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RegressionInClassSolutions

# RegressionInClassSolutions - ESI 6321 APPLIED PROBABILITY...

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ESI 6321 APPLIED PROBABILITY METHODS IN ENGINEERING OEM 2009 January 13, 2008 IN-CLASS ASSIGNMENT - SOLUTIONS Problem 1: (a) S XY = x i y i i ! " n x y = 1083.67 " 20 1478 20 # \$ % & ( 12.75 20 # \$ % & ( = 141.445 S XX = x i 2 i ! " n x 2 = 143,215.8 " 20 1478 20 # \$ % & ( 2 = 33,991.6 ˆ ! 1 = S XY S XX = 141.445 33991.6 = 0.00416 ˆ ! 0 = y " ˆ ! 1 x = 12.75 20 " 0.00416 1478 20 # \$ % & ( = 0.33 (b) ˆ y = ˆ ! 0 + ˆ ! 1 x = 0.33 + 0.00416 " 85 = 0.684 . (c) ˆ μ = ˆ ! 0 + ˆ ! 1 x = 0.33 + 0.00416 " 90 = 0.704 . (d) 1 ˆ 0.00416 ! = . (e) Denote the temperature in ° C by c , i.e., x = 9 5 c + 32 . This means that the new regression line becomes: ˆ y = ˆ ! 0 + ˆ ! 1 x = ˆ ! 0 + ˆ ! 1 9 5 c + 32 " # \$ % & = ˆ ! 0 + 32 ˆ ! 1 ( ) + 9 5 ˆ ! 1 c = 0.46 + 0.00749 c . (f) 1 9 ˆ 0.00749 5 ! = .

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(g) The coefficient of determination is equal to R 2 = 1 ! SSE SST = 1 ! y i 2 i " ! ˆ # 0 y i i " ! ˆ # 1 x i y i i " y i 2 i " ! n y 2 = 1 ! 8.86 ! 0.33 \$ 12.75 ! 0.00416 \$ 1083.67 8.86 ! 20(12.75 / 20) 2 = 1 ! 0.1444 0.7319 = 0.803 . Problem 2: (a) The residual is equal to e = y ! ˆ " 0 + ˆ " 1 x ( ) = 46.1 ! 48.013 ! 2.33 # 3.7 ( ) = 46.1 ! 39.39 = 6.71 . (b) SSE = y i 2 i ! " ˆ # 0 y i i ! " ˆ # 1 x i y i i ! = 23530 " 48.013 \$ 572 + 2.33 \$ 1697.8 = 22.438 ˆ ! 2 = SSE n " 2 = 22.438 14 " 2 = 1.8698 (c) S XX = x i 2 i ! " n x 2 = 157.42 " 14 43 14 # \$ % & ( 2 = 25.35 se ˆ ! 1 ( ) = ˆ " 2 S XX = 1.8698 25.35 = 0.2716 se ( ˆ ! 0 ) = ˆ " 2 1 n + x 2 S XX # \$ % & ( = 1.8698 1 14 + 43/ 14 ( ) 2 25.35 # \$ % % & ( ( = 0.9107 (d) The critical t -value of a t- distribution with n -2 = 12 degrees of freedom at the 95%- significance level is equal to 2.179. The test statistic is equal to ˆ ! 1 / se ( ˆ ! 1 ) = " 2.33/ 0.2716 = " 8.57
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RegressionInClassSolutions - ESI 6321 APPLIED PROBABILITY...

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