chapter 1

Chapter 12: Simple Linear Regression and Correlation73.22a. n = 9, Σx = 228, Σx= 5958, Σy= 93.76, Σy= 982. 2932 andiiiΣ y− 243.93xi i= 2348.15 , giving βˆ1= = − . 148919 , βˆ0= 14. 1903921638the equation yˆ= 14.19 − (.1489)x.i, andb. β1is the expected increase in load associated with a one-day age increase (so a negativevalue of β1corresponds to a decrease). We wish to test H0: β1= −.10 vs.H β < .10 (the alternative contradicts prior belief). H o will be rejected at level.05 if0:1−( − .10)βˆ1−t = ≤ −tsβˆ1.05.7= −1.895.4608− .1489 + 1sˆ= = .0342 . Thus t = = −1.43β1182.0342≤ −1.895 , do not reject H o .x so ( − x )c. Σ = 306, Σx2 = 7946,ii∑contrasted with 182 for the given 9x values is larger, the original set of values is preferable.d. ( t )( s)( 28 − 25.33)2. With SSE = 1.4862, s = .4608, and. Because –1.43 is not( 306)x i= 7946 − = 143 here, as12x i' s . Even though the sample size for the proposed2. 025,7==1 9+( 2.365)( .4608)(.3877) . 429 1638β ˆ + βˆso the 95% CI is 10 .02 ± .42 = ( 9.60, 10.44).( 28) 10.02,0 1=2, and74.3.5979a. βˆ1= = . 0805 , βˆ0=1. 6939 , yˆ 1.69 + (.0805)x44.713= .3.5979b. βˆ1= = 12. 2254 , βˆ0= −20. 4046 , yˆ −20.40+ ( 12. 2254)x.2943c. r = .992, so r 2 = .984 for either regression.= .386