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Solutions to assignment #5 of ba 303, focusing on regression analysis. It includes estimates of regression equations, predictions, and interpretations of coefficients. Students are expected to turn in their completed assignments by december 8.

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Download BA 303 Assignment #5: Regression Analysis and more Assignments Introduction to Business Management in PDF only on Docsity! 1 BA 303 Assignment #5 The due date for this assignment is Thursday December 8 Instructions: Complete the following questions and turn in your solutions on or before the date due. Remember to include your name and student number on your submitted document. Chapter 15 1. ASW - page 642 Question # 2 a. The estimated regression equation is ŷ = 45.06 + 1.94x1 An estimate of y when x1 = 45 is ŷ = 45.06 + 1.94(45) = 132.36 b. The estimated regression equation is ŷ = 85.22 + 4.32x2 An estimate of y when x2 = 15 is ŷ = 85.22 + 4.32(15) = 150.02 c. The estimated regression equation is ŷ = -18.37 + 2.01x1 + 4.74x2 An estimate of y when x1 = 45 and x2 = 15 is ŷ = -18.37 + 2.01(45) + 4.74(15) = 143.18 2. ASW – page 644 Question # 7 a. The regression output is as follows: The regression equation is Price = 356 - 0.0987 Capacity + 123 Comfort Predictor Coef SE Coef T P Constant 356.1 197.2 1.81 0.114 Capacity -0.09874 0.04588 -2.15 0.068 Comfort 122.87 21.80 5.64 0.001 S = 51.14 R-Sq = 83.2% R-Sq(adj) = 78.4% Analysis of Variance 2 Source DF SS MS F P Regression 2 90548 45274 17.31 0.002 Residual Error 7 18304 2615 Total 9 108852 b. b1 = -.0987 is an estimate of the change in the price with respect to a 1 cubic inch change in capacity with the comfort rating held constant. b2 = 123 is an estimate of the change in the price with respect to a 1 unit change in the comfort rating with the capacity held constant. c. ŷ = 356 - .0987(4500) + 123 (4) = 404 3. ASW – page 649 Question # 14 a. 2 SSR 12,000 .75 SST 16,000 R = = = b. 2 2 1 91 (1 ) 1 .25 .68 1 7a nR R n p − = − − = − = − − c. The adjusted coefficient of determination shows that 68% of the variability has been explained by the two independent variables; thus, we conclude that the model does not explain a large amount of variability. 4. ASW – page 656 Question # 22 a. SSE = SST - SSR = 16000 - 12000 = 4000 2 SSE 4000 571.43 - -1 7 s n p = = = SSR 12000MSR 6000 2p = = = b. F = MSR/MSE = 6000/571.43 = 10.50 Using F table (2 degrees of freedom numerator and 7 denominator), p-value is less than .01 Because p-value ,α≤ we reject H0. There is a significant relationship among the variables. 5. ASW – page 665 Question # 33 a. two b. E(y) = β0 + β1 x1 + β2 x2 + β3 x3 where