Statistical Analysis of Sales Calls and Other Data Sets, Exams of Nursing

Download Statistical Analysis of Sales Calls and Other Data Sets and more Exams Nursing in PDF only on Docsity! 1 Week 8 : Final Exam - Final Exam Page: 1 2 3 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, 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) a. Ans: Mean, median, mode, and standard deviation, Q1, Q3, Min, and 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 2 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 and 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. 5 Ans: 0.00000000003+0.000000002+0.00000004+0.0000005+0.000005+0.00003+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 and 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 = (22000­27009)/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 and 23,000 gallons for a given day. Ans: z­ score for 20000 gallons = (20000­27009)/4530 = ­1.547 From the Standard Normal cumulative proportions p­ value = 0.0606 z­ score for 23000 gallons = (23000­27009)/4530 = ­ 0.885 From the Standard Normal cumulative proportions p­ value = 0.1880 probability that the demand for gasoline falls between 20,000 and 23,000 gallons for a given day = (0.1880­0.0606) = 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 6 day)? (Points : 18) Ans: For the demand to be met 90 % of the time, it means that p­value = 0.9 Z­score for (p­value = 0.9) = 1.28 gallons of gasoline should be on hand = 1.28*4530 + 27009 = 32807.4 gallons ~ approx 32808 gallons 7 5. (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 a. Compute the 90% confidence interval for the population mean refueling and baggage time. Ans: 90 % confidence interval means Z(Upper – 95%) = 1.645 Z ( Lower – 5%) = ­1.645 90 % Confidence interval lower limit = 24.2­1.645*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 and baggage handling time at a foreign airport location would lie between 23.049 minutes and 25.351 minutes for 90% of the times. c. How many refueling and baggage handling times should be sampled so that we may construct a 90% confidence interval with a sampling error of .5 minutes for the population mean refueling and baggage time? (Points : 18) Ans: Nos. of refueling and 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 10 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 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. a. Formulate the null and 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), and clearly show the rejection and nonrejection regions. Ans: New defect % = 45 /2000 = 0.0225 At 99 % level of significance (one­sided), the 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 Non­rejection region. 11 d. Compute the test statistic. Ans: Z ( Lower Value) = 0.021 e. Decide whether you can reject Ho and 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 and 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 p­value for the hypothesis test and interpret this value. What does this mean? Ans: New defect % = 45 /2000 = 0.0225 Observed p­value is 2.25 % which means that sample mean observed value recorded 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) 12 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% 15 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). Page: 1 2 3 Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/fontdata.js Week 8 : Final Exam - Final Exam Page: 1 2 3 Page 2 16 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 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 17 122 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 20 a. Analyze the above output to determine the regression equation. Ans: The regression equation is PRICE ($1,000s) = 2.26 + 2.89 ACREAGE b. Find and interpret in the context of this problem. Ans: Price of land depends on acreage and varies with the acreage. Price of the land can be predicted by multiplying the acreage of the land with 2.89 and then adding a constant value of 2.26. The land price arrived would show up in $ 1,000s (thousands of Dollars). c. Find and interpret the coefficient of determination (r­squared). Ans: 21 The coefficient of determination, r 2, is useful because it gives the proportion of the variance (fluctuation) of price of the land ( in ,000 $) that can be predicted from the acreage of the land. 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, R­Sq 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 and interpret coefficient of correlation. Ans: Coefficient of correlation. R, measures the strength and the direction of a linear relationship between two variables. In this case, the r value is 0.997 and so there is a string relationship between the acreage of the land and 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 two­tailed test. Find the observed p­value and 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: 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). g. Find the 95% prediction interval for the price of a single plot of farmland that is 50 acres. Interpret this interval. Ans: 22 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 = 25 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 26 0.000 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 and interpret the multiple index of determination (R­Sq). Ans: R-Sq = 93.1% So, the variation in the current monthly auto insurance premium (Y) is explained by driving experience (X1, in years) 27 and number of driving violations in the past three years (X2) to the extent of 93.1%. c. Perform the t­tests on and on (use two tailed test with (𝛼 = .05). Interpret your results. Ans: A t­stat 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 and beyond the sample. The greater the t­stat the greater the relative influence of the independent variable on the dependent variable. A t­stat of less than 1.96 with significance greater than 0.05 indicates that the independent variable is NOT a significant 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 and t-stats value for number of driving violations in the past three years (X2) is 6.16. t –stats value of X1 ( POSITIVE PART) and X2 values both are more than 1.96 which means that independent predictors “X1 driving experience (X1, in years)” and “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 and 1 driving violation during the past 3 years. Use both a point estimate and 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).