HW-05 - HOMEWORK FOR EXST 7036 (Chapter 5) Student Name:...

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Unformatted text preview: HOMEWORK FOR EXST 7036 (Chapter 5) Student Name: Tanza Erlambang 5.1) For the horseshoe crab data ( www.stat.ufl.edu/~aa/introcda/appendix.html .), fit model using weight and width as predictors. a. Report the prediction equation Answer : logit ( ) π = -9.36 + 0.83 (weight) + 0.31 (width) b. Conduct a likelihood-ratio test of Ho: β 1 = β 2 = 0. interpret. Answer : Likelihood-ratio= 32.87, with df= 2→ P-value < 0.0001. Thus,HO: β 1 =β 2 =0 is rejected → there is strongly evidence that at least one variable affects on response. c. Conduct separate likelihood-ratio tests for the partial effects of each variable. Why does neither test show evidence of an effect when the test in (b) shows very strong evidence? Answer : Wald test for predictor of weight : Z 2 =( β 1 / SE ) 2 = (0.83/0.67) 2 = 1.54, with df=1 → p-value> 0.1. Thus, HO: β 1 =0 is rejected. Wald test for predictor of width: Z 2 =( β 2 / SE ) 2 = (0.31/0.18) 2 = 2.96, with df=1 → p- value> 0.05. Thus, HO: β 2 =0 is rejected. Conclusion: Separately, each predictors (weight or width) showed no effects on response. 5.8) Refer to the classification table in Table 5.3 with π = 0.50. Table 5.3. Classification Table for Horseshoe crab data 1----------------------------------------------------------------------------------------------------- Prediction, π = 0.64 Prediction, π = 0.50---------------------------------------------- Actual y = 1 y = 0 y = 1 y = 0 Total------------------------------------------------------------------------------------------------------...
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This note was uploaded on 04/11/2010 for the course EXST 7036 taught by Professor Brianmarx during the Spring '10 term at LSU.

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HW-05 - HOMEWORK FOR EXST 7036 (Chapter 5) Student Name:...

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