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Unformatted text preview: DSC 330 HW3 SOLUTIONS 1. Textbook 4.18 (There is a typo in question (b). The null hypothesis should be u1D43B : u1D6FD 1 = ... = u1D6FD 4 = 0.) u1D6FC = 5% Solutions: a. u1D445 2 = . 912: 91.2% of the total variability of equivalent widths is explained by the model containing the four independent variables. u1D445 2 u1D44E = . 894. This statistic has a similar interpretation to that of u1D445 2 , but is adjusted for both the sample size u1D45B and the number of predictors in the model. The u1D445 2 u1D44E is the preferred measure of model fit because it may decrease as a useless predictor is introduced. b. u1D43B : u1D6FD 1 = u1D6FD 2 = u1D6FD 3 = u1D6FD 4 = 0 u1D43B u1D44E : At least one of them is nonzero. The test statistic is F = 51.720. RR: u1D439 > u1D439 4 , 20 , . 05 = 2 . 87. Conclusion: Reject u1D43B , so the model is useful. 2. Solutions: a. u1D445 2 = . 362. 36.2% of the variability in the AC scores can be explained by the model containing the variables selfesteem score, optimism score, and group cohesion score.containing the variables selfesteem score, optimism score, and group cohesion score....
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This note was uploaded on 01/29/2012 for the course DSC 330 taught by Professor Staff during the Spring '08 term at Oregon.
 Spring '08
 Staff

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