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solutions5 - Stat 665 (Spring 2011) Kaizar Solutions 5...

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Unformatted text preview: Stat 665 (Spring 2011) Kaizar Solutions 5 Exercises 20 points total 1. Problem 3.11, b and c (b) Using R, I found the following output: Estimate Std. Error z value Pr(>|z|) (Intercept) 1.6094 0.1414 11.380 < 2e-16 *** x 0.5878 0.1764 3.332 0.000861 *** The prediction equation is: log( ) = 1 . 6094 + 0 . 5878 x I estimate that the log expected number of imperfections on wafers treated with B is 0.5878 higher than the log expected number of imperfections on wafers treated with A. That is, I estimate that wafers treated with B are expected to have e . 5878 = 1 . 8 times as many imperfections than wafers treated with A. (c) From the above output, the p-value associated with the Wald test of the the null hy- pothesis that = 0 is 0.000861. The Likelihood Ratio Test statistic is 27.857-16.268 = 11.589. If the null hypothesis that = 0 is true, then the LRT has a 2 1 distribution. The p-value is 0.00066. Regardless of which test I use, I reject the null hypothesis and conclude at significance 0.05 that wafers treated with B have more imperfections than wafers treated with A. 2. Modified Problem 3.12 (a) Using R, I found the following output: Estimate Std. Error z value Pr(>|z|) (Intercept) 2.0541 0.1132 18.142 <2e-16 *** z-0.2296 0.1701-1.349 0.177 The prediction equation is: log( ) = 2 . 0541- . 2296 z I estimate that the log expected number of imperfections on wafers treated with high coats is 0.2296 lower than the log expected number of imperfections on wafers treatedcoats is 0....
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solutions5 - Stat 665 (Spring 2011) Kaizar Solutions 5...

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