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Second, I don't believe you. I say it's always smarter to use the partitioned data than the aggregate data. If you have a data set that includes the gender of the subject, you're always better off building two models (one for each gender) instead of one big model. Why throw away information?
If you believe the OP's assertion
Similarly, for just about any given set of data, you can find some partition which reverses the apparent correlation
then it is demonstrably false that your strategy always improves matters. Why do you believe that your strategy is better?
My first reading of this quote was essentially "the map loses to the terrain". I interpreted "theory" as "our beliefs" and "practice" as "reality".
Possibly, yes; but reading a discussion about a topic I don't know anything about is hard, so I'm less likely to get anything out of it, despite the fact that it is there in what you wrote. I'm claiming that the additional "distracting" material would actually serve as a hook to get the reader interested in putting effort into understanding the point of the post.
I'd need to read it again, with pen and paper, to gain an understanding of why the Student-t distribution is the right thing to compute. At the very least I can say this: the probability of one's vote tilting the election is certainly higher in very close elections (as measured beforehand by polls, say) than in an election such as Obama-McCain 2008. The article you quoted suggests the difference in probabilities is much higher than I anticipated. (Unless my calculation, which models the closest possible election, is incorrect.)
Edited to add: Okay, I've inc... (read more)
Jane estimates the probability of her vote tilting the presidential election at 1 in 1,000,000; Eric estimates the probability of his vote tilting the presidential election at 1 in 100,000,000. I find both of these estimates orders of magnitude too low.
Eric presumably is modeling the election by saying that with 100,000,000 voters (besides himself), there are 100,000,001 outcomes of their votes, only one of which is a tie which his vote will break. But his conclusion that the odds of deciding the election are about 1 in 100,000,000 assumes that all of thes... (read more)
A presentation critique: psychologically, we tend to compare the relative areas of shapes. Your ovals in Figure 1 are scaled so that their linear dimensions (width, for example) are in the ratio 2:5:3; however, what we see are ovals whose areas are in ratio 4:25:9, which isn't what you're trying to convey. I think this happens for later shapes as well, although I didn't check them all.
If my estimate is 1000, and someone else's is 300, that's too big a discrepancy to explain by minor variations. It casts doubt on the assumption of identical thermometers. Assuming that I only have the other people's estimates, and there's no opportunity for discussion, I'll search for reasons why we might have come up with completely different answers, but if I find no error in my own, I'll discard all such outliers.
What if everyone else's estimate is between 280 and 320? Do you discard your own estimate if it's an outlier? Does the answer depend on whether you can find an error in your reasoning?
Part of the output of your quizzes is a line of the form "Your chance of being well calibrated, relative to the null hypothesis, is 50.445538580926 percent." How is this number computed?
I chose "25% confident" for 25 questions and got 6 of them (24%) right. That seems like a pretty good calibration ... but 50.44% chance of being well calibrated relative to null doesn't seem that good. Does that sentence mean that an observer, given my test results, would assign a 50.44% probability to my being well calibrated and a 49.56% probability to my not being well calibrated? (or to my randomly choosing answers?) Or something else?