Math 533 Week 8 Final Exam Set 1, Exams of Health sciences

Download Math 533 Week 8 Final Exam Set 1 and more Exams Health sciences in PDF only on Docsity! MATH_533_Week_8_Final_Exam_Set_1.doc (1) 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 91.23529 412 Median 91 Mode 93 Standard Deviation 12.37704 232 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 MATH_533_Week_8_Final_Exam_Set_1.doc (1) 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. Urba n Subur ban Total Good 60 168 228 Borderl ine 36 72 108 Poor 24 40 64 Tot al 120 280 400 If you choose a customer at random, then find the probability that the customer a. is considered “borderline.” MATH_533_Week_8_Final_Exam_Set_1.doc (1) From the Standard Normal cumulative proportions p- value = 0.0606 z- score for 23000 gallons = (23000-27009)/4530 = - 0.885 From the Standard Normal cumulative proportions p- value = 0.1880 probability that the demand for gasoline falls between 20,000 & 23,000 gallons for a given day = (0.1880-0.0606) = 0.1274 c. How many gallons of gasoline should be on hand at the beginning of each day so that we can meet the demand 90% of the time (i.e., the station stands a 10% chance of running out of gasoline for that day)? (Points : 18) Ans: For the demand to be met 90 % of the time, it means that p- value = 0.9 Z-score for (p-value = 0.9) = 1.28 gallons of gasoline should be on hand = 1.28*4530 + 27009 = 32807.4 gallons ~ approx 32808 gallons MATH_533_Week_8_Final_Exam_Set_1.doc (1) 5. (TCO C) An operations analyst from an airline company has been asked to develop a fairly accurate estimate of the mean refueling & baggage handling time at a foreign airport. A random sample of 36 refueling & baggage handling times yields the following results. Sample Size = 36 Sample Mean = 24.2 minutes Sample Standard Deviation = 4.2 minutes a. Compute the 90% confidence interval for the population mean refueling & baggage time. Ans: 90 % confidence interval means Z(Upper – 95%) = 1.645 Z ( Lower – 5%) = -1.645 90 % Confidence interval lower limit = 24.2-1.645*4.2/6 = 23.049 90 % Confidence interval Upper limit = 24.2+1.645*4.2/6 = 25.351 b. Interpret this interval. Ans: Mean refueling & baggage handling time at a foreign airport location would lie between 23.049 minutes & 25.351 minutes for 90% of the times. c. How many refueling & baggage handling times should be sampled so that we may construct a 90% MATH_533_Week_8_Final_Exam_Set_1.doc (1) confidence interval with a sampling error of .5 minutes for the population mean refueling & baggage time? (Points : 18) Ans: Nos. of refueling & baggage handling times that should be sampled so that we may construct a 90% confidence interval = (4.2/0.5) ^ 2 = 70.56 ~approx. 71 6. (TCO C) The manufacturer of a certain brand of toothpaste claims that a highpercentage of dentists recommend the use of their toothpaste. A random sample of 400 MATH_533_Week_8_Final_Exam_Set_1.doc (1) lower limit = 0.021 So, more than 2.1 % of car(= more than 42 cars) with paint defects would clearly fall into rejection region. Anything less than 2.1 % of “car ( less than 42 cars) with paint defects” would mean Non- rejection region. MATH_533_Week_8_Final_Exam_Set_1.doc (1) d. Compute the test statistic. Ans: Z ( Lower Value) = 0.021 e. Decide whether you can reject Ho & accept Ha or not. Ans: New defect % = 45 /2000 = 0.0225 Since the new defect % 2.25 % is higher than the lower limit of 2.1%, we would reject the null Hypotheses (H0). We would accept the Alternative hypotheses: H1: proportion of paint defects after the implementation is now more than .03 f. Explain & interpret your conclusion in part e. What does this mean? Ans: It means that at 99 % level of significance, proportion of paint defects after the implementation of the defect reduction program in cars is now more than .03. So, the defect reduction program has not really worked for Ford Motor Company quality improvement team. g. Determine the observed p-value for the hypothesis test & interpret this value. What does this mean? Ans: New defect % = 45 /2000 = 0.0225 Observed p-value is 2.25 % which means that sample mean observed value recorded MATH_533_Week_8_Final_Exam_Set_1.doc (1) 2.25 % of the cars having paint defects. h. Does the sample data provide evidence to conclude that the proportion of paint defects is now less than .03 (with a = .01)? (Points : 24) Ans: Sample data doesn’t provide evidence to conclude that the proportion of paint defects is now less than .03 (with a = .01). 8. (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. a. Formulate the null & alternative hypotheses. Ans: Null Hypotheses: H0: dealership average profit on sales of new cars = 4.5% Alternative hypotheses: H1: dealership average profit on sales of new cars is more than 4.5 % MATH_533_Week_8_Final_Exam_Set_1.doc (1) 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). PRIC E ACR EAG E PRE DICT 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 MATH_533_Week_8_Final_Exam_Set_1.doc (1) 145 45.0 61 19.2 235 80.0 250 90.0 278 95.0 118 41.0 46 14.0 69 22.0 MATH_533_Week_8_Final_Exam_Set_1.doc (1) 220 81.5 235 78.0 50 16.0 25 10.0 290 100. 0 Correlations: PRICE, ACREAGE Pearson correlation of PRICE & 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 MATH_533_Week_8_Final_Exam_Set_1.doc (1) The coefficient of determination is a measure of how well the regression line (PRICE ($1,000s) = 2.26 + 2.89 ACREAGE) represents the data. In this case, R-Sq value is 99.4% which means that variation in the price of the land can be explained/predicted to the extent of 99.4 % by the variation in the acreage of the land. d. Find & interpret coefficient of correlation. Ans: Coefficient of correlation. R, measures the strength & the direction of a linear relationship between two variables. In this case, the r value is 0.997 & so there is a string relationship between the acreage of the land & price of the land ( in ,000 $). e. Does the data provide significant evidence (𝛼 = .05) that the acreage can be used to predict the price? Test the utility of this model using a two-tailed test. Find the observed p- value & interpret. Ans: at 95 % level of significance, the observed p – value is 0.000 which is less than 0.05. So, reject the null hypothesis. So, the data provides significant evidence (at 𝛼 = .05) that the acreage can be used to predict the price. f. Find the 95% confidence interval for mean price of plots of farmland that are 50 acres. Interpret this interval. Ans: MATH_533_Week_8_Final_Exam_Set_1.doc (1) 95% confidence interval for mean price of plots of farmland that are 50 acres is (144.05, 149.66). So, at 95% confidence interval, the mean price of plots of farmland that are 50 acres would lie in the interval of (144.05, 149.66). MATH_533_Week_8_Final_Exam_Set_1.doc (1) 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 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): 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 MATH_533_Week_8_Final_Exam_Set_1.doc (1) 0.020 Cell Contents: Pearson correlation P-Value a. Analyze the above output to determine the multiple regression equation. Ans: The multiple Regression equation is The regression equation is Y = 55.1 - 1.37 X1 + 8.05 X2 b. Find & interpret the multiple index of determination (R-Sq). Ans: R-Sq = 93.1% So, the variation in the current monthly auto insurance premium (Y) is explained by driving experience (X1, in years) & number of driving violations in the past three years (X2) to the extent of 93.1%. c. Perform the t-tests on & on (use two tailed test with (𝛼 = .05). Interpret your results. Ans: A t-stat of greater than 1.96 with a significance less than 0.05 indicates that the independent variable is a significant predictor of the dependent variable within & beyond the sample. The greater the t-stat the greater the relative influence of the independent variable on the dependent variable. A t-stat of less than1.96 with significance greater than 0.05 indicates that the independent variable is NOT a significant MATH_533_Week_8_Final_Exam_Set_1.doc (1) predictor of the dependent variable. Predictor Coef SE Coef T P Constant 55.1387.309 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 MATH_533_Week_8_Final_Exam_Set_1.doc (1) 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