFICULTY: Easy
KEYWORDS: mean, sampling distribution, proportion, central limit theorem
171. Referring to Table 7-7, the standard error of all possible sample proportions is ______.
ANSWER:
0.0342
TYPE: FI DIFFICULTY: Easy
KEYWORDS: standard error, sampling distribution, proportion, central limit theorem
172. Referring to Table 7-7, ____ % of the samples are likely to have between 35% and 40% who
take advantage of online customer service.
ANSWER:
53.48 using Excel or 53.46 using Table E.2
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
173. Referring to Table 7-7, ____ % of the samples are likely to have less than 37.5% who take
advantage of online customer service.
ANSWER:
50
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
220
Sampling Distributions
174. Referring to Table 7-7, 90% of the samples proportions symmetrically around the population
proportion will have between _____% and _____% of the customers who take advantage of
online customer service.
ANSWER:
31.87 and 43.13
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, proportion, central limit theorem
175. Referring to Table 7-7, 95% of the samples proportions symmetrically around the population
proportion will have between _____% and _____% of the customers who take advantage of
online customer service.
ANSWER:
30.79 and 44.21
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: sampling distribution, proportion, central limit theorem