chapter 1

Chapter 13: Nonlinear and Multiple Regression6.a. : β = 10H vs. H : β ≠ 10oa. The test statistic isβˆ1t =st ≥ t . 025,4 = 2.776 or if t ≤ −2. 776 .s 7.265s ˆ = = = . 565 , andβ 112.869S xxβˆ1, and we will reject H o if6.19268t = = 10.97 . Since 10.97≥ 2. 776 , we reject H o and conclude that the model.565is useful.ˆ7 . 0=− yˆ 7.0= 1046 −1051.49= −5.b. y ( ) = 1008.14 + 6.19268( 7.0) 1051. 49, from which the residual isy ( ) 49 . Similarly, the other residuals are -.73, 4.11,7.91, 3.58, and –9.38. The plot of the residuals vs x follows:10RESI10-101015x20Because a curved pattern appears, a linear regression function may be inappropriate.c. The standardized residuals are calculated as− 5.49e * == −1.0741, and similarly the others are -.123,21 ( 7.0 −14.48)7.265 1++6 165.5983.624, 1.208, .587, and –1.841. The plot of e* vs x follows :1SRES10-1-210x1520This plot gives the same information as the previous plot. No values are exceptionallylarge, but the e* of –1.841 is close to 2 std deviations away from the expected value of 0.396