chapter 1

Chapter 12: Simple Linear Regression and Correlation69.a. The test statistic value istβˆ1−1= , and H o will be rejected if eithersβ ˆ 1t ≥ t . 025,11= 2.201 or t ≤ −2. 201. With22Σx = 243, Σx= 5965, Σy= 241, Σy= 5731 and Σ = 5805iiiix ,βˆ1= .913819 , βˆ0= 1. 457072 , SSE = 75. 126 , s = 2. 613 , and . 0693y i is ,ˆ =β1.9138 −1t = = −1.24. Because –1.24 is neither ≤ −2. 201 nor ≥ 2. 201.0693be rejected. It is plausible that β 1.1=, H o cannotb.r=16,902136( )( 128.15)= .97070.a. sample size = 8b. yˆ= 326.976038 − ( 8. 403964)x. When x = 35.5, ˆ = 28. 64y .c. Yes, the model utility test is statistically significant at the level .01.2d. r = r = 0.9134 = 0. 9557e. First check to see if the value x = 40 falls within the range of x values used to generatethe least-squares regression equation. If it does not, this equation should not be used.Furthermore, for this particular model an x value of 40 yields a g value of –9.18, which isan impossible value for y.71.a. r2 = . 5073r (positive because ˆβ1is positive.)2b. = + r = .5073 = . 7122c. We test test β 0H vs β 00:1=H . The test statistic t = 3.93 gives p-value =0:1≠.0013, which is < .01, the given level of significance, therefore we reject H o and concludethat the model is useful.384