chapter 1

Chapter 13: Nonlinear and Multiple Regressione. A computer generated residual analysis:Residual Model DiagnosticsNormal Plot of ResidualsI Chart of ResidualsRe sidua l210-1Resid ual3210-1-23. 0SL=2.820X=-0. 06771-2-1.5 - 1.0 -0.5 0. 0 0.5 1.0 1. 5No rmal Sco re-31 2 3 4 5 6 7 8Observati on Nu mbe r-3 .0SL=-2 .956Histogram of ResidualsResiduals vs. Fits32Freq uen cy21Residua l10- 10-1.5 - 1.0 -0. 5 0.0 0.5 1.0 1.5 2.0Resid ual- 20 1 2 3FitLooking at the residual vs. fits (bottom right), one standardized residual, corresponding tothe third observation, is a bit large. There are only two positive standardized residuals,but two others are essentially 0. The patterns in the residual plot and the normalprobability plot (upper left) are marginally acceptable.16.Σx i, Σy i′ = 313. 10 , Σx 2 i= 8. 0976 , Σy ′ 2 i= 288, 013 ,Σx ′iyi= 255.11 , (all from computer printout, where y i= ln ( L 178)a. = 9. 72βˆ1= 6.6667 andˆ0= 20. 6917′ ), from whichβ (again from computer output). Thusβ ˆ = β ˆ = 16.6667 and ˆ 0α = e βˆ= 968,927, 163 .b. We first predict y′ using the linear model and then exponentiate:( ) 25. 6917y ′ = 20 .6917 + 6.6667 .75 =25.6917ˆ = Lˆ= e = 1.438051363×y .1117810c. We first compute a prediction interval for the transformed data and then exponentiate.With t 2. 228 , s = .5946, and. 025,10=prediction interval for y′ is112, so(.95− x )2− ( Σx)21++2Σx= 1.082/12, the( 2.228)( .5496)( 1.082) = 27.0251±1.4334 ( 25.5917,28.4585)27 .0251=± .25.5917 28.4585The P.I. for y is then ( ,e )e .404