Solutions to Exam 1, Math 382, Spr05

Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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1. (66 pts.) Forced expiratory volume (FEV) - the amount of air, in liters, expelled during 1 second during a forceful exhalation, was fit to the following variables . age, m years . height, in inches . male?, 1 if male and 0 if female . smoke?, 1 if a smoker, 0 if a non-smoker for 654 young people. Below is the associated (edited) Minitab output for fitting the underlying model FEV = Po +Pl (age) + P2{height) +P3{male?) + P4{smoke?) Regression Analysis The regression equation is FEV = - 4.46 + 0.0655 age + 0.104 height + 0.157 male? - 0.0872 . smoke? ... Predictor Coef StDev. T P 4 Constant -4.4570 0.2228 A 0.000 age 0.065509 0.009489 * 0.000 height 0.104199 0.004758 * 0.000 male? 0.15710 B * * smoke? -0.08725 * -1.47 C S = 0.4122 R-Sq = 77.5% R-Sq(adj) = 77.4% Analysis of Variance Source DF SS MS F P Regression 4 380.640 95.160 560.02 0.000 Residual Error 649 110.280 0.170 Total 653 490.920 a. What is the typical size, in liters, of the difference between the predicted and actual FEV s for the young people in this study? 5.::: c> . 1/2 2.. / i j.e,,:s b. One of the 654 young people in the study is a 14 year-old girl who smokes and is 66 inchestall. Find her residual ifher FEV was recorded as2.236 liters. Show your work. A F£v :: - t. tb .J- 0. 06"!i!i( If) + 0, loft bt::) -t D./.>?( 0) -. () 372-(1) ~ J..2fD (,'M A. . fL~5IPc)~ '::: f 6V - FE: V ': 1. .2."'; ~ - ), 'l-t; 0 = - I.plt , (~ 2

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c. (True o~ We need to assume that the errors are independent normals with the same variance to conclude that Po = -4.4570, PI = 0.065509, A ::: 0.104199"P3 = 0.15710,P4 = -0.08725 arethe least squares estimates of Po, PI' P2' P3' P4.
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