Chapter13 - CHAPTER 13 Section 13.1 1. a. 2 ^i is 10 1 - 1...

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393 CHAPTER 13 Section 13.1 1. a. 15 = x and ( 29 250 2 = - x x j , so s.d. of i i Y Y ˆ - is ( 29 = - - - 250 15 5 1 1 10 2 i x 6.32, 8.37, 8.94, 8.37, and 6.32 for i = 1, 2, 3, 4, 5. b. Now 20 = x and ( 29 1250 2 = - x x i , giving standard deviations 7.87, 8.49, 8.83, 8.94, and 2.83 for i = 1, 2, 3, 4, 5. c. The deviation from the estimated line is likely to be much smaller for the observation made in the experiment of b for x = 50 than for the experiment of a when x = 25. That is, the observation (50, Y) is more likely to fall close to the least squares line than is (25, Y). 2. The pattern gives no cause for questioning the appropriateness of the simple linear regression model, and no observation appears unusual. 3. a. This plot indicates there are no outliers, the variance of ε is reasonably constant, and the ε are normally distributed. A straight-line regression function is a reasonable choice for a model. 200 150 100 1 0 -1 filtration rate residuals
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Chapter 13: Nonlinear and Multiple Regression 394 b. We need S xx = ( 29 ( 29 8295 . 886 , 18 20 9 . 2817 85 . 914 , 415 2 2 = - = - x x i . Then each * i e can be calculated as follows: ( 29 8295 . 886 , 18 895 . 140 20 1 1 4427 . 2 * - + + = i i i x e e . The table below shows the values: standardized residuals * / i e e standardized residuals * / i e e -0.31064 0.644053 0.6175 0.64218 -0.30593 0.614697 0.09062 0.64802 0.4791 0.578669 1.16776 0.565003 1.2307 0.647714 -1.50205 0.646461 -1.15021 0.648002 0.96313 0.648257 0.34881 0.643706 0.019 0.643881 -0.09872 0.633428 0.65644 0.584858 -1.39034 0.640683 -2.1562 0.647182 0.82185 0.640975 -0.79038 0.642113 -0.15998 0.621857 1.73943 0.631795 Notice that if * i e ˜ e / s, then * / i e e ˜ s . All of the * / i e e ’s range between .57 and .65, which are close to s. c. This plot looks very much the same as the one in part a. 100 150 200 -2 -1 0 1 2 filtration rate standardized residuals
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Chapter 13: Nonlinear and Multiple Regression 395 4. a. The (x, residual) pairs for the plot are (0, -.335), (7, -.508), (17. -.341), (114, .592), (133, .679), (142, .700), (190, .142), (218, 1.051), (237, -1.262), and (285, -.719). The plot shows substantial evidence of curvature. b. The standardized residuals (in order corresponding to increasing x) are -.50, -.75, -.50, .79, .90, .93, .19, 1.46, -1.80, and -1.12. A standardized residual plot shows the same pattern as the residual plot discussed in the previous exercise. The z percentiles for the normal probability plot are –1.645, -1.04, -.68, -.39, -.13, .13, .39, .68, 1.04, 1.645. The plot follows. The points follow a linear pattern, so the standardized residuals appear to have a normal distribution. 2 1 0 -1 -2 1 0 -1 -2 percentile std resid Normal Probability Plot for the Standardized Residuals 5. a.
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Chapter13 - CHAPTER 13 Section 13.1 1. a. 2 ^i is 10 1 - 1...

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