## Description

**INSTANT DOWNLOAD COMPLETE TEST BANK WITH ANSWERS**

**Business Statistics for Contemporary Decision Making 7th Edition by Black – Test Bank**

File: ch02, Chapter 2: Charts and Graphs

True/False

- A summary of data in which raw data are grouped into different intervals and the number of items in each group is listed is called a frequency distribution.

Ans: True

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- A graphical representation of a frequency distribution is called a pie chart.

Ans: False

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Easy

Learning Objective: 2.3: Construct different types of qualitative data graphs, including pie charts, bar graphs, and Pareto charts, in order to interpret the data being graphed.

- If the individual class frequency is divided by the total frequency, the result is the median frequency.

Ans: False

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- A cumulative frequency polygon is also called an ogive.

Ans: True

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

Learning Objective: 2.2: Construct different types of quantitative data graphs, including histograms, frequency polygons, ogives, dot plots, and stem-and-leaf plots, in order to interpret the data being graphed.

- A histogram can be described as a type of vertical bar chart.

Ans: True

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

Learning Objective: 2.2: Construct different types of quantitative data graphs, including histograms, frequency polygons, ogives, dot plots, and stem-and-leaf plots, in order to interpret the data being graphed.

- For any given data set, a frequency distribution with a larger number of classes will always be better than the one with a smaller number of classes.

Ans: False

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- One advantage of a stem and leaf plot over a frequency distribution is that the values of the original data are retained.

Ans: True

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

Learning Objective: 2.2: Construct different types of quantitative data graphs, including histograms, frequency polygons, ogives, dot plots, and stem-and-leaf plots, in order to interpret the data being graphed.

- One rule that must always be followed in constructing frequency distributions is that the adjacent classes must overlap.

Ans: False

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- For a company in gardening supplies business, the best graphical way to show the percentage of a total budget that is spent on each of a number of different expense categories is the stem and leaf plot.

Ans: False

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Hard

- A cumulative frequency distribution provides a running total of the frequencies in the classes.

Ans: True

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The difference between the highest number and the lowest number in a set of data is called the differential frequency.

Ans: False

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- In a histogram, the tallest bar represents the class with the highest cumulative frequency.

Ans: False

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- A scatter plot shows how the numbers in a data set are scattered around their average.

Ans: False

Response: See section 2.4 Charts and Graphs for Two Variables.

Difficulty: Medium

Learning objective: 2.4: Recognize basic trends in two-variable scatter plots of numerical data.

- A scatter plot is a two-dimensional graph plot of data containing pairs of observations on two numerical variables.

Ans: True

Response: See section 2.4 Charts and Graphs for Two Variables

Difficulty: Medium

Learning objective: 2.4: Recognize basic trends in two-variable scatter plots of numerical data.

- A scatter plot is useful for examining the relationship between two numerical variables.

Ans: True

Response: See section 2.4 Charts and Graphs for Two Variables

Difficulty: Medium

Learning objective: 2.4: Recognize basic trends in two-variable scatter plots of numerical data.

- Dot Plots are mainly used to display a large data set.

Ans: False

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

Multiple Choice

- Consider the following frequency distribution:

Class Interval Frequency

10-under 20 15

20-under 30 25

30-under 40 10

What is the midpoint of the first class?

- a) 10
- b) 20
- c) 15
- d) 30
- e) 40

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the following frequency distribution:

Class Interval Frequency

10-under 20 15

20-under 30 25

30-under 40 10

What is the relative frequency of the first class?

- a) 0.15
- b) 0.30
- c) 0.10
- d) 0.20
- e) 0.40

Ans: b

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the following frequency distribution:

Class Interval Frequency

10-under 20 15

20-under 30 25

30-under 40 10

What is the cumulative frequency of the second class interval?

- a) 25
- b) 40
- c) 15
- d) 50

Ans: b

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The number of phone calls arriving at a switchboard each hour has been recorded, and the following frequency distribution has been developed.

Class Interval Frequency

20-under 40 30

40-under 60 45

60-under 80 80

80-under 100 45

What is the midpoint of the last class?

- a) 80
- b) 100
- c) 95
- d) 90
- e) 85

Ans: d

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The number of phone calls arriving at a switchboard each hour has been recorded, and the following frequency distribution has been developed.

Class Interval Frequency

20-under 40 30

40-under 60 45

60-under 80 80

80-under 100 45

What is the relative frequency of the second class?

- a) 0.455
- b) 900
- c) 0.225
- d) 0.750
- e) 0.725

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The number of phone calls arriving at a switchboard each hour has been recorded, and the following frequency distribution has been developed.

Class Interval Frequency

20-under 40 30

40-under 60 45

60-under 80 80

80-under 100 45

What is the cumulative frequency of the third class?

- a) 80
- b) 0.40
- c) 155
- d) 75
- e) 105

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the following stem and leaf plot:

Stem Leaf

1 0, 2, 5, 7

2 2, 3, 4, 4

3 0, 4, 6, 6, 9

4 5, 8, 8, 9

5 2, 7, 8

Suppose that a frequency distribution was developed from this, and there were 5 classes (10-under 20, 20-under 30, etc.). What would the frequency be for class 30-under 40?

- a) 3
- b) 4
- c) 6
- d) 7
- e) 5

Ans: e

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- Consider the following stem and leaf plot:

Stem Leaf

1 0, 2, 5, 7

2 2, 3, 4, 8

3 0, 4, 6, 6, 9

4 5, 8, 8, 9

5 2, 7, 8

Suppose that a frequency distribution was developed from this, and there were 5 classes (10-under 20, 20-under 30, etc.). What would be the relative frequency of the class 20-under 30?

- a) 0.4
- b) 0.25
- c) 0.20
- d) 4
- e) 0.50

Ans: c

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- Consider the following stem and leaf plot:

Stem Leaf

1 0, 2, 5, 7

2 2, 3, 4, 8

3 0, 4, 6, 6, 9

4 5, 8, 8, 9

5 2, 7, 8

Suppose that a frequency distribution was developed from this, and there were 5 classes (10-under 20, 20-under 30, etc.). What was the highest number in the data set?

- a) 50
- b) 58
- c) 59
- d) 78
- e) 98

Ans: b

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- Consider the following stem and leaf plot:

Stem Leaf

1 0, 2, 5, 7

2 2, 3, 4, 8

3 0, 4, 6, 6, 9

4 5, 8, 8, 9

5 2, 7, 8

Suppose that a frequency distribution was developed from this, and there were 5 classes (10-under 20, 20-under 30, etc.). What was the lowest number in the data set?

- a) 0
- b) 10
- c) 7
- d) 2
- e) 1

Ans: b

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- Consider the following stem and leaf plot:

Stem Leaf

1 0, 2, 5, 7

2 2, 3, 4, 8

3 0, 4, 6, 6, 9

4 5, 8, 8, 9

5 2, 7, 8

Suppose that a frequency distribution was developed from this, and there were 5 classes (10-under 20, 20-under 30, etc.). What is the cumulative frequency for the 30-under 40 class interval?

- a) 5
- b) 9
- c) 13
- d) 14
- e) 18

Ans: c

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- An instructor has decided to graphically represent the grades on a test. The instructor uses a plus/minus grading system (i.e. she gives grades of A-, B+, etc.). Which of the following would provide the most information for the students?
- a) A histogram
- b) bar chart
- c) A cumulative frequency distribution
- d) A frequency distribution
- e) A scatter plot

Ans: b

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Medium

Learning Objective: 2.3: Construct different types of qualitative data graphs, including pie charts, bar graphs, and Pareto charts, in order to interpret the data being graphed.

- The following represent the ages of students in a class:

19, 23, 21, 19, 19, 20, 22, 31, 21, 20

If a stem and leaf plot were to be developed from this, how many stems would there be?

- a) 2
- b) 3
- c) 4
- d) 5
- e) 10

Ans: b

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- A person has decided to construct a frequency distribution for a set of data containing 60 numbers. The lowest number is 23 and the highest number is 68. If 5 classes are used, the class width should be approximately _______.
- a) 4
- b) 12
- c) 8
- d) 5
- e) 9

Ans: e

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- A person has decided to construct a frequency distribution for a set of data containing 60 numbers. The lowest number is 23 and the highest number is 68. If 7 classes are used, the class width should be approximately _______.
- a) 5
- b) 7
- c) 9
- d) 11
- e) 12

Ans: b

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- A frequency distribution was developed. The lower endpoint of the first class is 9.30, and the midpoint is 9.35. What is the upper endpoint of this class?
- a) 9.50
- b) 9.60
- c) 9.70
- d) 9.40
- e) 9.80

Ans: d

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The cumulative frequency for a class is 27. The cumulative frequency for the next (non-empty) class will be _______.
- a) less than 27
- b) equal to 27
- c) next class frequency minus 27
- d) 27 minus the next class frequency
- e) 27 plus the next class frequency

Ans: e

Response: See section 2.1 Frequency Distributions

Difficulty: Hard

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The following class intervals for a frequency distribution were developed to provide information regarding the starting salaries for students graduating from a particular school:

Salary Number of Graduates

($1,000s)

18-under 21 –

21-under 25 –

24-under 27 –

29-under 30 –

Before data was collected, someone questioned the validity of this arrangement. Which of the following represents a problem with this set of intervals?

- a) There are too many intervals.
- b) The class widths are too small.
- c) Some numbers between 18,000 and 30,000 would fall into two different intervals.
- d) The first and the second interval overlap.
- e) There are too few intervals.

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The following class intervals for a frequency distribution were developed to provide information regarding the starting salaries for students graduating from a particular school:

Salary Number of Graduates

($1,000s)

18-under 21 –

21-under 25 –

24-under 27 –

29-under 30 –

Before data was collected, someone questioned the validity of this arrangement. Which of the following represents a problem with this set of intervals?

- a) There are too many intervals.
- b) The class widths are too small.
- c) Some numbers between 18,000 and 30,000 would not fall into any of these intervals.
- d) The first and the second interval overlap.
- e) There are too few intervals.

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Hard

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The following class intervals for a frequency distribution were developed to provide information regarding the starting salaries for students graduating from a particular school:

Salary Number of Graduates

($1,000s)

18-under 21 –

21-under 25 –

24-under 27 –

29-under 30 –

Before data was collected, someone questioned the validity of this arrangement. Which of the following represents a problem with this set of intervals?

- a) There are too many intervals.
- b) The class widths are too small.
- c) The class widths are too large.
- d) The second and the third interval overlap.
- e) There are too few intervals.

Ans: d

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Abel Alonzo, Director of Human Resources, is exploring employee absenteeism at the Harrison Haulers Plant during the last operating year. A review of all personnel records indicated that absences ranged from zero to twenty-nine days per employee. The following class intervals were proposed for a frequency distribution of absences.

Absences Number of Employees

(Days)

0-under 5 –

5-under 10 –

10-under 15 –

20-under 25 –

25-under 30 –

Which of the following represents a problem with this set of intervals?

- a) There are too few intervals.
- b) Some numbers between 0 and 29, inclusively, would not fall into any interval.
- c) The first and second interval overlaps.
- d) There are too many intervals.
- e) The second and the third interval overlap.

Ans: b

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Abel Alonzo, Director of Human Resources, is exploring employee absenteeism at the Harrison Haulers Plant during the last operating year. A review of all personnel records indicated that absences ranged from zero to twenty-nine days per employee. The following class intervals were proposed for a frequency distribution of absences.

Absences Number of Employees

(Days)

0-under 10 –

10-under 20 –

20-under 30 –

Which of the following might represent a problem with this set of intervals?

- a) There are too few intervals.
- b) Some numbers between 0 and 29 would not fall into any interval.
- c) The first and second interval overlaps.
- d) There are too many intervals.
- e) The second and the third interval overlap.

Ans: a

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the relative frequency distribution given below:

Class Interval Relative Frequency

20-under 40 0.2

40-under 60 0.3

60-under 80 0.4

80-under 100 0.1

There were 60 numbers in the data set. How many numbers were in the interval 20-under 40?

- a) 12
- b) 20
- c) 40
- d) 10
- e) 15

Ans: a

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the relative frequency distribution given below:

Class Interval Relative Frequency

20-under 40 0.2

40-under 60 0.3

60-under 80 0.4

80-under 100 0.1

There were 60 numbers in the data set. How many numbers were in the interval 40-under 60?

- a) 30
- b) 50
- c) 18
- d) 12
- e) 15

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the relative frequency distribution given below:

Class Interval Relative Frequency

20-under 40 0.2

40-under 60 0.3

60-under 80 0.4

80-under 100 0.1

There were 60 numbers in the data set. How many of the number were less than 80?

- a) 90
- b) 80
- c) 0.9
- d) 54
- e) 100

Ans: d

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the following frequency distribution:

Class Interval Frequency

100-under 200 25

200-under 300 45

300-under 400 30

What is the midpoint of the first class?

- a) 100
- b) 150
- c) 25
- d) 250
- e) 200

Ans: b

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the following frequency distribution:

Class Interval Frequency

100-under 200 25

200-under 300 45

300-under 400 30

What is the relative frequency of the second class interval?

- a) 0.45
- b) 0.70
- c) 0.30
- d) 0.33
- e) 0.50

Ans: a

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the following frequency distribution:

Class Interval Frequency

100-under 200 25

200-under 300 45

300-under 400 30

What is the cumulative frequency of the second class interval?

- a) 25
- b) 45
- c) 70
- d) 100
- e) 250

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Consider the following frequency distribution:

Class Interval Frequency

100-under 200 25

200-under 300 45

300-under 400 30

What is the midpoint of the last class interval?

- a) 15
- b) 350
- c) 300
- d) 200
- e) 400

Ans: b

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Pinky Bauer, Chief Financial Officer of Harrison Haulers, Inc., suspects irregularities in the payroll system and orders an inspection of “each and every payroll voucher issued since January 1, 2000.” Each payroll voucher was inspected and the following frequency distribution was compiled.

Errors per Voucher Number of Vouchers

0-under 2 500

2-under 4 400

4-under 6 300

6-under 8 200

8-under 10 100

The relative frequency of the first class interval is _________.

- a) 0.50
- b) 0.33
- c) 0.40
- d) 0.27
- e) 0.67

Ans: b

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Pinky Bauer, Chief Financial Officer of Harrison Haulers, Inc., suspects irregularities

in the payroll system and orders an inspection of “each and every payroll voucher issued since January 1, 2000.” Each payroll voucher was inspected and the following frequency distribution was compiled.

Errors per Voucher Number of Vouchers

0-under 2 500

2-under 4 400

4-under 6 300

6-under 8 200

8-under 10 100

The cumulative frequency of the second class interval is _________.

- a) 1,500
- b) 500
- c) 900
- d) 1,000
- e) 1,200

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Medium

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- Pinky Bauer, Chief Financial Officer of Harrison Haulers, Inc., suspects irregularities in the payroll system and orders an inspection of “each and every payroll voucher issued since January 1, 2000.” Each payroll voucher was inspected and the following frequency distribution was compiled.

Errors per Voucher Number of Vouchers

0-under 2 500

2-under 4 400

4-under 6 300

6-under 8 200

8-under 10 100

The midpoint of the first class interval is _________.

- a) 500
- b) 2
- c) 1.5
- d) 1
- e) 250

Ans: d

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The 1999 and 2000 market share data of the three competitors (A, B, and C) in an oligopolistic industry are presented in the following pie charts.

Which of the following is true?

- a) Only company B gained market share.
- b) Only company C lost market share.
- c) Company A lost market share.
- d) Company B lost market share.
- e) All companies lost market share

Ans: b

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Medium

Learning Objective: 2.3: Construct different types of qualitative data graphs, including pie charts, bar graphs, and Pareto charts, in order to interpret the data being graphed.

- The 1999 and 2000 market share data of the three competitors (A, B, and C) in an oligopolistic industry are presented in the following pie charts. Total sales for this industry were $1.5 billion in 1999 and $1.8 billion in 2000. Company C’s sales in 2000 were ___________.

- a) $342 million
- b) $630 million
- c) $675 million
- d) $828 million
- e) $928 million

Ans: a

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Medium

- The 1999 and 2000 market share data of the three competitors (A, B, and C) in an oligopolistic industry are presented in the following pie charts. Total sales for this industry were $1.5 billion in 1999 and $1.8 billion in 2000.

Company B’s sales in 1999 were ___________.

- a) $342 million
- b) $630 million
- c) $675 million
- d) $828 million
- e) $928 million

Ans: c

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Medium

- The 1999 and 2000 market share data of the three competitors (A, B, and C) in an oligopolistic industry are presented in the following pie charts.

Which of the following may be a false statement?

- a) Sales revenues declined at company C.
- b) Only company C lost market share.
- c) Company A gained market share.
- d) Company B gained market share.
- e) Both Company A and Company B gained market share

Ans: a

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Hard

- Each day, the office staff at Oasis Quick Shop prepares a frequency distribution and an ogive of sales transactions by dollar value of the transactions. Saturday’s cumulative frequency ogive follows.

The total number of sales transactions on Saturday was _____________.

- a) 200
- b) 500
- c) 300
- d) 100
- e) 400

Ans: b

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- Each day, the office staff at Oasis Quick Shop prepares a frequency distribution and an ogive of sales transactions by dollar value of the transactions. Saturday’s cumulative frequency ogive follows.

The percentage of sales transactions on Saturday that were under $100 each was _____________.

- a) 100
- b) 10
- c) 80
- d) 20
- e) 15

Ans: d

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- Each day, the office staff at Oasis Quick Shop prepares a frequency distribution and an ogive of sales transactions by dollar value of the transactions. Saturday’s cumulative frequency ogive follows.

The percentage of sales transactions on Saturday that were at least $100 each was _____________.

- a) 100
- b) 10
- c) 80
- d) 20
- e) 15

Ans: c

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

The percentage of sales transactions on Saturday that were between $100 and $150 was _____________.

- a) 20%
- b) 40%
- c) 60%
- d) 80%
- e) 10%

Ans: c

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Hard

- Each day, the office staff at Oasis Quick Shop prepares a frequency distribution and a histogram of sales transactions by dollar value of the transactions. Friday’s histogram follows.

On Friday, the approximate number of sales transactions in the 125-under 150 category was _____________.

- a) 50
- b) 100
- c) 150
- d) 200
- e) 85

Ans: d

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- Each day, the office staff at Oasis Quick Shop prepares a frequency distribution and a histogram of sales transactions by dollar value of the transactions. Friday’s histogram follows.

On Friday, the approximate number of sales transactions between $100 and $150 was _____________.

- a) 100
- b) 200
- c) 300
- d) 400
- e) 500

Ans: c

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- The staff of Mr. Wayne Wertz, VP of Operations at Portland Peoples Bank, prepared a cumulative frequency ogive of waiting time for walk-in customers.

The total number of walk-in customers included in the study was _________.

- a) 100
- b) 250
- c) 300
- d) 450
- e) 500

Ans: d

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- The staff of Mr. Wayne Wertz, VP of Operations at Portland Peoples Bank, prepared a cumulative frequency ogive of waiting time for walk-in customers.

The percentage of walk-in customers waiting one minute or less was _________.

- a) 22%
- b) 11%
- c) 67%
- d) 10%
- e) 5%

Ans: a

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- The staff of Mr. Wayne Wertz, VP of Operations at Portland Peoples Bank, prepared a cumulative frequency ogive of waiting time for walk-in customers.

The percentage of walk-in customers waiting more than 6 minutes was ______.

- a) 22%
- b) 11%
- c) 67%
- d) 10%
- e) 75%

Ans: b

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

The percentage of walk-in customers waiting between 1 and 6 minutes was ___.

- a) 22%
- b) 11%
- c) 37%
- d) 10%
- e) 67%

Ans: e

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- The staff of Mr. Wayne Wertz, VP of Operations at Portland Peoples Bank, prepared a frequency histogram of waiting time for walk-in customers.

Approximately _____ walk-in customers waited less than 2 minutes.

- a) 20
- b) 30
- c) 100
- d) 180
- e) 200

Ans: d

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- The staff of Mr. Wayne Wertz, VP of Operations at Portland Peoples Bank, prepared a frequency histogram of waiting time for walk-in customers.

Approximately ____ walk-in customers waited at least 7 minutes.

- a) 20
- b) 30
- c) 100
- d) 180
- e) 200

Ans: b

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- The staffs of the accounting and the quality control departments rated their respective supervisor’s leadership style as either (1) authoritarian or (2) participatory. Sixty-eight percent of the accounting staff rated their supervisor “authoritarian,” and thirty-two percent rated him “participatory.” Forty percent of the quality control staff rated their supervisor “authoritarian,” and sixty percent rated her “participatory.” The best graphic depiction of these data would be two ___________________.
- a) histograms
- b) frequency polygons
- c) ogives
- d) pie charts
- e) scatter plots

Ans: d

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Hard

- The staff of Ms. Tamara Hill, VP of Technical Analysis at Blue Sky Brokerage, prepared a frequency histogram of market capitalization of the 937 corporations listed on the American Stock Exchange in January 2003.

Approximately ________ corporations had capitalization exceeding $200,000,000.

- a) 50
- b) 100
- c) 700
- d) 800
- e) 890

Ans: b

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- The staff of Ms. Tamara Hill, VP of Technical Analysis at Blue Sky Brokerage, prepared a frequency histogram of market capitalization of the 937 corporations listed on the American Stock Exchange in January 2003.

Approximately ________ corporations had capitalizations of $200,000,000 or less.

- a) 50
- b) 100
- c) 700
- d) 800
- e) 900

Ans: d

Response: See section 2.2 Quantitative Data Graphs

Difficulty: Medium

- The following graphic of PCB Failures is a _____________.

- a) Scatter Plot
- b) Pareto Chart
- c) Pie Chart
- d) Cumulative Histogram Chart
- e) Line diagram

Ans: b

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Medium

- According to the following graphic, the most common cause of PCB Failures is a _____________.

- a) Cracked Trace
- b) Bent Pin
- c) Missing Part
- d) Solder Bridge
- e) Wrong Part

Ans: a

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Medium

Learning Objective: 2.3: Construct different types of qualitative data graphs, including pie charts, bar graphs, and Pareto charts, in order to interpret the data being graphed

- According to the following graphic, “Bent Pins” account for ____% of PCB Failures.

- a) 10
- b) 20
- c) 30
- d) 40
- e) 50

Ans: b

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Hard

Learning Objective: 2.3: Construct different types of qualitative data graphs, including pie charts, bar graphs, and Pareto charts, in order to interpret the data being graphed

- The following graphic of residential housing data (selling price and size in square feet) is a _____________.

- a) scatter plot
- b) Pareto chart
- c) pie chart
- d) cumulative histogram
- e) cumulative frequency distribuion

Ans: a

Response: See section 2.4 Graphical Depiction of Two-Variable Numerical Data: Scatter Plots

Difficulty: Medium

Learning objective: 2.4: Recognize basic trends in two-variable scatter plots of numerical data.

- The following graphic of residential housing data (selling price and size in square feet) indicates _____________.

- a) an inverse relation between the two variables
- b) no relation between the two variables
- c) a direct relation between the two variables
- d) a negative exponential relation between the two variables
- e) a sinusoidal relationship between the two variables

Ans: c

Response: See section 2.4 Graphical Depiction of Two-Variable Numerical Data: Scatter Plots

Difficulty: Medium

Learning objective: 2.4: Recognize basic trends in two-variable scatter plots of numerical data.

- An instructor made a frequency table of the scores his students got on a test

Score Frequency

30-under 40 1

40-under 50 4

50-under 60 5

60-under 70 10

70-under 80 20

80-under 90 10

90-under 100 5

The midpoint of the last class interval is _________.

- a) 90
- b) 5
- c) 95
- d) 100
- e) 50

Ans: c

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- An instructor made a frequency table of the scores his students got on a test

Score Frequency

30-under 40 1

40-under 50 4

50-under 60 5

60-under 70 10

70-under 80 20

80-under 90 10

90-under 100 5

Approximately what percent of students got more than 70?

- a) 36
- b) 20
- c) 50
- d) 10
- e) 64

Ans: e

Response: See section 2.1 Frequency Distributions

Difficulty: Easy

Learning Objective: 2.1: Construct a frequency distribution from a set of data.

- The following is a bar chart of the self-reported race for 189 pregnant women.

Approximately _____ percent of pregnant women are African-American

- a) 20
- b) 14
- c) 5
- d) 35
- e) 50

Ans: b

Response: See section 2.3 Qualitative Data Graphs

Difficulty: Medium

- The following graphic of cigarettes smoked (sold) per capita (CIG) and deaths per 100K population from lung cancer (LUNG) indicates _________

- a) a weak relation between the two variables
- b) a pretty strong relation between the two variables
- c) when the number of cigarettes smoked (sold) per capita (CIG) increases the deaths per 100K population from lung cancer (LUNG)decreases
- d) a negative relation between the two variables
- e) no relation between the two variables

Ans: b

Response: See section 2.4 Graphical Depiction of Two-Variable Numerical Data: Scatter Plots

Difficulty: Medium

Learning objective: 2.4: Recognize basic trends in two-variable scatter plots of numerical

File: ch04, Chapter 4: Probability

True/false

- Inferring the value of a population parameter from the statistic on a random sample drawn from the population is an inferential process under uncertainty.

Ans: True

Response: See section 4.1 Introduction to Probability

Difficulty: Easy

Learning Objective: 4.1: Describe what probability is and when one would use it.

- The method of assigning probabilities to uncertain outcomes based on laws and rules is called the classical method.

Ans: True

Response: See section 4.2 Methods of Assigning Probabilities

Difficulty: Easy

Learning Objective: 4.2: Differentiate among three methods of assigning probabilities: the classical method, relative frequency of occurrence, and subjective probability.

- Assigning probabilities by dividing the number of ways that an event can occur by the total number of possible outcomes in an experiment is called the relative frequency of occurrence method.

Ans: False

Response: See section 4.2 Methods of Assigning Probabilities

Difficulty: Easy

Learning Objective: 4.2: Differentiate among three methods of assigning probabilities: the classical method, relative frequency of occurrence, and subjective probability.

- Assigning probabilities to uncertain events based on one’s beliefs or intuitions is called classical method.

Ans: False

Response: See section 4.2 Methods of Assigning Probabilities

Difficulty: Easy

Learning Objective: 4.2: Differentiate among three methods of assigning probabilities: the classical method, relative frequency of occurrence, and subjective probability.

- An experiment is a process that produces outcomes.

Ans: True

Response: See section 4.3 Structure of Probability

Difficulty: Easy

Learning Objective: 4.3: Deconstruct the elements of probability by defining experiments, sample spaces, and events, classifying events as mutually exclusive, collectively exhaustive, complementary, or independent, and counting possibilities.

- An event that cannot be broken down into other events is called a certainty outcome.

Ans: False

Response: See section 4.3 Structure of Probability

Difficulty: Easy

Learning Objective: 4.3: Deconstruct the elements of probability by defining experiments, sample spaces, and events, classifying events as mutually exclusive, collectively exhaustive, complementary, or independent, and counting possibilities.

- The list of all elementary events for an experiment is called the sample space.

Ans: True

Response: See section 4.3 Structure of Probability

Difficulty: Easy

Learning Objective: 4.3: Deconstruct the elements of probability by defining experiments, sample spaces, and events, classifying events as mutually exclusive, collectively exhaustive, complementary, or independent, and counting possibilities.

- If the occurrence of one event does not affect the occurrence of another event, then the two events are mutually exclusive.

Ans: False

Response: See section 4.3 Structure of Probability

Difficulty: Hard

- If the occurrence of one event precludes the occurrence of another event, then the two events are independent.

Ans: False

Response: See section 4.3 Structure of Probability

Difficulty: Hard

- If two events are mutually exclusive, then the two events are also independent.

Ans: False

Response: See section 4.3 Structure of Probability

Difficulty: Hard

- If two events are mutually exclusive, then their joint probability is always zero.

Ans: True

Response: See section 4.5 Addition Laws

Difficulty: Medium

Learning Objective: 4.5: Calculate probabilities using the general law of addition, along with a joint probability table, the complement of a union, or the special law of addition if necessary.

- Given two events, A and B, if the probability of either A or B occurring is 0.8, then the probability of neither A nor B occurring is -0.8.

Ans: False

Response: See section 4.5 Addition Laws

Difficulty: Easy

Learning Objective: 4.5: Calculate probabilities using the general law of addition, along with a joint probability table, the complement of a union, or the special law of addition if necessary.

- Given two events, A and B, if the probability of A is 0.6, the probability of B is 0.4, and the joint probability of A and B is 0.24, then the two events are independent.

Ans: True

Response: See section 4.6 Multiplication Laws

Difficulty: Medium

Learning Objective: 4.6 Calculate joint probabilities of both independent and dependent events using the general and special laws of multiplication.

- Given that two events, A and B, are independent, if the marginal probability of A is 0.6, the conditional probability of A given B will be 0.4.

Ans: False

Response: See section 4.7 Conditional Probability

Difficulty: Medium

Learning Objective: 4.7: Calculate conditional probabilities with various forms of the law of conditional probability, and use them to determine if two events are independent.

- Given two events A and B each with a non-zero probability, if the conditional probability of A given B is zero, it implies that the events A and B are independent.

Ans: False

Response: See section 4.7 Conditional Probability

Difficulty: Hard

Learning Objective: 4.7: Calculate conditional probabilities with various forms of the law of conditional probability, and use them to determine if two events are independent

- Given two events A and B each with a non-zero probability, if the conditional probability of A given B is zero, it implies that the events A and B are mutually exclusive.

Ans: True

Response: See section 4.7 Conditional Probability

Difficulty: Hard

Learning Objective: 4.7: Calculate conditional probabilities with various forms of the law of conditional probability, and use them to determine if two events are independent.

- Bayes’ rule is a rule to assign probabilities under the classical method.

Ans: False

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Easy

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

- Bayes’ rule is an extension of the law of conditional probabilities to allow revision of original probabilities with new information.

Ans: True

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Easy

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

- The probability of A È B where A is receiving a state grant and B is receiving a federal grant is the probability of receiving no more than one of the two grants.

Ans: False

Response: See section 4.4 Marginal, Union, Joint, and Conditional Probabilities

Difficulty: Easy

Learning Objective: 4.4: Compare marginal, union, joint, and conditional probabilities by defining each one.

- Probability is used to develop knowledge of the fundamental mathematical tools for quantitatively assessing risk.

Ans: True

Response: See section 4.1 Introduction to Probability

Difficulty: Easy

Learning Objective: 4.1: Describe what probability is and when one would use it.

Multiple Choice

- Belinda Bose is reviewing a newly proposed advertising campaign. Based on her 15 years experience, she believes the campaign has a 75% chance of significantly increasing brand name recognition of the product. This is an example of assigning probabilities using the ________________ method.
- a) subjective probability
- b) relative frequency
- c) classical probability
- d)
*a**priori*probability - e)
*a posterior*probability

Ans: a

Response: See section 4.2 Methods of Assigning Probabilities

Difficulty: Medium

- Which of the following is
**not**a legitimate probability value? - a) 0.67
- b) 15/16
- c) 0.23
- d) 4/3
- e) 0.98

Ans: d

Response: See section 4.2 Methods of Assigning Probabilities

Difficulty: Easy

- Which of the following is
**not**a legitimate probability value? - a) 0.67
- b) 15/16
- c) 0.23
- d) 2/3
- e) -0.28

Ans: e

Response: See section 4.2 Methods of Assigning Probabilities

Difficulty: Easy

- The list of all elementary events for an experiment is called _______.
- a) the sample space
- b) the exhaustive list
- c) the population space
- d) the event union
- e) a frame

Ans: a

Response: See section 4.3 Structure of Probability

Difficulty: Easy

- In a set of 15 aluminum castings, two castings are defective (D), and the remaining thirteen are good (G). A quality control inspector randomly selects three of the fifteen castings without replacement, and classifies each as defective (D) or good (G). The sample space for this experiment contains ____________ elementary events.
- a) 3,375
- b) 2,730
- c) 455
- d) 15
- e) 3

Ans: c

Response: See section 4.3 Structure of Probability

Difficulty: Medium

- In a set of 10 aluminum castings, two castings are defective (D), and the remaining eight are good (G). A quality control inspector randomly selects three of the ten castings with replacement, and classifies each as defective (D) or good (G). The sample space for this experiment contains __________ elementary events.
- a) 1,000
- b) 720
- c) 100
- d) 10
- e) 3

Ans: a

Response: See section 4.3 Structure of Probability

Difficulty: Medium

- If X and Y are mutually exclusive events, then if X occurs _______.
- a) Y must also occur
- b) Y cannot occur
- c) X and Y are independent
- d) X and Y are complements
- e) A and Y are collectively exhaustive

Ans: b

Response: See section 4.3 Structure of Probability

Difficulty: Easy

- Consider the following sample space, S, and several events defined on it. S = {Albert, Betty, Abel, Jack, Patty, Meagan}, and the events are: F = {Betty, Patty, Meagan}, H = {Abel, Meagan}, and P = {Betty, Abel}. F Ç H is ___________.
- a) {Meagan}
- b) {Betty, Patty, Abel, Meagan}
- c) empty, since F and H are complements
- d) empty, since F and H are independent
- e) empty, since F and H are mutually exclusive

Ans: a

Response: See section 4.3 Structure of Probability

Difficulty: Easy

- Consider the following sample space, S, and several events defined on it. S = {Albert, Betty, Abel, Jack, Patty, Meagan}, and the events are: F = {Betty, Patty, Meagan}, H = {Abel, Meagan}, and P = {Betty, Abel}. F È H is ___________.
- a) {Meagan}
- b) {Betty, Abel, Patty, Meagan}
- c) empty, since F and H are complements
- d) empty, since F and H are independent
- e) empty, since F and H are mutually exclusive

Ans: b

Response: See section 4.3 Structure of Probability

Difficulty: Easy

- Consider the following sample space, S, and several events defined on it. S = {Albert, Betty, Abel, Jack, Patty, Meagan}, and the events are: F = {Betty, Patty, Meagan}, H = {Abel, Meagan}, and P = {Betty, Abel}. The complement of F is ___________.
- a) {Albert, Betty, Jack, Patty}
- b) {Betty, Patty, Meagan}
- c) {Albert, Abel, Jack}
- d) {Betty, Abel}
- e) {Meagan}

Ans: c

Response: See section 4.3 Structure of Probability

Difficulty: Easy

- If X and Y are mutually exclusive, then _______.
- a) the probability of the union is zero
- b) P(X) = 1 – P(Y)
- c) the probability of the intersection is zero
- d) the probability of the union is one
- e) P(Y) = P(X)

Ans: c

Response: See section 4.3 Structure of Probability

Difficulty: Easy

- Let A be the event that a student is enrolled in an accounting course, and let S be the event that a student is enrolled in a statistics course. It is known that 30% of all students are enrolled in an accounting course and 40% of all students are enrolled in statistics. Included in these numbers are 15% who are enrolled in both statistics and accounting. Find P(S).
- a) 0.15
- b) 0.30
- c) 0.40
- d) 0.55
- e) 0.60

Ans: c

Response: See section 4.4 Marginal, Union, Joint, and Conditional Probabilities

Difficulty: Easy

Learning Objective: 4.4: Compare marginal, union, joint, and conditional probabilities by defining each one.

- Let A be the event that a student is enrolled in an accounting course, and let S be the event that a student is enrolled in a statistics course. It is known that 30% of all students are enrolled in an accounting course and 40% of all students are enrolled in statistics. Included in these numbers are 15% who are enrolled in both statistics and accounting. Find the probability that a student is in accounting and is also in statistics.
- a) 0.15
- b) 0.70
- c) 0.55
- d) 0.12
- e) 0.60

Ans: a

Response: See section 4.4 Marginal, Union, Joint, and Conditional Probabilities

Difficulty: Easy

Learning Objective: 4.4: Compare marginal, union, joint, and conditional probabilities by defining each one.

- Let A be the event that a student is enrolled in an accounting course, and let S be the event that a student is enrolled in a statistics course. It is known that 30% of all students are enrolled in an accounting course and 40% of all students are enrolled in statistics. Included in these numbers are 15% who are enrolled in both statistics and accounting. A student is randomly selected, and it is found that the student is enrolled in accounting. What is the probability that this student is also enrolled in statistics?
- a) 0.15
- b) 0.75
- c) 0.375
- d) 0.50
- e) 0.80

Ans: d

Response: See section 4.7 Conditional Probability

Difficulty: Medium

Learning Objective: 4.7: Calculate conditional probabilities with various forms of the law of conditional probability, and use them to determine if two events are independent

- Let A be the event that a student is enrolled in an accounting course, and let S be the event that a student is enrolled in a statistics course. It is known that 30% of all students are enrolled in an accounting course and 40% of all students are enrolled in statistics. Included in these numbers are 15% who are enrolled in both statistics and accounting. A student is randomly selected, what is the probability that the student is enrolled in either accounting or statistics or both?
- a) 0.15
- b) 0.85
- c) 0.70
- d) 0.55
- e) 0.90

Ans: d

Response: See section 4.5 Addition Laws

Difficulty: Medium

Learning Objective: 4.5: Calculate probabilities using the general law of addition, along with a joint probability table, the complement of a union, or the special law of addition if necessary.

- Abel Alonzo, Director of Human Resources, is exploring employee absenteeism at the Plano Power Plant. Ten percent of all plant employees work in the finishing department; 20% of all plant employees are absent excessively; and 7% of all plant employees work in the finishing department and are absent excessively. A plant employee is selected randomly; F is the event “works in the finishing department;” and A is the event “is absent excessively.” P(A È F) = _____________.
- a) 0.07
- b) 0.10
- c) 0.20
- d) 0.23
- e) 0.37

Ans: d

Response: See section 4.5 Addition Laws

Difficulty: Medium

- Abel Alonzo, Director of Human Resources, is exploring employee absenteeism at the Plano Power Plant. Ten percent of all plant employees work in the finishing department; 20% of all plant employees are absent excessively; and 7% of all plant employees work in the finishing department and are absent excessively. A plant employee is selected randomly; F is the event “works in the finishing department;” and A is the event “is absent excessively.” P(A|F) = _____________.
- a) 0.37
- b) 0.70
- c) 0.13
- d) 0.35
- e) 0.80

Ans: b

Response: See section 4.7 Conditional Probability

Difficulty: Medium

Learning Objective: 4.7: Calculate conditional probabilities with various forms of the law of conditional probability, and use them to determine if two events are independent.

- Abel Alonzo, Director of Human Resources, is exploring employee absenteeism at the Plano Power Plant. Ten percent of all plant employees work in the finishing department; 20% of all plant employees are absent excessively; and 7% of all plant employees work in the finishing department and are absent excessively. A plant employee is selected randomly; F is the event “works in the finishing department;” and A is the event “is absent excessively.” P(F|A) = _____________.
- a) 0.35
- b) 0.70
- c) 0.13
- d) 0.37
- e) 0.10

Ans: a

Response: See section 4.7 Conditional Probability

Difficulty: Medium

Learning Objective: 4.7: Calculate conditional probabilities with various forms of the law of conditional probability, and use them to determine if two events are independent

- Max Sandlin is exploring the characteristics of stock market investors. He found that sixty percent of all investors have a net worth exceeding $1,000,000; 20% of all investors use an online brokerage; and 10% of all investors a have net worth exceeding $1,000,000 and use an online brokerage. An investor is selected randomly, and E is the event “net worth exceeds $1,000,000” and O is the event “uses an online brokerage.” P(O È E) = _____________.
- a) 0.17
- b) 0.50
- c) 0.80
- d) 0.70
- e) 0.10

Ans: d

Response: See section 4.5 Additional Laws

Difficulty: Medium

- Max Sandlin is exploring the characteristics of stock market investors. He found that sixty percent of all investors have a net worth exceeding $1,000,000; 20% of all investors use an online brokerage; and 10% of all investors a have net worth exceeding $1,000,000 and use an online brokerage. An investor is selected randomly, and E is the event “net worth exceeds $1, 000, 000,” and O is the event “uses an online brokerage.” P(O|E) = _____________.
- a) 0.17
- b) 0.50
- c) 0.80
- d) 0.70
- e) 0.88

Ans: a

Response: See section 4.7 Conditional Probability

Difficulty: Medium

- Given P(A) = 0.40, P(B) = 0.50, P(A Ç B) = 0.15. Find P(A È B).
- a) 0.90
- b) 1.05
- c) 0.75
- d) 0.65
- e) 0.60

Ans: c

Response: See section 4.5 Additional Laws

Difficulty: Easy

- Given P (A) = 0.40, P (B) = 0.50, P (A Ç B) = 0.15. Find P (A|B).
- a) 0.20
- b) 0.80
- c) 0.30
- d) 0.375
- e) 0.15

Ans: c

Response: See section 4.7 Conditional Probability

Difficulty: Medium

43 Given P (A) = 0.40, P (B) = 0.50, P (A Ç B) = 0.15. Find P (B|A).

- a) 0.20
- b) 0.80
- c) 0.30
- d) 0.375
- e) 0.15

Ans: d

Response: See section 4.7 Conditional Probability

Difficulty: Medium

- Given P (A) = 0.40, P (B) = 0.50, P (A Ç B) = 0.15. Which of the following is true?
- a) A and B are independent
- b) A and B are mutually exclusive
- c) A and B are collectively exhaustive
- d) A and B are not independent
- e) A and B are complimentary

Ans: d

Response: See section 4.7 Conditional Probability

Difficulty: Hard

- An automobile dealer wishes to investigate the relation between the gender of the buyer and type of vehicle purchased. Based on the following joint probability table that was developed from the dealer’s records for the previous year, P (Male) = ________

Type of | Buyer Gender | ||

Vehicle | Female | Male | Total |

SUV | |||

Not SUV | .32 | .48 | |

Total | .40 | 1.00 |

- a) 0.48
- b) 0.50
- c) 0.20
- d) 0.02
- e) 0.60

Ans: e

Response: See section 4.5 addition Laws

Difficulty: Medium

- An automobile dealer wishes to investigate the relation between the gender of the buyer and type of vehicle purchased. Based on the joint probability table below that was developed from the dealer’s records for the previous year, P (Female Ç SUV) = _______.

Type of | Buyer Gender | |||

Vehicle | Female | Male | Total | |

SUV | ||||

Not SUV | .30 | .40 | ||

Total | .60 | 1.00 | ||

- a) 0.30
- b) 0.40
- c) 0.12
- d) 0.10
- e) 0.60

Ans: d

Response: See section 4.6 Multiplication Laws

Difficulty: Medium

Learning Objective: 4.6 Calculate joint probabilities of both independent and dependent events using the general and special laws of multiplication.

- An automobile dealer wishes to investigate the relation between the gender of the buyer and type of vehicle purchased. Based on the following joint probability table that was developed from the dealer’s records for the previous year, P (Female) = __________.

Type of | Buyer Gender | ||

Vehicle | Female | Male | Total |

SUV | |||

Not SUV | .30 | .40 | |

Total | .60 | 1.00 |

- a) 0.30
- b) 0.40
- c) 0.12
- d) 0.10
- e) 0.60

Ans: b

Response: See section 4.5 Addition Laws

Difficulty: Medium

- An automobile dealer wishes to investigate the relation between the gender of the buyer and type of vehicle purchased. Based on the following joint probability table that was developed from the dealer’s records for the previous year, P (SUV) = ___________.

Type of | Buyer Gender | ||

Vehicle | Female | Male | Total |

SUV | |||

Not SUV | .30 | .40 | |

Total | .60 | 1.00 |

- a) 0.30
- b) 0.40
- c) 0.12
- d) 0.10
- e) 0.60

Ans: a

Response: See section 4.5 Addition Laws

Difficulty: Medium

- Meagan Dubean manages a portfolio of 200 common stocks. Her staff classified the portfolio stocks by ‘industry sector’ and ‘investment objective.’

Investment | Industry Sector | |||

Objective | Electronics | Airlines | Healthcare | Total |

Growth | 100 | 10 | 40 | 150 |

Income | 20 | 20 | 10 | 50 |

Total | 120 | 30 | 50 | 200 |

If a stock is selected randomly from Meagan’s portfolio, P (Growth) = _______.

- a) 50
- b) 0.83
- c) 0.67
- d) 0.75
- e) 0.90

Ans: d

Response: See section 4.5 Addition Laws

Difficulty: Medium

- Meagan Dubean manages a portfolio of 200 common stocks. Her staff classified the portfolio stocks by ‘industry sector’ and ‘investment objective.’

Investment | Industry Sector | |||

Objective | Electronics | Airlines | Healthcare | Total |

Growth | 100 | 10 | 40 | 150 |

Income | 20 | 20 | 10 | 50 |

Total | 120 | 30 | 50 | 200 |

If a stock is selected randomly from Meagan’s portfolio, P (Healthcare È Electronics) = _______.

- a) 0.25
- b) 0.85
- c) 0.60
- d) 0.75
- e) 0.90

Ans: b

Response: See section 4.5 Addition laws

Difficulty: Medium

- Meagan Dubean manages a portfolio of 200 common stocks. Her staff classified the portfolio stocks by ‘industry sector’ and ‘investment objective.’

Investment | Industry Sector | |||

Objective | Electronics | Airlines | Healthcare | Total |

Growth | 100 | 10 | 40 | 150 |

Income | 20 | 20 | 10 | 50 |

Total | 120 | 30 | 50 | 200 |

If a stock is selected randomly from Meagan’s portfolio, P (Airlines|Income) = _______.

- a) 10
- b) 0.40
- c) 0.25
- d) 0.67
- e) 0.90

Ans: b

Response: See section 4.7 Conditional Probability

Difficulty: Medium

Investment | Industry Sector | |||

Objective | Electronics | Airlines | Healthcare | Total |

Growth | 100 | 10 | 40 | 150 |

Income | 20 | 20 | 10 | 50 |

Total | 120 | 30 | 50 | 200 |

Which of the following statements is **not** true?

- a) Growth and Income are complementary events.
- b) Electronics and Growth are dependent.
- c) Electronics and Healthcare are mutually exclusive.
- d) Airlines and Healthcare are collectively exhaustive.
- e) Growth and Income are collectively exhaustive.

Ans: d

Response: See section 4.3 Structure of Probability

Difficulty: Hard

Investment | Industry Sector | |||

Objective | Electronics | Airlines | Healthcare | Total |

Growth | 84 | 21 | 35 | 140 |

Income | 36 | 9 | 15 | 60 |

Total | 120 | 30 | 50 | 200 |

Which of the following statements is true?

- a) Growth and Healthcare are complementary events.
- b) Electronics and Growth are independent.
- c) Electronics and Growth are mutually exclusive.
- d) Airlines and Healthcare are collectively exhaustive.
- e) Electronics and Healthcare are collectively exhaustive.

Ans: b

Response: See section 4.6 Multiplication Laws

Difficulty: Hard

- The table below provides summary information about students in a class. The sex of each individual and the major is given.

Male | Female | Total | |

Accounting | 12 | 18 | 30 |

Finance | 10 | 8 | 18 |

Other | 26 | 26 | 52 |

Total | 48 | 52 | 100 |

If a student is randomly selected from this group, what is the probability that the student is male?

- a) 12
- b) 0.48
- c) 0.50
- d) 0.52
- e) 0.68

Ans: b

Response: See section 4.5 Addition Laws

Difficulty: Easy

- The table below provides summary information about students in a class. The sex of each individual and the major is given.

Male | Female | Total | |

Accounting | 12 | 18 | 30 |

Finance | 10 | 8 | 18 |

Other | 26 | 26 | 52 |

Total | 48 | 52 | 100 |

If a student is randomly selected from this group, what is the probability that the student is a female who majors in accounting?

- a) 18
- b) 0.60
- c) 0.35
- d) 0.40
- e) 0.78

Ans: a

Response: See Section 4.5 Addition Laws

Difficulty: Easy

- The table below provides summary information about students in a class. The sex of each individual and the major is given.

Male | Female | Total | |

Accounting | 12 | 18 | 30 |

Finance | 10 | 8 | 18 |

Other | 26 | 26 | 52 |

Total | 48 | 52 | 100 |

A student is randomly selected from this group, and it is found that the student is majoring in finance. What is the probability that the student is a male?

- a) 21
- b) 0.10
- c) 0.56
- d) 0.48
- e) 0.78

Ans: c

Response: See section 4.7 Conditional Probability

Difficulty: Hard

- A market research firm is investigating the appeal of three package designs. The table below gives information obtained through a sample of 200 consumers. The three package designs are labeled A, B, and C. The consumers are classified according to age and package design preference.

A | B | C | Total | |

Under 25 years | 22 | 34 | 40 | 96 |

25 or older | 54 | 28 | 22 | 104 |

Total | 76 | 62 | 62 | 200 |

If one of these consumers is randomly selected, what is the probability that the person prefers design A?

- a) 76
- b) 0.38
- c) 0.33
- d) 0.22
- e) 0.39

Ans: b

Response: See section 4.5 Addition Laws

Difficulty: Easy

- A market research firm is investigating the appeal of three package designs. The table below gives information obtained through a sample of 200 consumers. The three package designs are labeled A, B, and C. The consumers are classified according to age and package design preference.

A | B | C | Total | |

Under 25 years | 22 | 34 | 40 | 96 |

25 or older | 54 | 28 | 22 | 104 |

Total | 76 | 62 | 62 | 200 |

If one of these consumers is randomly selected, what is the probability that the person prefers design A and is under 25?

- a) 0.22
- b) 0.11
- c) 0.18
- d) 0.54
- e) 0.78

Ans: b

Response: See section 4.5 Addition Laws

Difficulty: Easy

- A market research firm is investigating the appeal of three package designs. The table below gives information obtained through a sample of 200 consumers. The three package designs are labeled A, B, and C. The consumers are classified according to age and package design preference.

A | B | C | Total | |

Under 25 years | 22 | 34 | 40 | 96 |

25 or older | 54 | 28 | 22 | 104 |

Total | 76 | 62 | 62 | 200 |

If one of these consumers is randomly selected and is under 25, what is the probability that the person prefers design A?

- a) 22
- b) 0.23
- c) 0.29
- d) 0.18
- e) 0.78

Ans: b

Response: See section 4.7 Conditional Probability

Difficulty: Hard

A | B | C | Total | |

Under 25 years | 22 | 34 | 40 | 96 |

25 or older | 54 | 28 | 22 | 104 |

Total | 76 | 62 | 62 | 200 |

If one of these consumers is randomly selected and prefers design B, what is the probability that the person is 25 or older?

- a) 28
- b) 0.14
- c) 0.45
- d) 0.27
- e) 0.78

Ans: c

Response: See section 4.7 Conditional Probability

Difficulty: Hard

A | B | C | Total | |

Under 25 years | 22 | 34 | 40 | 96 |

25 or older | 54 | 28 | 22 | 104 |

Total | 76 | 62 | 62 | 200 |

Are “B” and “25 or older” independent and why or why not?

- a) No, because P (25 or over | B) ¹ P (B)
- b) Yes, because P (B) = P(C)
- c) No, because P (25 or older | B) ¹ P (25 or older)
- d) Yes, because P (25 or older Ç B) ¹ 0
- e) No, because age and package design are different things

Ans: c

Response: See section 4.7 Conditional Probability

Difficulty: Hard

- An analysis of personal loans at a local bank revealed the following facts: 10% of all personal loans are in default (D), 90% of all personal loans are not in default (D΄), 20% of those in default are homeowners (H | D), and 70% of those not in default are homeowners (H | D΄). If a personal loan is selected at random P (H Ç D’) = ___________.
- a) 0.20
- b) 0.63
- c) 0.90
- d) 0.18
- e) 0.78

Ans: b

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Medium

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

- An analysis of personal loans at a local bank revealed the following facts: 10% of all personal loans are in default (D), 90% of all personal loans are not in default (D΄), 20% of those in default are homeowners (H | D), and 70% of those not in default are homeowners (H | D΄). If a personal loan is selected at random, P (D | H) = ___________.
- a) 0.03
- b) 0.63
- c) 0.02
- d) 0.18
- e) 0.78

Ans: a

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Hard

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

- A market research firms conducts studies regarding the success of new products. The company is not always perfect in predicting the success. Suppose that there is a 50% chance that any new product would be successful (and a 50% chance that it would fail). In the past, for all new products that ultimately were successful, 80% were predicted to be successful (and the other 20% were inaccurately predicted to be failures). Also, for all new products that were ultimately failures, 70% were predicted to be failures (and the other 30% were inaccurately predicted to be successes). What is the a priori probability that a new product would be a success?
- a) 0.50
- b) 0.80
- c) 0.70
- d) 0.60
- e) 0.95

Ans: a

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Medium

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

- A market research firms conducts studies regarding the success of new products. The company is not always perfect in predicting the success. Suppose that there is a 50% chance that any new product would be successful (and a 50% chance that it would fail). In the past, for all new products that ultimately were successful, 80% were predicted to be successful (and the other 20% were inaccurately predicted to be failures). Also, for all new products that were ultimately failures, 70% were predicted to be failures (and the other 30% were inaccurately predicted to be successes). For any randomly selected new product, what is the probability that the market research firm would predict that it would be a success?
- a) 80
- b) 0.50
- c) 0.45
- d) 0.55
- e) 0.95

Ans: d

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Hard

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

- A market research firm conducts studies regarding the success of new products. The company is not always perfect in predicting the success. Suppose that there is a 50% chance that any new product would be successful (and a 50% chance that it would fail). In the past, for all new products that ultimately were successful, 80% were predicted to be successful (and the other 20% were inaccurately predicted to be failures). Also, for all new products that were ultimately failures, 70% were predicted to be failures (and the other 30% were inaccurately predicted to be successes). If the market research predicted that the product would be a success, what is the probability that it would actually be a success?
- a) 27
- b) 0.73
- c) 0.80
- d) 0.24
- e) 1.00

Ans: b

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Hard

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

- Assigning probability 1/3 of rolling a 4 or 5 on a single die is an example of assigning probabilities using the ________________ method
- a) subjective probability
- b) relative frequency
- c) classical probability
- d)
*a**priori*probability - e)
*a posterior*probability

Ans: c

Response: See section 4.2 Methods of Assigning Probabilities

Difficulty: Medium

- The number of different committees of 2 students that can be chosen from a group of 5 students is
- a) 20
- b) 2
- c) 5
- d) 10
- e) 1

Ans: d

Response: See section 4.3 Structure of Probability

Difficulty: Medium

- It is known that 20% of all students in some large university are overweight, 20% exercise regularly and 2% are overweight and exercise regularly. What is the probability that a randomly selected student is either overweight or exercises regularly or both?
- a) 0.40
- b) 0.38
- c) 0.20
- d) 0.42
- e) 0.10

Ans: b

Response: See section 4.5 addition laws

Difficulty: Medium

Learning Objective: 4.5: Calculate probabilities using the general law of addition, along with a joint probability table, the complement of a union, or the special law of addition if necessary

- Suppose 5% of the population have a certain disease. A laboratory blood test gives a positive reading for 95% of people who have the disease. What is the probability of testing positive and having the disease?
- a) 0.0475
- b) 0.95
- c) 0.05
- d) 0.9

e)0.02

Ans: a

Response: See section 4.6 Multiplication Laws

Difficulty: Hard

Learning Objective: 4.6 Calculate joint probabilities of both independent and dependent events using the general and special laws of multiplication.

- Suppose that 5% of all TVs made by some Company in 2010 are defective. If 2 of these TVs are randomly selected what is the probability that both are defective?
- a) 0.95
- b) 0.0025
- c) 0.9025
- d) 0.0475
- e) 0.1

Ans: b

Response: See section 4.6 Multiplication Laws

Difficulty: Medium

- It is known that 20% of all students in some large university are overweight, 20% exercise regularly and 2% are overweight and exercise regularly. What is the probability that a randomly selected student is overweight given that this student exercises regularly?
- a) 0.40
- b) 0.38
- c) 0.20
- d) 0.42
- e) 0.10

Ans: e

Response: See section 4.7 Conditional Probability

Difficulty: Medium

- Suppose 5% of the population have a certain disease. A laboratory blood test gives a positive reading for 95% of people who have the disease and 10% positive reading of people who do not have the disease. . What is the probability of testing positive?
- a) 0.0475
- b) 0.1425
- c) 0.95
- d) 9
- e) 0.3333

Ans: b

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Hard

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

- Suppose 5% of the population have a certain disease. A laboratory blood test gives a positive reading for 95% of people who have the disease and 10% positive reading of people who do not have the disease. . What is the probability that a randomly selected person has the disease given that this person is testing positive?
- a) 0.0475
- b) 0.1425
- c) 0.95
- d) 9
- e) 0.3333

Ans: e

Response: See section 4.8 Revision of Probabilities: Bayes’ Rule

Difficulty: Hard

Learning Objective: 4.8: Calculate conditional probabilities using Bayes’ rule.

File: Ch09, Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations

True/False

- Hypotheses are tentative explanations of a principle operating in nature.

Ans: True

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Easy

Learning Objective: 9.1: Develop both one- and two-tailed null and alternative hypotheses that can be tested in a business setting by examining the rejection and non-rejection regions in light of Type I and Type II errors

- The first step in testing a hypothesis is to establish a true null hypothesis and a false alternative hypothesis.

Ans: False

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Easy

Learning Objective: 9.1: Develop both one- and two-tailed null and alternative hypotheses that can be tested in a business setting by examining the rejection and non-rejection regions in light of Type I and Type II errors

- In testing hypotheses, the researcher initially assumes that the alternative hypothesis is true and uses the sample data to reject it.

Ans: False

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Easy

Learning Objective: 9.1: Develop both one- and two-tailed null and alternative hypotheses that can be tested in a business setting by examining the rejection and non-rejection regions in light of Type I and Type II errors

- The null and the alternative hypotheses must be mutually exclusive and collectively exhaustive.

Ans: True

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Medium

- Generally speaking, the hypotheses that business researchers want to prove are stated in the alternative hypothesis.

Ans: True

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Medium

- The probability of committing a Type I error is called the power of the test.

Ans: False

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Medium

- When a true null hypothesis is rejected, the researcher has made a Type I error.

Ans: True

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Easy

- When a false null hypothesis is rejected, the researcher has made a Type II error.

Ans: False

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Medium

- When a researcher fails to reject a false null hypothesis, a Type II error has been committed.

Ans: True

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Easy

- Power is equal to (1 –
*b*), the probability of a test rejecting the null hypothesis that is indeed false.

Ans: True

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Medium

- The rejection region for a hypothesis test becomes smaller if the level of significance is changed from 0.01 to 0.05.

Ans: False

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Hard

- Whenever hypotheses are established such that the alternative hypothesis is “μ>8”, where μ is the population mean, the hypothesis test would be a two-tailed test.

** **

Ans: False

Response: See section 9.1 Introduction to Hypothesis Testing Difficulty: Easy

Learning Objective:

9.1: Develop both one- and two-tailed null and alternative hypotheses that can be tested in a business setting by examining the rejection and non-rejection regions in light of Type I and Type II errors

** **

- Whenever hypotheses are established such that the alternative hypothesis is “μ≠8”, where μ is the population mean, the hypothesis test would be a two-tailed test.

Ans: True

Response:

See section 9.1 Introduction to Hypothesis Testing

Difficulty: Easy

Learning Objective:

9.1: Develop both one- and two-tailed null and alternative hypotheses that can be tested in a business setting by examining the rejection and non-rejection regions in light of Type I and Type II errors

- Whenever hypotheses are established such that the alternative hypothesis is “μ>8”, where μ is the population mean, the p-value is the probability of observing a sample mean greater than the observed sample mean assuming that μ=8.

Ans: True

Response: See section 9.2 Testing Hypotheses about a Population Mean using the *z* Statistic (*s* Known)

Difficulty: Hard

Learning Objective: 9.2: Reach a statistical conclusion in hypothesis testing problems about a population mean with a known population standard deviation using the z statistic.

- If a null hypothesis was not rejected at the 0.10 level of significance, it will be rejected at a 0.05 level of significance based on the same sample results.

Ans: False

Response: See section 9.2 Testing Hypotheses about a Population Mean using the *z* Statistic (*s* Known)

Difficulty: Medium

Learning Objective: 9.2: Reach a statistical conclusion in hypothesis testing problems about a population mean with a known population standard deviation using the z statistic

- The rejection and nonrejection regions are divided by a point called the critical value.

Ans: True

Response:

See section 9.1 Introduction to Hypothesis Testing

Difficulty: Medium

Learning Objective:

9.1: Develop both one- and two-tailed null and alternative hypotheses that can be tested in a business setting by examining the rejection and non-rejection regions in light of Type I and Type II errors

- The probability of type II error becomes bigger if the level of significance is changed from 0.01 to 0.05.

Ans: False

Response: See section 9.1 Introduction to Hypothesis Testing

Difficulty: Hard

- If a null hypothesis is not rejected at the 0.05 level of significance, the p-value is bigger than 0.05

Ans: True

Response: See section 9.2 Testing Hypotheses about a Population Mean using the *z* Statistic (*s* Known)

Difficulty: Hard

Learning Objective: 9.2: Reach a statistical conclusion in hypothesis testing problems about a population mean with a known population standard deviation using the z statistic

Multiple Choice

- Consider the following null and alternative hypotheses.

Ho: *m* £ 67

Ha: *m* > 67

These hypotheses _______________.

- a) indicate a one-tailed test with a rejection area in the right tail
- b) indicate a one-tailed test with a rejection area in the left tail
- c) indicate a two-tailed test
- d) are established incorrectly
- e) are not mutually exclusive

Ans: a

*z* Statistic (*s* Known)

Difficulty: Easy

Learning Objective: 9.2: Reach a statistical conclusion in hypothesis testing problems about a population mean with a known population standard deviation using the z statistic.

- Consider the following null and alternative hypotheses.

Ho: *m* ³ 67

Ha: *m* < 67

These hypotheses _______________.

- a) indicate a one-tailed test with a rejection area in the right tail
- b) indicate a one-tailed test with a rejection area in the left tail
- c) indicate a two-tailed test
- d) are established incorrectly
- e) are not mutually exclusive

Ans: b

*z* Statistic (*s* Known)

Difficulty: Easy

Learning Objective: 9.2: Reach a statistical conclusion in hypothesis testing problems about a population mean with a known population standard deviation using the z statistic.

- Consider the following null and alternative hypotheses.

Ho: *m* = 67

Ha: *m* ¹ 67

These hypotheses ______________.

- a) indicate a one-tailed test with a rejection area in the right tail
- b) indicate a one-tailed test with a rejection area in the left tail
- c) indicate a two-tailed test
- d) are established incorrectly
- e) are not mutually exclusive

Ans: c

*z* Statistic (*s* Known)

Difficulty: Easy

- Consider the following null and alternative hypotheses.

Ho: *p* £ 0.16

Ha: *p* > 0.16

These hypotheses _______________.

- a) indicate a one-tailed test with a rejection area in the right tail
- b) indicate a one-tailed test with a rejection area in the left tail
- c) indicate a two-tailed test
- d) are established incorrectly
- e) are not mutually exclusive

Ans: a

Response: See section 9.4 Testing Hypotheses about a Population Proportion

Difficulty: Easy

Learning Objective: 9.4: Reach a statistical conclusion in hypothesis testing problems about a population proportion using the z statistic.

- Consider the following null and alternative hypotheses.

Ho: *p* ³ 0.16

Ha: *p* < 0.16

These hypotheses _______________.

- a) indicate a one-tailed test with a rejection area in the right tail
- b) indicate a one-tailed test with a rejection area in the left tail
- c) indicate a two-tailed test
- d) are established incorrectly
- e) are not mutually exclusive

Ans: b

Response: See section 9.4 Testing Hypotheses about a Population Proportion

Difficulty: Easy

Learning Objective: 9.4: Reach a statistical conclusion in hypothesis testing problems about a population proportion using the z statistic.

- Consider the following null and alternative hypotheses.

Ho: *p* = 0.16

Ha: *p* ¹ 0.16

These hypotheses _______________.

- a) indicate a one-tailed test with a rejection area in the right tail
- b) indicate a one-tailed test with a rejection area in the left tail
- c) indicate a two-tailed test
- d) are established incorrectly
- e) are not mutually exclusive

Ans: c

Response: See section 9.4 Testing Hypotheses about a Population Proportion

Difficulty: Easy

Learning Objective: 9.4: Reach a statistical conclusion in hypothesis testing problems about a population proportion using the z statistic.

- Suppose the alternative hypothesis in a hypothesis test is: “the population mean is less than 60”. If the sample size is 50, s is known, and alpha =.05, the critical value of
*z*is _______.

- a) 1.645
- b) -1.645
- c) 1.96
- d) -1.96
- e) 2.05

Ans: b

*z* Statistic (*s* Known)

Difficulty: Medium

- Suppose the alternative hypothesis in a hypothesis test is “the population mean is greater than 60”. If the sample size is 80, s is known, and alpha = .01, the critical value of
*z*is _______.

- a) 2.575
- b) -2.575
- c) 2.33
- d) -2.33
- e) 2.45

Ans: c

*z* Statistic (*s* Known)

Difficulty: Medium

- In a two-tailed hypothesis about a population mean with a sample size of 100, s is known, and alpha = 0.10, the rejection region would be _______.

- a)
*z*> 1.64 - b)
*z*> 1.28 - c)
*z*< -1.28 and*z*> 1.28 - d)
*z*< -1.64 and*z*> 1.64 - e)
*z*< -2.33 and*z*> 2.33

Ans: d

*z* Statistic (*s* Known)

Difficulty: Medium

- In a two-tailed hypothesis about a population mean with a sample size of 100, s is known, and
*α*= 0.05, the rejection region would be _______.

- a)
*z*> 1.64 - b)
*z*> 1.96 - c)
*z*< -1.96 and*z*> 1.96 - d)
*z*< -1.64 and*z*> 1.64 - e)
*z*< -2.33 and*z*> 2.33

Ans: c

*z* Statistic (*s* Known)

Difficulty: Medium

- Suppose you are testing the null hypothesis that a population mean is less than or equal to 80, against the alternative hypothesis that the population mean is greater than 80. If the sample size is 49, s is known, and alpha = .10, the critical value of
*z*is _______.

- a) 1.645
- b) -1.645
- c) 1.28
- d) -1.28
- e) 2.33

Ans: c

Response: See section 9.2 Testing Hypotheses about a Population Mean using the *z* Statistic (s Known)

Difficulty: Medium

- Suppose you are testing the null hypothesis that a population mean is less than or equal to 80, against the alternative hypothesis that the population mean is greater than 80. The sample size is 49 and alpha =.05. If the sample mean is 84 and the population standard deviation is 14, the observed
*z*value is _______.

- a) 2
- b) -2
- c) 14
- d) -14
- e) 49

Ans: a

*z* Statistic (*s* Known)

Difficulty: Medium

- Suppose you are testing the null hypothesis that a population mean is greater than or equal to 60, against the alternative hypothesis that the population mean is less than 60. The sample size is 64 and
*a*= .05. If the sample mean is 58 and the population standard deviation is 16, the observed*z*value is _______.

- a) -1
- b) 1
- c) -8
- d) 8
- e) 58

Ans: a

*z* Statistic (*s* Known)

Difficulty: Medium

- A researcher is testing a hypothesis of a single mean. The critical
*z*value for

*a* = .05 and an one‑tailed test is 1.645. The observed *z* value from sample data is 1.13. The decision made by the researcher based on this information is to ______ the null hypothesis.

- a) reject
- b) not reject
- c) redefine
- d) change the alternate hypothesis into
- e) restate the null hypothesis

Ans: b

*z* Statistic (*s* Known)

Difficulty: Medium

- A researcher is testing a hypothesis of a single mean. The critical
*z*value for

*a* = .05 and a two‑tailed test is __+__1.96. The observed *z* value from sample data is ‑1.85. The decision made by the researcher based on this information is to _____ the null hypothesis.

- a) reject
- b) not reject
- c) redefine
- d) change the alternate hypothesis into
- e) restate the null hypothesis

Ans: b

*z* Statistic (*s* Known)

Difficulty: Medium

- A researcher is testing a hypothesis of a single mean. The critical
*z*value for

*a* = .05 and a two‑tailed test is __+__1.96. The observed *z* value from sample data is 2.85. The decision made by the researcher based on this information is to _____ the null hypothesis.

- a) reject
- b) not reject
- c) redefine
- d) change the alternate hypothesis into
- e) restate the null hypothesis

Ans: a

*z* Statistic (*s* Known)

Difficulty: Medium

- A researcher is testing a hypothesis of a single mean. The critical
*z*value for*a*= .05 and a two‑tailed test is__+__1.96. The observed*z*value from sample data is -2.11. The decision made by the researcher based on this information is to _____ the null hypothesis.

- a) reject
- b) not reject
- c) redefine
- d) change the alternate hypothesis into
- e) restate the null hypothesis

Ans: a

*z* Statistic (*s* Known)

Difficulty: Medium

- A company produces an item that is supposed to have a six inch hole punched in the center. A quality control inspector is concerned that the machine which punches the hole is “out‑of‑control” (hole is too large or too small). In an effort to test this, the inspector is going to gather a sample punched by the machine and measure the diameter of the hole. The alternative hypothesis used to statistical test to determine if the machine is out‑of‑control is

- a) the mean diameter is > 6 inches
- b) the mean diameter is < 6 inches
- c) the mean diameter is = 6 inches
- d) the mean diameter is ≠ 6 inches
- e) the mean diameter is ≥ 6 inches

Ans: d

*z* Statistic (*s* Known)

Difficulty: Medium

- Jennifer Cantu, VP of Customer Services at Tri-State Auto Insurance, Inc., monitors the claims processing time of the claims division. Her standard includes “a mean processing time of 5 days or less.” Each week, her staff checks for compliance by analyzing a random sample of 60 claims. Jennifer’s null hypothesis is ________.

- a)
*m*> 5 - b)
*s*> 5 - c)
*n*= 60 - d)
*m*< 5 - e)
*m**=*5

Ans: e

*z* Statistic (*s* Known)

Difficulty: Medium

- In performing a hypothesis test where the null hypothesis is that the population mean is 23 against the alternative hypothesis that the population mean is not equal to 23, a random sample of 17 items is selected. The sample mean is 24.6 and the sample standard deviation is 3.3. It can be assumed that the population is normally distributed. The degrees of freedom associated with this are _______.

- a) 17
- b) 16
- c) 15
- d) 2
- e) 1

Ans: b

Response: See section 9.3 Testing Hypotheses about a Population Mean using the *t* Statistic (*s* Unknown)

Difficulty: Easy

Learning Objective: 9.3: Reach a statistical conclusion in hypothesis testing problems about a population mean with an unknown population standard deviation using the t statistic.

- In performing a hypothesis test where the null hypothesis is that the population mean is 4.8 against the alternative hypothesis that the population mean is not equal to 4.8, a random sample of 25 items is selected. The sample mean is 4.1 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed. The degrees of freedom associated with this are _______.

- a) 25
- b) 24
- c) 26
- d) 2
- e) 1

Ans: b

Response: See section 9.3 Testing Hypotheses about a Population Mean using the *t* Statistic (*s* Unknown)

Difficulty: Easy

Learning Objective: 9.3: Reach a statistical conclusion in hypothesis testing problems about a population mean with an unknown population standard deviation using the t statistic.

- In performing a hypothesis test where the null hypothesis is that the population mean is 4.8 against the alternative hypothesis that the population mean is not equal to 4.8, a random sample of 25 items is selected. The sample mean is 4.1 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed. The level of significance is selected to be 0.10. The table “
*t*” value for this problem is _______.

- a) 318
- b) 1.711
- c) 2.492
- d) 2.797
- e) 3.227

Ans: b

Response: See section 9.3 Testing Hypotheses about a Population Mean using the *t* Statistic (*s* Unknown)

Difficulty: Medium

Learning Objective: 9.3: Reach a statistical conclusion in hypothesis testing problems about a population mean with an unknown population standard deviation using the t statistic.

- In performing a hypothesis test where the null hypothesis is that the population mean is 4.8 against the alternative hypothesis that the population mean is not equal to 4.8, a random sample of 25 items is selected. The sample mean is 4.1 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed. The observed “
*t*” value for this problem is _______.

- a) -12.5
- b) 12.5
- c) -2.5
- d) -0.7
- e) 0.7

Ans: c

*t* Statistic (*s* Unknown)

Difficulty: Medium

- In performing a hypothesis test where the null hypothesis is that the population mean is 6.9 against the alternative hypothesis that the population mean is not equal to 6.9, a random sample of 16 items is selected. The sample mean is 7.1 and the sample standard deviation is 2.4. It can be assumed that the population is normally distributed. The observed “
*t*” value for this problem is _______.

- a) 0.05
- b) 0.20
- c) 0.33
- d) 1.33
- e) 1.43

Ans: c

*t* Statistic (*s* Unknown)

Difficulty: Medium

- In performing a hypothesis test where the null hypothesis is that the population mean is 6.9 against the alternative hypothesis that the population mean is not equal to 6.9, a random sample of 16 items is selected. The sample mean is 7.1 and the sample standard deviation is 2.4. It can be assumed that the population is normally distributed. The level of significance is selected as 0.05. The table “
*t*” value for this problem is _______.

- a) 753
- b) 2.947
- c) 2.120
- d) 2.131
- e) 2.311

Ans: d

*t* Statistic (*s* Unknown)

Difficulty: Medium

- In performing a hypothesis test where the null hypothesis is that the population mean is 6.9 against the alternative hypothesis that the population mean is not equal to 6.9, a random sample of 16 items is selected. The sample mean is 7.1 and the sample standard deviation is 2.4. It can be assumed that the population is normally distributed. The level of significance is selected as 0.05. The decision rule for this problem is to reject the null hypothesis if the observed “
*t*” value is _______.

- a) less than -2.131 or greater than 2.131
- b) less than -1.761 or greater than 1.761
- c) less than -1.753 or greater than 1.753
- d) less than -2.120 or greater than 2.120
- e) less than -3.120 or greater than 3.120

Ans: a

*t* Statistic (*s* Unknown)

Difficulty: Medium

- The diameter of 3.5 inch diskettes is normally distributed. Periodically, quality control inspectors at Dallas Diskettes randomly select a sample of 16 diskettes. If the mean diameter of the diskettes is too large or too small the diskette punch is shut down for adjustment; otherwise, the punching process continues. The last sample showed a mean and standard deviation of 3.49 and 0.08 inches, respectively. Using
*a*= 0.05, the critical “*t*” values are _______.

- a) -2.120 and 2.120
- b) -2.131 and 2.131
- c) -1.753 and 1.753
- d) -1.746 and 1.746
- e) -2.567 and 2.567

Ans: b

*t* Statistic (*s* Unknown)

Difficulty: Medium

- The diameter of 3.5 inch diskettes is normally distributed. Periodically, quality control inspectors at Dallas Diskettes randomly select a sample of 16 diskettes. If the mean diameter of the diskettes is too large or too small the diskette punch is shut down for adjustment; otherwise, the punching process continues. The null hypothesis is ______.

- a)
*n*¹ 16 - b)
*n*= 16 - c)
*m*= 3.5 - d)
*m*¹5 - e)
*m*≥5

Ans: c

*t* Statistic (*s* Unknown)

Difficulty: Easy

- The diameter of 3.5 inch diskettes is normally distributed. Periodically, quality control inspectors at Dallas Diskettes randomly select a sample of 16 diskettes. If the mean diameter of the diskettes is too large or too small the diskette punch is shut down for adjustment; otherwise, the punching process continues. The last sample showed a mean and standard deviation of 3.49 and 0.08 inches, respectively. Using
*a*= 0.05, the appropriate decision is _______.

- a) reject the null hypothesis and shut down the punch
- b) reject the null hypothesis and do not shut down the punch
- c) do not reject the null hypothesis and shut down the punch
- d) do not reject the null hypothesis and do not shut down the punch
- e) do nothing

Ans: d

*t* Statistic (*s* Unknown)

Difficulty: Hard

- The diameter of 3.5 inch diskettes is normally distributed. Periodically, quality control inspectors at Dallas Diskettes randomly select a sample of 16 diskettes. If the mean diameter of the diskettes is too large or too small the diskette punch is shut down for adjustment; otherwise, the punching process continues. The last sample showed a mean and standard deviation of 3.55 and 0.08 inches, respectively. Using
*a*= 0.05, the appropriate decision is _______.

- a) reject the null hypothesis and shut down the punch
- b) reject the null hypothesis and do not shut down the punch
- c) do not reject the null hypothesis and shut down the punch
- d) do not reject the null hypothesis and do not shut down the punch
- e) do nothing

Ans: a

*t* Statistic (*s* Unknown)

Difficulty: Hard

- In performing hypothesis tests about the population mean, if the population standard deviation is not known, a
*t*test can be used to test the mean if _________________.

- a) n is small
- b) the sample is random
- c) the population mean is known
- d) the population is normally distributed
- e) the population is chi-square distributed

Ans: d

*t* Statistic (*s* Unknown)

Difficulty: Medium

- Suppose a researcher is testing a null hypothesis that
*m*= 61. A random sample of*n*= 36 is taken resulting in a sample mean of 63 and*s*= 9. The observed*t*value is _______.

- a) -0.22
- b) 0.22
- c) 1.33
- d) 8.08
- e) 7.58

Ans: c

*t* Statistic (*s* Unknown)

Difficulty: Medium

- A political scientist wants to prove that a candidate is currently carrying more than 60% of the vote in the state. She has her assistants randomly sample 200 eligible voters in the state by telephone and only 90 declare that they support her candidate. The observed
*z*value for this problem is _______.

- a) -4.33
- b) 4.33
- c) 0.45
- d) -.31
- e) 2.33

Ans: a

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Medium

- A company believes that it controls more than 30% of the total market share for one of its products. To prove this belief, a random sample of 144 purchases of this product is contacted. It is found that 50 of the 144 purchased this company’s brand of the product. If a researcher wants to conduct a statistical test for this problem, the alternative hypothesis would be _______.

- a) the population proportion is less than 0.30
- b) the population proportion is greater than 0.30
- c) the population proportion is not equal to 0.30
- d) the population mean is less than 40
- e) the population mean is greater than 40

Ans: b

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Easy

- A company believes that it controls more than 30% of the total market share for one of its products. To prove this belief, a random sample of 144 purchases of this product is contacted. It is found that 50 of the 144 purchased this company’s brand of the product. If a researcher wants to conduct a statistical test for this problem, the observed
*z*value would be _______.

- a) 05
- b) 0.103
- c) 0.35
- d) 1.24
- e) 1.67

Ans: d

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Medium

- A company believes that it controls more than 30% of the total market share for one of its products. To prove this belief, a random sample of 144 purchases of this product is contacted. It is found that 50 of the 144 purchased this company’s brand of the product. If a researcher wants to conduct a statistical test for this problem, the test would be _______.

- a) a one-tailed test
- b) a two-tailed test
- c) an alpha test
- d) a finite population test
- e) a finite sample test

Ans: a

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Easy

- Ophelia O’Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is “no more than 5% of personal loans should be in default.” On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday’s sample contained 30 defaulted loans. Ophelia’s null hypothesis is _______.

- a)
*p*> 0.05 - b)
*p*= 0.05 - c)
*n*= 30 - d)
*n*= 500 - e)
*n*= 0.05

Ans: b

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Easy

- Ophelia O’Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is “no more than 5% of personal loans should be in default.” On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday’s sample contained 30 defaulted loans. Using
*a*= 0.10, the critical*z*value is _______.

- a) 645
- b) -1.645
- c) 1.28
- d) -1.28
- e) 2.28

Ans: c

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Medium

- Ophelia O’Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is “no more than 5% of personal loans should be in default.” On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday’s sample contained 30 defaulted loans. Using
*a*= 0.10, the observed*z*value is _______.

- a) 1.03
- b) -1.03
- c) 0.046
- d) -0.046
- e) 1.33

Ans: a

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Medium

- Ophelia O’Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is “no more than 5% of personal loans should be in default.” On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday’s sample contained 30 defaulted loans. Using
*a*= 0.10, the appropriate decision is _______.

- a) reduce the sample size
- b) increase the sample size
- c) reject the null hypothesis
- d) do not reject the null hypothesis
- e) do nothing

Ans: d

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Hard

- Ophelia O’Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is “no more than 5% of personal loans should be in default.” On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday’s sample contained 38 defaulted loans. Using
*a*= 0.10, the appropriate decision is _______.

- a) reduce the sample size
- b) increase the sample size
- c) reject the null hypothesis
- d) do not reject the null hypothesis
- e) do nothing

Ans: c

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Hard

- The executives of CareFree Insurance, Inc. feel that “a majority of our employees perceive a participatory management style at CareFree.” A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance. Eighty employees rate the management as participatory. The null hypothesis is __________.

- a)
*n*= 30 - b)
*n*= 200 - c)
*p*= 0.50 - d)
*p*< 0.50 - e)
*n*> 200

Ans: c

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Easy

- The executives of CareFree Insurance, Inc. feel that “a majority of our employees perceive a participatory management style at CareFree.” A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance. Eighty employees rate the management as participatory. The appropriate decision is __________.

- a) do not reject the null hypothesis
- b) reject the null hypothesis
- c) reduce the sample size
- d) increase the sample size
- e) do nothing

Ans: b

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Hard

- The executives of CareFree Insurance, Inc. feel that “a majority of our employees perceive a participatory management style at CareFree.” A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance. Ninety employees rate the management as participatory. The appropriate decision is __________.

- a) do not reject the null hypothesis
- b) reject the null hypothesis
- c) reduce the sample size
- d) increase the sample size
- e) maintain
*status quo*

Ans: a

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Hard

- Elwin Osbourne, CIO at GFS, Inc., suspects that at least 25% of e-mail messages sent by GFS employees are not business related. A random sample of 300 e-mail messages was selected to test this hypothesis at the 0.01 level of significance. Fifty-four of the messages were not business related. The null hypothesis is ____.

- a)
*b*= 30 - b)
*n*= 300 - c)
*p*< 25 - d)
*p*≠ 0.25 - e)
*p*= 0.25

Ans: e

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Easy

- Elwin Osbourne, CIO at GFS, Inc., suspects that at least 25% of e-mail messages sent by GFS employees are not business related. A random sample of 300 e-mail messages was selected to test this hypothesis at the 0.01 level of significance. Fifty-four of the messages were not business related. The appropriate decision is _______.

- a) increase the sample size
- b) gather more data
- c) reject the null hypothesis
- d) do not reject the null hypothesis
- e) maintain
*status quo*

Ans: c

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Hard

- Elwin Osbourne, CIO at GFS, Inc., suspects that at least 25% of e-mail messages sent by GFS employees are not business related. A random sample of 300 e-mail messages was selected to test this hypothesis at the 0.01 level of significance. Sixty of the messages were not business related. The appropriate decision is _______.

- a) increase the sample size
- b) gather more data
- c) maintain
*status quo* - d) do not reject the null hypothesis
- e) reject the null hypothesis

Ans: d

Response: See section 9.4 Testing Hypotheses about a Proportion

Difficulty: Hard

- Discrete Components, Inc. manufactures a line of electrical resistors. Presently, the carbon composition line is producing 100 ohm resistors. The population variance of these resistors “must not exceed 4” to conform to industry standards. Periodically, the quality control inspectors check for conformity by randomly selecting 10 resistors from the line, and calculating the sample variance. The last sample had a variance of 4.36. Assume that the population is normally distributed. Using
*a*= 0.05, the null hypothesis is _________________.

- a)
*m*= 100 - b)
*s*£ 10 - c)
*s*2³ 4 - d)
*s*2= 4 - e)
*n*= 100

Ans: d

Response: See section 9.5 Testing Hypotheses about a Variance

Difficulty: Easy

Learning Objective: 9.5: Reach a statistical conclusion in hypothesis testing problems about a population variance using the chi-square statistic.

- Discrete Components, Inc. manufactures a line of electrical resistors. Presently, the carbon composition line is producing 100 ohm resistors. The population variance of these resistors “must not exceed 4” to conform to industry standards. Periodically, the quality control inspectors check for conformity by randomly select 10 resistors from the line, and calculating the sample variance. The last sample had a variance of 4.36. Assume that the population is normally distributed. Using
*a*= 0.05, the critical value of chi-square is _________________.

- a) 18.31
- b) 16.92
- c) 3.94
- d) 3.33
- e) 19.82

Ans: b

Response: See section 9.5 Testing Hypotheses about a Variance

Difficulty: Medium

Learning Objective: 9.5: Reach a statistical conclusion in hypothesis testing problems about a population variance using the chi-square statistic.

- Discrete Components, Inc. manufactures a line of electrical resistors. Presently, the carbon composition line is producing 100 ohm resistors. The population variance of these resistors “must not exceed 4” to conform to industry standards. Periodically, the quality control inspectors check for conformity by randomly select 10 resistors from the line, and calculating the sample variance. The last sample had a variance of 4.36. Assume that the population is normally distributed. Using
*a*= 0.05, the observed value of chi-square is _________________.

- a) 1.74
- b) 1.94
- c) 10.90
- d) 9.81
- e) 8.91

Ans: d

Response: See section 9.5 Testing Hypotheses about a Variance

Difficulty: Medium

Learning Objective: 9.5: Reach a statistical conclusion in hypothesis testing problems about a population variance using the chi-square statistic.

- Discrete Components, Inc. manufactures a line of electrical resistors. Presently, the carbon composition line is producing 100 ohm resistors. The population variance of these resistors “must not exceed 4” to conform to industry standards. Periodically, the quality control inspectors check for conformity by randomly select 10 resistors from the line, and calculating the sample variance. The last sample had a variance of 4.36. Assume that the population is normally distributed. Using
*a*= 0.05, the appropriate decision is _________________.

- a) increase the sample size
- b) reduce the sample size
- c) reject the null hypothesis
- d) do not reject the null hypothesis
- e) maintain
*status quo*

Ans: d

Response: See section 9.5 Testing Hypotheses about a Variance

Difficulty: Hard

- David Desreumaux, VP of Human Resources of American First Banks (AFB), is reviewing the employee training programs of AFB banks. Based on a recent census of personnel, David knows that the variance of teller training time in the Southeast region is 8, and he wonders if the variance in the Southwest region is the same number. His staff randomly selected personnel files for 15 tellers in the Southwest Region, and determined that their mean training time was 25 hours and that the standard deviation was 4 hours. Assume that teller training time is normally distributed. Using
*a*= 0.10, the null hypothesis is ________.

- a)
*m*= 25 - b)
*s*2= 8 - c)
*s*2= 4 - d)
*s*2£ 8 - e)
*s*^{2}= 16

Ans: b

Response: See section 9.5 Testing Hypotheses about a Variance

Difficulty: Easy

- David Desreumaux, VP of Human Resources of American First Banks (AFB), is reviewing the employee training programs of AFB banks. Based on a recent census of personnel, David knows that the variance of teller training time in the Southeast region is 8, and he wonders if the variance in the Southwest region is the same number. His staff randomly selected personnel files for 15 tellers in the Southwest Region, and determined that their mean training time was 25 hours and that the standard deviation was 4 hours. Assume that teller training time is normally distributed. Using
*a*= 0.10, the critical values of chi-square are ________.

- a) 7.96 and 26.30
- b) 6.57 and 23.68
- c) -1.96 and 1.96
- d) -1.645 and 1.645
- e) -6.57 and 23.68

Ans: b

Response: See section 9.5 Testing Hypotheses about a Variance

Difficulty: Medium

- David Desreumaux, VP of Human Resources of American First Banks (AFB), is reviewing the employee training programs of AFB banks. Based on a recent census of personnel, David knows that the variance of teller training time in the Southeast region is 8, and he wonders if the variance in the Southwest region is the same number. His staff randomly selected personnel files for 15 tellers in the Southwest Region, and determined that their mean training time was 25 hours and that the standard deviation was 4 hours. Assume that teller training time is normally distributed. Using
*a*= 0.10, the observed value of chi-square is ________.

- a) 28.00
- b) 30.00
- c) 56.00
- d) 60.00
- e) 65.00

Ans: a

Response: See section 9.5 Testing Hypotheses about a Variance

Difficulty: Medium

- David Desreumaux, VP of Human Resources of American First Banks (AFB), is reviewing the employee training programs of AFB banks. Based on a recent census of personnel, David knows that the variance of teller training time in the Southeast region is 8, and he wonders if the variance in the Southwest region is the same number. His staff randomly selected personnel files for 15 tellers in the Southwest Region, and determined that their mean training time was 25 hours and that the standard deviation was 4 hours. Assume that teller training time is normally distributed. Using
*a*= 0.10, the appropriate decision is ________.

- a) increase the sample size
- b) reduce the sample size
- c) do not reject the null hypothesis
- d) maintain
*status quo* - e) reject the null hypothesis

Ans: e

Response: See section 9.5 Testing Hypotheses about a Variance

Difficulty: Hard

- A coffee-dispensing machine is supposed to deliver 8 ounces of liquid into each paper cup, but a consumer believes that the actual mean amount is less. The consumer obtained a sample of 49 cups of the dispensed liquid with average of 7.75 ounces. If the variance of the dispensed liquid per cup is 0.81 ounces, and α=0.05, the p-value is approximately

- 05
- 025
- 06
- 015
- 10

Ans: b

*z* Statistic (*s* Known)

Difficulty: Hard

- A coffee-dispensing machine is supposed to deliver 8 ounces of liquid into each paper cup, but a consumer believes that the actual mean amount is less. The consumer obtained a sample of 49 cups of the dispensed liquid with average of 7.75 ounces. If the variance of the dispensed liquid delivered per cup is 0.81 ounces, and α=0.05, the appropriate decision is ________.

- a) increase the sample size
- b) reduce the sample size
- c) do not reject the 8-ounces claim
- d) maintain
*status quo* - e) reject the 8-ounces claim

Ans: e

*z* Statistic (*s* Known)

Difficulty: Hard

- A coffee-dispensing machine is supposed to deliver 8 ounces of liquid into each paper cup, but a consumer believes that the actual mean amount is less. The consumer obtained a sample of 16 cups of the dispensed liquid with sample mean of 7.75 ounces and sample variance of 0.81. If the dispensed liquid delivered per cup is normally distributed, the appropriate decision at α=0.05 is

- a) increase the sample size
- b) reduce the sample size
- c) do not reject the 8-ounces claim
- d) maintain
*status quo* - e) reject the 8-ounces claim

Ans: c

*t* Statistic (*s* Unknown)

Difficulty: Hard

- The lifetime of a squirrels follows a normal distribution with mean μ months and standard deviation s=7 months. To test the hypothesis that μ>40, 25 squirrels are randomly selected. The null hypothesis is rejected when the sample mean is bigger than 43. Assuming that μ=45, the probability of type II error is approximately

- 076

- 016

- 05

- 011

- 983

Ans: a

Response: See section 9.6 Solving for Type II Errors.

Difficulty: Hard

Learning Objective: 9.6: Solve for possible Type II errors when failing to reject the null hypothesis.

- The lifetime of a squirrels follows a normal distribution with mean μ months and standard deviation s=7 months. To test the hypothesis that μ>40, 25 squirrels are randomly selected. Assuming that μ=45, and α=0.1, the probability of type II error is approximately

- 076
- 016
- 05
- 011
- 983

Ans: d

Response: See section 9.6 Solving for Type II Errors.

Difficulty: Hard

Learning Objective: 9.6: Solve for possible Type II errors when failing to reject the null hypothesis.

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