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Statistics, section 2, Fall 2019
My Assignments

Anderson et al-Stats for Business & Economics 14/e (Homework)

James Finch

Statistics, section 2, Fall 2019

Instructor: Dr. Friendly

Current Score : 30 / 114

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	10
8/12	14/17	0/5	1/2	3/4	0/4	2/14	–/14	2/9	–/33

Total

30/114 (26.3%)

Instructions

Drawing from the authors' unmatched experience as professors and consultants, Anderson/Sweeney/Williams/Camm/Cochran/Fry/Ohlmann's Statistics for Business and Economics, 14th 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 links to an Excel data file.

Question 7 guides the student through a hypothesis test.

Question 8 showcases a full ANOVA table the student completes and then uses to perform a hypothesis test.

Question 9 displays grading for a multiple regression equation that is then used to make predictions.

Question 10 guides students through the decision analysis process. 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. 8/12 points | Previous Answers ASWSBE14 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 57 Yes, 43 No, and 20 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
	=	57 + 43 + 20
	=	(No Response) 120

Given that there are 57 Yes answers, find the corresponding relative frequency, leaving your answer as a fraction.

relative frequency = (No Response)

19/40

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

120

360 degrees

171

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

129

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 171, 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 −

171 + 129

129

degrees

Step 6

The pie chart for this data will have a sector of 60 degrees representing the No Opinion answer, a sector of 129 degrees representing the No answer, and a third sector of 171 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, 16.7%.

The second piece is labeled No, 47.5%.

The third piece is labeled No opinion, 35.8%.

A circle graph has 3 uneven pieces.

The first piece is labeled Yes, 47.5%.

The second piece is labeled No, 16.7%.

The third piece is labeled No opinion, 35.8%.

A circle graph has 3 uneven pieces.

The first piece is labeled Yes, 35.8%.

The second piece is labeled No, 16.7%.

The third piece is labeled No opinion, 47.5%.

A circle graph has 3 uneven pieces.

The first piece is labeled Yes, 47.5%.

The second piece is labeled No, 35.8%.

The third piece is labeled No opinion, 16.7%.

(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 bars.

The height of each bar will correspond to the frequency of each group. There were 57 Yes answers, 43 No answers, and 20 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: 57 units tall

No: 43 units tall

No opinion: 20 units tall

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2. 14/17 points | Previous Answers aswsbe14 4.e.033.defective 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	352	199	249	800
Intended Enrollment Status	Part-Time	148	159	193	500
Totals		500	358	442	1,300

(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	0.271	0.153	0.192	0.615
Intended Enrollment Status	Part-Time	0.114	0.122	0.148	0.385
Totals		0.385	0.275	0.340	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.)

0.249

(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.)

0.704

(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)

0.237

and

P(A ∩ B)

0.271

, so the events

are not

independent.

Solution or Explanation

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

(a)

		Undergraduate Major			Totals
		Business	Engineering	Other	Totals
Intended Enrollment Status	Full-Time	0.271	0.153	0.192	0.615
Intended Enrollment Status	Part-Time	0.114	0.122	0.148	0.385
Totals		0.385	0.275	0.340	1.000

(b)

P(business) = 0.385,

P(engineering) = 0.275,

and

P(other) = 0.340,

so business is the undergraduate major that produces the most potential MBA students.

(c)

P(engineering | full-time) =

P(engineering ∩ full-time)

P(full-time)

0.153

0.615

= 0.249

(d)

P(full-time | business) =

P(full-time ∩ business)

P(business)

0.271

0.385

= 0.704

(e)

For independence, we must have that

P(A)P(B) = P(A ∩ B);

from the joint probability table in part (a) of this problem, we have

P(A)	=	0.615
P(B)	=	0.385.

P(A)P(B) = (0.615)(0.385) = 0.237,

but

P(A ∩ B) = 0.271.

Because

P(A)P(B) ≠ P(A ∩ B),

the events are not independent.

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3. 0/5 points | Previous Answers aswsbe14 6.e.033.defective My Notes

Consider the following exponential probability density function.

f(x) =

e^−x/3 for x ≥ 0

(a)

Write the formula for

P(x ≤ x₀).

Σ13e−(x3)

$webMathematica generated answer key$

(b)

Find

P(x ≤ 2).

(Round your answer to four decimal places.)

0.4866

(c)

Find

P(x ≥ 3).

(Round your answer to four decimal places.)

0.3679

(d)

Find

P(x ≤ 4).

(Round your answer to four decimal places.)

0.7364

(e)

Find

P(2 ≤ x ≤ 4).

(Round your answer to four decimal places.)

0.2498

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^−x₀/3

(b)

P(x ≤ 2) = 1 − e^−2/3 = 1 − 0.5134 = 0.4866

(c)

P(x ≥ 3) = 1 − P(x ≤ 3) = 1 − (1 − e^−3/3) = e⁻¹ = 0.3679

(d)

P(x ≤ 4) = 1 − e^−4/3 = 1 − 0.2636 = 0.7364

(e)

P(2 ≤ x ≤ 4) = P(x ≤ 4) − P(x ≤ 2) = 0.7364 − 0.4866 = 0.2498

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4. 1/2 points | Previous Answers aswsbe14 8.aq.502.defective 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?

1,821

(b)

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

The confidence interval becomes more narrow than the population proportion. The confidence interval becomes wider than the population proportion. The width of the confidence interval becomes the same as the standard error. correct

The confidence interval becomes wider because the standard error becomes larger. The confidence interval becomes more narrow because the sampling distribution becomes larger.

Solution or Explanation

(a)

To determine the required sample size to obtain a margin of error of 0.023, move the sample size slider until the interval on either side of the interval reads 0.023. In this case, a sample size of 1,821 generates a margin of error equaling 0.023.

(b)

In the formula everything under and including the square root sign is called the standard error. As sample size decreases, the denominator under the square root becomes smaller causing the overall fraction to become larger and the square root to become larger. By continuing with the multiplication of 1.96 and then addition/subtraction of 0.5 leads to a wider interval as the sample size decreases.

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5. 3/4 points | Previous Answers aswsbe14 9.jmp.011.defective 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
is
statistically different than the statewide proportion.
(c)

Which of the following correctly describes the p-value?

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 gives the probability that any other sample will have a proportion equal to the given sample's proportion. 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 is the probability that the given proportion is equal to the population proportion.

Solution or Explanation

(a)

To determine the p-value for the hypothesis test of proportions, refer to the applet. The p-value is listed in the Test Probabilities table as the

"Prob > |z|".

From the applet, see that this value is 0.0111.

(b)

Recall that the null hypothesis is rejected when the p-value is less than α. For a test with 5% significance,

α = 0.05.

From before, the p-value is 0.0111. Since

0.0111 < 0.05,

the null hypothesis should be rejected. The null hypothesis stated that the proportion of students with cell phones at the local college was equal to the proportion of students with cell phones statewide. Since this is rejected, the proportion of students at the college with cell phones is statistically different than the statewide proportion.

(c)

The p-value is a value between 0 and 1 that represents the probability that another sample would behave similarly. That is, if another sample of students at the college were taken in the same way as the first, the p-value is the probability that the proportion of the new sample would be as extreme as the first. Thus, assuming p = 0.75, the p-value gives the probability of a similar sample being as extreme as the given sample 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?

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6. 0/4 points | Previous Answers aswsbe14 10.e.006.defective My Notes

DATAfile: Hotel

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

Suppose that you are responsible for making arrangements for a business convention and that you have been charged with choosing a city for the convention that has the least expensive hotel rooms. You have narrowed your choices to Atlanta and Houston. The file named Hotel contains samples of prices for rooms in Atlanta and Houston that are consistent with a SmartMoney survey conducted by Smith Travel Research. Because considerable historical data on the prices of rooms in both cities are available, the population standard deviations for the prices can be assumed to be $20 in Atlanta and $25 in Houston. Based on the sample data, can you conclude that the mean price of a hotel room in Atlanta is lower than one in Houston?

State the hypotheses. (Let μ₁ = mean hotel price in Atlanta and let μ₂ = mean hotel price in Houston.)

H₀: μ₁ − μ₂ ≥ 0

H_a: μ₁ − μ₂ < 0

H₀: μ₁ − μ₂ < 0

H_a: μ₁ − μ₂ ≥ 0

H₀: μ₁ − μ₂ = 0

H_a: μ₁ − μ₂ ≠ 0

H₀: μ₁ − μ₂ ≠ 0

H_a: μ₁ − μ₂ = 0

H₀: μ₁ − μ₂ > 0

H_a: μ₁ − μ₂ = 0

Calculate the test statistic. (Round your answer to two decimal places.)

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

p-value =

What is your conclusion?

Do not reject H₀. There is insufficient evidence to conclude that the mean price of a hotel room in Atlanta is lower than the mean price of a hotel room in Houston. Do not reject H₀. There is sufficient evidence to conclude that the mean price of a hotel room in Atlanta is lower than the mean price of a hotel room in Houston. correct

Reject H₀. There is sufficient evidence to conclude that the mean price of a hotel room in Atlanta is lower than the mean price of a hotel room in Houston. Reject H₀. There is insufficient evidence to conclude that the mean price of a hotel room in Atlanta is lower than the mean price of a hotel room in Houston.

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7. 2/14 points | Previous Answers aswsbe14 12.e.004.defective 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	10	15	35
Good	490	485	465

(a)

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

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

(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.)

18.229

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

p-value =

0.0001

State your conclusion.

Do not reject H₀. We 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. Do not reject H₀. We cannot 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.

(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

A vs. C

B vs. C

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

Supplier A Supplier B correct

Supplier C none

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8. –/14 points aswsbe14 13.e.008.defective 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
	85	71	59
	76	76	63
	83	72	61
	77	73	69
	70	69	76
	89	89	74
Sample mean	80	75	67
Sample variance	48.0	52.4	50.0
Sample standard deviation	6.93	7.24	7.07

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: 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: Not all the population means are equal.
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.

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

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9. 2/9 points | Previous Answers ASWSBE14 15.E.009. My Notes

Spring is a peak time for selling houses. The table below contains the selling price, number of bathrooms, square footage, and number of bedrooms of 26 homes in a certain city.

Selling Price	Baths	Sq Ft	Beds
160,000	1.5	1,786	3
170,000	2	1,768	3
178,000	1	1,219	3
182,500	1	1,578	2
195,100	1.5	1,125	4
212,500	2	1,196	2
245,900	2	2,128	3
250,000	3	1,280	2
255,000	2	1,596	3
258,000	2.5	2,374	4
267,000	2.5	2,439	3
268,000	2	1,470	4
275,000	2	1,668	4

Selling Price	Baths	Sq Ft	Beds
295,000	2.5	1,860	3
325,000	2	2,056	4
325,000	3.5	2,776	4
328,400	2	1,408	4
331,000	1.5	1,972	3
344,500	2.5	1,736	3
365,000	2.5	1,990	4
385,000	2.5	3,640	4
395,000	2.5	1,908	4
399,000	2	2,108	3
430,000	2	2,462	4
430,000	2	2,615	4
454,000	3.5	3,700	4

(b)

Develop an estimated regression equation that can be used to predict the selling price given the three independent variables (number of baths, square footage, and number of bedrooms). (Round your numerical values to two decimal places. Let

x₁

represent the number of baths,

x₂

represent the square footage,

x₃

represent the number of bedrooms, and

represent the selling price.)

ŷ =

(c)

It is argued that we do not need both number of baths and number of bedrooms. Develop an estimated regression equation that can be used to predict selling price given square footage and the number of bedrooms. (Round your numerical values to two decimal places. Let

x₁

represent the square footage,

x₂

represent the number of bedrooms, and

represent the selling price.)

ŷ =

(d)

Suppose your house has three bedrooms and is 2,450 square feet. What is the predicted selling price (in $) using the model developed in part (c). (Round your answer to the nearest cent.)

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10. –/33 points ASWSBE14 19.E.006. My Notes

Question Part

Points

Submissions Used

–/1

0/100

Total

–/33

A survey ranked the Finger Lakes region of New York state as the top wine region to visit. Finger Lakes vineyards typically specialize in growing grapes for white wines such as Chardonnay and Riesling. Seneca Hill Winery recently purchased land for the purpose of establishing a new vineyard. Management is considering two varieties of white grapes for the new vineyard: Chardonnay and Riesling. The Chardonnay grapes would be used to produce a dry Chardonnay wine, and the Riesling grapes would be used to produce a semidry Riesling wine. It takes approximately four years from the time of planting before new grapes can be harvested. This length of time creates a great deal of uncertainty concerning future demand and makes the decision concerning the type of grapes to plant difficult. Three possibilities are being considered: Chardonnay grapes only; Riesling grapes only; and both Chardonnay and Riesling grapes. Seneca management decided that for planning purposes it would be adequate to consider only two demand possibilities for each type of wine: strong or weak. With two possibilities for each type of wine it was necessary to assess four probabilities. With the help of some forecasts in industry publications management made the following probability assessments.

Chardonnay Demand	Riesling Demand
Chardonnay Demand	Weak	Strong
Weak	0.05	0.50
Strong	0.25	0.20

Revenue projections show an annual contribution to profit of $30,000 if Seneca Hill only plants Chardonnay grapes and demand is weak for Chardonnay wine, and $80,000 if the company only plants Chardonnay grapes and demand is strong for Chardonnay wine. If the company only plants Riesling grapes, the annual profit projection is $35,000 if demand is weak for Riesling grapes and $55,000 if demand is strong for Riesling grapes. If Seneca plants both types of grapes, the annual profit projections are as shown in the following table.

Chardonnay Demand	Riesling Demand
Chardonnay Demand	Weak	Strong
Weak	$32,000	$50,000
Strong	$36,000	$70,000

(a)

What is the decision to be made, what is the chance event, and what is the consequence?

The decision to be made is . The chance event is . The consequence is .

Identify the alternatives for the decisions and the possible outcomes for the chance events.

The alternatives for the decisions are . The possible outcomes for the chance events are .

(b)

Develop a decision tree. (Enter monetary values in thousands and percentages in decimal form.)

Decision Tree

Description

A decision tree begins at decision node 1 and an upper, middle, and lower branch extend from this node to the right. The upper branch, labeled Plant Chardonnay, stops at chance node 2 and an upper and lower branch extend from this node to the right. The middle branch, labeled Plant both grapes, stops at chance node 3 and four branches extend from this node to the right. The lower branch, labeled Plant Riesling, stops at chance node 4 and an upper and lower branch extend from this node to the right.
The branches extending from decision node 1 from top to bottom are labeled Plant Chardonnay, Plant both grapes, and Plant Riesling. The next set of branches from top to bottom are labeled Weak for Chardonnay, Strong for Chardonnay, Weak Chardonnay Weak Riesling, Weak Chardonnay Strong Riesling, Strong Chardonnay Weak Riesling, Strong Chardonnay Strong Riesling, Weak for Riesling, and Strong for Riesling. There is an answer blank at the end of each branch as well as below each branch.

(c)

Use the expected value approach to recommend which alternative Seneca Hill Winery should follow in order to maximize expected annual profit.

EV(Plant Chardonnay) EV(Plant both grapes) EV(Plant Riesling) The best decision is to plant grapes.

(d)

Suppose management is concerned about the probability assessments when demand for Chardonnay wine is strong. Some believe it is likely for Riesling demand to also be strong in this case. Suppose the probability of strong demand for Chardonnay and weak demand for Riesling is 0.05 and that the probability of strong demand for Chardonnay and strong demand for Riesling is 0.40. How does this change the recommended decision? Assume that the probabilities when Chardonnay demand is weak are still 0.05 and 0.50.

EV(Plant Chardonnay) EV(Plant both grapes) EV(Plant Riesling) The best decision is to plant grapes.

(e)

Other members of the management team expect the Chardonnay market to become saturated at some point in the future, causing a fall in prices. Suppose that the annual profit projections fall to $50,000 when demand for Chardonnay is strong and Chardonnay grapes only are planted. Using the original probability assessments, determine how this change would affect the optimal decision.

EV(Plant Chardonnay) EV(Plant both grapes) EV(Plant Riesling) The best decision is to plant grapes.

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Anderson et al-Stats for Business & Economics 14/e (Homework)

Current Score : 30 / 114

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

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(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)