chapter 1

Chapter 13: Nonlinear and Multiple Regression19.a. No, there is definite curvature in the plot.′ x′b. = β + β ( ) + εY10where1x′ = and y ′ = ln( lifetime). Plotting y′ vs.tempx′ gives a plot which has a pronounced linear appearance (and in fact r 2 = .954 for thestraight line fit).c. x ′ = . 082273 , y ′ = 123. 64 , x ′ 2 = . 00037813 , ′ 2 = 879. 88Σ iΣ iΣ iΣ iy ,Σx ′ ′iyi= .57295 , from which βˆ1= 3735. 4485 andˆ0= −10. 2045from computer output). With x = 220, x ′ = . 00445 soˆy ˆ ′ = −10.2045+3735.4485 .00445 = 6. and thus ˆ = ey′= 875. 50( ) 7748d. For the transformed data, SSE = 1.39857, and = n = n 6,y ′ 6.83157 , ′ 5. 32891= 2 .= 3 .1 2 3=β (values ready .n ′ 8. 44695y ,= 1 .y , from which SSPE = 1.36594, SSLF = .02993,.02993 /1f == .33 . Comparing this to F. 01,1,15= 8. 681.36594 /15be rejected., it is clear that H o cannot20. After examining a scatter plot and a residual plot for each of the five suggested models as wellas for y vs. x, I felt that the power model Y = αxβ ⋅ ε ( y ′ = ln( y)vs.x ′ = ln( x)) provided the bet fit. The transformation seemed to remove most of the curvaturefrom the scatter plot, the residual plot appeared quite random, e * < 1. 65 for every i, therewas no indication of any influential observations, and r 2 = .785 for the transformed data.′i21.410Y ′0 1where x′ = . The summaryxx ′ = 159. , y =121. 50 , x ′ 2 = 4058. 8 , 2 = 1865. 2a. The suggested model is = β + β ( x ) + εΣ iquantities are 01Σ iΣ iΣx ′ y = 2281.6 , from which βˆ1= −. 1485 and βˆ0=18. 1391iiregression function isy148518.1391−x= .Σ iy ,, and the estimatedˆ1485b. x = 500 ⇒ y = 18.1391 − = 15. 17 .500406