chapter 1

Chapter 13: Nonlinear and Multiple Regressionc. When x 1 = 8.0 and x 2 = 33.1 the residual is e = 2.71 and the standardized residual is e* =.44; since e* = e/(sd of the residual), sd of residual = e/e* = 6.16. Thus the estimatedvariance of Yˆ is ( 6.99) − ( 6.16) = 10. 915y ˆ = 24.29 and t 2. 3062. 025,8=24 .29 2.306 3.304 =d. 4. 07. 05,3,8=2, the desired C.I. is( ) ( 16.67,31.91)± ., so the estimated sd is 3.304. SinceF , so H β = β = β 0 will be rejected if f ≥ 4. 070:3 4 5=SSE 8,2 k= s = 390.88, and f =≥ 4.07. With( 894.95−390.88)3( )= 3. 44 , and since 3.44 is not390.88, H o cannot be rejected and the quadratic terms should all be deleted. (n.b.: this isnot a modification which would be suggested by a residual plot.852.a. The complete 2 nd order model obviously provides a better fit, so there is a need toaccount for interaction between the three predictors.b. A 95% CI for y when x 1 =x 2 =30 and x 3 =10 is. 66573 ± 2.120 .01785 =( ) (.6279,.7036)53. Some possible questions might be:Is this model useful in predicting deposition of poly-aromatic hydrocarbons? A test of modelutility gives us an F = 84.39, with a p-value of 0.000. Thus, the model is useful.Is x 1 a significant predictor of y while holding x 2 constant? A test of H β 0 vs the0:1=two-tailed alternative gives us a t = 6.98 with a p-value of 0.000., so this predictor issignificant.A similar question, and solution for testing x 2 as a predictor yields a similar conclusion: Witha p-value of 0.046, we would accept this predictor as significant if our significance levelwere anything larger than 0.046.54.a. For = x = x = x = 1x , ˆ = 84.67 + .650 − .258 + ... + .050 = 85. 3901 2 3 4+The single y corresponding to thesey − yˆ = 85.4 − 85.390 = .010 .y .xivalues is 85.4, sox 1,...,x ′ ′ denote the uncoded variables, x ′1= . 1x1+ .3,x ′2= . 1x2+ .3,x ′3= x3+ 2.5, and x 4′ = 15x4+ 160 ; Substitution of x 1= 10x′1−3,4+ 160x 2= 10x′2−3,x3= x′3− 2.5,and4= x′15b. Letting4x yields the uncoded function.423