chapter 1

Chapter 12: Simple Linear Regression and Correlation17. Note: n = 23 in this study.a. For a one (mg /cm 2 ) increase in dissolved material, one would expect a .144 (g/l) increasein calcium content. Secondly, 86% of the observed variation in calcium content can beattributed to the simple linear regression relationship between calcium content anddissolved material.b. µ = 3.678 + .144( 50) 10. 878y⋅50 =c.r2SSE= .86 = 1 −SSTSSEs = = n − 2Then = 1. 46, so = ( SST )( 1 −.86) = ( 320.398)( .14) = 44. 85572SSE .44.855722118.a.( 987.645) − ( 1425)( 10.68)2( ) − ( 1425)− ( − .00736023)( 1425)ˆ 15− 404.3250β1=== −.0073602315 139,037.2554,933.750010.68βˆ0== 1.41122185 , y 1.4112− . 007360x15= .b. βˆ1 = −. 00736023c. With x now denoting temperature in C⎛ 9 ⎞ο , y = β ˆ + ˆ0 β1⎜x + 32⎟ ⎝ 5 ⎠( ˆ 9+ 32βˆ) + ˆ x = 1.175695 . 0132484xβ , so the new ˆβ1is -.0132484 and=1β5ˆ0=1.0 1−the new β 175695 .d. Using the equation of a, predicted y = β + β ( 200) = . 0608factor cannot be negative.ˆˆ0 1−, but the deflection362