chapter 1

Chapter 13: Nonlinear and Multiple Regression− .04304Ha, with t = = −2.428.01773f. We test : β ≠ 30. The p-value is approximately2(.012) = .024. At significance level .01 we do not reject H o . The interaction termshould not be retained.45.a. The appropriate hypotheses are β = β = β = β 0βi≠ 0. The test statistic isfH vs. :0:1 2 3 4=2R=−n−k−1.9464H at least onek2( )= = 87.6 ≥ 7.10 = F1−R( 1 .946) .001,4, 20( )smallest available significance level from Table A.9), so we can reject H o at anysignificance level. We conclude that at least one of the four predictor variables appearsto provide useful information about tenacity.20a(theb. The adjusted R 2 value is241 −=20( 1 − .946) . 935n1−Rn − ⎝ ⎠−1⎛ SSE ⎞ n −1( ) ( ) ( 2⎜ ⎟ = 1 − 1 − )k + 1 SST n − k + 1= , which does not differ much from R 2 = .946.c. The estimated average tenacity when x 1 = 16.5, x 2 = 50, x 3 = 3, and x 4 = 5 isyˆ= 6.121 − .082x+ .113x+ .256 x − . 219 xy ˆ = 6.121 − .082( 16.5) + .113( 50) + .256( 3) − .219( 5) = 10. 091 . For a 99% C.I.,2.84510 .091 ± 2.845 .350 = 9.095,11.087 . Therefore,t , so the interval is ( ) ( ). 005,20 =when the four predictors are as specified in this problem, the true average tenacity isestimated to be between 9.095 and 11.087.46.a. Yes, there does appear to be a useful linear relationship between repair time and the twomodel predictors. We determine this by conducting a model utility test:H β = β 0 vs. H : at least one β ≠ 0. We reject H o if f ≥ F . = 4. 26 .0:1 2=The calculated statistic isfa=SSRkSSE( n−k−1)=iMSRMSE=10.632( 20.9)9=05,2,95.315= 22.91. Since.23222.91≥ 4.26 , we reject H o and conclude that at least one of the two predictor variablesis useful.b. We will reject H β 0 in favor of H : β 0 if t ≥ t 3.. = 25 . The test: =0 2astatistic is 1.250t = = 4. 01 which is ≥ 3. 25, so we reject H o and conclude that the “type of.312repair” variable does provide useful information about repair time, given that the“elapsed time since the last service” variable remains in the model.2 ≠005,9419