mont4e_sm_ch11_sec09 - Section 11-9 11-75 a b c d Yes No...

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Unformatted text preview: Section 11-9 11-75 a) b) c) d) Yes, No Yes, Yes, ln y = ln 0 + 1 ln x + ln ln y = ln 0 + x ln 1 + ln 1 1 = 0 + 1 + y x 11-76 a) There is curvature in the data. 800 Vapor Pressure (mm Hg) 700 600 500 400 300 200 100 0 280 330 380 Tem p erature (K) 11-45 15 10 y 5 0 0 500 1000 1500 2000 2500 3000 3500 x b) y = - 1956.3 + 6.686 x c) Source Regression Residual Error Total d) There is a curve in the residuals. Residuals Versus the Fitted Values (response is Vapor Pr) DF 1 9 10 SS 491662 124403 616065 MS 491662 13823 F 35.57 P 0.000 200 100 Residual 0 -100 -200 -100 0 100 200 300 400 500 600 Fitted Value e) The data are linear after the transformation to y* = ln y and x* = 1/x. 7 6 5 Ln(VP) 4 3 2 1 0 .0 0 2 7 0 .0 03 2 0 .0 0 3 7 1/T ln y = 20.6 - 5201(1/x) 11-46 Analysis of Variance Source Regression Residual Error Total DF 1 9 10 SS 28.511 0.004 28.515 MS 28.511 0.000 F 66715.47 P 0.000 Residuals Versus the Fitted Values (response is y*) 0.02 0.01 Residual 0.00 -0.01 -0.02 -0.03 1 2 3 4 5 6 7 Fitted Value There is still curvature in the data, but now the plot is convex instead of concave. 11-77 a) 15 10 y 5 0 0 500 1000 1500 2000 2500 3000 3500 x b) c) ^ y = -0.8819 + 0.00385 x H 0 : 1 = 0 H1 : 1 0 = 0.05 f 0 = 122.03 f 0 > f 0.05,1, 48 Reject H0. Conclude that regression model is significant at = 0.05 d) No, it seems the variance is not constant, there is a funnel shape. 11-47 Residuals Versus the Fitted Values (response is y) 3 2 1 0 Residual -1 -2 -3 -4 -5 0 5 10 Fitted Value e) ^ y = 0.5967 + 0.00097 x . Yes, the transformation stabilizes the variance. 11-48 ...
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