chapter 1

Chapter 12: Simple Linear Regression and Correlation37.a. n = 10, = 2615Σ ix , y = 39. 20 , x 2 = 860, 675 , 2 = 161. 94Σ iΣ iΣ iy ,Σ y12,027xi i= 11,453.5 , so βˆ1= = . 00680058 , βˆ0= 2. 141647701,768,525which SSE = .09696713, s = .11009492 s = . 11009492 = &.110= σˆ ,.110σ ˆ =.000262ˆ 1176,852=β, fromb. We wish to test : β = . 10060either ≥ t .= 1. 86005,8H vs H : β ≠ . 10060ot or t ≤ −t= 1. 860a. 05,8−.0068 − .0060t == 3.06 ≥ 1.1860.000262. Since, H o is rejected.. At level .10, H o is rejected if38.a. From Exercise 23, which also refers to Exercise 19, SSE = 16.205.45, so36.75s 2 = 1350.454 , s = 36. 75 , and sˆ= = . 0997β1368.6361.711t = 17.2 > 4.318 = t.0005,14.0997: β = 0 1. Thus= , so p-value < .001. Because the p-value < .01,H is rejected at level .01 in favor of the conclusion that the model is usefulo( )1≠ 0β .b. The C.I. for1the C.I. for1β is 1 .711 ( 2.179)( .0997) = 1.711±.217 = ( 1.494,1.928)10β is ( 14 .94,19.28).± . Thus39. SSE = 124,039.58– (72.958547)(1574.8) – (.04103377)(222657.88) = 7.9679, and SST =39.828Source df SS MS fRegr 1 31.860 31.860 18.0Error 18 7.968 1.77Total 19 39.828F , so : β = 10Let’s use α = .001. Then = 15.38 18. 0. 001,1,18