Sampling Distributions 7-1
CHAPTER 7: 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
KEYWORDS: unbiased
Copyright ©2015 Pearson Education
7-2
Sampling Distributions
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 right
skewed 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 right skewed with mean = 10 minutes and standard error = 0.8 minutes.
b) Distribution is right skewed 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 µ .
d) The standard deviation of the sampling distribution of the sample mean is equal to σ .
ANSWER:
d
TYPE: MC DIFFICULTY: Easy
KEYWORDS: sampling distribution, properties, mean
7. Which of the following is true about the sampling distribution of the sample mean?
a) The mean of the sampling distribution is always µ .
b) The standard deviation of the sampling distribution is always σ .
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, mean
Copyright ©2015 Pearson Education
Sampling Distributions 7-3
8. True or False: The amount of time it takes to complete an examination has a left skewed
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:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, central limit theorem
9. Suppose the ages of students in Statistics 101 follow a right skewed 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
sample.
collected is referred to as a(n)
ANSWER:
biased
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: unbiased
Copyright ©2015 Pearson Education
7-4
Sampling Distributions
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 selected 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 µ = 6 ounces and σ = 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?
a) The mean of the sampling distribution is 6 ounces.
b) The standard deviation of the sampling distribution is 2.5 ounces.
c) The shape of the sampling distribution is approximately normal.
d) All of the above are correct.
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, unbiased
14. The mean score of all pro golfers for a particular course has a mean of 70 and a standard
deviation of 3.0. Suppose 36 pro golfers played the course today. Find the probability that the
mean score of the 36 pro 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 week 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 weeks has been selected. What is the approximate probability that the mean number of
loaves sold in the sampled weeks exceeds 7,895 loaves?
ANSWER:
Approximately 0
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
Copyright ©2015 Pearson Education
Sampling Distributions 7-5
16. Sales prices of baseball cards from the 1960s are known to possess a right skewed 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) Right skewed 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) Right skewed 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 $3.26 million with a standard deviation of $1.2 million
in a certain year in the past. Suppose a sample of 100 major league players was taken. What was
the standard error for the sample mean salary?
a) $0.012 million
b) $0.12 million
c) $12 million
d) $1,200.0 million
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: standard error, mean
18. Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million
in a certain year in the past. Suppose a sample of 100 major league players was taken. Find the
approximate probability that the mean salary of the 100 players exceeded $3.5 million.
a) Approximately 0
b) 0.0228
c) 0.9772
d) Approximately 1
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
Copyright ©2015 Pearson Education
7-6
Sampling Distributions
19. Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million
in a certain year in the past. Suppose a sample of 100 major league players was taken. Find the
approximate probability that the mean salary of the 100 players exceeded $4.0 million.
a) Approximately 0
b) 0.0228
c) 0.9772
d) Approximately 1
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
20. Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million
in a certain year in the past. Suppose a sample of 100 major league players was taken. Find the
approximate probability that the mean salary of the 100 players was no more than $3.0 million.
a) Approximately 0
b) 0.0151
c) 0.9849
d) Approximately 1
ANSWER:
b
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
21. Major league baseball salaries averaged $3.26 million with a standard deviation of $1.2 million
in a certain year in the past. Suppose a sample of 100 major league players was taken. Find the
approximate probability that the mean salary of the 100 players was less than $2.5 million.
a) Approximately 0
b) 0.0151
c) 0.9849
d) Approximately 1
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability, central limit theorem
Copyright ©2015 Pearson Education
Sampling Distributions 7-7
22. At a computer manufacturing company, the actual size of a particular type 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
23. At a computer manufacturing company, the actual size of a particular type 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
24. At a computer manufacturing company, the actual size of a particular type 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
25. At a computer manufacturing company, the actual size of a particular type 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
Copyright ©2015 Pearson Education
7-8
Sampling Distributions
26. 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
b) 0.050
c) 0.200
d) 0.800
ANSWER:
c
TYPE: MC DIFFICULTY: Easy
KEYWORDS: standard error, mean
27. 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
28. 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
Copyright ©2015 Pearson Education
Sampling Distributions 7-9
29. 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
30. 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
31. 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
32. 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
Copyright ©2015 Pearson Education
7-10
Sampling Distributions
33. For sample sizes greater than 30, the sampling distribution of the mean will be approximately
normally distributed
a) regardless of the shape of the population.
b) only if the shape of the population is symmetrical.
c) only if the standard deviation of the samples are known.
d) only if the population is normally distributed.
ANSWER:
a
TYPE: MC DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, central limit theorem
34.
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
35.
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
36.
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
37.
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
Copyright ©2015 Pearson Education
Sampling Distributions 7-11
38.
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.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, central limit theorem
39.
True or False: If the amount of gasoline purchased per car at a large service station has a
population mean of 15 gallons and a population standard deviation of 4 gallons, then 99.73% of
all cars will purchase between 3 and 27 gallons.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
40.
True or False: If the amount of gasoline purchased per car at a large service station has a
population mean of 15 gallons and a population standard deviation of 4 gallons 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 gallons.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
EXPLANATION: The sample is too small for the normal approximation.
KEYWORDS: sampling distribution, mean, probability
41.
True or False: If the amount of gasoline purchased per car at a large service station has a
population mean of 15 gallons and a population standard deviation of 4 gallons 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 gallons.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
42.
True or False: If the amount of gasoline purchased per car at a large service station has a
population mean of 15 gallons and a population standard deviation of 4 gallons 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 gallons.
ANSWER:
True
TYPE: TF DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, probability
Copyright ©2015 Pearson Education
7-12
Sampling Distributions
43.
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
KEYWORDS: sampling distribution, mean
44.
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
45.
True or False: A sampling distribution is a distribution for a statistic.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution
46.
True or False: Suppose µ = 50 and σ = 10 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.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
47.
True or False: Suppose µ = 80 and σ = 20 for a population. In a sample where n = 100 is
randomly taken, 95% of all possible sample means will fall above 76.71.
ANSWER:
True
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
48.
True or False: Suppose µ = 50 and σ = 10 for a population. In a sample where n = 100 is
randomly taken, 90% of all possible sample means will fall between 49 and 51.
ANSWER:
False
TYPE: TF DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
Copyright ©2015 Pearson Education
Sampling Distributions 7-13
49.
True or False: The Central Limit Theorem ensures that the sampling distribution of the sample
mean approaches a normal distribution as the sample size increases.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: central limit theorem
50.
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
51.
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
52.
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
53.
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
54.
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
Copyright ©2015 Pearson Education
7-14
Sampling Distributions
55.
The amount of tea leaves in a can from a particular production line is normally distributed
with µ = 110 grams and σ = 25 grams. A sample of 25 cans 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
56.
The amount of tea leaves in a can from a particular production line is normally distributed with
µ = 110 grams and σ = 25 grams. A sample of 25 cans 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
57.
The amount of tea leaves in a can from a particular production line is normally distributed with
µ = 110 grams and σ = 25 grams. A sample of 25 cans 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
58.
The amount of tea leaves in a can from a particular production line is normally distributed with
µ = 110 grams and σ = 25 grams. A sample of 25 cans 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
59.
The amount of tea leaves in a can from a particular production line is normally distributed with
µ = 110 grams and σ = 25 grams. A sample of 25 cans 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
Copyright ©2015 Pearson Education
Sampling Distributions 7-15
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. What is the standard error of the mean?
ANSWER:
2.5 minutes
TYPE: PR DIFFICULTY: Easy
KEYWORDS: standard error, mean
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. What is the probability that the sample mean is between 45 and 52
minutes?
ANSWER:
0.4974
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: sampling distribution, mean, probability
62.
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
63.
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. 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
64.
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. 90% of the sample means will be greater than what value?
ANSWER:
41.8 minutes
TYPE: PR DIFFICULTY: Difficult
KEYWORDS: sampling distribution, mean, probability
Copyright ©2015 Pearson Education
7-16
Sampling Distributions
65.
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
66.
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
67.
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 will be approximately normal only if the
population sampled is normal.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, mean, 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 exceeds 36.01 oz. is __________.
ANSWER:
0.3446
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. 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
Copyright ©2015 Pearson Education
Sampling Distributions 7-17
70.
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
71.
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
72.
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,
the middle 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
KEYWORDS: sampling distribution, mean, value, central limit theorem
73.
A manufacturer of power tools claims that the mean 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
74.
A manufacturer of power tools claims that the mean 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
Copyright ©2015 Pearson Education
7-18
Sampling Distributions
75.
A manufacturer of power tools claims that the mean 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
76.
A manufacturer of power tools claims that the mean 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
77.
A manufacturer of power tools claims that the mean 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
78.
A manufacturer of power tools claims that the mean 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, the middle 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
79.
To use the normal distribution to approximate the binomial distribution, we need ______ and
______ to be at least 5.
ANSWER:
nπ and n (1 − π )
TYPE: FI DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, central limit theorem
Copyright ©2015 Pearson Education
Sampling Distributions 7-19
80.
True or False: The sample mean is an unbiased estimate of the population mean.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: mean, unbiased
81.
True or False: The sample proportion is an unbiased estimate of the population proportion.
ANSWER:
True
TYPE: TF DIFFICULTY: Difficult
KEYWORDS: proportion, unbiased
82.
True or False: The mean of the sampling distribution of a sample proportion is the population
proportion, π .
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion
83.
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
84.
True or False: The standard deviation of the sampling distribution of a sample proportion is
π (1 − π )
n
where π is the population proportion.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion
Copyright ©2015 Pearson Education
7-20
Sampling Distributions
85.
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, mean
86.
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, mean
87.
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
88.
True or False: The sample proportion is an unbiased estimator for the population proportion.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: unbiased, proportion
89.
True or False: The sample mean is an unbiased estimator for the population mean.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: unbiased, mean
90.
Assume that house prices in a neighborhood are normally distributed with a standard deviation
of $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
Copyright ©2015 Pearson Education
Sampling Distributions 7-21
SCENARIO 7-1
The time spent studying by students in the week before final exams follows a normal distribution
with a standard deviation of 8 hours. A random sample of 4 students was taken in order to estimate
the mean study time for the population of all students.
91.
Referring to Scenario 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
92.
Referring to Scenario 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
93.
Referring to Scenario 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
94.
Referring to Scenario 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
Copyright ©2015 Pearson Education
7-22
Sampling Distributions
SCENARIO 7-2
The mean selling price of new homes in a small town 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.
95.
Referring to Scenario 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
96.
Referring to Scenario 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
KEYWORDS: sampling distribution, mean, probability, central limit theorem
97.
Referring to Scenario 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
98.
Referring to Scenario 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
Copyright ©2015 Pearson Education
Sampling Distributions 7-23
SCENARIO 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.
99.
Referring to Scenario 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
100. Referring to Scenario 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
101. Referring to Scenario 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
SCENARIO 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.
102. Referring to Scenario 7-4, the mean 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
Copyright ©2015 Pearson Education
7-24
Sampling Distributions
103. Referring to Scenario 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
104. True or False: Referring to Scenario 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
105. Referring to Scenario 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
106. Referring to Scenario 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
107. Referring to Scenario 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
TYPE: PR DIFFICULTY: Easy
KEYWORDS: sampling distribution, proportion, probability
108. Referring to Scenario 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
Copyright ©2015 Pearson Education
Sampling Distributions 7-25
109. Referring to Scenario 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
110. A study at a college in the west coast reveals that, historically, 45% of the students are
minority students. The expected percentage of minority students in their next group of freshmen
is _______.
ANSWER:
45%
TYPE: FI DIFFCULTY: Moderate
KEYWORDS: sampling distribution, proportion, mean
111. A study at a college in the west coast reveals that, historically, 45% of the students are
minority students. If random samples of size 75 are selected, the standard error of the proportion
of students in the samples who are minority students is _________.
ANSWER:
0.05745
TYPE: FI DIFFCULTY: Moderate
KEYWORDS: sampling distribution, proportion, standard error
112. A study at a college in the west coast reveals that, historically, 45% of the 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
113. A study at a college in the west coast reveals that, historically, 45% of the 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:
0.1920 using Excel or 0.1922 using Table E.2
TYPE: FI DIFFCULTY: Easy
KEYWORDS: sampling distribution, proportion, probability
Copyright ©2015 Pearson Education
7-26
Sampling Distributions
114. A study at a college in the west coast reveals that, historically, 45% of the 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
115. A study at a college in the west coast reveals that, historically, 45% of the 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
SCENARIO 7-5
According to an article, 19% of the entire population in a developing country have high-speed access
to the Internet. Random samples of size 200 are selected from the country’s population.
116. Referring to Scenario 7-5, the population mean of all the sample proportions is ______.
ANSWER:
19% or 0.19
TYPE: FI DIFFICULTY: Easy
KEYWORDS: proportion, mean, sampling distribution, central limit theorem
117. Referring to Scenario 7-5, the standard error of all the sample proportions is ______.
ANSWER:
0.0277
TYPE: FI DIFFICULTY: Easy
KEYWORDS: proportion, standard error, sampling distribution, central limit theorem
118. Referring to Scenario 7-5, 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
Copyright ©2015 Pearson Education
Sampling Distributions 7-27
119. Referring to Scenario 7-5, 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
120. Referring to Scenario 7-5, 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
121. Referring to Scenario 7-5, 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
122. Referring to Scenario 7-5, 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
123. Referring to Scenario 7-5, 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
Copyright ©2015 Pearson Education
7-28
Sampling Distributions
SCENARIO 7-6
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.
124. Referring to Scenario 7-6, the population mean of all possible sample proportions is ______.
ANSWER:
0.375 or 37.5%
TYPE: FI DIFFICULTY: Easy
KEYWORDS: proportion, mean, sampling distribution, central limit theorem
125. Referring to Scenario 7-6, the standard error of all possible sample proportions is ______.
ANSWER:
0.0342
TYPE: FI DIFFICULTY: Easy
KEYWORDS: proportion, standard error, sampling distribution, central limit theorem
126. Referring to Scenario 7-6, ____ % 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
127. Referring to Scenario 7-6, ____ % 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
128. Referring to Scenario 7-6, 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
Copyright ©2015 Pearson Education
Sampling Distributions 7-29
129. Referring to Scenario 7-6, 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
Copyright ©2015 Pearson Education