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Regression Analysis Solutions - Assignment #3, Assignments of Introduction to Business Management

Solutions to assignment #3 questions related to simple linear regression analysis from asw textbook. Includes calculations for sum of squares, regression coefficients, and hypothesis testing.

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Pre 2010

Uploaded on 08/19/2009

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Download Regression Analysis Solutions - Assignment #3 and more Assignments Introduction to Business Management in PDF only on Docsity! 1 BA 303 Assignment #3 – Solution Key Chapter 14 __ 1. ASW - page 573 - Question # 15 a. The estimated regression equation and the mean for the dependent variable are: $ . .y x yi i= + =0 2 2 6 8 The sum of squares due to error and the total sum of squares are SSE SST= ∑ − = = ∑ − =( $ ) . ( )y y y yi i i2 212 40 80 Thus, SSR = SST - SSE = 80 - 12.4 = 67.6 b. r2 = SSR/SST = 67.6/80 = .845 The least squares line provided a very good fit; 84.5% of the variability in y has been explained by the least squares line. c. r = = +. .845 9192 __ 2. ASW - page 574 - Question # 16 a. The estimated regression equation and the mean for the dependent variable are: ˆ 30.33 1.88 23.2iy x y= − = The sum of squares due to error and the total sum of squares are 2 2ˆSSE ( ) 6.33 SST ( ) 114.80i i iy y y y= ∑ − = = ∑ − = Thus, SSR = SST - SSE = 114.80 - 6.33 = 108.47 b. r2 = SSR/SST = 108.47/114.80 = .945 The least squares line provided an excellent fit; 94.5% of the variability in y has been explained by the estimated regression equation. c. r = = −. .945 9721 Note: the sign for r is negative because the slope of the estimated regression equation is negative. (b1 = -1.88) 2 __ 3. ASW - page 575 - Question # 20 a. Let x = income and y = home price. Summations needed to compute the slope and y-intercept are: 21424 2455.5 ( )( ) 4011 ( ) 1719.618i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − = 1 2 ( )( ) 4011 2.3325 1719.618( ) i i i x x y yb x x Σ − − = = = Σ − 0 1 136.4167 (2.3325)(79.1111) 48.11b y b x= − = − = − ˆ 48.11 2.3325y x= − + b. The sum of squares due to error and the total sum of squares are 2 2ˆSSE ( ) 2017.37 SST ( ) 11,373.09i i iy y y y= ∑ − = = ∑ − = Thus, SSR = SST - SSE = 11,373.09 – 2017.37 = 9355.72 r2 = SSR/SST = 9355.72/11,373.09 = .82 We see that 82% of the variability in y has been explained by the least squares line. .82 .91r = = + c. ˆ 48.11 2.3325(95) 173.5y = − + = or approximately $173,500 __ 4. ASW - page 567 - Question # 12 (this question not required – assigned in assignment #2) __ 5. ASW - page 585 - Question # 23 a. s2 = MSE = SSE / (n - 2) = 12.4 / 3 = 4.133 b. s = = =MSE 4 133 2 033. . c. 2( ) 10ix xΣ − = 1 2 2.033 0.643 10( ) b i ss x x = = = Σ − 5 F = MSR / MSE = 9836.74/310.96 = 31.63 F.05 = 5.32 (1 degree of freedom numerator and 8 denominator) Since F = 31.63 > F.05 = 5.32 we reject H0: β1 = 0. Upper support and price are related. c. r2 = SSR/SST = 9,836.74/12,324.4 = .80 The estimated regression equation provided a good fit; we should feel comfortable using the estimated regression equation to estimate the price given the upper support rating. d. ŷ = 19.93 + 31.21(4) = 144.77 __ 8. ASW - page 592 - Question # 33 a. s = 1.453 b. 23.8 ( ) 30.8ix x x= Σ − = p 2 2 p ˆ 2 ( )1 1 (3 3.8)1.453 .068 5 30.8( )y i x x s s n x x − − = + = + = Σ − $ . . . . ( ) .y x= − = − =30 33 188 30 33 188 3 24 69 $ / $y t syp p± α 2 24.69 ± 3.182 (.68) = 24.69 ± 2.16 or 22.53 to 26.85 c. 2 2 p ind 2 ( )1 1 (3 3.8)1 1.453 1 1.61 5 30.8( )i x x s s n x x − − = + + = + + = Σ − d. $ /y t sp ind± α 2 24.69 ± 3.182 (1.61) = 24.69 ± 5.12 or 19.57 to 29.81 __ 9. ASW - page 592 - Question # 35 a. s = 145.89 23.2 ( ) 0.74ix x x= Σ − = 6 p 2 2 p ˆ 2 ( )1 1 (3 3.2)145.89 68.54 6 0.74( )y i x x s s n x x − − = + = + = Σ − ˆ 1790.5 581.1 1790.5 581.1(3) 3533.8y x= + = + = $ / $y t syp p± α 2 3533.8 ± 2.776 (68.54) = 3533.8 ± 190.27 or $3343.53 to $3724.07 b. 2 2 p ind 2 ( )1 1 (3 3.2)1 145.89 1 161.19 6 0.74( )i x x s s n x x − − = + + = + + = Σ − $ /y t sp ind± α 2 3533.8 ± 2.776 (161.19) = 3533.8 ± 447.46 or $3086.34 to $3981.26