ch06 - CHAPTER 6 Note to Instructor For computer exercises...

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CHAPTER 6 Note to Instructor: For computer exercises, the procedure ‘Regression’ under ‘Stat’ in Minitab can be used for the regression analysis except for computing confidence intervals on the regressor variables. Sections 6-2 6-1. a) The regression equation is Thermal = 0.0249 + 0.129 Density Predictor Coef StDev T P Constant 0.024934 0.001786 13.96 0.000 Density 0.128522 0.007738 16.61 0.000 S = 0.0005852 R-Sq = 98.6% R-Sq(adj) = 98.2% Analysis of Variance Source DF SS MS F P Regression 1 0.000094464 0.000094464 275.84 0.000 Residual Error 4 0.000001370 0.000000342 Total 5 0.000095833 x 129 . 0 0249 . 0 y ˆ + = b) 0.0005747 -0.0007088 0.0001486 -0.0004799 -0.0000644 0.0005298 c) SSE = 0.000001370 2 ˆ σ = 0.000000342 d) se( 1 ˆ β ) = 0.007738, se( 0 ˆ β ) = 0.001786 e) SST = 0.000095833 SSR = 0.000094464, SSE = 0.000001370, and SSR + SSE = 0.000095834 SST = SSR + SSE f) R 2 = 98.6%. This is interpreted as 98.6% of the total variability in thermal conductivity can be explained by the fitted regression model. g) See the Minitab output given in part a. Based on the t-tests, we conclude that the slope and intercept are nonzero. h) See the Minitab output in part a. Based on the analysis of variance, we can reject the null hypothesis and conclude that the regression is significant. i) β 0 : 0.024934 ± 2.776(0.001786); 0.02, 0.03 β 1 : 0.128522 ± 2.776(0.007738); 0.107, 0.15 j) Residual Plots 1
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0.28 0.23 0.18 0.0005 0.0000 -0.0005 Density Residual Residuals Versus Density (response is Conduct) 0.060 0.055 0.050 0.0005 0.0000 -0.0005 Fitted Value Residual Residuals Versus the Fitted Values (response is Conduct) 2
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0.0005 0.0000 -0.0005 1 0 -1 Normal Score Residual Normal Probability Plot of the Residuals (response is Conduct) k) r = 0.993, P-value = 0, Therefore, we conclude there is a significant correlation between density and conductivity. 6-2. a) The regression equation is Usage = - 6.34 + 9.21 Temp Predictor Coef StDev T P Constant -6.336 1.668 -3.80 0.003 Temp 9.20836 0.03377 272.64 0.000 S = 1.943 R-Sq = 100.0% R-Sq(adj) = 100.0% Analysis of Variance Source DF SS MS F P Regression 1 280583 280583 74334.36 0.000 Residual Error 10 38 4 Total 11 280621 x 21 . 9 34 . 6 y ˆ + - = b) -1.25010 -0.19519 -0.30208 -1.61751 0.49740 2.07214 1.71688 -0.02329 -2.55294 -1.15260 -1.25734 4.06464 3
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c) SSE = 38 2 ˆ σ = 4 d) se( 1 ˆ β ) = 1.668, se( 0 ˆ β ) = 0.03377 e) SST = 280621 SSR = 280583, SSE = 38, and SSR + SSE = 280621 SST = SSR + SSE f) R 2 = 100%. This is interpreted as 100% of the total variability in Usage can be explained by the fitted regression model. g) See the Minitab output given in part a. Based on the t-tests, we conclude that the slope and intercept are nonzero. h) See the Minitab output in part a. Based on the analysis of variance, we can reject the null hypothesis and conclude that the regression is significant. i) β 0 : -6.34 ± 2.228(1.668); 2.62,10.06 β 1 : 9.21 ± 2.228(0.03377); 9.13,9.29 j) Residual Plots. 80 70 60 50 40 30 20 4 3 2 1 0 -1 -2 -3 Temp Residual Residuals Versus Temp (response is Usage) 700 600 500 400 300 200 4 3 2 1 0 -1 -2 -3 Fitted Value Residual Residuals Versus the Fitted Values (response is Usage) 4
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4 3 2 1 0 -1 -2 -3 2 1 0 -1 -2 Normal Score Residual Normal Probability Plot of the Residuals (response is Usage) k) r = 1.000, P-value = 0, Therefore, we conclude there is a significant correlation between temperature and usage. 6-3. a) The regression equation is Deflect = 0.393 + 0.00333 Temp Predictor Coef SE Coef T P Constant 0.39346 0.04258 9.24 0.000 Temp 0.0033285 0.0005815 5.72 0.000 S = 0.006473 R-Sq = 64.5% R-Sq(adj) = 62.6% Analysis of Variance Source DF SS MS F P Regression 1 0.0013727 0.0013727 32.76 0.000 Residual Error 18 0.0007542 0.0000419 Total 19 0.0021270 x y 00333 . 0 393 . 0 ˆ + = b) -0.0054488 0.0072519 0.0065614 -0.0127685 0.0069249 -0.0004269 -0.0027627 -0.0044372 0.0002431 -0.0074196 0.0015643 0.0078738 0.0035833 -0.0077656 -0.0011131 -0.0024386 0.0105570 -0.0054342 -0.0014357 0.0068913 c) SSE = 0.0007542 2 ˆ σ = 0.0000419 5
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d) se( 1 ˆ β ) = 0.0005815, se( 0 ˆ β ) = 0.04258 e) SST = 0.0020402 SSR = 0.0013149, SSE = 0.0007253, and SSR + SSE = 0.0020402 SST = SSR + SSE f) R 2 = 64.5%. This is interpreted as 64.5% of the total variability in deflection can be
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