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# BOOKCH11ANSWERS - Chapter 12 Simple Regression 12.1 a...

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Chapter 12: Simple Regression 12.1 a. Compute the sample correlation x y ( ) i x x - 2 ( ) i x x - ( ) i y y - 2 ( ) i y y - ( )( ) i i x x y y - - 2 5 -1.8 3.24 -2.4 5.76 4.32 5 8 1.2 1.44 0.6 0.36 0.72 3 7 -0.8 0.64 -0.4 0.16 0.32 1 2 -2.8 7.84 -5.4 29.16 15.12 8 15 4.2 17.64 7.6 57.76 31.92 19 37 30.8 93.2 52.4 x = 19/5 = 3.8, y = 37/5 = 7.4, 2 ( ) 1 i x x x s n - = - = 30.8 4 = 2.7749, 2 ( ) 1 i y y y s n - = - = 93.2 4 = 4.827, ( )( ) 1 i i xy x x y y s n - - = - = 52.4/4 = 13.1 xy x y s r s s = = 13.1/(2.7749)(4.827) = .97802 b. Compute the sample correlation x y ( ) i x x - 2 ( ) i x x - ( ) i y y - 2 ( ) i y y - ( )( ) i i x x y y - - 7 5 -1.8 3.24 -2.4 5.76 4.32 10 8 1.2 1.44 0.6 0.36 0.72 8 7 -0.8 0.64 -0.4 0.16 0.32 6 2 -2.8 7.84 -5.4 29.16 15.12 13 15 4.2 17.64 7.6 57.76 31.92 44 37 0 30.8 0 93.2 52.4 x = 44/5 = 8.8, y = 37/5 = 7.4, 2 ( ) 1 i x x x s n - = - = 30.8 4 = 2.7749, 2 ( ) 1 i y y y s n - = - = 93.2 4 = 4.827, ( )( ) 1 i i xy x x y y s n - - = - = 52.4/4 = 13.1 xy x y s r s s = = 13.1/(2.7749)(4.827) = .97802 c. Compute the sample correlation

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229 Statistics for Business & Economics, 6 th edition x y ( ) i x x - 2 ( ) i x x - ( ) i y y - 2 ( ) i y y - ( )( ) i i x x y y - - 12 4 -3.6 12.96 -1.8 3.24 6.48 15 6 -0.6 0.36 0.2 0.04 -0.12 16 5 0.4 0.16 -0.8 0.64 -0.32 21 8 5.4 29.16 2.2 4.84 11.88 14 6 -1.6 2.56 0.2 0.04 -0.32 78 29 45.2 8.8 17.6 x = 78/5 = 15.6, y = 29/5 = 5.8, 2 ( ) 1 i x x x s n - = - = 45.2 4 = 3.36155, 2 ( ) 1 i y y y s n - = - = 8.8 4 = 1.48324, ( )( ) 1 i i xy x x y y s n - - = - = 17.6/4 = 4.4 xy x y s r s s = = 4.4/(3.36155)(1.48324) = .88247 d. Compute the correlation coefficient x y ( ) i x x - 2 ( ) i x x - ( ) i y y - 2 ( ) i y y - ( )( ) i i x x y y - - 2 8 -1.8 3.24 -5 25 9 5 12 1.2 1.44 -1 1 -1.2 3 14 -0.8 0.64 1 1 -0.8 1 9 -2.8 7.84 -4 16 11.2 8 22 4.2 17.64 9 81 37.8 19 65 0 30.8 0 124 56 x = 19/5 = 3.8, y = 65/5 = 13, 2 ( ) 1 i x x x s n - = - = 30.8 4 = 2.77488, 2 ( ) 1 i y y y s n - = - = 124 4 = 5.56776, ( )( ) 1 i i xy x x y y s n - - = - = 56/4 = 14 xy x y s r s s = = 14/(2.77488)(5.56776) = .90615
Chapter 12: Simple Regression 230 12.2 a. 1 : 0, : 0 o H H ρ ρ = , 2 ( 2) (1 ) r n t r - = - = 2 .35 38 2.303 (1 .35 ) t = = - , t 38,.05 2.021, t 38,.01 2.704. Reject H 0 at the 5% level. Insufficient evidence to reject H 0 at the 1% level. b. 1 : 0, : 0 o H H ρ ρ = , 2 ( 2) (1 ) r n t r - = - = 2 .5 58 4.397 (1 .5 ) t = = - t 58,.05 2.000, t 58,.01 2.660. Therefore, reject H 0 at the 1% level. c. 1 : 0, : 0 o H H ρ ρ = , 2 ( 2) (1 ) r n t r - = - = 2 .62 43 5.182 (1 .62 ) t = = - t 43,.05 2.021, t 43,.01 2.704. Therefore, reject H 0 at the 1% level. d. 1 : 0, : 0 o H H ρ ρ = , 2 ( 2) (1 ) r n t r - = - = 2 .60 23 3.597 (1 .60 ) t = = - t 23,.05 2.069, t 23,.01 2.807. Therefore, reject H 0 at the 1% level. 12.3 Let x = Examination and y = Project x y ( ) i x x - 2 ( ) i x x - ( ) i y y - 2 ( ) i y y - ( )( ) i i x x y y - - 81 76 2.4 5.76 -0.7 0.49 -1.68 62 71 -16.6 275.56 -5.7 32.49 94.62 74 69 -4.6 21.16 -7.7 59.29 35.42 78 76 -0.6 0.36 -0.7 0.49 0.42 93 87 14.4 207.36 10.3 106.09 148.32 69 62 -9.6 92.16 -14.7 216.09 141.12 72 80 -6.6 43.56 3.3 10.89 -21.78 83 75 4.4 19.36 -1.7 2.89 -7.48 90 92 11.4 129.96 15.3 234.09 174.42 84 79 5.4 29.16 2.3 5.29 12.42 786 767 824.4 668.1 575.8 x = 78.6, y = 76.7, 2 ( ) 1 i x x x s n - = - = 824.4 9 = 9.5708, 2 ( ) 1 i y y y s n - = - = 668.1 9 = 8.6159, ( )( ) 1 i i xy x x y y s n - - = - = 575.8/9 = 63.97778, xy x y s r s s = = 63.97778/(9.5708)(8.6159) = .7759 12.4 1 : 0, : 0 o H H ρ ρ = , 2 ( 2) (1 ) r n t r - = - = 2 .75 47 7.7736 (1 .75 ) t = = -

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231 Statistics for Business & Economics, 6 th edition t 47,.05 1.684, t 47,.01 2.423. Therefore, reject H 0 at all common levels of alpha 12.5 1 : 0, : 0 o H H ρ ρ = , 2 ( 2) (1 ) r n t r - = - = 2 .11 351 2.073 1 (.11 ) t = = - Therefore, reject H 0 at the 2.5% level since 2.073 > 1.96 = z .025 t 351,.025 12.6 1 : 0, : 0 o H H ρ ρ = , 2 .51 66 4.8168 (1 .51 ) t = = - Therefore, reject H 0 at the 5% level since 4.8168 > 1.671 t 66,.05 12.7 a. Using the computer, calculate the sample correlation Correlations: Dow5day, Dow1Yr Pearson correlation of Dow5day and Dow1Yr = -0.407 P-Value = 0.168 b. 1 : 0, : 0 o H H ρ ρ = , 2 .4066 11 1.4761 1 ( .4066) t - = = - - - t 11,.05 = ±1.796, p-value of .168 > alpha of .10. Do not reject H 0 at the 10% level 12.8 a. Let x = Instructor Rating and y = Expected Grade.
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