 # George Mason University Business Statistics Worksheet

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## Full Question

Sampling Methods and
the Central Limit Theorem
Chapter 8
LEARNING OBJECTIVES

So What? Why do we sample?

Methods to select a sample (Simple, Systematic, Stratified
and Cluster random sampling)

Other Important Aspects: Applying the central limit
theorem to find probabilities of selecting possible sample
means from a specified population.

Sampling Error
Sampling distribution of the sample Mean.
Central limit theorem.
Standard error of the mean.
Exercise
On making Decisions on Important Matters:
1.
What does “winging it” mean to you?
2.
List the reasons you and other may choose to “winging it”?
3.
What are the risks and costs of “winging it”, if any (please
list)?
4.
What is a superior approach (what are wise steps) in making
important decisions?
Factors that may drive firms to “Winging it”:

Eliminate or Reduce upfront costs
Get to the market faster
Competitive Pressures (e.g. profitability, defend market
share)
Other: human factors:

Wanting to move on (get closure)
Inexperience (lack of insights or limited knowledge)
Arrogance or misplaced self confidence
Laziness and lack of discipline
Potential Costs and Risks of “Winging it” on
important Decisions
What are potential costs and/or risks?

Financial Loss
Job Loss
Harming others
Reputational Loss, Shame, Embarrassment, Loss of status
Loss of loved ones
Loss of freedom (incarceration)
Loss of health or even life
Use Data: Do not wing it
It is a Capital Mistake to theorize
before one has data.
Sir Arthur Ignatius Conan Doyle
(1859 – 1930, Scottish physician and writer,
author of Sherlock Holmes
Use Data: Do not “wing it”
are much less
than those
using no data at all.
Charles Babbage
(1791 – 1871, English polymath,
inventor and mechanical engineer)
On important matters:
On Important Matters; some prudent steps to consider:

objective)

Collect a Reasonable and Accurate Data Sample

Perform a proper Analysis (be analytical)

Reality test your conclusions with respected and more
Why Sample the Population?
1.
2.
3.
4.
5.
To contact the whole population would be timeconsuming.
The cost of studying all the items in a population may be
prohibitive.
The physical impossibility of checking all items in the
population.
The destructive nature of some tests.
Probability Sampling
A probability sample is a sample selected such that each item or
person in the population being studied has a known likelihood of
being included in the sample.
Most Commonly Used Probability Sampling
Methods

Simple Random Sample
Systematic Random Sampling
Stratified Random Sampling
Cluster Sampling
Excel: 25 tips – Including Vlookup

Simple Random Sample
Simple Random Sample: A sample selected so that each
item or person in the population has the same chance of
being included.
Example: Simple Random Sample:
Using Tables of Random Numbers

A population consists of 845 employees. A sample of 52 employees is to
be selected from that population.

A more convenient method of selecting a random sample is to use the
identification number of each employee and a table of random
numbers such as the one in Appendix B.6.
Simple Random Sample: Using Excel

A bed and breakfast has eight rooms available for rent.
Listed below is the number of these eight rooms rented each day
during the Month of June.
Let’s select a sample of five nights.
Simple Random Sample:
Using Excel

At a Bed and Breakfast, There
are eight rooms available for
rent .
Listed are the number of the
eight rooms rented each day
during June.
Use Excel to select a sample
of five nights during the
month of June.
Note: Data Analysis add-in must be
installed
Systematic Random Sampling
Systematic Random Sampling: The items or individuals of the
population are arranged in some order. A random starting point is
selected and then every kth member of the population is selected for
the sample.
EXAMPLE
A population consists of 845 employees of Nitra Industries. A sample of 52
employees is to be selected from that population.

Use the systematic random sampling to select the samples.
Systematic Random Sampling
EXAMPLE

A population consists of 845 employees at a company.
A sample of 52 employees is to be selected from that population.

Use the systematic random sampling to select the samples.

Systematic Random Sampling
Data Given:
N=845 and n=52
Steps:
Step 1: Calculate k
k = N/n
= 845/52 (round down)
= 16
Step 2: Use simple random sampling to select the first sample
Step 3: Select every 16th element on the list after Step 2
Stratified Random Sampling
Stratified Random Sampling: A population is first divided into
subgroups, called strata, and a sample is selected from each
stratum.
Question: When should we Use this method?
Useful when a population can be clearly divided in groups
based on some characteristics.
Stratified Random Sampling
Suppose we want to study the advertising expenditures for the 352 largest companies
in the United States to determine whether firms with high returns on equity (a measure of
profitability) spent more of each sales dollar on advertising than firms with a low return or
deficit.
To make sure that the sample is a fair representation of the 352 companies, the
companies are grouped on percent return on equity and a sample proportional to the
relative size of the group is randomly selected.
Cluster Sampling
Cluster Sampling: A population is divided into clusters using naturally
occurring geographic or other boundaries.
Then, clusters are randomly selected and a sample is collected by
randomly selecting from each cluster.
Cluster Sampling
Suppose you want to determine the views of residents in Virginia, about state
and federal environmental protection policies.
Step 1: Subdivide the state into small units—either counties or regions,
Step 2: Select at random, say 4 regions, then
Step 3: Take samples of the residents in each of these regions and
interview them.
Sampling Error
The sampling error is the difference between a Sample statistic and its
corresponding Population parameter.
X 
s 
s 
p 
2
2
Sampling Distribution of the Sample Mean
The sampling distribution of the sample mean is a probability
distribution consisting of all possible sample means of a given
sample size selected from a population.
Example : Sampling Distribution of the
Sample Means

A Company has seven production employees (considered the population).
The hourly earnings of each employee are given in the table below.
Questions:
1. What is the population mean?
2. What is the sampling distribution of the sample mean for samples of size 2?
3. What is the mean of the sampling distribution?
4. What observations can be made about the population and the sampling
distribution?
Example: Sampling Distribution of the
Sample Means
Note: Applying the Combination Counting Rule – (see next slide for reference)
Recall – Combination Counting Rules
A combination is the number of ways to choose r
objects from a group of n objects without regard to
order.
Example – Sampling Distribution of
the Sample Means
Example – Sampling Distribution of
the Sample Means
Sampling Distribution of the
Sample Means – Example
Example – Sampling Distribution of the
Sample Means
Conclusions:
A. The mean of the distribution of the sample mean (\$7.71) is equal to the mean of the
population.
B. The spread in the distribution of the sample mean is less than the spread in the population
values.
C. As the size of the sample is increased, the spread of the distribution of the sample mean
becomes smaller.
D. The shape of the sampling distribution of the sample mean and the shape of the frequency
distribution of the population values are different. The distribution of the sample mean tends
to be more bell-shaped and to approximate the normal probability distribution.
Central Limit Theorem
CENTRAL LIMIT THEOREM If all samples of a particular size are selected
from any population, the sampling distribution of the sample mean is
approximately a normal distribution.
This approximation improves with larger samples.
Central Limit Theorem
Samples Sizes and Characteristic are based on:

For any sample size, the sampling distribution of the sample
mean will also be normal if the population follows a normal
probability distribution.
If the population distribution is symmetrical (but not normal),
the normal shape of the distribution of the sample mean
emerges with samples as small as 10.
If a distribution is skewed or has thick tails, it may require
samples of 30 or more to observe the normality feature.
The mean of the sample distribution (X bar) is equal to μ and
the sample variance is equal to σ2/n.
Sampling Methods and the
Central Limit Theorem
Central Limit Theorem – Example

Spence Sprockets, Inc. employs 40 people and faces some major decisions regarding health care for
these employees.
Before making a final decision on what health care plan to purchase, Ed decides to form a committee of
five representative employees. The committee will be asked to study the health care issue carefully and
make a recommendation as to what plan best fits the employees’ needs. Ed feels the views of newer
employees toward health care may differ from those of more experienced employees.
1.
2.
3.
If Ed randomly selects this committee, what can he expect in terms of the mean years with
Spence Sprockets for those on the committee?
How does the shape of the distribution of years of experience of all employees (the population)
compare with the shape of the sampling distribution of the mean?
The lengths of service (rounded to the nearest year) of the 40 employees currently on the Spence
Sprockets, Inc., payroll are as follows.
Central Limit Theorem – Example
25 Samples of Five Employees
25 Samples of 20 Employees
Standard Error of the Mean
1. The mean of the distribution of sample means will be exactly
equal to the population mean if we are able to select all possible
samples of the same size from a given population.
2. There will be less dispersion in the sampling distribution of the
sample mean than in the population. As the sample size
increases, the standard error of the mean decreases
Using the Sampling
Distribution of the Sample Mean (Sigma Known)

If a population follows the normal distribution, the sampling
distribution of the sample mean will also follow the normal
distribution.
If the shape is known to be non-normal, but the sample contains at
least 30 observations, the central limit theorem guarantees the
sampling distribution of the mean follows a normal distribution.
To determine the probability a sample mean falls within a particular
region, use:
X 
z
 n
If “z” is know, but x̅ is unknown
Apply algebra to convert the z formula:
X 
z 

n
x̅ = µ + (z* σ/√n)
Note: Important in solving some of the homework and exam problems.
Using the Sampling Distribution of the Sample
Mean (Sigma Unknown)

If the population does not follow the normal distribution, but
the sample is of at least 30 observations, the distribution of
the sample means will follow the normal distribution.
To determine the probability a sample mean falls within a
particular region, use:
X 
t
s n
Using the Sampling Distribution of the Sample Mean (Sigma
Known) – Example

The Quality Assurance Department for Cola, Inc., maintains records regarding the
amount of cola in its Jumbo bottle. The actual amount of cola in each bottle is
critical, but varies a small amount from one bottle to the next. Cola, Inc., does not
wish to under-fill the bottles. On the other hand, it cannot overfill each bottle. Its
records indicate that the amount of cola follows the normal probability
distribution.
The mean amount per bottle is 31.2 ounces and the population standard deviation
is 0.4 ounces.
At 8 A.M. today the quality technician randomly selected 16 bottles from the filling
line. The mean amount of cola contained in the bottles is 31.38 ounces.
Questions:
1.
Is this an unlikely result?
2.
Is it likely the process is putting too much soda in the bottles? To put it another
way, is the sampling error of 0.18 ounces unusual?
Using the Sampling Distribution of the Sample Mean
(Sigma Known) – Example
Step 1: Find the z value corresponding to the sample mean of 31.38.
X   31.38  31.20
z

 1.80
 n
\$0.4 16
Using the Sampling Distribution of the Sample Mean
(Sigma Known) – Example
Step 2: Find the probability of observing a z equal to or greater
than 1.80
Find P(z  1.80)  ?
Using the Standard Normal Distribution Table in
Finding Probability – Example
Find P(z  1.80)  ?
Step 3: 0.5 – 0.4641
= 0.0359
Using the Sampling Distribution of the Sample
Mean (Sigma Known) – Example
What do we conclude?
The process is putting too much cola in the bottles.
Why: It is unlikely, less than a 4 (3.59) percent chance, we
could select a sample of 16 observations from a normal
population with a mean of 31.2 ounces and a population
standard deviation of 0.4 ounces and find the sample
mean equal to or greater than 31.38 ounces.
APPENDIX – EXERCISES
1
5
9
13
17
21
25
29
33
37
41
45
45. Exercise – Example
45. Nike’s annual report says that the average American buys 6.5
pairs of sports shoes per year. Suppose the population standard
deviation is 2.1 and that a sample of 81 customers will be examined
next year.
a) What is the standard error of the mean in this experiment?
b) What is the probability that the sample mean is between 6 and 7
pairs of sports shoes?
c) What is the probability that the difference between the sample
mean and the population mean is less than 0.25 pairs?
d) What is the likelihood the sample mean is greater than 7 pairs?
45. Exercise – Example
Data:
µ = 6.5 σ = 2.1 N = 81
a) What is the standard error of the mean in this experiment?
Answer: σ error of the Mean = 2.1 / √81 = 0.23
45. Exercise – Example
Data:
µ = 6.5 σ = 2.1 N = 81
b) What is the probability that the sample mean is between 6 and 7 pairs of
sports shoes?
Obtain z for 6 = – 2.14 (area = – 0.4838)
Obtain z for 7 = 2.14 (area = 0.4838)
P = 0.4838 + 0.4838
= 0.9676
45. Exercise – Example
Data:
c)
µ = 6.5 σ = 2.1 N = 81
What is the probability that the difference between the sample mean and
the population mean is less than 0.25 pairs?
First:
Subtract 0.25 from the Mean = 6.25
Add 0.25 to Mean = 6.75
Next:
Obtain z for 6.25 = – 1.07 (area = – 0.3577)
Obtain z for 6.75 = 1.07 (area = 0.3577)
Finally: P = 0.3577 + 0.3577
= 0.7154
45. Exercise – Example
Data:
µ = 6.5 σ = 2.1 N = 81
d) What is the likelihood the sample mean is greater than 7 pairs?
Recall z for 7 = 2.14 (area = 0.4838)
P = 0.5000 – 0.4838 =
0.0162
41. Exercise – Example
41.Following is a list of the 50 states (see next page) with the numbers 0
through 49 assigned to them.
a) You wish to select a sample of eight from this list. The selected random
numbers are 45, 15, 81, 09, 39, 43, 90, 26, 06, 45, 01, and 42. Which
states are included in the sample?
b) You wish to use a systematic sample of every sixth item and the digit
02 is chosen as the starting point. Which states are included?
Number
0
State
Alabama
1
2
Arizona
3
Arkansas
4
California
5
6
Connecticut
7
Delaware
8
Florida
9
Georgia
10
Hawaii
11
Idaho
12
Illinois
13
Indiana
14
Iowa
15
Kansas
16
Kentucky
17
Louisiana
18
Maine
19
Maryland
20
Massachusetts
21
Michigan
22
Minnesota
23
Mississippi
24
Missouri
25
Montana
26
27
28
New Hampshire
29
New Jersey
30
New Mexico
31
New York
32
North Carolina
33
North Dakota
34
Ohio
35
Oklahoma
36
Oregon
37
Pennsylvania
38
Rhode Island
39
South Carolina
40
South Dakota
41
Tennessee
42
Texas
43
Utah
44
Vermont
45
Virginia
46
Washington
47
West Virginia
48
Wisconsin
49
Wyoming
41. Exercise – Example
a) You wish to select a sample of eight from this list. The selected random
numbers are 45, 15, 81, 09, 39, 43, 90, 26, 06, 45, 01, and 42. Which states
are included in the sample?
Virginia, Utah
41. Exercise – Example
a) You wish to select a sample of eight from this list. The selected random
numbers are 45, 15, 81, 09, 39, 43, 90, 26, 06, 45, 01, and 42. Which states
are included in the sample?
Observe: Only eight of these are between 0 and 49
Next: Remove the two that are out of range, namely 81 and 90
45, 15, 81, 09, 39, 43, 90, 26, 06, 45, 01, and 42
Next: Remove duplicate, namely 45
Finally: Select 8 states associated with the 9 numbers provided
Virginia, Utah
41. Exercise – Example
b) You wish to use a systematic sample of every sixth item and the digit 02 is
chosen as the starting point. Which states are included?
Observe: Only eight of these are between 0 and 49
Next: Remove the two that are out of range, namely 81 and 90
45, 15, 81, 09, 39, 43, 90, 26, 06, 45, 01, and 42
Next: Remove duplicate, namely 45
Finally: Select 8 states associated with the 9 numbers provided
Virginia, Utah
41. Exercise – Example
b) You wish to use a systematic sample of every sixth item and the digit 02
is chosen as the starting point. Which states are included?
Number
State
2
Arizona
8
Florida
14
Iowa
20
Massachusetts
26
32
North Carolina
38
Rhode Island
44
Vermont
37. Exercise – Example
37. Crossett Trucking Company claims that the mean weight of its
delivery trucks when they are fully loaded is 6,000 pounds and
the standard deviation is 150 pounds.
Assume that the population follows the normal distribution.
Forty trucks are randomly selected and weighed. Within what
limits will 95% of the sample means occur?
Data Given: µ= 6,000 σ = 150 n = 40 a – 1 = 95%
If “z” is know, but x̅ is unknown
Apply algebra to convert the z formula:
X 
z 

n
x̅ = µ + (z* σ/√n)
Note: Important in solving some of the homework and exam problems.
37. Exercise – Example
Assume that the population follows the normal distribution. Forty trucks are
randomly selected and weighed. Within what limits will 95% of the sample
means occur?
Data Given: µ = 6,000 σ = 150 n = 40 Confidence Level = 95%
X 
Formula – apply algebra: x̅ = µ + (z* σ/√n)
z

First: Derive the standard deviation of the sample mean:
s = σ/√n = 150/√40 = 23.72
Next: Find z for 0.95
.95/2 = 47.50
z = 1.96
Next: 6,000 + – (1.96 * 23.72)
Answer: 95% of the sample means will be between 5,953.51 lbs and 6,046.49 lbs
n
33. Exercise – Example
33. Recent studies indicate that the typical 50-year-old woman spends \$350 per
year for personal-care products. The distribution of the amounts spent
follows a normal distribution with a standard deviation of \$45 per year. We
select a random sample of 40 women. The mean amount spent for those
sampled is \$335. What is the likelihood of finding a sample mean this large or
larger from the specified population?
Data Given: µ = 350 σ = 45 n = 40 sample mean = 335
33. Exercise – Example
33. What is the likelihood of finding a sample mean this large or larger from
the specified population?
Data Given: µ = 350 x̅ = 335 σ = 45 n = 40
z = (335 – 350)/(45/√40)
= (-15)/7.1151
= -2.11
area = 0.4826
P = 0.5000 + 0.4826
= 0.9826
z
X 
 n
29. Exercise – Example
29. The Quality Control Department employs five technicians during the day
shift. Listed below is the number of times each technician instructed the
production foreman to shut down the manufacturing process last week.
a) How many different samples of two technicians are possible from this
population?
b) List all possible samples of two observations each and compute the mean
of each sample.
c) Compare the mean of the sample means with the population mean.
d) Compare the shape of the population distribution with the shape of the
distribution of the sample means.
Technician
Shutdowns
Taylor
4
Hurley
3
Gupta
5
Rousche
3
Huang
2
29. Exercise – Example
Data Given: n = 5 r = 2
a)
How many different samples of two technicians are possible from this
population?
5
C2 = 10
29. Exercise – Example
b) List all possible samples of two observations each and compute the mean of
each sample.
Shutdowns Mean Shutdowns Mean
4, 3 3.5 3, 3 3.0
4, 5 4.5 3, 2 2.5
4, 3 3.5 5, 3 4.0
4, 2 3.0 5, 2 3.5
3, 5 4.0 3, 2 2.5
Mean Frequency Probability
2.5
2
.20
3.0
2
.20
3.5
3
.30
4.0
2
.20
4.5
1
.10
10
1.00
29. Exercise – Example
c) Compare the mean of the sample means with the population mean.
Mean of the sample means = Sum of the sample means/Number of samples
Sample Mean = (3.5 + 4.5 + · · · 2.5)/10 = 3.4 shutdowns
Population mean = (4 + 3 + 5 + 3 + 2)/5 = 3.4 shutdowns
 Mean of the sample means = Population mean
d) Compare the shape of the population distribution with the shape of the
distribution of the sample means.
 The population values are relatively uniform in shape
 The distribution of sample means tends toward normality
25. Exercise – Example
25. There are 25 motels in Goshen, Indiana. The number of rooms in each motel
follows:
90
72
75
60
75
72
84
72
88
74
105 115 68
74
80
64
104 82
48
58
60
80
48
58
a) Using a table of random numbers (Appendix B.6), select a random sample of
five motels from this population.
b) Obtain a systematic sample by selecting a random starting point among the
first five motels and then select every fifth motel.
c) Suppose the last five motels are “cut-rate” motels. Describe how you would
select a random sample of three regular motels and two cut-rate motels.
100
25. Exercise – Example
90
72
75
60
75
72
84
72
88
74
105 115 68
74
80
64
104 82
48
58
60
80
48
58
100
a) Using a table of random numbers (Appendix B.6), select a random sample of
five motels from this population.
b) Obtain a systematic sample by selecting a random starting point among the
first five motels and then select every fifth motel.
Answers will vary. Selected the third observation. So the sample consists of
75, 72, 68, 82, 48. Answers will vary.
c.
Suppose the last five motels are “cut-rate” motels. Describe how you would
select a random sample of three regular motels and two cut-rate motels.
Select three numbers. Then number the last five numbers 20 to 24.
Randomly select two numbers from that group.
21. Exercise – Example
21. A population consists of the following three values: 1, 2, and 3.
a) List all possible samples of size 2 (including possible repeats) and
compute the mean of every sample.
b) Find the means of the distribution of the sample mean and the
population mean. Compare the two values.
c) Compare the dispersion of the population with that of the sample
mean.
d) Describe the shapes of the two distributions.
21. Exercise – Example
b) Find the means of the distribution of the sample mean and the population
mean. Compare the two values.
They are equal
Mean of sample means is (1.0+1.5 +2.0 …… + 3.0)/9 = 18/9 = 2.0
The population mean is (1+2+3)3 = 6/3 = 2.0
c) Find the means of the distribution of the sample mean and the population
mean. Compare the two values
The Variance of the population is twice as large (more dispersed) as that of the
sample means.
Variance of sample means is (1.0 + 0.25 +0.0 …. +1.0)/9= 3/9 = 1/3
Variance of the population values is (1+0 +1)/3 = 2/3
d) Describe the shapes of the two distributions
Sample means follow a triangular shape peaking at 2.
The population is uniform between 1 and 3.
17. Exercise – Example
17. In a certain section of Southern California, the distribution of
monthly rent for a one-bedroom apartment has a mean of \$2,200
and a standard deviation of \$250. The distribution of the monthly
rent does not follow the normal distribution. In fact, it is positively
skewed.
What is the probability of selecting a sample of 50 one-bedroom
apartments and finding the mean to be at least \$1,950 per month?
17. Exercise – Example
Answer: According to the Central Limit Theorem, the distribution of sample
means will have a normal distribution even though the population distribution
is positively skewed
Central Limit Theorem: If a distribution is skewed or has thick tails, it may
require samples of 30 or more to observe the normality feature.
Data Provided: μ = 2,200, σ = 260, n = 50
Formula
z 
X 

n
z = (1,950 – 2,200)/ (250/ √50)
= 7.07
P = virtually 1 or 100%
13. Exercise – Example
13. Consider all of the coins (pennies, nickels, quarters, etc.) in your pocket or
purse as a population.
a) Make a frequency table beginning with the current year and counting
backward to record the ages (in years) of the coins. For example, if the
current year is 2011, then a coin with 2007 stamped on it is 4 years old.
b) Draw a histogram or other graph showing the population distribution.
c) Randomly select five coins and record the mean age of the sampled coins.
Repeat this sampling process 20 times. Now draw a histogram or other
graph showing the distribution of the sample means.
d) Compare the shapes of the two histograms.
9. Exercise – Example
9. In the law firm Tybo and Associates, there are six partners. Listed next is the
number of cases each associate actually tried in court last month.
Ruud…….. 3
Wu…….. 6
Sass…….. 3
Flores…….. 3
Wilhelms…….. 0
Schueller…….. 1
a) How many different samples of 3 are possible?
b) List all possible samples of size 3, and compute the mean number of cases
in each sample.
c) Compare the mean of the distribution of sample means to the population
mean.
d) On a chart similar to CHART 8–1, compare the dispersion in the population
with that of the sample means.
9. Exercise – Example
Data Given: 6 Partners
a) How many different samples of 3 are possible?
6C3 = 20
c)
Compare the mean of the distribution of sample means to the population
mean.
They are the same:
Sample Mean = 53.33/ 20 = 2.67
µ = (3 + 6 + 3 + 3 + 0 + 1)/6 = 2.67
d) The population has more dispersion than the sample means.
Answer: The sample means vary from 1.33 to 4.0. The population varies from 0
to 6.
5. Exercise – Example
5.
A population consists of the following four values: 12, 12, 14, and 16.
a) List all samples of size 2, and compute the mean of each sample.
b) Compute the mean of the distribution of the sample mean and the
population mean. Compare the two values.
Answer: Sample Mean = 12 + 13 + 14 + 13 + 14 + 15)/6 = 13.5
µ = (12 + 12 + 14 + 16)/4 = 13.5
a)
Compare the dispersion in the population with that of the sample mean.
Answer: More dispersion with population data compared to the sample
means.
 Population varies from 12 to 16
 Sample Means vary from 12 to 15
3. Exercise – Example
Listed below are the 35 members of the Metro Toledo Automobile Dealers
Association. We would like to estimate the mean revenue from dealer service
departments.
a) We want to select a random sample of five dealers. The random numbers
are: 05, 20, 59, 21, 31, 28, 49, 38, 66, 08, 29, and 02. Which dealers
would be included in the sample?
b) Use the table of random numbers to select your own sample of five
dealers.
c) A sample is to consist of every seventh dealer. The number 04 is selected
as the starting point. Which dealers are included in the sample?
https://www.chegg.com/homework-help/listed-35-members-metro-toledo-automobiledealers-associatio-chapter-8-problem-3e-solution-9780077388621exc?trackid=06d09b51&strackid=461d375a&ii=1
1. Exercise – Example
1.
a)
The following is a list of Marco’s Pizza stores in Lucas County. Also noted is
whether the store is corporate-owned (C) or manager-owned (M). A sample of
four locations is to be selected and inspected for customer convenience,
safety, cleanliness, and other features.
The random numbers selected are 08, 18, 11, 54, 02, 41, and 54. Which stores
are selected?
b) Use the table of random numbers to select your own sample of locations.
c)
A sample is to consist of every seventh location. The number 03 is the starting
point. Which locations will be included in the sample?
d) Suppose a sample is to consist of three locations, of which two are corporateowned and one is manager-owned. Select a sample accordingly.
1. Exercise – Example
NAME: ___________________________________
1) When all the items in a population have an equal chance of being selected for a sample,
the process is called ________.
1)
A) Sampling error B) Nonprobability sampling
C) Simple random sampling D) z-score
2) For a given population, the mean of all the sample means (X) of sample size n, and the
mean of all (N) population observations (X) are ________.
2)
A) Equal to the population mean μ B) Not equal
C) Equal to (x – ) D) Equal to (X)
3) What is the difference between a sample mean and the population mean called? 3)
A) Point estimate B) Standard error of the mean
C) Interval estimate D) Sampling error
4) The Office of Student Services at a large western state university maintains information
on the study habits of its full-time students. Their studies indicate that the mean amount
of time undergraduate students study per week is 20 hours. The hours studied follows
the normal distribution with a standard deviation of six hours. Suppose we select a
random sample of 144 current students. What is the standard error of the mean?
4)
A) 0.25 B) 2.00 C) 0.50 D) 6.00
5) The mean of all possible sample means is equal to ________. 5)
A) The sample variance B)
n
C) The population mean D) The population variance
6) All possible samples of size n are selected from a population and the mean of each
sample is determined. What is the mean of the sample means?
6)
A) It is larger than the population mean.
B) It is smaller than the population mean.
C) The population mean.
D) It cannot be estimated in advance.
1
7) The Intelligence Quotient (IQ) test scores for adults are normally distributed with a
mean of 100 and a standard deviation of 15. What is the probability we could select a
sample of 50 adults and find that the mean of this sample exceeds 104?
7)
A) 0.9706 B) 0.0294
C) Approximately one D) 0.9412
8) Which of the following is the standard error of the mean? 8)
A) x/n B)
n
C) s D)
9) A statewide sample survey is to be made. First, the state is subdivided into counties.
Seven counties are selected at random and further sampling is concentrated on these
seven counties. What type of sampling is this?
9)
A) Simple random B) Cluster sampling
C) Stratified sampling D) Systematic random sampling
10) Mileage tests were conducted on a randomly selected sample of 100 newly developed
automobile tires. The mean tread wear was 50,000 miles, with a standard deviation of
3,500 miles. What is the best estimate of the average tread life in miles for the entire
population of these tires?
10)
A) 500 B) 35 C) 50,000 D) 3,500
11) Sampling error is the difference between a sample statistic and its corresponding
________.
11)
A) Trend B) Variance
C) Sample mean D) Population parameter
12) The Intelligence Quotient (IQ) test scores for adults are normally distributed with a
mean of 100 and a standard deviation of 15. What is the probability we could select a
sample of 50 adults and find the mean of this sample is between 95 and 105?
12)
A) Very unlikely B) 1.00
C) 0.9818 D) 0.0182
13) Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones
Brothers has maintained careful records regarding the time to prepare a return. The
mean time to prepare a return is 90 minutes and the standard deviation of this
distribution is 14 minutes. Suppose 100 returns from this year are selected and analyzed
regarding the preparation time. What is the standard error of the mean?
13)
A) 1.4 minutes B) 140 minutes C) 90 minutes D) 14 minutes
2
14) The weight of trucks traveling on a particular section of I-475 has a population mean of
15.8 tons and a population standard deviation of 4.2 tons. What is the probability a state
highway inspector could select a sample of 49 trucks and find the sample mean to be
14.3 tons or less?
14)
A) 0.1368 B) 0.0062 C) 0.3632 D) 0.4938
15) When dividing a population into subgroups so that a random sample from each
subgroup can be collected, what type of sampling is used?
15)
A) Stratified random sampling B) Simple random sampling
C) Systematic sampling D) Cluster sampling
16) Sampling error is defined as ________. 16)
A) (x – ) B) C) N – n D) /n
17) The tread life of tires mounted on light-duty trucks follows the normal probability
distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles.
Suppose we select a sample of 40 tires and use a simulator to determine the tread life.
What is the standard error of the mean?
17)
A) 40 B) 632.46
C) 4000 D) Cannot be determined.
18) The Office of Student Services at a large western state university maintains information
on the study habits of its full-time students. Their studies indicate that the mean amount
of time undergraduate students study per week is 20 hours. The hours studied follows
the normal distribution with a standard deviation of six hours. Suppose we select a
random sample of 144 current students. What is the probability that the mean of this
sample is between 19.25 hours and 21.0 hours?
18)
A) 0.0160 B) 0.9104 C) 0.9544 D) 0.0986
19) Suppose a research firm conducted a survey to determine the mean amount steady
smokers spend on cigarettes during a week. A sample of 100 steady smokers revealed
that the sample mean is \$20 and the sample standard deviation is \$5. What is the
probability that a sample of 100 steady smokers spend between \$19 and \$21?
19)
A) 1.0000 B) 0.4772 C) 0.9544 D) 0.0228
20) As the size of the sample increases, what happens to the shape of the distribution of
sample means?
20)
A) It cannot be predicted in advance. B) It is negatively skewed.
C) It is positively skewed. D) It approaches a normal distribution.
3
21) When testing the safety of cars using crash tests, a sample of one or two cars is used
because ________.
21)
A) It is quicker B) The population is very large
C) Sampling is more accurate D) Cars are destroyed
22) A university has 1,000 computers available for students to use. Each computer
has a 250-gigabyte hard drive. The university wants to estimate the space
occupied on the hard drives. A random sample of 100 computers showed a mean
of 115 gigabytes used with a standard deviation of 20 gigabytes. What is the
probability that a sample mean is between 111 and 119 gigabytes?
22)
23) A population consists of 14 values. How many samples of size five are
possible?
23)
24) LongLast Inc. produces car batteries. The mean life of these batteries is 60
months. The distribution of the battery life closely follows the normal
probability distribution with a standard deviation of eight months. As a part of
its testing program, LongLast tests a sample of 25 batteries. What proportion of
the samples will have a mean useful life less than 56 months?
24)
25) A university has 1,000 computers available for students to use. Each computer
has a 250-gigabyte hard drive. The university wants to estimate the space
occupied on the hard drives. A random sample of 100 computers showed a mean
of 115 gigabytes used with a standard deviation of 20 gigabytes. What is the
standard error of the mean?
25)
4

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