mont4e_sm_ch12_supplemental

# mont4e_sm_ch12_supplemental - Supplemental Exercises 12-88...

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Supplemental Exercises 12-88. a) 5 4 3 6887 . 1 485 . 21 231 . 0 4203 ˆ x x x y + + = b) H 03 4 5 0 : βββ === H j 1 0 : β≠ for at least one j α = 0.01 38 . 4 25 . 1651 36 , 3 , 01 . 0 = = f f Reject H 0 and conclude that regression is significant. P-value < 0.00001 c) All at α = 0.01 t ., . 005 36 272 = 0 : 3 0 = β H H 04 0 : β = H 05 0 : β = H 13 0 : H 14 0 : β H 15 0 : β 06 . 2 0 = t 91 . 22 0 = t 00 . 3 0 = t || /, tt 02 /> α 3 6 3 6 3 6 > α > α Do not reject H 0 Reject H 0 Reject H 0 d) R Adj. R 2 0 993 = . 2 0 9925 = . e) Normality assumption appears reasonable. However there is a gap in the line. -100 0 100 -2 -1 0 1 2 Normal Score Residual Normal Probability Plot of the Residuals (response is y) f) Plot is satisfactory. 3000 4000 5000 -100 0 100 Fitted Value Residual Residuals Versus the Fitted Values (response is y) 12-61

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g) Slight indication that variance increases as x 3 increases. 30000 29500 29000 100 0 -100 x3 Residual Residuals Versus x3 (response is y) h) 89 . 3862 ) 1589 ( 6887 . 1 ) 170 ( 485 . 21 ) 28900 ( 231 . 0 4203 ˆ = + + = y 12-89. a) H 03 4 5 0 : ββ β === H j 1 0 : β≠ for at least one j α = 0.01 36 , 3 , 0 36 , 3 , 01 . 0 38 . 4 39 . 1321 α f f f f >> = = Reject H 0 and conclude that regression is significant. P-value < 0.00001 b) α = 0.01 t ., . 005 36 272 = H 0 : β = H 04 0 : β = H 05 0 : β = H 13 0 : β H 14 0 : β H 15 0 : β 45 . 1 0 = t 95 . 19 0 = t 53 . 2 0 = t || /, tt 02 /> α 3 6 3 6 3 6 > α α Do not reject H 0 Reject H 0 Do not reject H 0 c) Curvature is evident in the residuals plots from this model. Residual Percent 0.03 0.02 0.01 0.00 -0.01 -0.02 -0.03 99 95 90 80 70 60 50 40 30 20 10 5 1 Normal Probability Plot of the Residuals (response is y*) 12-62
Fitted Value Residual 8.5 8.4 8.3 8.2 8.1 8.0 0.03 0.02 0.01 0.00 -0.01 -0.02 Residuals Versus the Fitted Values (response is y*) x3* 10.32 10.31 10.30 10.29 10.28 10.27 10.26 0.03 0.02 0.01 0.00 -0.01 -0.02 Residuals Versus x3* (response is y*) 12-90. From data in table 12-6 a) 5 4 3 2 1 005 . 0 594 . 0 454 . 0 2206 . 0 291 . 0 86 . 2 ˆ x x x x x y + + + = 0 : 5 4 3 2 1 0 = = = = = β H 0 one least at : 1 j H α = 0.01 19 , 5 , 0 19 , 5 , 01 . 0 17 . 4 81 . 4 α f f f f > = = Reject H 0 . P -value = 0.005 b) α = 0.05 093 . 2 19 , 025 . = t 0 : 1 0 = H H 02 0 : β= H 03 0 : β = H 04 0 : β = H 05 0 : β = H 11 0 : β≠ H 12 0 : H 13 0 : β H 14 0 : β H 15 0 : β 47 . 2 0 = t 74 . 2 0 = t 42 . 2 0 = t 80 . 2 0 = t 26 . 0 0 = t || /, tt 1 > α 1 9 > α 1 9 1 9 1 9 9 > α > α /> α Reject H 0 Reject H 0 Reject H 0 Reject H 0 Do not reject H 0 c) 4 3 2 1 609 . 0 455 . 0 19919 . 0 290 . 0 148 . 3 ˆ x x x x y + + = 0 : 4 3 2 1 0 = = = = H 0 : 1 j H for at least one j α = 0.05 20 , 4 , 0 20 , 4 , 05 . 0 87 . 2 28 . 6 f f f f > = = Reject H 0 . α = 0.05 t ., . 025 20 2 086 = H 01 0 : H 0 : H 0 : β = H 0 : β = H 0 : H 0 : H 0 : β H 0 : β 53 . 2 0 = t 89 . 2 0 = t 49 . 2 0 = t 05 . 3 0 = t 2 > α 2 0 > α 2 0 2 0 0 > α > α Reject H 0 Reject H 0 Reject H 0 Reject H 0 d) The addition of the 5 th regressor results in a loss of one degree of freedom in the denominator and the reduction in SS E is not enough to compensate for this loss.

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mont4e_sm_ch12_supplemental - Supplemental Exercises 12-88...

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