chapter 1

Chapter 13: Nonlinear and Multiple Regression72.2 SSE .80017a. R = 1−= 1 − = . 9506SST 16.18555weld strength can be attributed to the given model., or 95.06% of the observed variation inb. The complete second order model consists of nine predictors and nine correspondingcoefficients. The hypotheses are H β = ... = β 0 vs. H : at least one of theβi' s ≠ 0 . The test statistic is0:19=2Rkf =2( 1−R ) ( n−k− 1)region is f ≥ F .= 2. 25 . The calculated statistic is05,9,27a, where k = 9, and n = 37.The rejectionf= −.95069( 1 .9506)27= 57.68which is ≥ 2. 25 , so we reject the null hypothesis. The complete second order model isuseful.c. To test β 0H vs : β ≠ 700:7=predictor), = f = 2 .32 = 1. 52H (the coefficient corresponding to the wc*wtat . With df = 27, the p-value ≈ 2 (.073) = . 146(from Table A.8). With such a large p-value, this predictor is not useful in the presenceof all the others, so it can be eliminated.2d. The point estimate is y ˆ = 3.352 + .098( 10) + .222( 12) + .297( 6) −.0102( 10 )2− .037( 6 ) + .0128( 10)( 12) = 7. 962 . With t 2. 052. 025,27=( ) = 7.962 ± .154 ( 7.808,8.116)7 .962 2.052 .0750=, the 95% P.I. would be± . Because of thenarrowness of the interval, it appears that the value of strength can be accuratelypredicted.73.a. We wish to test β = β 0is2RH vs. :0:1 2=Haeither1β or β 02≠kf =2( 1−R ) ( n−k− 1)2 .29f ≥ Fα , k , n−k−1= F.01,2,5= 13.27 . R = 1−= . 9986202.88. The test statistic, where k = 2 for the quadratic model. The rejection region isdoubt about it, folks – the quadratic model is useful!, giving f = 1783. Nob. The relevant hypotheses are β 0βˆβ ˆ 2H vs. H : β 00:2=a2≠2t = , and H o will be rejected at level .001 if either ≥ 6. 869s− .00163141. The test statistic value ist or ≤ −6. 869t (df= n – 3 = 5). Since t == −48.1≤ −6.869 , H o is rejected. The.00003391quadratic predictor should be retained.434