chapter 1

Chapter 12: Simple Linear Regression and Correlation32. Let β1denote the true average change in runoff for each 1 m 3 increase in rainfall. To test thehypotheses H : β = 10 vs. H : β ≠ 10 , the calculated t statistic isoaβˆ1.82697t = = = 22.64 which (from the printout) has an associated p-value of P =s .03652βˆ10.000. Therefore, since the p-value is so small, H o is rejected and we conclude that there is auseful linear relationship between runoff and rainfall.A confidence interval for β1is based on n – 2 = 15 – 2 = 13 degrees of freedom.t. 025,13=2.160, so the interval estimate is( 2.160)( .03652) (.748,.906)βˆ1t.025,13⋅ s ˆ = .82697 ±=±β1. Therefore, we can beconfident that the true average change in runoff, for each 1 m 3 increase in rainfall, issomewhere between .748 m 3 and .906 m 3 .33.a. From the printout in Exercise 15, the error d.f. = n – 2 = 25, t 2. 060 . Theconfidence interval is then. 025,25=( 2.060)( .01280) (.081,.134)βˆ1t.025,25⋅sˆ = .10748 ±=±β1. Therefore, weestimate with a high degree of confidence that the true average change in strengthassociated with a 1 Gpa increase in modulus of elasticity is between .081 MPa and .134MPa.b. We wish to test : β = . 11Hovs. H : β > .a 11βˆ1− .1 .10748 − .1t = = = .58s .01280βˆ1. The calculated t statistic is, which yields a p-value of .277. A large p-valuesuch as this would not lead to rejecting H o , so there is not enough evidence to contradictthe prior belief.34.a. : β = 10H ; H : β ≠ 10ot > tα nor t > 3. 106t = 5.29RR:/ 2, −2a: Reject H o . The slope differs significantly from 0, and the model appears tobe useful.b. At the level α = 0. 01, reject h o if the p-value is less than 0.01. In this case, the reportedp-value was 0.000, therefore reject H o . The conclusion is the same as that of part a.c. : β = 1. 15H ; H : β