# Chpt11 - Chapter 11 Inferential Methods in Regression and...

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Chapter 11 Inferential Methods in Regression and Correlation Section 11.1 1. (a) The slope of the estimated regression line ( β = .095) is the expected change of in the response variable y for each one-unit increase in the x variable. This, of course, is just the usual interpretation of the slope of a straight line. Since x is measured in inches, a one-unit increase in x corresponds to a one- inch increase in pressure drop. Therefore, the expected change in flow rate is .095 m 3 /min. (b) When the pressure drop, x, changes from 10 inches to 15 inches, then a 5 unit increase in x has occurred. Therefore, using the definition of the slope from (a), we expect about a 5(.095) = .475 m 3 /min. increase in flow rate (it is an increase since the sign of β = .095 is positive ). (c) For x = 10, μ y.10 = -.12 + .095(10) = .830. For x = 15, μ y.15 = -.12 + .095(15) = 1.305. (d) When x = 10, the flow rate y is normally distributed with a mean value of μ y.10 = .830 and a standard deviation of σ y.10 = σ = .025. Therefore, we standardize and use the z table to find: P(y > .835) = P(z > 025 . 830 . 835 . ) = P(z > .20) = 1 - P(z .20) = 1- .5793 = .4207 (using Table I). 2. (a) The slope of the estimated regression line ( β = -.01) is the expected change in reaction time for a one degree Fahrenheit increase in the temperature of the chamber. So, with a one degree Fahrenheit increase in temperature, the true average reaction time will decrease by .01 hours. With a 10 degree increase in temperature, the true average reaction time will decrease by (10)(.01) = 1 hour. (b) When x = 200, 3 ) 200 ( 01 . 5 200 = = y μ When x = 250, 5 . 2 ) 250 ( 01 . 5 250 = = y (c) = = < < ) 250 when 6 . 2 4 . 2 ( x y P 8164 . 0918 . 9082 . ) 33 . 1 ( ) 33 . 1 ( ) 33 . 1 33 . 1 ( 075 . 5 . 2 6 . 2 075 . 5 . 2 4 . 2 = = < < = < < = < < z P z P z P z P Next, the probability that all five observed reaction times are between 2.4 and 2.6 is (.8164) 5 = .3627 3. (a) The simple linear regression model states that y = α + β x. Here, y = ln(V) and x = 1/T, so the model becomes ln(V) = α + β (1/T) + e. Exponentiating both sides gives: exp(ln(V)) = exp( α + β (1/T)+e), or, V = ε = ε = γ T e e / βα T e e / 1 ) ( 0 ( γ 1 ) 1/T ε , where ε is the antilog of the error term e, γ 0 = e α , and γ 1 =

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Chapter 11 Inferential Methods in Regression and Correlation 2 e β . To summarize, the model is V = γ 0 ( γ 1 ) 1/T ε , which is the model for a multiplicative relationship between the response and predictor variables. (b) For estimation, we normally set the error term equal to its expected value. For a multiplicative model, the expected value of the error term is 1 (for additive models the expected value is 0). So, for α = 20.607, β = -5200.762 and a temperature of T = 300, we predict a vapor pressure of about V = γ 0 ( γ 1 ) 1/T ε = (e 20.607 )(e -5200.762 ) 1/300 (1) = 26.341. 4. (a) 100 50 0 100 90 80 70 60 50 40 30 20 10 0 rainfall volume runoff volume The scatterplot appears linear, so a simple linear regression model seems reasonable. (b) The following quantities are needed: () ( ) 1278 . 1 ) 2 . 53 )( 82697 (. 867 . 42 82697 . 4 . 20586 4 . 17024 7 . 14435 15 643 41999 4 . 20586 15 798 63040 4 . 17024 15 643 798 51232 867 . 42 200 . 53 2 2 = = = = = = = = = = = = = = x b y a S S b S S S y x xx xy yy xx
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## This note was uploaded on 02/23/2009 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

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Chpt11 - Chapter 11 Inferential Methods in Regression and...

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