Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

MATH 533 Week 8 Final Exam(100% CORRECT Solution) | Already graded A+

Typology: Exams

2022/2023

1 / 22

Download MATH 533 Week 8 Final Exam(100% CORRECT Solution) | Already graded A+ and more Exams Mathematics in PDF only on Docsity! Week 8 : Final Exam - Final Exam Page 1 1. (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 a. Compute the mean, median, mode, & standard deviation, Q1, Q3, Min, & Max for the above sample data on number of sales calls per month. b. In the context of this situation, interpret the Median, Q1, & Q3. (Points : 33) a. Ans: Mean, median, mode, & standard deviation, Q1, Q3, Min, & Max for the above sample data on number of sales calls per month Mean 91.23529412 Median 91 Mode 93 Standard Deviation 12.37704232 1st quartile 82.00 3rd quartile 100.00 Minimum 72 Maximum 119 b. Median of the above sales calls means that if all the sales calls data points are arranged in an ascending order, then 91 Nos. of calls made would fall in the middle. So, there are as 8 sales calls data point above this median & 8 sales calls data point below this median point. Q1 is the first quartile points which is 82 nos. of calls made. It means that there are 25 % of sales calls data point which lie below this point. Q3 is the third quartile points which is 100 nos. of calls made. It means that there are 75 % of sales calls data point which lie below this point. 2. (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 a. is considered “borderline.” Page 1 of 22 Ans: P(Customer is considered “borderline) = 108/400 = 27/100 b. is considered “good” & resides in an urban location. Ans: P(is considered “good” & resides in an urban location) = 60/400 = 3/20 c. is suburban, given that customer is considered “poor.” (Points : 18) Ans: P(is suburban, given that customer is considered “poor”) = 40/64 =5/8 3. (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 a. exactly 14 customers will pay for their purchases using credit cards. Ans: This is a binomial distribution with p = 0.70, so q = 1- p = 0.30 n = 20 Probability distribution: P(exactly 14 customers will pay for their purchases using credit cards) = 20 C 14 *(0.70)^14 *(0.30)^6 = 0.19164 b. at least 10 customers will pay for their purchases using credit cards. Ans : 0.0308+0.0653+0.1144+0.1643+0.1916+0.1788+0.1304+0.0716+0.02 78+0.0068+0.0 008 = 0.9826 c. at most 12 customers will pay for their purchases using credit cards. (Points : 18) Ans: 0.00000000003+0.000000002+0.00000004+0.0000005+0.000005+0.00 003+0.0002+0 .0010+0.0038+0.0120+0.0308+0.0653+0.1144 = 0.2275 4. (TCO B) The demand for gasoline at a local service station is normally distributed with a mean of 27,009 gallons per day & a standard deviation of 4,530 gallons per day. a. Find the probability that the demand for gasoline exceeds 22,000 gallons for a given day. Ans: z- score = (22000-27009)/4530 = -1.1057395 From the Standard Normal cumulative proportions p- value = 0.1346 probability that the demand for gasoline exceeds 22,000 gallons for a given day = 1- 0.1346 = 0.8654 b. Find the probability that the demand for gasoline falls between 20,000 & 23,000 gallons for a given day. Page 2 of 22 ~approx. 71 6. (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 Page 5 of 22 dentists results in 310 recommending their toothpaste. a. Compute the 99% confidence interval for the population proportion of dentists who recommend the use of this toothpaste. Ans: p value = 310 /400 = 0.775 Sample Size = 40072.1% of 99% confidence interval for the population proportion of dentists = 99% lower population proportion of dentists = 0.721 = 72.1% 99% Upper population proportion of dentists = 0.829 = 82.9 % b. Interpret this confidence interval. Ans: So, for 99 % of the cases/times, More than 72.1% of population proportion of dentists & less than 82.9 % of population proportion of dentists would recommend the use of their toothpaste. c. How large a sample size will need to be selected if we wish to have a 99% confidence interval that is accurate to within 3%? (Points : 18) Ans: Sample size = 1286 7. (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 & 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. a. Formulate the null & alternative hypotheses. Ans: Null Hypotheses: H0: proportion of paint defects after the implementation = .03 Alternative hypotheses: H1: proportion of paint defects after the implementation is now more than .03 b. State the level of significance. Ans: Level of significance is = 1- 0.01 = 0.99 = 99 % c. Find the critical value (or values), & clearly show the rejection & nonrejection regions. Ans: New defect % = 45 /2000 = 0.0225 At 99 % level of significance (one-sided), the lower limit = 0.021 So, more than 2.1 % of car(= more than 42 cars) with paint defects would Page 6 of 22 clearly fall into rejection region. Anything less than 2.1 % of “car ( less than 42 cars) with paint defects” would mean Non- rejection region. Page 7 of 22 b. State the level of significance. Ans: Level of significance = 1-0.10 = 0.9 = 90 % c. Find the critical value (or values), & clearly show the rejection & nonrejection regions. Ans: Critical Value = 4.5 + 1.28 * 0.2 = 4.756 So, less than 4.756% of dealership average profit on sales of new cars would be rejection region. More than 4.756% of dealership average profit on sales of new cars would be non- rejection region d. Compute the test statistic. Ans: Z – higher value = 4.756 % e. Decide whether you can reject Ho & accept Ha or not. Ans: Since, the Sample Mean = 4.97% lie above the critical value of 4.756 %, we would reject H0 & accept Ha. f. Explain & interpret your conclusion in part e. What does this mean? Ans: It means that at 90 % level of significance, the dealership averages more than 4.5% profit on sales of new cars. g. Determine the observed p-value for the hypothesis test & interpret this value. What does this mean? Ans: P – value = Sample Mean = 4.97% so, it means that random sample of 81 cars gave profit of 4.97 % on sales of new car. h. 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)? (Points : 24) Ans: Yes, the sample data does provide evidence to conclude that the dealership averages more than 4.5% profit on sales of new cars (using a = .10). Week 8 : Final Exam - Final Exam Page 2 1. (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 Page 10 of 22 Page 11 of 22 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 & 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 & Y=Price ($1,000s). PRICE ACREAG E PREDIC T 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 Page 12 of 22 lie in the interval of (144.05, 149.66). Page 15 of 22 g. Find the 95% prediction interval for the price of a single plot of farmland that is 50 acres. Interpret this interval. Ans: 95% prediction interval for the price of a single plot of farmland that are 50 acres (131.82, 161.90. So, at 95% prediction interval, the price of a single plot of farmland that is 50 acres would lie in the interval of (131.82, 161.90. h. What can we say about the price for a plot of farmland that is 250 acres? (Points : 48) Ans: Price for a plot of farmland that is 250 acres = PRICE ($1,000s) = 2.26 + 2.89 ACREAGE = 2.26 + 2.89 * 250 = 724.760 ( In ,000 $ ) Week 8 : Final Exam - Final Exam 4 Page 16 of 22 Y X1 X2 Predict X1 Predict X2 74 5 2 8 1 38 14 0 50 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 0 60 13 3 1. (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), & current monthly auto insurance premium (Y). A sample of 12 insured drivers is selected at random. The data is given below (in MINITAB): Regression Analysis: Y versus X1, X2 The regression equation is Y = 55.1 - 1.37 X1 + 8.05 X2 Page 17 of 22 predictor of the dependent variable. 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 t-stats value for X1 driving experience (X1, in years) is -2.81 & t-stats value for number of driving violations in the past three years (X2) is 6.16. t –stats value of X1 ( POSITIVE PART) & X2 values both are more than 1.96 which means that independent predictors “X1 driving experience (X1, in years)” & “number of driving violations in the past three years (X2)” both are significant predictors of the current monthly auto insurance premium (Y). d. Predict the monthly premium for an individual having 8 years of driving experience & 1 driving violation during the past 3 years. Use both a point estimate & the appropriate interval estimate. (Points : 31) Ans: The regression equation is Y = 55.1 - 1.37 X1 + 8.05 X2 Point Estimate = 55.1 – 1.37 *8 + 8.05 *1 = 55.1 – 10.96 + 8.05 = $ 52.19 Interval Estimate would be (45.62, 58.79). 17) Of the decision styles listed below, this one should be used for a non-urgent, yet critical & unusual problem affecting multiple stakeholders. A. Programmed decision making B. Autocratic C. Collaborative/participative D. Leader decides 18) While this decision style offers an opportunity for stakeholder input, it most lessens accountability for decision making. A. Consensus B. Collaborative/participative C. Democratic D. Leader decides E. Autocratic 19) A manager may properly choose to ignore a problem A. when a SWOT analysis indicates an environmental opportunity . when the costs of doing nothing are less than the cost of addressing the problem C. when the problem is surfaced through a positive trigger point . when a Pareto chart has properly clarified the causes of a problem . because of a bias against change 20) Jan is the owner of a retail store, & is evaluating the need for changes to the store’s commission structure for sales personnel. She has decided to exclude the sales personnel affected from the decision making process. Jan is probably seeking to minimize this force of influence on the decision making process. . Lack of political support Page 20 of 22 Page 21 of 22 . Resource availability . Resistance from external stakeholders . Ethical considerations Self serving bias 21) Urgent crisis conditions might call for a decision making style that is collaborative B. democratic . autocratic . participative . consensus 22) Upon review of the customer comment cards dropped in the box in his convenience store, Bob sees that 61% of the customers filling out a card have expressed dissatisfaction with aiting times. Bob decides to add an additional cashier during the day. Bob is using the inking style in order to make this decision. A. metaphorical . deductive logical . inductive logical . creative Page 22 of 22