f07finalsoln - Stat 571 Final Exam Solution 1(a 37 24/66 =...

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Stat 571 Final Exam - Solution December 19, 2007 1.(a) 37 * 24 / 66 = 13 . 45 (b) X 2 = 1 . 5365+0 . 055+ · · · +2 . 943 = 8 . 8705. Using df=2 we get a p-value slightly above 0 . 01. We reject the null hypothesis: there is evidence that mole rats from different colonies choose their tubers differently. Note: from the small contributions of colony B cells, mole rats from colony B seem to have average tuber preferences. Colonies A and C have the largest contributions, so they seem to differ more markedly from the average tuber preference profile. 2.(a) ˆ b 1 = 26 , 198 . 4 / 1006 . 6 = 26 . 02662. ˆ b 0 = 471 . 4 - 26 . 02 * 16 . 1 = 52 . 37. For each increase of 1kg in female’s weight, there is an av- erage increase of 26,027 eggs. (b) r = 26 , 198 . 4 / 1006 . 6 * 811 , 738 = 0 . 916 so r 2 = 0 . 84. It means that 84% of the variability in the number of eggs is explained by the female’s weight. (c) SSError= (1 - r 2 ) * 811 , 738 = 129 , 878 so s 2 e = MSEr- ror = SSError / (15 - 2) = 9990 . 93, and s e = 99 . 954 (which seems correct from the graph provided). Then the standard error of ˆ b 1 is 9990 . 93 / 1006 . 6 = 9 . 925 = 3 . 15.
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