chapter 1

Chapter 12: Simple Linear Regression and CorrelationΣx i, Σy i= 5. 37 , Σx 2 i=12. 000 , Σy 2 i= 9. 3501,2Σ x y( )i i= 333 . 200Sxx= 12,000 − = 2000 ,42( 5.37)( 200)( 5.37)Syy= 9.3501 − = 2.140875 , andxy= 333 −= 64.544S 64.5ˆ1 xyβ = = .03225S 2000=5.37 200and βˆ0= − (.03225) = −.27000 .4413. For this data, n = 4, = 200xxS .(.03225)( 64.5) . 060750SSE S −ˆ1 = 2.14085 −=yyβ S.r2=xySSE= 1 −SST.060750= 1 − = .9722.14085. This is a very high value of2r , which confirmsthe authors’ claim that there is a strong linear relationship between the two variables.14.a. n = 24, Σx i= 4308 , Σy i= 40. 09 , Σx 2 i= 773, 790 , Σy 2 i= 76. 8823 ,2Σ x y( )i i= 7,243.65 . 4308Sxx= 773,790 − = 504.0 ,242( 40.09)S yy = 76.8823 − = 9.9153 , and24( 4308 )( 40.09 )SSxy= 7 ,243.65 −= 45.8246 .45.8246. 0909224ˆ1 xyβ = =S 504= andxx40.09 4308βˆ0= − (.09092) = −14.6497 . The equation of the estimated regression2424line is yˆ = −14.6497+ . 09092x.b. When x = 182, y ˆ = −14.6497+ .09092( 182) = 1. 8997temperature is 182, we would predict an efficiency ratio of 1.8997.359. So when the tankc. The four observations for which temperature is 182 are: (182, .90), (182, 1.81), (182,1.94), and (182, 2.68). Their corresponding residuals are: . 90 − 1.8997 = −0.9977 ,1.81− 1.8997 = −0.0877, 1 .94 − 1.8997 = 0. 0423 , 2 .68 − 1.8997 = 0. 7823 .These residuals do not all have the same sign because in the cases of the first two pairs ofobservations, the observed efficiency ratios were smaller than the predicted value of1.8997. Whereas, in the cases of the last two pairs of observations, the observedefficiency ratios were larger than the predicted value.d. S ˆ1 9.9153 (.09092)( 45.8246) =yy− S = −5. 7489SSE β .r2=xySSE= 1 −SST5.7489= 1 − = .42029.9153. (42.02% of the observed variation inefficiency ratio can be attributed to the approximate linear relationship between theefficiency ratio and the tank temperature.)