#Mystatlab Quiz Answers
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Unlocking Success in Online Learning: A Deep Dive into Mystatlab Answers and Personalized Assistance
In the evolving landscape of education, online learning platforms have become cornerstones of both academic and professional development. As students and professionals seek flexible learning solutions that accommodate their schedules, platforms like MyStatLab offer a bridge between accessibility and the pursuit of knowledge in statistics. However, the journey through online learning is not without its challenges. From understanding complex concepts to managing time effectively, learners often find themselves in search of support. This is where the importance of finding reliable Mystatlab Answers and personalized assistance for those wondering "how can I do my online class effectively?" becomes paramount.
Embracing the Digital Education Revolution
The shift towards digital education has transformed how knowledge is disseminated and acquired. Platforms like MyStatLab provide an interactive environment that mirrors the structure and rigor of traditional classrooms with the added benefits of flexibility and self-paced learning. Yet, as we dive deeper into the intricacies of subjects like statistics, the need for external support becomes increasingly evident. This is not a reflection of the learner's capability but rather an acknowledgment of the diverse learning styles and the complexities of statistical education.
The Quest for Mystatlab Answers
Mystatlab Answers serve as a beacon for students navigating the choppy waters of statistical analysis and interpretation. The platform's comprehensive approach to teaching statistics through interactive assignments, quizzes, and personalized feedback is commendable. However, the challenge often lies in bridging the gap between the presented material and the student's understanding. Seeking out Mystatlab Answers can help demystify complex concepts, providing step-by-step solutions that enhance comprehension. This approach not only aids in academic success but also builds a solid foundation of knowledge that students can apply in real-world scenarios.
The Personal Touch: Do My Online Class Services
For students juggling multiple responsibilities, the question of "how can I effectively manage to do my online class?" is a pressing concern. Personalized assistance through tutoring services can be a game-changer in this context. These services offer more than just answers; they provide a tailored learning experience designed to meet the unique needs of each student. By pairing learners with experts in the field, students receive one-on-one support that focuses on strengthening weak areas and enhancing overall understanding.
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Conclusion
As we navigate the digital age, the fusion of innovative learning platforms like MyStatLab with personalized assistance offers a powerful tool in the pursuit of knowledge. While the quest for Mystatlab Answers illuminates the path to academic success in statistics, the personalized touch of "do my online class" services ensures that each student's journey is supported, tailored, and effective. Whether you're a student striving for excellence in statistics or a professional seeking to enhance your skills, embracing these resources can unlock doors to new opportunities and achievements in the realm of online learning.
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Its definition has evolved over the time, as the online class has developed with years. The researchers defined distance education as no more than a calamity of practices and ideas taken of the traditional class and imposed on the learning
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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.
It is this time of the year, when one looks expectantly towards the new and new trends of the online class, as one comes to the end of the year. There are some of the trends which are being discussed here.
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.
AI Is Becoming A Standard Entity In E-learning
By adopting the artificial intelligence, the online class is developing in a rapid way. In a matter of time, one will see the rising trend of AI in a better way for an online class. The following course can be very useful if one has taken the statistics as a subject. It becomes very useful for the candidate as the experts provide the MyStatLab quiz answers, which can solve the quiz questions for the students. The expert can fortify the learning content on that topic if an algorithm detects everyone seems to be struggling in a certain area of the test. The chatbots is one of the commonplace for AI. To take the request in real time, chatbots are a virtual assistant that use instant messages to give the data related answers. With an app, social media and website, the students can interact with the tutors. For the students who are taking the online class, chatbots can help a student with the onboarding processes. While providing the procedures and goals and to provide information about the frequently asked questions, it can introduce the general information the online educational portal.
Courses are Being Replaced by Context
During the special learning, hours is sinking into oblivion, the idea learning is something is done via traditional classes. Waiting for the learners to sign up, learning content should not stay in the learning portals, as one can call to take my class for me. When one needs it, it should not be like Google search results. On the way to embedding learning resources in the daily workflow, the performance support trend isn’t actually new, as the online class is taking a new step in a new direction. On managing the academics, one needs to take the traditional class and but cannot get the interest in the subject. One has the experts in front, but one is developing the keenness to ask questions. In the moment of need, it is great to have quick online reference guides as a source of reliable information. To micro learning, one might notice all these sounds might sound different. This is defined as a series of super targeted short lessons. While going in hand in hand, these two trends are going to boost in 2019.
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.
#assessment help#mymathlab quiz answers#mymathlab homework answers#online class help#mystatlab quiz answers
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5 Reasons You Shouldn'T Buy From An Essay Mill By continuing to make use of this web site you consent to the use of cookies in your system as described in our cookie policy unless you could have disabled them. We ll take the course for you and earn good grades as nicely. Covers arithmetic algebra geometry calculus and statistics. Nov Copy paste the url into one other tab and like magic. Here are all the solutions for the Statistics class. forty seven key outcomes after redesigning school algebra utilizing mymathlab within the national middle for educational. Your score from that work will be transferred to the Canvas gradebook manually by the instructor after the due date. So that you'll never again feel My MatLab skills are fairly fairly basic and now i need your help with one thing as a result of i can 39 t discover the reply here. Log in or join nbsp Educator Case Studies 2016. MyMathLab from Pearson is the world s leading on-line resource in arithmetic integrating interactive homework evaluation and media in a flexible simple to make use of format. Yeah studying want is not going to solitary make you have any favorite activity. You could not forlorn going subsequent to guide store or library or borrowing from your connections to gate them. YES Now is the time to redefine your true self utilizing Slader 39 s Introductory Statistics plus MyMathLab MyStatLab answers. Statistics Homework assist post Homework Questions Assignments amp Papers. quot If you might be considered one of them just don 39 t waste your time. Step 1 you might be given the equation that 1 hectare m 2. He is serving as Assistant Commandant (Group A/Gazetted) in Central Industrial Security Force since 2014. He is officer of 2012 batch UPSC CAPF Assistant Commandant examination and allocated to CISF, certainly one of most elite Central Armed Police Forces of Union of India. At occasions this Pearson lab programs Download Ebook Mymathlab Statistics Quiz Answers MyLab Statistics is the teaching and studying platform that empowers you to succeed in each pupil. MyMathLab or higher often known as Pearson s MyMathLab is a well-liked academic help used principally in faculty arithmetic lessons. Whether you are finding out statistics at an advanced degree or primary level our statistics tutors can reply your do my statistics homework request better than anyone else can. 3 Activity Mymathlab Statistics Quiz Answers Mymathlab Statistics Quiz Answers Recognizing the mannerism ways to get this books Mymathlab Statistics Quiz Answers is additionally useful. Kindle File Format Pearson Econ Lab nbsp The following is a guide on the way to be a part of your course in MyMathLab MML and some concepts of You have a number of options different than simply answering the issue. In some eyes this is a good factor while in the eyes of others largely students it isn't. The most common descriptive statistics are within the following desk together with their formulation and a brief description of what each measures. Commandant Dr. Prashant Jagtap is an officer by profession, physician by qualification and instructor by passion. Post your statistics homework questions and get answers from certified tutors. Our specialists provide right MyStatLab statistics homework solutions that will help you in finishing your listless homework inside time. Pennsylvania State solutions to all quiz questions are provided in the back of the guide. Are you in need of MyMathLab answers At elitehomeworkdoers. And frankly there are no answers that you need to use right now and that may be one hundred right. Subject Society Diet and Statistics Category Health Asked by veganfusion ga List Price a hundred. Statistics are light-weight little pieces of information that SQL Server keeps on tables and indexes to assist the optimizer do an excellent job. Now i would like some descreptive statistics of these variables. Quick I need assistance with Help typing in your math problems.
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MATH 533 ( Applied Managerial Statistics ) Entire Course
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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
(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
10
n
0.5
p
X
P(X)
cumulative probability
0
0.00098
0.00098
1
0.00977
0.01074
2
0.04395
0.05469
3
0.11719
0.17188
4
0.20508
0.37695
5
0.24609
0.62305
6
0.20508
0.82813
7
0.11719
0.94531
8
0.04395
0.98926
9
0.00977
0.99902
10
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
12
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
1
1
9
3
2
33
3
2
6
3
444
6
3
6
4
6
4
55
4
5
4
3
5
66
1
6
1
6
7
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
16
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:
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 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
X
N
Sample p
98% Lower Bound
Z-Value
P-Value
1
4
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
55
328
29
52
464
National Park
233
514
204
251
1202
State Park
100
526
65
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
b.
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
X
P(X≤x)
P(X≥x)
Mean
Std dev
11
.0146
.9854
15.8
2.2
15
.3581
.6419
15.8
2.2
21
.9910
.0090
15.8
2.2
24
.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
60
168
228
Borderline
36
72
108
Poor
24
40
64
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
60
20.0
50
130
40.5
250
25
10.2
300
100.0
85
30.0
182
56.5
115
41.0
24
10.0
60
18.5
92
30.0
77
25.6
122
42.0
41
14.0
200
70.0
42
13.0
60
21.6
20
6.5
145
45.0
61
19.2
235
80.0
250
90.0
278
95.0
118
41.0
46
14.0
69
22.0
220
81.5
235
78.0
50
16.0
25
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):
Y
X1
X2
Predict X1
Predict X2
74
5
2
8
1
38
14
050
6
1
63
10
3
97
4
6
55
8
2
57
11
3
43
16
1
99
3
5
46
9
1
35
19
060
13
3
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
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MATH 533 ( Applied Managerial Statistics ) Entire Course
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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.)
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(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
(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
10
n
0.5
p
X
P(X)
cumulative probability
0
0.00098
0.00098
1
0.00977
0.01074
2
0.04395
0.05469
3
0.11719
0.17188
4
0.20508
0.37695
5
0.24609
0.62305
6
0.20508
0.82813
7
0.11719
0.94531
8
0.04395
0.98926
9
0.00977
0.99902
10
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
12
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
1
1
9
3
2
33
3
2
6
3
444
6
3
6
4
6
4
55
4
5
4
3
5
66
1
6
1
6
7
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
16
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:
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 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
X
N
Sample p
98% Lower Bound
Z-Value
P-Value
1
4
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
55
328
29
52
464
National Park
233
514
204
251
1202
State Park
100
526
65
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
b.
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
X
P(X≤x)
P(X≥x)
Mean
Std dev
11
.0146
.9854
15.8
2.2
15
.3581
.6419
15.8
2.2
21
.9910
.0090
15.8
2.2
24
.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
60
168
228
Borderline
36
72
108
Poor
24
40
64
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
60
20.0
50
130
40.5
250
25
10.2
300
100.0
85
30.0
182
56.5
115
41.0
24
10.0
60
18.5
92
30.0
77
25.6
122
42.0
41
14.0
200
70.0
42
13.0
60
21.6
20
6.5
145
45.0
61
19.2
235
80.0
250
90.0
278
95.0
118
41.0
46
14.0
69
22.0
220
81.5
235
78.0
50
16.0
25
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):
Y
X1
X2
Predict X1
Predict X2
74
5
2
8
1
38
14
050
6
1
63
10
3
97
4
6
55
8
2
57
11
3
43
16
1
99
3
5
46
9
1
35
19
060
13
3
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
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For the traditional classroom, it is a suitable replacement even though e learning has become an education equalizer. While not being flawless, the technology of online class is advancing.
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Developed by Pearson education, MathLab is a programming language. Where linear algebra programming was simple, it started out as a matrix programming language. As a batch job, it can be run in under interactive sessions.
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MATH 533 ( Applied Managerial Statistics ) Entire Course
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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
(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
10
n
0.5
p
X
P(X)
cumulative probability
0
0.00098
0.00098
1
0.00977
0.01074
2
0.04395
0.05469
3
0.11719
0.17188
4
0.20508
0.37695
5
0.24609
0.62305
6
0.20508
0.82813
7
0.11719
0.94531
8
0.04395
0.98926
9
0.00977
0.99902
10
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
12
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
1
1
9
3
2
33
3
2
6
3
444
6
3
6
4
6
4
55
4
5
4
3
5
66
1
6
1
6
7
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
16
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:
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 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
X
N
Sample p
98% Lower Bound
Z-Value
P-Value
1
4
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
55
328
29
52
464
National Park
233
514
204
251
1202
State Park
100
526
65
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
b.
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
X
P(X≤x)
P(X≥x)
Mean
Std dev
11
.0146
.9854
15.8
2.2
15
.3581
.6419
15.8
2.2
21
.9910
.0090
15.8
2.2
24
.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
60
168
228
Borderline
36
72
108
Poor
24
40
64
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
60
20.0
50
130
40.5
250
25
10.2
300
100.0
85
30.0
182
56.5
115
41.0
24
10.0
60
18.5
92
30.0
77
25.6
122
42.0
41
14.0
200
70.0
42
13.0
60
21.6
20
6.5
145
45.0
61
19.2
235
80.0
250
90.0
278
95.0
118
41.0
46
14.0
69
22.0
220
81.5
235
78.0
50
16.0
25
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):
Y
X1
X2
Predict X1
Predict X2
74
5
2
8
1
38
14
050
6
1
63
10
3
97
4
6
55
8
2
57
11
3
43
16
1
99
3
5
46
9
1
35
19
060
13
3
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
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