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Homework5__solutions

# Homework5__solutions - 1 a 62 The Minitab output is shown...

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1. 62. a. The Minitab output is shown below: The regression equation is Defects = 22.2 - 0.148 Speed Predictor Coef SE Coef T P Constant 22.174 1.653 13.42 0.000 Speed -0.14783 0.04391 -3.37 0.028 S = 1.489 R-Sq = 73.9% R-Sq(adj) = 67.4% Analysis of Variance Source DF SS MS F P Regression 1 25.130 25.130 11.33 0.028 Residual Error 4 8.870 2.217 Total 5 34.000 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 14.783 0.896 ( 12.294, 17.271) ( 9.957, 19.608) b. Since the p -value corresponding to F = 11.33 = .028 < α = .05, the relationship is significant. c. = .739; a good fit. The least squares line explained 73.9% of the variability in the number of defects. d. Using the Minitab output in part (a), the 95% confidence interval is 12.294 to 17.271. 21. A microcomputer manufacturer has developed a regression model relating his sales (Y in \$10,000s) with three independent variables. The three independent variables are price per unit (Price in \$100s), advertising (ADV in \$1,000s) and the number of product lines (Lines). Part of the regression results is shown below. Coefficient Standard Error Intercept 1.0211 22.8752 Price -0.1524 0.1411 ADV 0.8849 0.2886 Lines -0.1463 1.5340 Analysis of Variance Source of Variation Degrees

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