chapter 1

Chapter 12: Simple Linear Regression and Correlation22.− 404.325a. βˆ1= = −.00736023, βˆ0=1. 41122185 ,54.933.75SSE = 7 .8518 − ( 1.41122185)( 10.68) − ( −.00736023)( 987.654) = . 049245 ,2 .049245s = = .003788 , and σ ˆ = s = . 0615513b.( 10.68)22 .049245SST = 7.8518 − = .24764 so r = 1 − = 1 − .199 = . 80115.2476423.a. Using the y i' s given to one decimal place accuracy is the answer to Exercise 19,SSE =formula gives22( 150 − 125.6) + ... + ( 670 − 639.0) = 16,213. 64. The computation( − 45.55190543)( 5010) ( 1.71143233)( 1,413,500)SSE = 2,207,100−−= 16,205.45b.( 5010)22 16,205.45SST = 2,207,100 − = 414,235.71 so = 1−= . 96114414,235.71r .24.a.1200y7002000 50 100364xAccording to the scatter plot of the data, a simple linear regression model does appear tobe plausible.b. The regression equation is y = 138 + 9.31 xc. The desired value is the coefficient of determination, r 2 = 99.0%.d. The new equation is y* = 190 + 7.55 x*. This new equation appears to differsignificantly. If we were to predict a value of y * for x * = 50, the value would be 567.9,where using the original data, the predicted value for x = 50 would be 603.5.