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Unformatted text preview: Stat 511 Midterm II - Answers 7 April 2011 1. Echinaceae variance components (a) Source df Plant 24 Tissue 2 Plant*Tissue 48 Extract(P*T) 75 Clarify(E*P*T) 150 error 300 Notes: Tissues and plants are crossed. Each combination is a biological sample. Extracts are nested in Samples, Clarifications are nested in Extracts, and Measurements are nested in Clarification (b) AIC = -2lnL + 2k. For REML, k is the number of variance components: that is either 5 or 4, depending on whether you include the random error or not. I accepted either. So, lnL = (105.1 - 8)/(-2) = -48.55 (c) The set of 8 models where Plants are a random factor give you the best understanding of which variance components should be included in a model. Model 2 has the lowest AIC among that set. Note: You can NOT compare models 9 and 10 to the other 8 because they have different fixed effects (plants are fixed in 9 and 10, but not in 1-8). I took off points if you tried to compare the two sets. (d) Model 1 is also possible because it has an AIC value within 2 of the best model 2. Ratio of reaction times (a) (0.89, 1.312) or perhaps adjacent values in the sorted lists. There is no correlation between the estimate and its se, so use a ordinary bootstrap. For a 99% ci, want the 0.005 and 0.995 quantiles. Those are given by the 5’th and 995’th values in the sorted vector of boostrap values. Those are (0.89, 1.312) (b) The observed value, 1.091 is somewhere in the middle of the randomization distribution. There are at least 20 values more extreme than the observed value, so you CAN tell me that p > (20 + 1) / (999 + 1), i.e. that p > . 021....
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This note was uploaded on 02/11/2012 for the course STAT 511 taught by Professor Staff during the Spring '08 term at Iowa State.
- Spring '08