chapter 1

Chapter 12: Simple Linear Regression and Correlation8.µ , and = 350a. = 1800 + 1.3( 2000) 4400Y⋅2000 =⎛ 5000 − 4400 ⎞= P ⎜Z>⎟ = P Z⎝ 350 ⎠( > 1.71) = . 0436σ , so P ( Y > 5000)b. Now E(Y) = 5050, so P ( Y > 5000 ) = P( Z > .14) = . 4443c. E Y −Y) = E(Y ) − E(Y ) = 5050 − 4400 650V(2 12 1=22( Y2 − Y1) = V ( Y2) + V ( Y1) = 350 + 350 =2−Y1= 494.97Y .⎛, and( ) ( ) 245, 000100 − 650 ⎞⎟494.97 ⎠Thus P Y − Y > 0) = P z > = P( Z > .71) . 2389( 2 1 ⎜=d. The standard deviation of Y −Y 494. 97 (from c), and⎝2 1=2− Y1)= 1800 + 1.3x2−3( 1800 + 1.3x1) = 1. ( x2−1)−1.3( x − x )E( Yx, so the s.d. of. Thus⎛2 1 ⎞P ( Y2 > Y1) = P(Y2− Y1> 0) = P⎜z >⎟ = .95 implies that⎝ 494.97 ⎠−1.3( x2 − x )− 1.645 =1, so x2− x1= 626. 33 .494.979.a. β1= expected change in flow rate (y) associated with a one inch increase in pressuredrop (x) = .095.b. We expect flow rate to decrease by 5β 1= . 475 .c. µ = −.12+ .095( 10) .83,and = −.12+ .095( 15) 1. 305Y⋅10 =356µ .Y⋅15 =⎛ .835 − .830 ⎞⎜⎟⎝ .025 ⎠> . 840⎛ .840 − .830 ⎞= P⎜Z > ⎟ = P Z > .40⎝ .025 ⎠= .d. P ( Y > . 835) = P Z >= P( Z > .20) = . 4207P( Y ) ( ) 3446e. Let Y 1 and Y 2 denote pressure drops for flow rates of 10 and 11, respectively. Thenµ so Y 1 - Y 2 has expected value .830 - .925 = -.095, and s.d.PY⋅11 = .925,22(.025) (.025) = . 035355+ . Thus⎛ + .095 ⎞> Y2) = P(Y1− Y > 0) = P⎜z > ⎟ = P Z⎝ .035355 ⎠( > 2.69) . 0036( Y1 2=