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# Stat- Anova - Melanie Hollis October 5 2011 Assignment 5...

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Melanie Hollis October 5, 2011 Assignment 5 2.1 Usefulness of Model (using ANOVA): Hypothesis: : Decision Rule: Reject in favor of if the p-value is less than 0.05. Test Statistic: F===4633.721378 P-value: Approximately 0 Conclusion: Reject in favor of . Model is Useful. ===0.9966 > anova(lm(current~last)) Analysis of Variance Table Response: current Df Sum Sq Mean Sq F value Pr(>F) last 1 1.5026e+12 1.5026e+12 4633.7 < 2.2e-16 *** Residuals 16 5.1884e+09 3.2427e+08 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 2.2 Usefulness of Model (using ANOVA): Hypothesis: : Decision Rule: Reject in favor of if the p-value is less than 0.05. Test Statistic: F===6.19596 P-value:0.0242 Conclusion: Reject in favor of . Model is Useful. ===0.2791 > anova(lm(price~loan))

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Melanie Hollis October 5, 2011 Analysis of Variance Table Response: price Df Sum Sq Mean Sq F value Pr(>F) loan 1 96.87 96.870 6.1961 0.02419 * Residuals 16 250.15 15.634 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 2.3 Usefulness of Model (using ANOVA): Hypothesis: : Decision Rule: Reject in favor of if the p-value is less than 0.05.
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Stat- Anova - Melanie Hollis October 5 2011 Assignment 5...

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