Unformatted text preview: Section 119 1175 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 1176 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) 1145 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) 1146 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. 1177 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. 1147 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. 1148 ...
View
Full Document
 Spring '09
 KAILASHKAPUR
 Linear Regression, Normal Distribution, Regression Analysis, Errors and residuals in statistics, Residual Error Total

Click to edit the document details