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?

Why do business wing it?

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”

Errors [even] using inadequate data

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:

Start with an Unbiased Mind-Set (must be clinically

objective)

Collect a Reasonable and Accurate Data Sample

Perform a proper Analysis (be analytical)

Reality test your conclusions with respected and more

experienced advisors (keep your ego in check)

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.

The sample results are adequate.

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

Tutorial: https://www.youtube.com/watch?v=PdXDLNNXPik

Excel: 25 tips – Including Vlookup

& Vlookup: https://www.youtube.com/watch?v=y8ygx1Zkcgs

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.

Tutorial: https://www.youtube.com/watch?v=JNm3M9cqWyc

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

Alaska

2

Arizona

3

Arkansas

4

California

5

Colorado

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

Nebraska

27

Nevada

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?

Answer: Alaska, Connecticut, Georgia, Kansas, Nebraska, South Carolina,

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

Answer: Alaska, Connecticut, Georgia, Kansas, Nebraska, South Carolina,

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

Answer: Alaska, Connecticut, Georgia, Kansas, Nebraska, South Carolina,

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

Nebraska

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?

Answer:

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.

Answers will vary

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.

Answers (a-d) will vary depending on the coins in your possession.

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?

Answers and steps:

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

Answers and steps:

https://www.chegg.com/homework-help/basic-statistics-for-business-and-economics-9th-editionchapter-8-problem-1e-solution-9781260501674?trackid=65a00ebd&strackid=1a664d65&ii=1

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

Purchase answer to see full

attachment

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?

Why do business wing it?

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”

Errors [even] using inadequate data

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:

Start with an Unbiased Mind-Set (must be clinically

objective)

Collect a Reasonable and Accurate Data Sample

Perform a proper Analysis (be analytical)

Reality test your conclusions with respected and more

experienced advisors (keep your ego in check)

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.

The sample results are adequate.

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

Tutorial: https://www.youtube.com/watch?v=PdXDLNNXPik

Excel: 25 tips – Including Vlookup

& Vlookup: https://www.youtube.com/watch?v=y8ygx1Zkcgs

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.

Tutorial: https://www.youtube.com/watch?v=JNm3M9cqWyc

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

Alaska

2

Arizona

3

Arkansas

4

California

5

Colorado

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

Nebraska

27

Nevada

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?

Answer: Alaska, Connecticut, Georgia, Kansas, Nebraska, South Carolina,

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

Answer: Alaska, Connecticut, Georgia, Kansas, Nebraska, South Carolina,

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

Answer: Alaska, Connecticut, Georgia, Kansas, Nebraska, South Carolina,

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

Nebraska

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?

Answer:

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.

Answers will vary

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.

Answers (a-d) will vary depending on the coins in your possession.

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?

Answers and steps:

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

Answers and steps:

https://www.chegg.com/homework-help/basic-statistics-for-business-and-economics-9th-editionchapter-8-problem-1e-solution-9781260501674?trackid=65a00ebd&strackid=1a664d65&ii=1

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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