WebAssign is not supported for this browser version. Some features or content might not work. System requirements

WebAssign

Welcome, demo@demo

(sign out)

Tuesday, April 1, 2025 05:57 EDT

Home My Assignments Grades Communication Calendar My eBooks

My Cengage Home Notifications Help My Account

Statistics, section 2, Fall 2019
My Assignments

Anderson - Essen. of Stats for Business & Econ 9/e (Homework)

James Finch

Statistics, section 2, Fall 2019

Instructor: Dr. Friendly

Current Score : 32 / 82

Due : Sunday, January 27, 2030 23:30 EST

Last Saved : n/a Saving... ()

Ask Your Teacher Extension Requests Print Assignment

Question

Points

1	2	3	4	5	6	7	8	9
12/12	12/17	0/5	0/2	1/4	6/14	–/14	1/6	–/8

Total

32/82 (39.0%)

Instructions

Drawing from the authors' unmatched experience as professors and consultants, Anderson/Sweeney/Williams/Camm/Cochran/Fry/Ohlmann's Essentials of Statistics for Business and Economics, 9th edition, published by Cengage Learning, delivers sound statistical methodology, a proven problem-scenario approach, and meaningful applications that clearly demonstrate how statistical information impacts decisions in actual business practice. More than 350 real business examples, relevant cases, and hands-on exercises present the latest statistical data and business information with unwavering accuracy. Choose optional coverage of popular commercial statistical software programs JMP® Student Edition 14 and Excel® 2016. An all new WebAssign online course management system is available with this powerful business statistics solution.

Question 1 is a multipart question that steps the student through the construction of a pie chart and frequency bar chart.

Question 2 features multiple question types and guides students through the process of interpreting values.

Question 3 asks students to derive a formula and then use that formula to calculate probabilities.

Question 4 includes an interactive applet to determine the sample size necessary for a given margin of error.

Question 5 is a simulation question utilizing a JMP applet.

Question 6 guides the student through a hypothesis test and includes a link to calculate a precise p-value.

Question 7 showcases a full ANOVA table to complete and then use to perform a hypothesis test.

Question 8 links to a data set and offers multiple question types for linear regression analysis.

Question 9 displays grading for multiple regression equations and interpretation of an analysis. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

Assignment Scoring

Your last submission is used for your score.

1. 12/12 points | Previous Answers ASWESBE9 2.E.003.MI.SA. My Notes

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

A questionnaire provides 67 Yes, 43 No, and 10 No Opinion answers.

(a)

In the construction of a pie chart, how many degrees would be in the section of the pie showing the Yes answers?

(b)

How many degrees would be in the section of the pie showing the No answers?

(c)

Construct a pie chart.

(d)

Construct a frequency bar chart.

Step 1

(a) In the construction of a pie chart, how many degrees would be in the section of the pie showing the Yes answers?

A pie chart displays categorical data as sections of a circle, called sectors. Pie charts depict the frequency, relative frequency, or percent frequency distribution of a set of data. Each sector will cover an area that corresponds to the relative frequency for each class in the data. Recall that the relative frequencies will always sum to 1. Since there are 360 degrees in a circle, each section will cover an area that is 360(relative frequency) degrees.

Therefore, the relative frequency for the Yes answers is needed here. The relative frequency of the Yes answers is the ratio of the Yes answers to the total sample size. The total sample size can be found by taking the sum of the answers from this questionnaire.

total sample size	=	Yes + No + No Opinion
	=	67 + 43 + 10
	=	120 120

Given that there are 67 Yes answers, find the corresponding relative frequency, leaving your answer as a fraction.

relative frequency = 67/120

67/120

Step 2

Now that the relative frequency has been found, we can find the number of degrees in the Yes answers portion of the pie chart by multiplying this value,

120

by the number of degrees in a circle, 360.

degrees in Yes portion

120

360 degrees

201

degrees

Step 3

(b) How many degrees would be in the section of the pie showing the No answers?

As in part (a), first find the relative frequency of the No answers. The sample size was found to be 120, so the relative frequency will be the ratio of the number of No answers to the total sample size. There were 43 No answers. Find this ratio, leaving your answer as a fraction.

relative frequency of No answers

number of No answers

total number of answers

43/120

Step 4

Now that the relative frequency has been found, we can find the number of degrees in the No answers portion of the pie chart by multiplying this value,

120

by the number of degrees in a circle, 360.

degrees in No portion

120

360 degrees

129

degrees

Step 5

(c) Construct a pie chart.

To create a pie chart, the angle for each sector is needed. Recall there are 360 degrees in a circle. The angle corresponding to a Yes answer was 201, and the angle corresponding to a No answer was 129. Since there are three responses in this questionnaire, the sum of these angles can be subtracted from 360.

degrees in the No Opinion sector

360 −

201 + 129

129

degrees

Step 6

The pie chart for this data will have a sector of 30 degrees representing the No Opinion answer, a sector of 129 degrees representing the No answer, and a third sector of 201 degrees representing the Yes answer. Construct a pie chart that corresponds to this data.

A circle graph has 3 uneven pieces.

The first piece is labeled Yes, 35.8%.

The second piece is labeled No, 8.3%.

The third piece is labeled No opinion, 55.8%.

A circle graph has 3 uneven pieces.

The first piece is labeled Yes, 8.3%.

The second piece is labeled No, 55.8%.

The third piece is labeled No opinion, 35.8%.

A circle graph has 3 uneven pieces.

The first piece is labeled Yes, 55.8%.

The second piece is labeled No, 8.3%.

The third piece is labeled No opinion, 35.8%.

A circle graph has 3 uneven pieces.

The first piece is labeled Yes, 55.8%.

The second piece is labeled No, 35.8%.

The third piece is labeled No opinion, 8.3%.

Step 7

(d) Construct a frequency bar chart.

A bar chart is another way to display categorical data. As with a pie chart, a bar chart can depict frequency, relative frequency, or percent frequency distribution data.

Place each answer type along the horizontal axis and use frequency along the vertical axis. Each answer type will have its own bar. Since there are 3 types of answers on this questionnaire, there will be 3

bars.

The height of each bar will correspond to the frequency of each group. There were 67 Yes answers, 43 No answers, and 10 No Opinion answers. Construct a frequency bar chart corresponding to this data.

A bar chart has 3 class labels along the horizontal axis. The horizontal axis is labeled: Response, and the vertical axis is labeled: Frequency. The vertical bars are centered over their labels and moving left to right their heights are as follows.

Yes: 67 units tall

No: 10 units tall

No opinion: 43 units tall

You have now completed the Master It.

Viewing Saved Work Revert to Last Response

2. 12/17 points | Previous Answers ASWESBE9 4.E.033. My Notes

Students taking a test were asked about their undergraduate major and intent to pursue their MBA as a full-time or part-time student. A summary of their responses follows.

		Undergraduate Major			Totals
		Business	Engineering	Other	Totals
Intended Enrollment Status	Full-Time	351	196	253	800
Intended Enrollment Status	Part-Time	149	159	193	501
Totals		500	355	446	1,301

(a)

Develop a joint probability table for these data. (Round your answers to three decimal places.)

		Undergraduate Major			Totals
		Business	Engineering	Other	Totals
Intended Enrollment Status	Full-Time	0.270	0.151	0.194	0.615
Intended Enrollment Status	Part-Time		0.122	0.148	0.385
Totals		0.384	0.273	0.343	1.000

(b)

Use the marginal probabilities of undergraduate major (business, engineering, or other) to comment on which undergraduate major produces the most potential MBA students.

From the marginal probabilities, we can tell that

business

majors produce the most potential MBA students.

(c)

If a student intends to attend classes full-time in pursuit of an MBA degree, what is the probability that the student was an undergraduate engineering major? (Round your answer to three decimal places.)

0.245

(d)

If a student was an undergraduate business major, what is the probability that the student intends to attend classes full-time in pursuit of an MBA degree? (Round your answer to three decimal places.)

0.702

(e)

Let A denote the event that the student intends to attend classes full-time in pursuit of an MBA degree, and let B denote the event that the student was an undergraduate business major. Are events A and B independent? Justify your answer. (Round your answers to three decimal places.)

P(A)P(B)

0.236

and

P(A ∩ B)

0.270

, so the events

are not

independent.

Viewing Saved Work Revert to Last Response

3. 0/5 points | Previous Answers ASWESBE9 6.E.033. My Notes

Consider the following exponential probability density function.

f(x) =

e^−x/5 for x ≥ 0

(a)

Write the formula for

P(x ≤ x₀).

676

$webMathematica generated answer key$

(b)

Find

P(x ≤ 2).

(Round your answer to four decimal places.)

0.3297

(c)

Find

P(x ≥ 5).

(Round your answer to four decimal places.)

0.3679

(d)

Find

P(x ≤ 6).

(Round your answer to four decimal places.)

0.6988

(e)

Find

P(2 ≤ x ≤ 6).

(Round your answer to four decimal places.)

0.3691

Solution or Explanation

Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

(a)

P(x ≤ x₀) = 1 − e^−x₀/5

(b)

P(x ≤ 2) = 1 − e^−2/5 = 1 − 0.6703 = 0.3297

(c)

P(x ≥ 5) = 1 − P(x ≤ 5) = 1 − (1 − e^−5/5) = e⁻¹ = 0.3679

(d)

P(x ≤ 6) = 1 − e^−6/5 = 1 − 0.3012 = 0.6988

(e)

P(2 ≤ x ≤ 6) = P(x ≤ 6) − P(x ≤ 2) = 0.6988 − 0.3297 = 0.3691

Viewing Saved Work Revert to Last Response

4. 0/2 points | Previous Answers ASWESBE9 8.AQ.502. My Notes

Use the applet "Sample Size and Interval Width when Estimating Proportions" to answer the following questions.

This applet illustrates how sample size is related to the width of a 95% confidence interval estimate for a population proportion.

(a)

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.023?

(b)

As the sample size decreases for any given confidence level, what happens to the confidence interval?

The confidence interval becomes wider than the population proportion. correct

The confidence interval becomes wider because the standard error becomes larger. The confidence interval becomes more narrow than the population proportion. The width of the confidence interval becomes the same as the standard error. The confidence interval becomes more narrow because the sampling distribution becomes larger.

Viewing Saved Work Revert to Last Response

5. 1/4 points | Previous Answers ASWESBE9 9.JMP.011. My Notes

In a certain state, 75% of college students own a cell phone. A poll is taken at a local college and 100 students are asked if they own a cell phone. Of those 100 students, 86 say that they do.

JMP Applet

Click here to access the JMP applet in a new window.
(a)

A company is interested in seeing if the proportion of students with cell phones at this college is statistically different than the proportion for the entire state. They perform the following hypothesis test.

H₀: p = 0.75
H₁: p ≠ 0.75

Determine the p-value for this test. (Round your answer to four decimal places.)

0.0111
(b)

The company needs to decide to reject or fail to reject the null hypothesis with a 5% significance level. Should they reject the null hypothesis?

The company should
reject
the null hypothesis. The proportion of students at the college with cell phones statistically different than the statewide proportion.
(c)

Which of the following correctly describes the p-value?

Assuming p = 0.75, the p-value describes the percent of the population that is larger than the proportion of the sample. Assuming p = 0.75, the p-value gives the probability of a similar sample having a sample proportion at least as extreme as the given sample proportion. Assuming p = 0.75, the p-value is the probability that the given proportion is equal to the population proportion. Assuming p = 0.75, the p-value gives the probability that any other sample will have a proportion equal to the given sample's proportion.

In JMP, how is the p-value labeled? What is the relationship between the p-value and the α-value in order to reject the null hypothesis?

Viewing Saved Work Revert to Last Response

6. 6/14 points | Previous Answers ASWESBE9 12.E.004. My Notes

You may need to use the appropriate technology to answer this question.

Benson Manufacturing is considering ordering electronic components from three different suppliers. The suppliers may differ in terms of quality in that the proportion or percentage of defective components may differ among the suppliers. To evaluate the proportion of defective components for the suppliers, Benson has requested a sample shipment of 500 components from each supplier. The number of defective components and the number of good components found in each shipment are as follows.

Component	Supplier
Component	A	B	C
Defective	17	22	42
Good	483	478	458

(a)

Formulate the hypotheses that can be used to test for equal proportions of defective components provided by the three suppliers.

H₀: Not all population proportions are equal.
H_a: p_A = p_B = p_C H₀: All population proportions are not equal.
H_a: p_A = p_B = p_C correct

H₀: p_A = p_B = p_C
H_a: Not all population proportions are equal. H₀: p_A = p_B = p_C
H_a: All population proportions are not equal.

(b)

Using a 0.05 level of significance, conduct the hypothesis test.

Find the value of the test statistic. (Round your answer to three decimal places.)

Find the p-value. (Round your answer to four decimal places.)

p-value =

0.0011

State your conclusion.

Do not reject H₀. We conclude that the suppliers do not provide equal proportions of defective components. correct

Reject H₀. We conclude that the suppliers do not provide equal proportions of defective components. Do not reject H₀. We cannot conclude that the suppliers do not provide equal proportions of defective components. Reject H₀. We cannot conclude that the suppliers do not provide equal proportions of defective components.

(c)

Conduct a multiple comparison test to determine if there is an overall best supplier or if one supplier can be eliminated because of poor quality. Use a 0.05 level of significance. (Round your answers for the critical values to four decimal places.)

Comparison

p_i − p_j

CV_ij

Significant
Diff > CV_ij

A vs. B

0.01

0.0300

A vs. C

0.05

0.0363

Yes

B vs. C

0.04

0.0378

Yes

Can any suppliers be eliminated because of poor quality? (Select all that apply.)

Supplier A Supplier B correct

Supplier C none

Viewing Saved Work Revert to Last Response

7. –/14 points ASWESBE9 13.E.008. My Notes

You may need to use the appropriate technology to answer this question.

A company manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. To measure how much employees at these plants know about quality management, a random sample of 6 employees was selected from each plant and the employees selected were given a quality awareness examination. The examination scores for these 18 employees are shown in the following table. The sample means, sample variances, and sample standard deviations for each group are also provided. Managers want to use these data to test the hypothesis that the mean examination score is the same for all three plants.

	Plant 1 Atlanta	Plant 2 Dallas	Plant 3 Seattle
	84	70	58
	74	76	65
	81	72	61
	76	74	68
	71	68	76
	82	84	74
Sample mean	78	74	67
Sample variance	26.0	32.0	50.4
Sample standard deviation	5.10	5.66	7.10

Set up the ANOVA table for these data. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments
Error
Total

Test for any significant difference in the mean examination score for the three plants. Use

α = 0.05.

State the null and alternative hypotheses.

H₀: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃ H₀: μ₁ = μ₂ = μ₃
H_a: μ₁ ≠ μ₂ ≠ μ₃ H₀: μ₁ = μ₂ = μ₃
H_a: Not all the population means are equal. H₀: μ₁ ≠ μ₂ ≠ μ₃
H_a: μ₁ = μ₂ = μ₃ H₀: At least two of the population means are equal.
H_a: At least two of the population means are different.

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to four decimal places.)

p-value =

State your conclusion.

Do not reject H₀. There is sufficient evidence to conclude that the means for the three plants are not equal. Reject H₀. There is not sufficient evidence to conclude that the means for the three plants are not equal. Reject H₀. There is sufficient evidence to conclude that the means for the three plants are not equal. Do not reject H₀. There is not sufficient evidence to conclude that the means for the three plants are not equal.

Viewing Saved Work Revert to Last Response

8. 1/6 points | Previous Answers ASWESBE9 14.E.006. My Notes

DATAfile: NFLPassing

The National Football League (NFL) records a variety of performance data for individuals and teams. To investigate the importance of passing on the percentage of games won by a team, the following data show the average number of passing yards per attempt (Yards/Attempt) and the percentage of games won (WinPct) for a random sample of 10 NFL teams for the 2011 season.†Source: NFL website, February 12, 2012

Team	Yards/Attempt	WinPct
Arizona Cardinals	6.5	50
Atlanta Falcons	7.1	63
Carolina Panthers	7.4	38
Chicago Bears	6.4	50
Dallas Cowboys	7.4	50
New England Patriots	8.3	81
Philadelphia Eagles	7.4	50
Seattle Seahawks	6.1	44
St. Louis Rams	5.2	13
Tampa Bay Buccaneers	6.2	25

(a)

Develop a scatter diagram with the number of passing yards per attempt on the horizontal axis and the percentage of games won on the vertical axis.

A scatter diagram has 9 points plotted on it. The horizontal axis ranges from 5 to 9 and is labeled: Yards/Attempt. The vertical axis ranges from 0 to 90 and is labeled: Win %. The points are plotted from left to right in an upward, diagonal direction starting in the lower left corner of the diagram and are between 5 to 8.5 on the horizontal axis and between 10 to 85 on the vertical axis.

A scatter diagram has 10 points plotted on it. The horizontal axis ranges from 5 to 9 and is labeled: Yards/Attempt. The vertical axis ranges from 0 to 90 and is labeled: Win %. The points appear to be distributed somewhat randomly and are between 5 to 8.5 on the horizontal axis. The 4 leftmost points are between 35 to 65 on the vertical axis and the 5 rightmost points are between 10 to 85 on the vertical axis.

(b)

What does the scatter diagram developed in part (a) indicate about the relationship between the two variables?

The scatter diagram indicates a negative linear relationship between average number of passing yards per attempt and the percentage of games won by the team. The scatter diagram indicates a positive linear relationship between average number of passing yards per attempt and the percentage of games won by the team. The scatter diagram indicates no noticeable linear relationship between average number of passing yards per attempt and the percentage of games won by the team.

(c)

Develop the estimated regression equation that could be used to predict the percentage of games won given the average number of passing yards per attempt. (Round your numerical values to three decimal places.)

ŷ =

·|(10σΧΨ)|

(d)

Provide an interpretation for the slope of the estimated regression equation.

The slope gives the percentage of games won when the average number of passes per attempt is 0. The slope gives the change in the average number of passes per attempt for every one percentage point decrease in the percentage of games won. The slope gives the change in the percentage of games won for every one yard increase in the average number of passes per attempt. The slope gives the average number of passes per attempt when the percentage of games won is 0%. The slope gives the change in the average number of passes per attempt for every one percentage point increase in the percentage of games won.

(e)

For the 2011 season, suppose the average number of passing yards per attempt for a certain NFL team was 6.2. Use the estimated regression equation developed in part (c) to predict the percentage of games won by that NFL team. (Note: For the 2011 season, suppose this NFL team's record was 7 wins and 9 losses. Round your answer to the nearest integer.)

Compare your prediction to the actual percentage of games won by this NFL team.

The predicted value is identical to the actual value. The predicted value is higher than the actual value. The predicted value is lower than the actual value.

Viewing Saved Work Revert to Last Response

9. –/8 points ASWESBE9 15.E.035. My Notes

A statistical program is recommended.

A company provides maintenance service for water-filtration systems throughout southern Florida. Customers contact the company with requests for maintenance service on their water-filtration systems. To estimate the service time and the service cost, the company's managers want to predict the repair time necessary for each maintenance request. Hence, repair time in hours is the dependent variable. Repair time is believed to be related to three factors, the number of months since the last maintenance service, the type of repair problem (mechanical or electrical), and the repairperson who performed the service. Data for a sample of 10 service calls are reported in the table below.

Repair Time in Hours	Months Since Last Service	Type of Repair	Repairperson
2.9	2	Electrical	Dave Newton
3.0	6	Mechanical	Dave Newton
4.8	8	Electrical	Bob Jones
1.8	3	Mechanical	Dave Newton
2.8	2	Electrical	Dave Newton
4.9	7	Electrical	Bob Jones
4.3	9	Mechanical	Bob Jones
4.8	8	Mechanical	Bob Jones
4.4	4	Electrical	Bob Jones
4.5	6	Electrical	Dave Newton

(a)

Ignore for now the months since the last maintenance service

(x₁)

and the repairperson who performed the service. Develop the estimated simple linear regression equation to predict the repair time

(y)

given the type of repair

(x₂).

Let

x₂ = 0

if the type of repair is mechanical and

x₂ = 1

if the type of repair is electrical. (Round your numerical values to three decimal places.)

ŷ =

(b)

Does the equation that you developed in part (a) provide a good fit for the observed data? Explain. (Round your answer to two decimal places.)

We see that % of the variability in the repair time has been explained by the type of repair. Since this is 55%, the estimated regression equation a good fit for the observed data.

(c)

Ignore for now the months since the last maintenance service and the type of repair associated with the machine. Develop the estimated simple linear regression equation to predict the repair time given the repairperson who performed the service. Let

x₃ = 0

if Bob Jones performed the service and

x₃ = 1

if Dave Newton performed the service. (Round your numerical values to three decimal places.)

ŷ =

(d)

Does the equation that you developed in part (c) provide a good fit for the observed data? Explain. (Round your answer to two decimal places.)

We see that % of the variability in the repair time has been explained by the repairperson. Since this is 55%, the estimated regression equation a good fit for the observed data.

Viewing Saved Work Revert to Last Response

Home My Assignments Extension Request

1	2	3	4	5	6	7	8	9	10	11	12
1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1
1/100	1/100	1/100	1/100	1/100	1/100	1/100	1/100	1/100	1/100	1/100	1/100

1	2	3	4	5
0/1	0/1	0/1	0/1	0/1
1/100	1/100	1/100	1/100	1/100

1	2
–/1	0/1
0/100	1/100

1	2	3	4
0/1	1/1	–/1	0/1
1/100	1/100	0/100	1/100

1	2	3	4	5	6
–/1	1/1	–/1	0/1	–/1	–/1
0/100	2/100	0/100	1/100	0/100	0/100

H₀: p	=	0.75
H₁: p	≠	0.75

WebAssign

Welcome, demo@demo

Anderson - Essen. of Stats for Business & Econ 9/e (Homework)

Current Score : 32 / 82

Due : Sunday, January 27, 2030 23:30 EST

Last Saved : n/a Saving... ()

Instructions

Assignment Submission

Assignment Scoring

(a) In the construction of a pie chart, how many degrees would be in the section of the pie showing the Yes answers?

(b) How many degrees would be in the section of the pie showing the No answers?

(c) Construct a pie chart.

(d) Construct a frequency bar chart.

JMP Applet

(a)

(b)

(c)