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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 self-esteem score, optimism score, and group cohesion score.containing the variables self-esteem 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