
#151




I agree that they should accept both answers for #38. When building the tree, it's true that the number of predictors m used are fewer than the total predictors available p. However, the final tree could very well be comprised of all p predictors. It is very ambiguously phrased indeed, which is why it may hard to argue for just one answer over another.

#152




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#153




I'm still working on written solutions, I'm hoping tomorrow. My only issues so far with the preliminary answer key are that the given information in #35 is impossible so I'm not sure yet how to proceed and I agree that the wording of I in #38 is bad and I can see both answers being reasonable.

#154




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#155




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#156




Since it is only asking about credibility for frequency and not aggregate it’s probably ok. The derivation in the text starts generally for frequency then only adds the Poisson assumption at the last step. And you can argue that a strong candidate could redo the derivation on the fly.
That doesn’t mean you shouldn’t appeal, just that it seems like a long shot. 
#157




I have to go pick up the kids so they can get into their Halloween costumes, here's the draft solutions that I have.
Of the controversial questions, I think #35 should be E. The given information in the problem isn't possible (if the sample mean is 2 and the prior is exponential, the posterior mode is below 2 and in particular not 2.1), so maybe the question will be thrown out. But choices II and III definitely have a lower posterior mode than the original prior, and choice I has a higher posterior mode than the original prior so long as the sample mean is below 7 (which it is), so I only, or E is still the best answer. For #38, I would have answer said I is false because it talked about individual trees, not splits. I don't know how the CAS will rule on that. And to add a new one to the mix, on #34, I think the answer can be anything from 0 to 0.5. I think the intention of the problem is that we have run for long enough to be roughly stationary, and our stationary distribution is symmetric, so the next accepted step of the chain is 5050 to be right or left. I agree with that part. But if sigma is extremely small, there is a high probability that the next proposal will be rejected, and the next iteration will remain where it is, making a small probability of moving right (or left) I'll work on cleaning these up some more tomorrow afternoon, but since people might be appealing things felt an early draft would be useful. 
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