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wk6_Checkpoint

Course: GENERAL MA GM533, Spring 2012
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6 Week : Simple Linear Regression - Checkpoint 1. Complete Exercise 13.8 (The Real Estate Case) on page 503 in your textbook. (Points : 9) 13.8 THE REAL ESTATE SALES PRICE CASE RealEst A real estate agency collects data concerning y = the sales price of a house (in thousands of dollars), and x = the home size (in hundreds of square feet). The data are given in the table below. The MINITAB output from fitting a...

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6 Week : Simple Linear Regression - Checkpoint 1. Complete Exercise 13.8 (The Real Estate Case) on page 503 in your textbook. (Points : 9) 13.8 THE REAL ESTATE SALES PRICE CASE RealEst A real estate agency collects data concerning y = the sales price of a house (in thousands of dollars), and x = the home size (in hundreds of square feet). The data are given in the table below. The MINITAB output from fitting a least squares regression line to the data is on the next page. a By using the formulas illustrated in Example 13.2 (see page 497) and the data provided, verify that (within rounding) b0 = 48.02 and b1 = 5.700, as shown on the MINITAB output. b Interpret the meanings of b0 and b1. Does the interpretation of b0 make practical sense? c Write the least squares prediction equation. Use the least squares line to obtain a point estimate of the mean sales price of all houses having 2,000 square d feet and a point prediction of the sales price of an individual house having 2,000 square feet. (Bowerman, Bruce.. Essentials of Business Statistics, 4th Edition. McGraw-Hill Learning Solutions, 2012. pp. 503 - 504). <vbk:0077587456#outline(13.3.2.4)> a. MINITAB output Regression Analysis: Sale Price versus Size The regression equation is Sale Price = 48.02 + 5.70 Size b1 = SSxy / SSxx = 1149.18 / 201.6 = 5.70. b0 = y-bar b1*x-bar = 155.19 5.70*18.8 = 48.03 which is within rounding 48.02. b. b1 is the estimated increase in mean sales price (5.700) for every hundred square foot increase in home size. b0 is the estimated mean sales price when square footage = 0. No, the interpretation of b0 makes no practical sense. c. y = 48.02 + 5.700x. d. y = 48.02 + 5.700 (20) = 162.02. That is, \$162,020. 2. Complete Exercise 13.21 (The Starting Salary) on page 511 in your textbook. (Points : 9) 13.21 THE STARTING SALARY CASE StartSal The MINITAB output of a simple linear regression analysis of the data set for this case (see Exercise 13.4 on page 501) is given in Figure 13.11. Recall that a labeled MINITAB regression output is on page 509. FIGURE 13.11: MINITAB Output of a Simple Linear Regression Analysis of the Starting Salary Data (Bowerman, Bruce.. Essentials of Business Statistics, 4th Edition. McGraw-Hill Learning Solutions, 2012. pp. 511 - 512). <vbk:0077587456#outline(13.5.2.1)> a. b0 = 14.816; b1 = 5.7066 b. SSE = 1.438; s = .5363 c. sb1 = .3953; t = 14.44 d. t > 2.571; reject H0. Yes, there is strong evidence that 1 0. e. t > 4.032; reject H0. Yes, there is very strong evidence that 1 0. f. p-value = .000. Can reject H0 at all s. Extremely strong evidence. g. [4.6903, 6.7229] h. [4.1128, 7.3004] i. sb = 1.235; t = 12.00 j. p-value = .000. Can reject H0 at all s. There is extremely strong evidence that 0 0. k. SSxx = 1.84069; sb = 1.235, sb = .3953 (Bowerman, Bruce.. Essentials of Business Statistics, 4th Edition. McGraw-Hill Learning Solutions, 2012. p. 655). <vbk:0077587456#outline(16.6.13)> a. b0 = 14.816 b1 = 5.7066 b. SSE = 1.438 s = .288 s = .5363 c. s = .3953 t = 14.44 t = b1/ sb1 = 5.7066 /.3953 = 14.44 d. df = 5 t.025 = 2.571 Reject H0 , Strong evidence of a significant relationship between x and y. e. t.005 = 4.032 Reject H0 , Very strong evidence of a significant relationship between x and y. f. p-value =.000 Reject at all , Extremely strong evidence of a significant relationship between x and y. g. 95% Cl: [b1 t.025 sb1 ] = 5.7066 (2.571)(.3953) = [4.690, 6.723] We are 95% confident that the mean starting salary increases by between \$4690 and \$6723 for each 1.0 increase in GPA. h. 99% Cl: [b1 t.005 sb1 ] = 5.7066 (4.032)(.3953) = [4.113, 7.300] We are 99% confident that the mean starting salary increases between by \$4113 and \$7300 for each 1.0 increase in GPA. i. sb0 = 1.235 t = 12.00 t = b0 / sb0 = 14.816 / 1.235 = 12.00 j. p-value = .000 Reject at all , Extremely strong evidence that the y-intercept is significant. sb1 = s/sqrt(SSxx) = .5363/sqrt(1.8407) = .3953 k. 2 b1 sb0 = s*sqrt(1/n + (x-bar)2/SSxx) = .5363*sqrt(1/7 + (.3.0814)2/1.8407) = 1.235 3. Complete Exercise 13.30 (The Fuel Consumption Case) on page 518 in your textbook. (Points : 9) 13.3 THE FUEL CONSUMPTION CASE FuelConl The following partial MINITAB regression output for the fuel consumption data relates to predicting the citys fuel consumption (in MMcf of natural gas) in a week that has an average hourly temperature of 40F. a Report (as shown on the computer output) a point estimate of and a 95 percent confidence interval for the mean fuel consumption for all weeks having an average hourly temperature of 40F. b Report (as shown on the computer output) a point prediction of and a 95 percent prediction interval for the fuel consumption in a single week that has an average hourly temperature of 40F. c Remembering that s = .6542; SSxx = 1,404.355;x=43.98 n = 8, hand calculate the distance value when x0 = 40. Remembering that the distance value equals (Sy^/S)2, use s and Sy^ from the computer output to calculate (within rounding) the distance value using this formula. Note that, because MINITAB rounds sy, the first hand calculation is the more accurate calculation of the distance value. d Remembering that for the fuel consumption data b0 = 15.84 and b1 = .1279, calculate (within rounding) the confidence interval of part a and the prediction interval of part b. e Suppose that next week the citys average hourly temperature will be 40F. Also, suppose that the citys natural gas company will use the point prediction y^=10.721 and order 10.721 MMcf of natural gas to be shipped to the city by a pipeline transmission system. The company will have to pay a fine to the transmission system if the citys actual gas useage y differs from the order of 10.721 MMcf by more than 10.5 percentthat is, is outside of the range [10.721 .105(10.721)] = [9.595, 11.847]. Discuss why the 95 percent prediction interval for y, [9.015, 12.427], says that y might be outside of the allowable range and thus does not make the company 95 percent confident that it will avoid paying a fine. Note: In the exercises of Chapter 14 we will use multiple regression analysis to predict y accurately enough that the company is likely to avoid paying a fine. (Bowerman, Bruce.. Essentials of Business Statistics, 4th Edition. McGraw-Hill Learning Solutions, 2012. p. 518). <vbk:0077587456#outline(13.6.3)> a. 10.721, [10.130, 11.312] b. 10.721, [9.015, 12.427] c. dv = 1/8 + (40-43.98) / 1404.355 = 0.1363; dv = (0.241 / 0.6542) = 0.1357 d. CI: 15.84 -0.1279 * 40 2.447*0.6542*sqrt(0.1363) = [10.13, 11.31] PI: 15.84 -0.1279 * 40 2.447*0.6542*sqrt(1.1363) = [9.01, 12.43] e. Since we are predicting fuel consumption for one day when the average temperature is 40 degrees we must use the prediction interval. Since 9.01 < 9.595 and 12.43 > 11.847 the city cannot be 95% confident it will not pay a fine. For the city to be at least 95% confident the PI would have to be inside the interval [9.595, 11.847]. 2 2 a. F = 106.30 b. F > 4.20. Reject H0 ; significant relationship at = .05 c. F > 7.64. Reject H0 ; significant relationship at = .01 d. p- value = .000. Can reject H0 at all levels of . There is extremely strong evidence of a regression relationship. (Bowerman, Bruce.. Essentials of Business Statistics, 4th Edition. McGraw-Hill Learning Solutions, 2012. p. 656). <vbk:0077587456#outline(16.6.13)>
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