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lecture7solutions - ISYE 6414 - Spring 2009 Solution...

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ISYE 6414 - Spring 2009 Solution Lecture 7. Problem 8.3. Pag. 335 1. The criticism is valid. Using to many parameters in the model can create a curve that is close to every point in the sample, however, for a new data, this model may not be valid any more. Overfitting implies a good fitting for the data point, but a unrealistic model for the population. 2. Yes. Using R 2 a penalizes the number of estimated parameters. Problem 8.4. Pag. 335 1. Fitted second order model: Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 82.935749 1.543146 53.745 <2e-16 *** x -1.183958 0.088633 -13.358 <2e-16 *** x2 0.014840 0.008357 1.776 0.0811 . --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 8.026 on 57 degrees of freedom Multiple R-squared: 0.7632, Adjusted R-squared: 0.7549 F-statistic: 91.84 on 2 and 57 DF, p-value: < 2.2e-16 The plot of the fitted curve is showed in Figure 1. The model seem to have a good fit. Figure 1: Data points and fitted second order model 2. H 0 : β 1 = β 11 = 0 vs. H a : [ β 1 = β 11 = 0] C . F * = MSR MSE = 5915 . 31 64 . 409 = 91 . 839 F 0 . 95 , 2 , 57 = 3 . 15884. Then, the null hypothesis is rejected. 1
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3. Confidence interval for ersponse mean when X = 48. Replacing in the model: x = - 11 . 9833, and ˆ Y h = 99 . 2546. s { ˆ Y h } = 1 . 4833 and t 0 . 975 , 57 = 2 . 00247. Then: 99 . 2546 ± 2 . 00247(1 . 4833) 96 . 2843 E ± ˆ Y h ² 102 . 2249 4. Prediction interval when X = 48.
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lecture7solutions - ISYE 6414 - Spring 2009 Solution...

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