ISM_chapter11 - Chapter 11: Linear Models and Estimation by...

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231 Chapter 11: Linear Models and Estimation by Least Squares 11.1 Using the hint, . ˆ ) ˆ ( ˆ ˆ ) ( ˆ 1 1 1 0 y x x y x x y = β + β = β + β = K 11.2 a. slope = 0, intercept = 1. SSE = 6. b. The line with a negative slope should exhibit a better fit. c. SSE decreases when the slope changes from .8 to .7. The line is pivoting around the point (0, 1), and this is consistent with ( y x , ) from part Ex. 11.1. d. The best fit is: y = 1.000 + 0.700 x . 11.3 The summary statistics are: x = 0, y = 1.5, S xy = –6, S xx = 10. Thus, y ˆ = 1.5 – .6 x . The graph is above. -2 -1 0 1 2 0.5 1.0 1.5 2.0 2.5 3.0 p11.3x p11.3y 11.4 The summary statistics are: x = 72, y = 72.1, S xy = 54,243, S xx = 54,714. Thus, y ˆ = 0.72 + 0.99 x . When x = 100, the best estimate of y is y ˆ = 0.72 + 0.99(100) = 99.72. 11.5 The summary statistics are: x = 4.5, y = 43.3625, S xy = 203.35, S xx = 42. Thus, y 21.575 + 4.842 x . Since the slope is positive, this suggests an increase in median prices over time. Also, the expected annual increase is $4,842. 11.6 a. intercept = 43.362, SSE = 1002.839. b. the data show an increasing trend, so a line with a negative slope would not fit well. c. Answers vary. d. Answers vary. e. (4.5, 43.3625) f. The sum of the areas is the SSE. 11.7 a. The relationship appears to be proportional to x 2 . b. No. c. No, it is the best linear model.
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232 Chapter 11: Linear Models and Estimation by Least Squares Instructor’s Solutions Manual 11.8 The summary statistics are: x = 15.505, y = 9.448, S xy = 1546.459, S xx = 2359.929. Thus, y ˆ = –0.712 + 0.655 x . When x = 12, the best estimate of y is y ˆ = –.712 + 0.655(12) = 7.148. 11.9 a. See part c . b. y ˆ = –15.45 + 65.17 x . c. The graph is above. 1 . 61 . 82 . 02 . 22 . 4 60 80 100 120 140 p11.9x p11.9y d. When x = 1.9, the best estimate of y is y ˆ = –15.45 + 65.17(1.9) = 108.373. 11.10 0 ) ( 2 ) ( 2 SSE 2 1 1 1 1 1 = β = β = β = = i n i i i i i n i i x y x x x y d d , so = = = β n i i n i i i x y x 1 2 1 1 ˆ . 11.11 Since = n i i i y x 1 = 134,542 and = n i i x 1 2 = 53,514, 1 ˆ β = 2.514. 11.12 The summary statistics are: x = 20.4, y = 12.94, S xy = –425.571, S xx = 1859.2. a. The least squares line is: y ˆ = 17.609 – 0.229 x .
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Chapter 11: Linear Models and Estimation by Least Squares 233 Instructor’s Solutions Manual b. The line provides a reasonable fit. 0 1 02 03 04 05 0 8 1 01 21 41 61 82 0 p11.12x p11.12y c. When x = 20, the best estimate of y is y ˆ = 17.609 – 0.229(20) = 13.029 lbs. 11.13 The summary statistics are: x = 6.177, y = 270.5, S xy = –5830.04, S xx = 198.29. a. The least squares line is: y ˆ = 452.119 – 29.402 x . b. The graph is above. 24681 0 1 2 200 300 400 500 p11.13x p11.13y 11.14 The summary statistics are: x = .325, y = .755, S xy = –.27125, S xx = .20625 a. The least squares line is: y ˆ = 1.182 – 1.315 x .
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234 Chapter 11: Linear Models and Estimation by Least Squares Instructor’s Solutions Manual 0.1 0.2 0.3 0.4 0.5 0.4 0.5 0.6 0.7 0.8 0.9 1.0 p11.14x p11.14y b. The graph is above. The line provides a reasonable fit to the data. 11.15 a. SSE ) ) ( )] ( ˆ [ ) ˆ ˆ ( 1 2 2 1 1 2 1 1 0 = = = = β = β β = n i i i n i i i n i i y y x x y y x y + = = β β n i i i n i i x x y y x x 1 1 1 2 2 1 ) )( ( ˆ 2 ) ( ˆ = = n i i y y 1 2 ) (+ xy xy S S 1 1 ˆ 2 ˆ β β = yy S xy S 1 ˆ β . b. Since xy yy S S 1 ˆ SSE β = , xx xy xy yy S S S S / ) ( SSE ˆ SSE 2 1 + = β + = . But, S xx > 0 and ( S xy ) 2 0. So, SSE yy S .
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This note was uploaded on 03/18/2009 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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ISM_chapter11 - Chapter 11: Linear Models and Estimation by...

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