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Anderson et al-Stats for Business & Economics 15/e (Homework)

James Finch

Statistics, section 2, Fall 2019

Instructor: Dr. Friendly

Current Score : 14 / 153

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

Last Saved : n/a Saving...  ()

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Question
Points
1 2 3 4 5 6 7 8 9 10
12/12 –/43 –/17 0/5 1/2 1/4 –/14 –/14 –/9 –/33
Total
14/153 (9.2%)

  • 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 is a case problem that uses summarization tools and techniques to gain insight into a dataset.

    Question 3 features multiple question types and guides students through the process of interpreting values.

    Question 4 asks students to derive a formula and then use that formula to calculate probabilities.

    Question 5 includes an interactive applet to determine the sample size necessary for a given margin of error.

    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.

    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 ASWSBE15 2.E.003.MI.SA. My Notes
Question Part
Points
Submissions Used
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 2/100 1/100
Total
12/12
 
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.

Tutorial Exercise
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
 = 120 Correct: Your answer is correct. seenKey

120
Given that there are 57 Yes answers, find the corresponding relative frequency, leaving your answer as a fraction.
relative frequency = 57/120 Correct: Your answer is correct. seenKey

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,
57
120
,
 by the number of degrees in a circle, 360.
degrees in Yes portion = 
57
120 Correct: Your answer is correct. seenKey

120
360 degrees
 = 171 Correct: Your answer is correct. seenKey

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
 
 = 43/120 Correct: Your answer is correct. seenKey

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,
43
120
,
 by the number of degrees in a circle, 360.
degrees in No portion = 
43
120 Correct: Your answer is correct. seenKey

120
360 degrees
 = 129 Correct: Your answer is correct. seenKey

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 Correct: Your answer is correct. seenKey

129
 degrees
 = 60 Correct: Your answer is correct. seenKey

60

 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.

Correct: Your answer is correct.
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 Correct: Your answer is correct. seenKey

3

 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.

Correct: Your answer is correct.
You have now completed the Master It.
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2. –/43 points ASWSBE15 3.CP.001. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43
–/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 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100
Total
–/43
 
  • Pelican Stores, a division of National Clothing, is a chain of women's apparel stores operating throughout the country. The chain recently ran a promotion in which discount coupons were sent to customers of other National Clothing stores. Data collected for a sample of 100 in-store credit card transactions at Pelican Stores during one day while the promotion was running are contained in the "Given Data" tab below. The proprietary card method of payment refers to charges made using a National Clothing charge card. Customers who made a purchase using a discount coupon are referred to as promotional customers and customers who made a purchase but did not use a discount coupon are referred to as regular customers. Because the promotional coupons were not sent to regular Pelican Stores customers, management considers the sales made to people presenting the promotional coupons as sales it would not otherwise make. Of course, Pelican also hopes that the promotional customers will continue to shop at its stores.
    Most of the variables shown in the data are self-explanatory, but two of the variables require some clarification.
    Items: the total number of items purchased
    Net Sales: the total amount ($) charged to the credit card
    Pelican's management would like to use this sample data to learn about its customer base and to evaluate the promotion involving discount coupons.
  • Use the methods of descriptive statistics presented in this chapter to summarize the data and comment on your findings.
    • Consider the complete data set.
      Customer Type of Customer Items Net Sales Method of Payment Gender Marital Status Age
      1 Regular 1 44.50 Discover Male Married 32
      2 Promotional 1 107.40 Proprietary Card Female Married 36
      3 Regular 1 26.50 Proprietary Card Female Married 32
      4 Promotional 5 104.40 Proprietary Card Female Married 28
      5 Regular 2 58.00 MasterCard Female Married 35
      6 Regular 1 49.50 MasterCard Female Married 45
      7 Promotional 2 83.00 Proprietary Card Female Married 30
      8 Regular 1 27.50 Visa Female Married 41
      9 Promotional 2 60.52 Proprietary Card Female Married 46
      10 Regular 1 48.50 Proprietary Card Female Married 37
      11 Regular 1 34.50 Proprietary Card Female Married 48
      12 Promotional 1 35.60 Proprietary Card Female Married 41
      13 Promotional 9 164.40 Visa Female Married 41
      14 Promotional 2 68.50 Visa Female Married 46
      15 Regular 1 53.50 Visa Male Single 25
      16 Promotional 2 75.40 Proprietary Card Male Single 37
      17 Promotional 3 98.00 Proprietary Card Female Single 23
      18 Regular 3 58.50 Discover Female Married 41
      19 Promotional 2 43.50 MasterCard Female Married 32
      20 Promotional 6 49.80 Proprietary Card Female Married 57
      21 Promotional 1 36.60 Proprietary Card Female Single 28
      22 Promotional 4 74.82 Proprietary Card Female Married 38
      23 Promotional 7 271.00 American Express Female Married 51
      24 Regular 2 79.00 Proprietary Card Female Married 42
      25 Promotional 2 43.50 Visa Male Married 48
      26 Promotional 1 34.02 Proprietary Card Female Married 60
      27 Regular 1 49.50 Proprietary Card Female Married 55
      28 Promotional 5 197.80 Proprietary Card Female Single 43
      29 Promotional 3 76.20 Proprietary Card Female Married 32
      30 Promotional 1 23.00 Proprietary Card Female Married 71
      31 Promotional 2 67.20 MasterCard Female Married 29
      32 Regular 1 80.00 Proprietary Card Female Married 52
      33 Promotional 3 68.20 Proprietary Card Female Married 45
      34 Regular 1 44.00 Proprietary Card Female Married 35
      35 Promotional 5 109.50 MasterCard Female Married 56
      36 Regular 1 33.50 MasterCard Male Single 37
      37 Regular 2 106.50 Visa Female Single 42
      38 Promotional 6 121.50 Proprietary Card Female Married 50
      39 Promotional 5 17.23 Proprietary Card Female Married 44
      40 Regular 2 57.50 Proprietary Card Female Married 59
      41 Promotional 13 202.80 Proprietary Card Female Married 43
      42 Promotional 4 24.50 Visa Female Married 46
      43 Regular 2 127.50 Proprietary Card Female Married 48
      44 Promotional 1 66.40 Proprietary Card Female Married 55
      45 Promotional 2 27.80 Proprietary Card Female Married 38
      46 Promotional 2 44.60 Proprietary Card Female Married 61
      47 Regular 1 30.00 MasterCard Female Married 46
      48 Promotional 3 68.64 Proprietary Card Female Married 30
      49 Promotional 1 18.82 Proprietary Card Female Married 33
      50 Promotional 9 149.20 MasterCard Female Married 46
      51 Promotional 6 181.62 Proprietary Card Female Married 38
      52 Promotional 5 122.80 Proprietary Card Male Married 68
      53 Regular 1 62.00 Discover Female Single 78
      54 Regular 2 78.00 Visa Female Single 20
      55 Regular 2 53.50 MasterCard Female Married 33
      56 Promotional 3 145.60 Proprietary Card Female Married 39
      57 Promotional 6 128.10 Proprietary Card Female Married 55
      58 Promotional 2 84.40 Proprietary Card Female Married 48
      59 Promotional 4 70.20 MasterCard Female Married 47
      60 Promotional 4 118.00 Proprietary Card Female Single 51
      61 Promotional 1 112.80 Proprietary Card Female Married 46
      62 Promotional 3 63.91 Proprietary Card Female Single 31
      63 Promotional 5 58.60 Proprietary Card Female Married 54
      64 Promotional 1 36.60 Proprietary Card Female Single 43
      65 Promotional 2 54.50 Proprietary Card Female Married 49
      66 Promotional 1 44.60 Proprietary Card Female Married 62
      67 Promotional 2 64.50 Proprietary Card Female Married 34
      68 Promotional 5 150.80 Proprietary Card Female Married 28
      69 Promotional 2 51.20 Proprietary Card Male Married 47
      70 Promotional 8 99.05 Proprietary Card Female Married 54
      71 Promotional 5 159.32 Proprietary Card Female Married 31
      72 Promotional 4 63.00 MasterCard Female Married 32
      73 Regular 1 74.00 Proprietary Card Female Single 22
      74 Promotional 2 50.50 Proprietary Card Female Married 32
      75 Promotional 2 49.22 Proprietary Card Female Married 74
      76 Promotional 4 89.74 Proprietary Card Female Married 63
      77 Regular 2 43.00 Proprietary Card Female Married 43
      78 Promotional 4 116.14 Proprietary Card Female Married 29
      79 Promotional 3 90.80 Proprietary Card Female Married 39
      80 Regular 2 94.00 Discover Female Married 54
      81 Promotional 2 83.00 MasterCard Female Married 68
      82 Promotional 6 58.20 Proprietary Card Female Single 31
      83 Promotional 4 62.50 Visa Female Married 36
      84 Promotional 3 51.00 Proprietary Card Female Married 45
      85 Regular 2 41.50 Visa Female Married 45
      86 Promotional 1 25.80 Proprietary Card Female Married 63
      87 Regular 6 148.00 MasterCard Female Single 49
      88 Regular 4 112.00 Proprietary Card Female Married 36
      89 Promotional 1 35.60 Proprietary Card Female Single 21
      90 Promotional 6 61.60 Proprietary Card Female Married 43
      91 Promotional 4 100.20 Proprietary Card Female Married 54
      92 Promotional 1 26.42 Proprietary Card Female Married 55
      93 Regular 5 164.75 Proprietary Card Female Married 73
      94 Promotional 17 233.50 Proprietary Card Female Married 31
      95 Regular 3 71.00 American Express Female Married 47
      96 Regular 1 43.50 MasterCard Female Married 44
      97 Promotional 9 257.00 Proprietary Card Female Married 31
      98 Promotional 10 292.59 Proprietary Card Female Married 53
      99 Promotional 2 52.60 Proprietary Card Female Married 31
      100 Promotional 1 32.44 Proprietary Card Female Married 44
    • Compute the mean, median, range, standard deviation, and skewness of net sales for all customers. (Round the mean, median, range, and standard deviation to two decimal places. Round the skewness to three decimal places.)
      mean $ median $ range standard deviation skewness
      What observations can be drawn from these descriptive statistics?
      Because the skewness is , the data are . This is also evidenced by the mean being the median.
    • Compute the mean, median, range, standard deviation, and skewness of net sales for married, single, regular, and promotional customers. (Round the mean, median, range, and standard deviation values to two decimal places. Round the skewness values to three decimal places.)
      Married Single Regular Promotional
      Mean $ $ $ $
      Median $ $ $ $
      Range
      Standard Deviation
      Skewness
      What observations can be drawn from these descriptive statistics?
      On average, the most amount was spent by customers and the least amount spent was by customers. Based on the skewness value, the distribution of married customers is . The skewness value of single customers indicates the distribution is . The skewness value of regular customers indicates the distribution is . The skewness value of promotional customers indicates the distribution is . The highest variability for net sales occurred for customers and the least variability for net sales occurred for customers.
    • Compute the sample covariance between age and net sales. (Round your answer to two decimal places.)
      Based on the sample covariance, what can be said about the relationship between age and net sales?
      Because the sample covariance is , there is a linear relationship between the age of a customer and the net sales.
      Compute the sample correlation coefficient between age and net sales. (Round your answer to three decimal places.)
      Based on the sample correlation coefficient, what can be said about the relationship between age and net sales?
      The value of the correlation coefficient implies there is linear relationship between the age of a customer and the net sales.
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3. –/17 points ASWSBE15 4.E.033. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
–/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100
Total
–/17
 
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
Intended
Enrollment
Status 		Full-Time 354 199 251 804
Part-Time 152 161 196 509
Totals 506 360 447 1,313
(a)
Develop a joint probability table for these data. (Round your answers to three decimal places.)
Undergraduate Major Totals
Business Engineering Other
Intended
Enrollment
Status 		Full-Time
Part-Time
Totals 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 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.)
(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.)
(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)
 = and
P(A ∩ B)
 = , so the events independent.
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4. 0/5 points  |  Previous Answers ASWSBE15 6.E.033. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5
–/1 0/1 –/1 –/1 –/1
0/100 1/100 0/100 0/100 0/100
Total
0/5
 
Consider the following exponential probability density function.
f(x) = 
1
4
e−x/4     for x ≥ 0
(a)
Write the formula for
P(x ≤ x0).
(b)
Find
P(x ≤ 3).
 (Round your answer to four decimal places.)
Incorrect: Your answer is incorrect.
(c)
Find
P(x ≥ 4).
 (Round your answer to four decimal places.)
(d)
Find
P(x ≤ 5).
 (Round your answer to four decimal places.)
(e)
Find
P(3 ≤ x ≤ 5).
 (Round your answer to four decimal places.)
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5. 1/2 points  |  Previous Answers ASWSBE15 8.AQ.502. My Notes
Question Part
Points
Submissions Used
1 2
1/1 0/1
2/100 1/100
Total
1/2
 
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?
Correct: Your answer is correct.
(b)
As the sample size decreases for any given confidence level, what happens to the confidence interval?
     Incorrect: Your answer is incorrect.
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6. 1/4 points  |  Previous Answers ASWSBE15 10.E.006. My Notes
Question Part
Points
Submissions Used
1 2 3 4
1/1 0/1 0/1 0/1
5/100 1/100 1/100 2/100
Total
1/4
 
Dataset Available: Download Hotel.xlsx
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.)
     Correct: Your answer is correct.
Calculate the test statistic. (Round your answer to two decimal places.)
Incorrect: Your answer is incorrect.
Calculate the p-value. (Round your answer to four decimal places.)
p-value = Incorrect: Your answer is incorrect.
What is your conclusion? (Use α = 0.05.)
     Incorrect: Your answer is incorrect.
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7. –/14 points ASWSBE15 12.E.004. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9 10 11 12 13 14
–/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100
Total
–/14
 
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
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.
    
(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 =
State your conclusion.
    
(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
pi − pj
CVij
 		Significant
Diff > CVij
A vs. B
A vs. C
B vs. C
Can any suppliers be eliminated because of poor quality? (Select all that apply.)

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8. –/14 points ASWSBE15 13.E.008. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9 10 11 12 13 14
–/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100
Total
–/14
 
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 59
76 76 64
81 74 63
76 73 68
71 69 76
86 82 66
Sample
mean 		79 74 66
Sample
variance 		32.0 22.0 33.2
Sample
standard
deviation 		5.66 4.69 5.76
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.
    
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.
    
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9. –/9 points ASWSBE15 15.E.009. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9
–/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100
Total
–/9
 
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 1,558 2
195,100 1.5 1,125 3
212,500 2 1,196 2
245,900 2 2,128 3
250,000 3 1,280 3
255,000 2 1,596 3
258,000 3.5 2,374 4
267,000 2.5 2,439 3
268,000 2 1,470 4
275,000 2 1,688 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 1,928 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
(a)
Develop scatter plots of selling price versus number of bathrooms.

What is the relationship between the selling price of a house and the number of bathrooms in it?
    
Develop a scatter plot of selling price versus square footage.

What is the relationship between the selling price of a house and its square footage?
    
Develop a scatter plot of selling price versus number of bedrooms.

What is the relationship between the selling price of a house and the number of bedrooms in it?
    
(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
y
 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
y
 represent the selling price.)
ŷ =
(d)
Suppose your house has three bedrooms and is 2,850 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 ASWSBE15 19.E.006. My Notes
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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
Weak Strong
Weak 0.05 0.50
Strong 0.25 0.20
Revenue projections show an annual contribution to profit of $10,000 if Seneca Hill only plants Chardonnay grapes and demand is weak for Chardonnay wine, and $60,000 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 $15,000 if demand is weak for Riesling grapes and $35,000 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
Weak Strong
Weak $12,000 $30,000
Strong $16,000 $50,000
(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.)
A decision tree has 16 answer blanks. Refer to the adjacent list for more details.
  • 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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