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domyonlineclass123 · 1 year ago

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mystatlabanswers · 2 years ago

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mikedonovanus · 6 years ago

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Mystatlab homework answers

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takeonlineclasshelp-blog · 6 years ago

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What Are The Trends Of Online Class For The Current Year

The trends of online class is changing rapidly and more and more students are joining to the class for completion of the education.

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VR and R Are On The Rise

In the training fields, virtual and augmented realities have been the buzzword for a couple of years, whether one is seeking for MyMathLab Answers. By hundreds of things, learners who are people can get distracted. This distraction can be done by newsfeeds, funny videos by the friends or even the constant IM notifications from the colleagues. For both entertainment and engagement, VR and AR are excellent options. Imagine a candidate is engaged in a lot of other activities, but at the same time, one needs to complete the academics. In that case, one can use the VR or AR technology, so that one can learn the subject in a proper way. Moreover, the learning can be much synchronized and proper way. While these two technologies are not the same, but the future possibilities of these technologies are very great. But it can be said, that AR is going to most likely to take the lead, while virtual reality can create a really immersive learning experience.

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Content Is Going Visual

The e-courses are the foundation of any e-learning. Since the traditional classroom has shifted to mobile on the go learning, the form of online courses are also changing. As a result, the use of animation, video, engaging scenario is making the online class more interesting. While consuming the mostly visual content, social media plays a significant role in habituating people. For learning whether recorded or live, the video is helping the students to pay attention to what is happening there. The students can develop a driving urge to learn the subject in a better way.

Self-Directed Learning

Self -directed learning is the approach of moving the onus of learning to the learners rather than teacher, in the general terms. As it maps to empowering learners to select how they want to learn, as an extension in the context of online and helping the students my assessment help. Through the LMS, the e-learning that pushed the self- directed learning is all about giving this control to the learners. When a student chooses the learn, one can select what one wants want and wishes to learn and over what time frame what wants to learn and so on. To decide the learning path, this approaches the control to the learners.

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markadi88-blog · 6 years ago

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studenteducare · 4 years ago

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whatdoesacoverpagelook758 · 5 years ago

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concludingparagraphessay821 · 5 years ago

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allhomeworkassignments · 6 years ago

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barbarapfister-blog · 8 years ago

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(MATH 533 Week 8 Final Exam Answers)

MATH 533 ( Applied Managerial Statistics ) Final Exam Answers

MATH 533 Final Exam Set 1

(TCO D) PuttingPeople2Work has a growing business placing out-of-work MBAs. They claim they can place a client in a job in their field in less than 36 weeks. You are given the following data from a sample. Sample size: 100 Population standard deviation: 5 Sample mean: 34.2 Formulate a hypothesis test to evaluate the claim. (Points : 10) Ho: µ = 36; Ha: µ ≠ 36 Ho: µ ≥ 36; Ha: µ < 36 Ho: µ ≤ 34.2; Ha: µ > 34.2 Ho: µ > 36; Ha: µ ≤ 36

Ans. b. H0 must always have equal sign, < 36 weeks 2. (TCO B) The Republican party is interested in studying the number of republicans that might vote in a particular congressional district. Assume that the number of voters is binomially distributed by party affiliation (either republican or not republican). If 10 people show up at the polls, determine the following: Binomial distribution

0.5

P(X)

cumulative probability

0.00098

0.00977

0.01074

0.04395

0.05469

0.11719

0.17188

0.20508

0.37695

0.24609

0.62305

0.20508

0.82813

0.11719

0.94531

0.04395

0.98926

0.00977

0.99902

0.00098

1.00000

What is the probability that no more than four will be republicans? (Points : 10) 38% 12% 21% 62%

Ans. a look at x=4, cumulative probability 3. (TCO A) Company ABC had sales per month as listed below. Using the Minitab output given, determine: (A) Range (5 points); (B) Median (5 points); and (C) The range of the data that would contain 68% of the results. (5 points). Raw data: sales/month (Millions of $) 23 45 34 34 56 67 54 34 45 56 23 19 Descriptive Statistics: Sales

Variable

Total Count

Mean

StDev

Variance

Minimum

Maximum

Range

Sales

40.83

15.39

236.88

19.00

67.00

48.00

Stem-and-Leaf Display: Sales Stem-and-leaf of Sales N = 12 Leaf Unit = 1.0

444

Reference: (TCO A) Company ABC had sales per month as listed below. Using the MegaStat output given, determine: (A) Range (5 points) (B) Median (5 points) (C) The range of the data that would contain 68% of the results. (5 points)

Raw data: sales/month (Millions of $) 19 34 23 34 56 45 35 36 46 47 19 23

count 12 mean 34.75 sample variance 146.20 sample standard deviation 12.09 minimum 19 maximum 56 range 37

Stem and Leaf plot for # 1 stem unit = 10 leaf unit = 1

count

12.00000

mean

34.75000

sample variance

146.20455

sample standard deviation

12.09151

minimum

19.00000

maximum

56.00000

range

37.00000

1st quartile

23.00000

median

34.50000

3rd quartile

45.25000

interquartile range

22.25000

mode

19.00000

4. (TCO C, D) Tesla Motors needs to buy axles for their new car. They are considering using Chris Cross Manufacturing as a vendor. Tesla’s requirement is that 95% of the axles are 100 cm ± 2 cm. The following data is from a test run from Chris Cross Manufacturing. Should Tesla select them as a vendor? Explain your answer. Descriptive statistics

count

mean

99.850

sample variance

4.627

sample standard deviation

2.151

minimum

96.9

maximum

104

range

7.1

population variance

4.338

population standard deviation

2.083

standard error of the mean

0.538

tolerance interval 95.45% lower

95.548

tolerance interval 95.45% upper

104.152

margin of error

4.302

1st quartile

98.850

median

99.200

3rd quartile

100.550

interquartile range

1.700

mode

103.000

(Points : 25) Reference: Chegg Tesla Motors needs to buy axles for their new car. They are considering using Chris Cross Manufacturing as a vendor. Tesla’s requirement is that 95% of the axles are 100 cm ± 5 cm. The following data is MegaStat output from a test run from Chris Cross Manufacturing.

Descriptive statistics count: 16 mean: 99.938 sample variance: 2.313 sample standard deviation: 1.521 minimum: 97 maximum: 102.9 range: 5.9 population variance: 2.169 population standard deviation: 1.473 standard error of the mean: 0.380 tollerance interval 95.45% lower: 96.896 tolerance interval 95.45% upper: 102.979 half-width: 3.042

1st quartile: 98.900 median: 99.850 3rd quartile: 100.475 interquartile range: 1.575 mode: 98.900

Question: Should Tesla select them as a vendor? Explain your answer.

Answers (1)

· Given that,

Tesla Motors needs to buy axles for their new car. They are considering using Chris Cross Manufacturing as a vendor. Tesla’s requirement is that 95% of the axles are 100 cm ± 5 cm. The following data is MegaStat output from a test run from Chris Cross Manufacturing:

1st quartile: 98.900 median: 99.850 3rd quartile: 100.475 interquartile range: 1.575 mode: 98.900 Now, we have to construct 95% confidence interval for the data from

the Chris Cross Manufacturing

(TCO D) A PC manufacturer claims that no more than 2% of their machines are defective. In a random sample of 100 machines, it is found that 4.5% are defective. The manufacturer claims this is a fluke of the sample. At a .02 level of significance, test the manufacturer’s claim, and explain your answer.

Test and CI for One Proportion

Test of p = 0.02 vs p > 0.02

Sample

Sample p

98% Lower Bound

Z-Value

P-Value

100

0.040000

0.000000

1.43

0.077

Reference:

Set up the hypotheses:

H0: p <= 0.02

Ha: p > 0.02

This is a one tailed test, since we will only reject for high proportions.

Since we are using a 0.02 level of significance (it’s just chance that the hypotheses happen to have the same value as this), we’ll reject the null hypothesis if our P Value is less than 0.02.

The computed P value from Megastat was 0.0371.

This is higher than the significance level.

Therefore, we do not reject H0:.

We can say that the proportion is still less than or equal to 2%, and this was a fluke.

Final Page 2

1. (TCO B) The following table gives the number of visits to recreational facilities by kind and geographical region. (Points : 30)

Ans.

East

South

Midwest

West

Totals

Local Park

328

464

National Park

233

514

204

251

1202

State Park

100

526

102

793

Totals

388

1368

298

405

2459

(A) Referring to the above table, if a visitor is chosen at random, what is the probability that he or she is either from the South or from the West? (15 points) (B) Referring to the above table, given that the visitor is from the Midwest, what is the probability that he or she visited a local park? (15 points)

a. Total people = 2459

South + West = 1368 + 405 = 1773

probability — divide these:

1773/2459 = approx 0.721

Total Midwest = 298

Midwest local park = 29

Divide:

(TCO B, F) The length of time Americans exercise each week is normally distributed with a mean of 15.8 minutes and a standard deviation of 2.2 minutes

P(X≤x)

P(X≥x)

Mean

Std dev

.0146

.9854

15.8

2.2

.3581

.6419

15.8

2.2

.9910

.0090

15.8

2.2

.9999

.0001

15.8

2.2

p(lower)

p(upper)

(A) Analyze the output above to determine what percentage of Americans will exercise between 11 and 21 minutes per week. (15 points) (B) What percentage of Americans will exercise less than 15 minutes? If 1000 Americans were evaluated, how many would you expect to have exercised less than 15 minutes? (15 points) (Points : 30)

MATH 533 Final Exam Set 2

(TCO A) Seventeen salespeople reported the following number of sales calls completed last month.

72 93 82 81 82 97 102 107 119 86 88 91 83 93 73 100 102

Compute the mean, median, mode, and standard deviation, Q1, Q3, Min, and Max for the above sample data on number of sales calls per month. b. In the context of this situation, interpret the Median, Q1, and Q3. (Points : 33)

(TCO B) Cedar Home Furnishings has collected data on their customers in terms of whether they reside in an urban location or a suburban location, as well as rating the customers as either “good,” “borderline,” or “poor.” The data is below.

Urban

Suburban

Total

Good

168

228

Borderline

108

Poor

Total

120

280

400

If you choose a customer at random, then find the probability that the customer

is considered “borderline.”

(TCO B) Historically, 70% of your customers at Rodale Emporium pay for their purchases using credit cards. In a sample of 20 customers, find the probability that

exactly 14 customers will pay for their purchases using credit cards.

(TCO C) An operations analyst from an airline company has been asked to develop a fairly accurate estimate of the mean refueling and baggage handling time at a foreign airport. A random sample of 36 refueling and baggage handling times yields the following results.

Sample Size = 36 Sample Mean = 24.2 minutes Sample Standard Deviation = 4.2 minutes

Compute the 90% confidence interval for the population mean refueling and baggage time.

(TCO C) The manufacturer of a certain brand of toothpaste claims that a high percentage of dentists recommend the use of their toothpaste. A random sample of 400 dentists results in 310 recommending their toothpaste.

Compute the 99% confidence interval for the population proportion of dentists who recommend the use of this toothpaste.

(TCO D) A Ford Motor Company quality improvement team believes that its recently implemented defect reduction program has reduced the proportion of paint defects. Prior to the implementation of the program, the proportion of paint defects was .03 and had been stationary for the past 6 months. Ford selects a random sample of 2,000 cars built after the implementation of the defect reduction program. There were 45 cars with paint defects in that sample. Does the sample data provide evidence to conclude that the proportion of paint defects is now less than .03 (with a = .01)? Use the hypothesis testing procedure outlined below.

Formulate the null and alternative hypotheses.

(TCO D) A new car dealer calculates that the dealership must average more than 4.5% profit on sales of new cars. A random sample of 81 cars gives the following result.

Sample Size = 81 Sample Mean = 4.97% Sample Standard Deviation = 1.8%

Does the sample data provide evidence to conclude that the dealership averages more than 4.5% profit on sales of new cars (using a = .10)? Use the hypothesis testing procedure outlined below.

Formulate the null and alternative hypotheses.

(TCO E) Bill McFarland is a real estate broker who specializes in selling farmland in a large western state. Because Bill advises many of his clients about pricing their land, he is interested in developing a pricing formula of some type. He feels he could increase his business significantly if he could accurately determine the value of a farmer’s land. A geologist tells Bill that the soil and rock characteristics in most of the area that Bill sells do not vary much. Thus the price of land should depend greatly on acreage. Bill selects a sample of 30 plots recently sold. The data is found below (in Minitab), where X=Acreage and Y=Price ($1,000s).

PRICE

ACREAGE

PREDICT

20.0

130

40.5

250

10.2

300

100.0

30.0

182

56.5

115

41.0

10.0

18.5

30.0

25.6

122

42.0

14.0

200

70.0

13.0

21.6

6.5

145

45.0

19.2

235

80.0

250

90.0

278

95.0

118

41.0

14.0

22.0

220

81.5

235

78.0

16.0

10.0

290

100.0

Correlations: PRICE, ACREAGE Pearson correlation of PRICE and ACREAGE = 0.997 P-Value = 0.000

Regression Analysis: PRICE versus ACREAGE The regression equation is PRICE = 2.26 + 2.89 ACREAGE

Predictor Coef SE Coef T P Constant 2.257 2.231 1.01 0.320 ACREAGE 2.89202 0.04353 66.44 0.000

S = 7.21461 R-Sq = 99.4% R-Sq(adj) = 99.3%

Analysis of Variance

Source DF SS MS F P Regression 1 229757 229757 4414.11 0.000 Residual Error 28 1457 52 Total 29 231215

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 146.86 1.37 (144.05, 149.66) (131.82, 161.90) 2 725.26 9.18 (706.46, 744.06) (701.35, 749.17)XX

XX denotes a point that is an extreme outlier in the predictors.

Values of Predictors for New Observations

New Obs ACREAGE 1 50 2 250

Analyze the above output to determine the regression equation.

(TCO E) An insurance firm wishes to study the relationship between driving experience (X1, in years), number of driving violations in the past three years (X2), and current monthly auto insurance premium (Y). A sample of 12 insured drivers is selected at random. The data is given below (in MINITAB):

Predict X1

Predict X2

050

060

Regression Analysis: Y versus X1, X2 The regression equation is Y = 55.1 – 1.37 X1 + 8.05 X2

Predictor Coef SE Coef T P Constant 55.138 7.309 7.54 0.000 X1 -1.3736 0.4885 -2.81 0.020 X2 8.053 1.307 6.16 0.000

S = 6.07296 R-Sq = 93.1% R-Sq(adj) = 91.6%

Analysis of Variance

Source DF SS MS F P Regression 2 4490.3 2245.2 60.88 0.000 Residual Error 9 331.9 36.9 Total 11 4822.3

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 52.20 2.91 (45.62, 58.79) (36.97, 67.44)

Values of Predictors for New Observations

New Obs X1 X2 1 8.00 1.00

Correlations: Y, X1, X2 Y X1 X1 -0.800 0.002

X2 0.933 -0.660 0.000 0.020

Cell Contents: Pearson correlation P-Value

Analyze the above output to determine the multiple regression equation.

MATH 533 Final Exam Set 3

MATH 533 Final Exam Set 4

#MATH 533 ( Applied Managerial Statistics ) Entire Course

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buyonlineclass-blog · 6 years ago

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How do the online portal in mathematics help in different sets of formats?

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takeonlineclasshelp-blog · 6 years ago

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How to remember to complete MyStatlab - Participation with timing

Online tests and their limited time will not worry about students anymore. Call and get professional help.

The calculation in MyStatLab is so lengthy and complex that without Stats, we would not be able to predict anything. Let us look into some tactics we can apply to make the subject easier:

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6. The sum may not have “significance”. Accept that: Devote some time thinking and accepting the fact. Sometimes interesting theories come from a finding that didn’t succeed.

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we-donaldpowers-blog · 8 years ago

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MATH 533 ( Applied Managerial Statistics ) Entire Course

Follow Below Link to Download Tutorial

http://devryfinalexam.com/downloads/math-533-applied-managerial-statistics-entire-course/

For More Information Visit Our Website ( https://homeworklance.com/ )

Email us At: [email protected] or [email protected]

(MATH 533 Applied Managerial Statistics – DeVry)

(MATH 533 Week 1) MATH 533 Week 1 Homework Problems (MyStatLab) MATH 533 Week 1 Graded Discussion Topics MATH 533 Week 1 Quiz

(MATH 533 Week 2) MATH 533 Week 2 Homework Problems (MyStatLab) MATH 533 Week 2 Graded Discussion Topics MATH 533 Week 2 Course Project – Part A (SALESCALL Inc.)

(MATH 533 Week 3) MATH 533 Week 3 Homework Problems (MyStatLab) MATH 533 Week 3 Graded Discussion Topics

(MATH 533 Week 4) MATH 533 Week 4 Homework Problems (MyStatLab) MATH 533 Week 4 Graded Discussion Topics

(MATH 533 Week 5) MATH 533 Week 5 Homework Problems (MyStatLab) MATH 533 Week 5 Quiz MATH 533 Week 5 Graded Discussion Topics

(MATH 533 Week 6) MATH 533 Week 6 Homework Problems (MyStatLab) MATH 533 Week 6 Graded Discussion Topics MATH 533 Week 6 Course Project – Part B (SALESCALL Inc.)

(MATH 533 Week 7) MATH 533 Week 7 Course Project – Part C (SALESCALL Inc.) MATH 533 Week 7 Graded Discussion Topics

(MATH 533 Week 8 Final Exam Answers)

MATH 533 ( Applied Managerial Statistics ) Final Exam Answers

MATH 533 Final Exam Set 1

0.5

P(X)

cumulative probability

0.00098

0.00977

0.01074

0.04395

0.05469

0.11719

0.17188

0.20508

0.37695

0.24609

0.62305

0.20508

0.82813

0.11719

0.94531

0.04395

0.98926

0.00977

0.99902

0.00098

1.00000

What is the probability that no more than four will be republicans? (Points : 10) 38% 12% 21% 62%

Variable

Total Count

Mean

StDev

Variance

Minimum

Maximum

Range

Sales

40.83

15.39

236.88

19.00

67.00

48.00

Stem-and-Leaf Display: Sales Stem-and-leaf of Sales N = 12 Leaf Unit = 1.0

444

Raw data: sales/month (Millions of $) 19 34 23 34 56 45 35 36 46 47 19 23

count 12 mean 34.75 sample variance 146.20 sample standard deviation 12.09 minimum 19 maximum 56 range 37

Stem and Leaf plot for # 1 stem unit = 10 leaf unit = 1

count

12.00000

mean

34.75000

sample variance

146.20455

sample standard deviation

12.09151

minimum

19.00000

maximum

56.00000

range

37.00000

1st quartile

23.00000

median

34.50000

3rd quartile

45.25000

interquartile range

22.25000

mode

19.00000

count

mean

99.850

sample variance

4.627

sample standard deviation

2.151

minimum

96.9

maximum

104

range

7.1

population variance

4.338

population standard deviation

2.083

standard error of the mean

0.538

tolerance interval 95.45% lower

95.548

tolerance interval 95.45% upper

104.152

margin of error

4.302

1st quartile

98.850

median

99.200

3rd quartile

100.550

interquartile range

1.700

mode

103.000

1st quartile: 98.900 median: 99.850 3rd quartile: 100.475 interquartile range: 1.575 mode: 98.900

Question: Should Tesla select them as a vendor? Explain your answer.

Answers (1)

· Given that,

1st quartile: 98.900 median: 99.850 3rd quartile: 100.475 interquartile range: 1.575 mode: 98.900 Now, we have to construct 95% confidence interval for the data from

the Chris Cross Manufacturing

Test and CI for One Proportion

Test of p = 0.02 vs p > 0.02

Sample

Sample p

98% Lower Bound

Z-Value

P-Value

100

0.040000

0.000000

1.43

0.077

Reference:

Set up the hypotheses:

H0: p <= 0.02

Ha: p > 0.02

This is a one tailed test, since we will only reject for high proportions.

Since we are using a 0.02 level of significance (it’s just chance that the hypotheses happen to have the same value as this), we’ll reject the null hypothesis if our P Value is less than 0.02.

The computed P value from Megastat was 0.0371.

This is higher than the significance level.

Therefore, we do not reject H0:.

We can say that the proportion is still less than or equal to 2%, and this was a fluke.

Final Page 2

1. (TCO B) The following table gives the number of visits to recreational facilities by kind and geographical region. (Points : 30)

Ans.

East

South

Midwest

West

Totals

Local Park

328

464

National Park

233

514

204

251

1202

State Park

100

526

102

793

Totals

388

1368

298

405

2459

a. Total people = 2459

South + West = 1368 + 405 = 1773

probability — divide these:

1773/2459 = approx 0.721

Total Midwest = 298

Midwest local park = 29

Divide:

(TCO B, F) The length of time Americans exercise each week is normally distributed with a mean of 15.8 minutes and a standard deviation of 2.2 minutes

P(X≤x)

P(X≥x)

Mean

Std dev

.0146

.9854

15.8

2.2

.3581

.6419

15.8

2.2

.9910

.0090

15.8

2.2

.9999

.0001

15.8

2.2

p(lower)

p(upper)

MATH 533 Final Exam Set 2

(TCO A) Seventeen salespeople reported the following number of sales calls completed last month.

72 93 82 81 82 97 102 107 119 86 88 91 83 93 73 100 102

Urban

Suburban

Total

Good

168

228

Borderline

108

Poor

Total

120

280

400

If you choose a customer at random, then find the probability that the customer

is considered “borderline.”

(TCO B) Historically, 70% of your customers at Rodale Emporium pay for their purchases using credit cards. In a sample of 20 customers, find the probability that

exactly 14 customers will pay for their purchases using credit cards.

Sample Size = 36 Sample Mean = 24.2 minutes Sample Standard Deviation = 4.2 minutes

Compute the 90% confidence interval for the population mean refueling and baggage time.

Compute the 99% confidence interval for the population proportion of dentists who recommend the use of this toothpaste.

Formulate the null and alternative hypotheses.

(TCO D) A new car dealer calculates that the dealership must average more than 4.5% profit on sales of new cars. A random sample of 81 cars gives the following result.

Sample Size = 81 Sample Mean = 4.97% Sample Standard Deviation = 1.8%

Does the sample data provide evidence to conclude that the dealership averages more than 4.5% profit on sales of new cars (using a = .10)? Use the hypothesis testing procedure outlined below.

Formulate the null and alternative hypotheses.

PRICE

ACREAGE

PREDICT

20.0

130

40.5

250

10.2

300

100.0

30.0

182

56.5

115

41.0

10.0

18.5

30.0

25.6

122

42.0

14.0

200

70.0

13.0

21.6

6.5

145

45.0

19.2

235

80.0

250

90.0

278

95.0

118

41.0

14.0

22.0

220

81.5

235

78.0

16.0

10.0

290

100.0

Correlations: PRICE, ACREAGE Pearson correlation of PRICE and ACREAGE = 0.997 P-Value = 0.000

Regression Analysis: PRICE versus ACREAGE The regression equation is PRICE = 2.26 + 2.89 ACREAGE

Predictor Coef SE Coef T P Constant 2.257 2.231 1.01 0.320 ACREAGE 2.89202 0.04353 66.44 0.000

S = 7.21461 R-Sq = 99.4% R-Sq(adj) = 99.3%

Analysis of Variance

Source DF SS MS F P Regression 1 229757 229757 4414.11 0.000 Residual Error 28 1457 52 Total 29 231215

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 146.86 1.37 (144.05, 149.66) (131.82, 161.90) 2 725.26 9.18 (706.46, 744.06) (701.35, 749.17)XX

XX denotes a point that is an extreme outlier in the predictors.

Values of Predictors for New Observations

New Obs ACREAGE 1 50 2 250

Analyze the above output to determine the regression equation.

Predict X1

Predict X2

050

060

Regression Analysis: Y versus X1, X2 The regression equation is Y = 55.1 – 1.37 X1 + 8.05 X2

Predictor Coef SE Coef T P Constant 55.138 7.309 7.54 0.000 X1 -1.3736 0.4885 -2.81 0.020 X2 8.053 1.307 6.16 0.000

S = 6.07296 R-Sq = 93.1% R-Sq(adj) = 91.6%

Analysis of Variance

Source DF SS MS F P Regression 2 4490.3 2245.2 60.88 0.000 Residual Error 9 331.9 36.9 Total 11 4822.3

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 52.20 2.91 (45.62, 58.79) (36.97, 67.44)

Values of Predictors for New Observations

New Obs X1 X2 1 8.00 1.00

Correlations: Y, X1, X2 Y X1 X1 -0.800 0.002

X2 0.933 -0.660 0.000 0.020

Cell Contents: Pearson correlation P-Value

Analyze the above output to determine the multiple regression equation.

MATH 533 Final Exam Set 3

MATH 533 Final Exam Set 4

#MATH 533 ( Applied Managerial Statistics ) Entire Course

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patricksaragosa-blog · 8 years ago

Text

MATH 533 ( Applied Managerial Statistics ) Entire Course

Follow Below Link to Download Tutorial

https://homeworklance.com/downloads/math-533-applied-managerial-statistics-entire-course/

For More Information Visit Our Website ( https://homeworklance.com/ )

Email us At: [email protected] or [email protected]

(MATH 533 Applied Managerial Statistics – DeVry)

(MATH 533 Week 1) MATH 533 Week 1 Homework Problems (MyStatLab) MATH 533 Week 1 Graded Discussion Topics MATH 533 Week 1 Quiz

(MATH 533 Week 2) MATH 533 Week 2 Homework Problems (MyStatLab) MATH 533 Week 2 Graded Discussion Topics MATH 533 Week 2 Course Project – Part A (SALESCALL Inc.)

(MATH 533 Week 3) MATH 533 Week 3 Homework Problems (MyStatLab) MATH 533 Week 3 Graded Discussion Topics

(MATH 533 Week 4) MATH 533 Week 4 Homework Problems (MyStatLab) MATH 533 Week 4 Graded Discussion Topics

(MATH 533 Week 5) MATH 533 Week 5 Homework Problems (MyStatLab) MATH 533 Week 5 Quiz MATH 533 Week 5 Graded Discussion Topics

(MATH 533 Week 6) MATH 533 Week 6 Homework Problems (MyStatLab) MATH 533 Week 6 Graded Discussion Topics MATH 533 Week 6 Course Project – Part B (SALESCALL Inc.)

(MATH 533 Week 7) MATH 533 Week 7 Course Project – Part C (SALESCALL Inc.) MATH 533 Week 7 Graded Discussion Topics

(MATH 533 Week 8 Final Exam Answers)

MATH 533 ( Applied Managerial Statistics ) Final Exam Answers

MATH 533 Final Exam Set 1

0.5

P(X)

cumulative probability

0.00098

0.00977

0.01074

0.04395

0.05469

0.11719

0.17188

0.20508

0.37695

0.24609

0.62305

0.20508

0.82813

0.11719

0.94531

0.04395

0.98926

0.00977

0.99902

0.00098

1.00000

What is the probability that no more than four will be republicans? (Points : 10) 38% 12% 21% 62%

Variable

Total Count

Mean

StDev

Variance

Minimum

Maximum

Range

Sales

40.83

15.39

236.88

19.00

67.00

48.00

Stem-and-Leaf Display: Sales Stem-and-leaf of Sales N = 12 Leaf Unit = 1.0

444

Raw data: sales/month (Millions of $) 19 34 23 34 56 45 35 36 46 47 19 23

count 12 mean 34.75 sample variance 146.20 sample standard deviation 12.09 minimum 19 maximum 56 range 37

Stem and Leaf plot for # 1 stem unit = 10 leaf unit = 1

count

12.00000

mean

34.75000

sample variance

146.20455

sample standard deviation

12.09151

minimum

19.00000

maximum

56.00000

range

37.00000

1st quartile

23.00000

median

34.50000

3rd quartile

45.25000

interquartile range

22.25000

mode

19.00000

count

mean

99.850

sample variance

4.627

sample standard deviation

2.151

minimum

96.9

maximum

104

range

7.1

population variance

4.338

population standard deviation

2.083

standard error of the mean

0.538

tolerance interval 95.45% lower

95.548

tolerance interval 95.45% upper

104.152

margin of error

4.302

1st quartile

98.850

median

99.200

3rd quartile

100.550

interquartile range

1.700

mode

103.000

1st quartile: 98.900 median: 99.850 3rd quartile: 100.475 interquartile range: 1.575 mode: 98.900

Question: Should Tesla select them as a vendor? Explain your answer.

Answers (1)

· Given that,

1st quartile: 98.900 median: 99.850 3rd quartile: 100.475 interquartile range: 1.575 mode: 98.900 Now, we have to construct 95% confidence interval for the data from

the Chris Cross Manufacturing

Test and CI for One Proportion

Test of p = 0.02 vs p > 0.02

Sample

Sample p

98% Lower Bound

Z-Value

P-Value

100

0.040000

0.000000

1.43

0.077

Reference:

Set up the hypotheses:

H0: p <= 0.02

Ha: p > 0.02

This is a one tailed test, since we will only reject for high proportions.

Since we are using a 0.02 level of significance (it’s just chance that the hypotheses happen to have the same value as this), we’ll reject the null hypothesis if our P Value is less than 0.02.

The computed P value from Megastat was 0.0371.

This is higher than the significance level.

Therefore, we do not reject H0:.

We can say that the proportion is still less than or equal to 2%, and this was a fluke.

Final Page 2

1. (TCO B) The following table gives the number of visits to recreational facilities by kind and geographical region. (Points : 30)

Ans.

East

South

Midwest

West

Totals

Local Park

328

464

National Park

233

514

204

251

1202

State Park

100

526

102

793

Totals

388

1368

298

405

2459

a. Total people = 2459

South + West = 1368 + 405 = 1773

probability — divide these:

1773/2459 = approx 0.721

Total Midwest = 298

Midwest local park = 29

Divide:

(TCO B, F) The length of time Americans exercise each week is normally distributed with a mean of 15.8 minutes and a standard deviation of 2.2 minutes

P(X≤x)

P(X≥x)

Mean

Std dev

.0146

.9854

15.8

2.2

.3581

.6419

15.8

2.2

.9910

.0090

15.8

2.2

.9999

.0001

15.8

2.2

p(lower)

p(upper)

MATH 533 Final Exam Set 2

(TCO A) Seventeen salespeople reported the following number of sales calls completed last month.

72 93 82 81 82 97 102 107 119 86 88 91 83 93 73 100 102

Urban

Suburban

Total

Good

168

228

Borderline

108

Poor

Total

120

280

400

If you choose a customer at random, then find the probability that the customer

is considered “borderline.”

(TCO B) Historically, 70% of your customers at Rodale Emporium pay for their purchases using credit cards. In a sample of 20 customers, find the probability that

exactly 14 customers will pay for their purchases using credit cards.

Sample Size = 36 Sample Mean = 24.2 minutes Sample Standard Deviation = 4.2 minutes

Compute the 90% confidence interval for the population mean refueling and baggage time.

Compute the 99% confidence interval for the population proportion of dentists who recommend the use of this toothpaste.

Formulate the null and alternative hypotheses.

(TCO D) A new car dealer calculates that the dealership must average more than 4.5% profit on sales of new cars. A random sample of 81 cars gives the following result.

Sample Size = 81 Sample Mean = 4.97% Sample Standard Deviation = 1.8%

Does the sample data provide evidence to conclude that the dealership averages more than 4.5% profit on sales of new cars (using a = .10)? Use the hypothesis testing procedure outlined below.

Formulate the null and alternative hypotheses.

PRICE

ACREAGE

PREDICT

20.0

130

40.5

250

10.2

300

100.0

30.0

182

56.5

115

41.0

10.0

18.5

30.0

25.6

122

42.0

14.0

200

70.0

13.0

21.6

6.5

145

45.0

19.2

235

80.0

250

90.0

278

95.0

118

41.0

14.0

22.0

220

81.5

235

78.0

16.0

10.0

290

100.0

Correlations: PRICE, ACREAGE Pearson correlation of PRICE and ACREAGE = 0.997 P-Value = 0.000

Regression Analysis: PRICE versus ACREAGE The regression equation is PRICE = 2.26 + 2.89 ACREAGE

Predictor Coef SE Coef T P Constant 2.257 2.231 1.01 0.320 ACREAGE 2.89202 0.04353 66.44 0.000

S = 7.21461 R-Sq = 99.4% R-Sq(adj) = 99.3%

Analysis of Variance

Source DF SS MS F P Regression 1 229757 229757 4414.11 0.000 Residual Error 28 1457 52 Total 29 231215

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 146.86 1.37 (144.05, 149.66) (131.82, 161.90) 2 725.26 9.18 (706.46, 744.06) (701.35, 749.17)XX

XX denotes a point that is an extreme outlier in the predictors.

Values of Predictors for New Observations

New Obs ACREAGE 1 50 2 250

Analyze the above output to determine the regression equation.

Predict X1

Predict X2

050

060

Regression Analysis: Y versus X1, X2 The regression equation is Y = 55.1 – 1.37 X1 + 8.05 X2

Predictor Coef SE Coef T P Constant 55.138 7.309 7.54 0.000 X1 -1.3736 0.4885 -2.81 0.020 X2 8.053 1.307 6.16 0.000

S = 6.07296 R-Sq = 93.1% R-Sq(adj) = 91.6%

Analysis of Variance

Source DF SS MS F P Regression 2 4490.3 2245.2 60.88 0.000 Residual Error 9 331.9 36.9 Total 11 4822.3

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 52.20 2.91 (45.62, 58.79) (36.97, 67.44)

Values of Predictors for New Observations

New Obs X1 X2 1 8.00 1.00

Correlations: Y, X1, X2 Y X1 X1 -0.800 0.002

X2 0.933 -0.660 0.000 0.020

Cell Contents: Pearson correlation P-Value

Analyze the above output to determine the multiple regression equation.

MATH 533 Final Exam Set 3

MATH 533 Final Exam Set 4

#MATH 533 ( Applied Managerial Statistics ) Entire Course

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MATH 533 ( Applied Managerial Statistics ) Full Course 2016

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(MATH 533 Applied Managerial Statistics – DeVry)

(MATH 533 Week 1) MATH 533 Week 1 Homework Problems (MyStatLab) MATH 533 Week 1 Graded Discussion Topics MATH 533 Week 1 Quiz (MATH 533 Week 2) MATH 533 Week 2 Homework Problems (MyStatLab) MATH 533 Week 2 Graded Discussion Topics MATH 533 Week 2 Course Project – Part A (SALESCALL Inc.) (MATH 533 Week 3) MATH 533 Week 3 Homework Problems (MyStatLab) MATH 533 Week 3 Graded Discussion Topics (MATH 533 Week 4) MATH 533 Week 4 Homework Problems (MyStatLab) MATH 533 Week 4 Graded Discussion Topics (MATH 533 Week 5) MATH 533 Week 5 Homework Problems (MyStatLab) MATH 533 Week 5 Quiz MATH 533 Week 5 Graded Discussion Topics (MATH 533 Week 6) MATH 533 Week 6 Homework Problems (MyStatLab) MATH 533 Week 6 Graded Discussion Topics MATH 533 Week 6 Course Project – Part B (SALESCALL Inc.) (MATH 533 Week 7) MATH 533 Week 7 Course Project – Part C (SALESCALL Inc.) MATH 533 Week 7 Graded Discussion Topics (MATH 533 Week 8 Final Exam Answers)

MATH 533 ( Applied Managerial Statistics ) Final Exam Answers

MATH 533 Final Exam Set 1

0.5

P(X)

cumulative probability

00.00098

0.00098

0.00977

0.01074

0.04395

0.05469

0.11719

0.17188

0.20508

0.37695

0.24609

0.62305

0.20508

0.82813

0.11719

0.94531

0.04395

0.98926

0.00977

0.99902

0.00098

1.00000

What is the probability that no more than four will be republicans? (Points : 10) 38% 12% 21% 62%Ans. a look at x=4, cumulative probability 3. (TCO A) Company ABC had sales per month as listed below. Using the Minitab output given, determine: (A) Range (5 points); (B) Median (5 points); and (C) The range of the data that would contain 68% of the results. (5 points). Raw data: sales/month (Millions of $) 23 45 34 34 56 67 54 34 45 56 23 19 Descriptive Statistics: Sales

Variable

Total Count

Mean

StDev

Variance

Minimum

Maximum

Range

Sales

40.83

15.39

236.88

19.00

67.00

48.00

Stem-and-Leaf Display: Sales Stem-and-leaf of Sales N = 12 Leaf Unit = 1.0

444

Raw data: sales/month (Millions of $) 19 34 23 34 56 45 35 36 46 47 19 23

count 12 mean 34.75 sample variance 146.20 sample standard deviation 12.09 minimum 19 maximum 56 range 37

Stem and Leaf plot for # 1 stem unit = 10 leaf unit = 1

count

12.00000

mean

34.75000

sample variance

146.20455

sample standard deviation

12.09151

minimum

19.00000

maximum

56.00000

range

37.00000

1st quartile

23.00000

median

34.50000

3rd quartile

45.25000

interquartile range

22.25000

mode

19.00000

count

mean

99.850

sample variance

4.627

sample standard deviation

2.151

minimum

96.9

maximum

104

range

7.1

population variance

4.338

population standard deviation

2.083

standard error of the mean

0.538

tolerance interval 95.45% lower

95.548

tolerance interval 95.45% upper

104.152

margin of error

4.302

1st quartile

98.850

median

99.200

3rd quartile

100.550

interquartile range

1.700

mode

103.000

1st quartile: 98.900 median: 99.850 3rd quartile: 100.475 interquartile range: 1.575 mode: 98.900

Question: Should Tesla select them as a vendor? Explain your answer.

Answers (1)

· Given that,

1st quartile: 98.900 median: 99.850 3rd quartile: 100.475 interquartile range: 1.575 mode: 98.900 Now, we have to construct 95% confidence interval for the data from

the Chris Cross Manufacturing

Test and CI for One Proportion

Test of p = 0.02 vs p > 0.02

Sample

Sample p

98% Lower Bound

Z-Value

P-Value

100

0.040000

0.000000

1.43

0.077

Reference:

Set up the hypotheses:

H0: p <= 0.02

Ha: p > 0.02

This is a one tailed test, since we will only reject for high proportions.

Since we are using a 0.02 level of significance (it’s just chance that the hypotheses happen to have the same value as this), we’ll reject the null hypothesis if our P Value is less than 0.02.

The computed P value from Megastat was 0.0371.

This is higher than the significance level.

Therefore, we do not reject H0:.

We can say that the proportion is still less than or equal to 2%, and this was a fluke.

Final Page 2

1. (TCO B) The following table gives the number of visits to recreational facilities by kind and geographical region. (Points : 30)Ans.

East

South

Midwest

West

Totals

Local Park

328

464

National Park

233

514

204

251

1202

State Park

100

526

102

793

Totals

388

1368

298

405

2459

a. Total people = 2459

South + West = 1368 + 405 = 1773

probability — divide these:

1773/2459 = approx 0.721

Total Midwest = 298

Midwest local park = 29

Divide:

(TCO B, F) The length of time Americans exercise each week is normally distributed with a mean of 15.8 minutes and a standard deviation of 2.2 minutes

P(X≤x)

P(X≥x)

Mean

Std dev

.0146

.9854

15.8

2.2

.3581

.6419

15.8

2.2

.9910

.0090

15.8

2.2

.9999

.0001

15.8

2.2

p(lower)

p(upper)

MATH 533 Final Exam Set 2

(TCO A) Seventeen salespeople reported the following number of sales calls completed last month.

72 93 82 81 82 97 102 107 119 86 88 91 83 93 73 100 102

Urban

Suburban

Total

Good

168

228

Borderline

108

Poor

Total

120

280

400

If you choose a customer at random, then find the probability that the customer

is considered “borderline.”

(TCO B) Historically, 70% of your customers at Rodale Emporium pay for their purchases using credit cards. In a sample of 20 customers, find the probability that

exactly 14 customers will pay for their purchases using credit cards.

Sample Size = 36 Sample Mean = 24.2 minutes Sample Standard Deviation = 4.2 minutes

Compute the 90% confidence interval for the population mean refueling and baggage time.

Compute the 99% confidence interval for the population proportion of dentists who recommend the use of this toothpaste.

Formulate the null and alternative hypotheses.

(TCO D) A new car dealer calculates that the dealership must average more than 4.5% profit on sales of new cars. A random sample of 81 cars gives the following result.

Sample Size = 81 Sample Mean = 4.97% Sample Standard Deviation = 1.8%

Does the sample data provide evidence to conclude that the dealership averages more than 4.5% profit on sales of new cars (using a = .10)? Use the hypothesis testing procedure outlined below.

Formulate the null and alternative hypotheses.

PRICE

ACREAGE

PREDICT

20.0

130

40.5

250

10.2

300

100.0

30.0

182

56.5

115

41.0

10.0

18.5

30.0

25.6

122

42.0

14.0

200

70.0

13.0

21.6

6.5

145

45.0

19.2

235

80.0

250

90.0

278

95.0

118

41.0

14.0

22.0

220

81.5

235

78.0

16.0

10.0

290

100.0

Correlations: PRICE, ACREAGE Pearson correlation of PRICE and ACREAGE = 0.997 P-Value = 0.000

Regression Analysis: PRICE versus ACREAGE The regression equation is PRICE = 2.26 + 2.89 ACREAGE

Predictor Coef SE Coef T P Constant 2.257 2.231 1.01 0.320 ACREAGE 2.89202 0.04353 66.44 0.000

S = 7.21461 R-Sq = 99.4% R-Sq(adj) = 99.3%

Analysis of Variance

Source DF SS MS F P Regression 1 229757 229757 4414.11 0.000 Residual Error 28 1457 52 Total 29 231215

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 146.86 1.37 (144.05, 149.66) (131.82, 161.90) 2 725.26 9.18 (706.46, 744.06) (701.35, 749.17)XX

XX denotes a point that is an extreme outlier in the predictors.

Values of Predictors for New Observations

New Obs ACREAGE 1 50 2 250

Analyze the above output to determine the regression equation.

Predict X1

Predict X2

050

060

Regression Analysis: Y versus X1, X2 The regression equation is Y = 55.1 – 1.37 X1 + 8.05 X2

Predictor Coef SE Coef T P Constant 55.138 7.309 7.54 0.000 X1 -1.3736 0.4885 -2.81 0.020 X2 8.053 1.307 6.16 0.000

S = 6.07296 R-Sq = 93.1% R-Sq(adj) = 91.6%

Analysis of Variance

Source DF SS MS F P Regression 2 4490.3 2245.2 60.88 0.000 Residual Error 9 331.9 36.9 Total 11 4822.3

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 52.20 2.91 (45.62, 58.79) (36.97, 67.44)

Values of Predictors for New Observations

New Obs X1 X2 1 8.00 1.00

Correlations: Y, X1, X2 Y X1 X1 -0.800 0.002

X2 0.933 -0.660 0.000 0.020

Cell Contents: Pearson correlation P-Value

Analyze the above output to determine the multiple regression equation.

MATH 533 Final Exam Set 3

MATH 533 Final Exam Set 4

#MATH 533 ( Applied Managerial Statistics ) Full Course 2016

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