Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Solutions: Assignment 10 November 23, 2005 12.4 a. x y 0 20 40 60 80 100 120 140 Figure 1: boxplot for 12.4 There are a couple of interesting features. First, we see that both x and y have outliers. 2 each. Also, y looks skewed. b. The scatterplot shows a possible linear relationship between x and y. As x increases, so does y. The two outliers from part a) are seen here, and they follow the linear trend. There is one value that looks to have a lower y than its x value would predict. This may merit further investigation. 1
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0 20 40 60 80 100 120 140 20 40 60 80 x y Figure 2: scatterplot for 12.4 12.16 a. 20 40 60 80 100 120 20 40 60 80 100 x y Figure 3: scatterplot for 12.16 There is a strong linear relationship between x and y, so a simple linear regression model should work well for this data. Regression Analysis: y: versus x: The regression equation is y: = - 1.13 + 0.827 x: 2
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Predictor Coef SE Coef T P Constant -1.128 2.368 -0.48 0.642 x: 0.82697 0.03652 22.64 0.000 S = 5.24046 R-Sq = 97.5% R-Sq(adj) = 97.3% b. Point estimates for the intercept and slope are -1.128 and 0.82697,respec- tivley, from the Minitab output. Could also use the equations in the book, ˆ β 1 = S xy S xx S xx = 63040 - (798) 2 / 15 = 20586 S xy = 51232 - (798)(643) / 15 = 17024 ˆ β 1 = 17024 / 20586 = . 82697 ˆ β 0 = ¯ y - ˆ β 1 * ¯ x = - 1 . 1278 From now on, I will use Minitab output to get answers, but realize that you can still solve using formulas from the book. c. Point estimate for runoff (y) when rainfall (x) is 50: Plug in 50 in the regression equation. y = β 0 + β 1 * x y = - 1 . 1278 + 0 . 82697 * 50 y = 40 . 221 d. From Minitab output, S=5.24 e. Proportion of variation in y accounted for by knowing x is R 2 R 2 in our model is 0.9753, so 97.5% of the variation in y is attributed to the relationship between x and y.
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