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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year, 6 months |
| seen | Dec 29 '11 at 3:49 | |
| stats | profile views | 50 |
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Nov 3 |
awarded | Yearling |
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Oct 2 |
awarded | Necromancer |
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Dec 18 |
comment |
How can I calculate a sample size for a ranked list of items across a population? Perhaps, my answer to "Sample size for a variable number of answers" is relevant. |
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Dec 7 |
answered | Are those two models equals? |
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Dec 7 |
answered | Application of Hidden Markov Model to CRM |
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Dec 1 |
comment |
Explaining two-tailed tests @chl I agree. However, for a person who is just being introduced to statistical ideas, re-writing the null for a one-tailed test may be a distraction when the focus is on how and why things change with respect to interpretation of the p-value. |
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Dec 1 |
revised |
Explaining two-tailed tests fixed phrasing |
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Dec 1 |
answered | Explaining two-tailed tests |
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Nov 30 |
comment |
Sample size for a variable number of answers @CarlBenson For a normal distribution the z-value associated with a 95% confidence interval is 1.96. See the wiki link (en.wikipedia.org/wiki/1.96) for an explanation. Let me know if the link does not help and I will add further details to my answer. |
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Nov 30 |
answered | Expected number of unseen cards when drawing $2n$ cards from a deck of size $n$ |
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Nov 30 |
comment |
What happens if a survival curve doesn't reach 0.5? @whuber Shouldn't the question be phrased as: "Does this mean that you can't estimate the median?". |
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Nov 30 |
revised |
Sample size for a variable number of answers added 1 characters in body |
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Nov 30 |
revised |
Sample size for a variable number of answers added 1138 characters in body |
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Nov 30 |
comment |
Sample size for a variable number of answers How did you get to 364? |
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Nov 30 |
awarded | Nice Answer |
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Nov 30 |
revised |
Sample size for a variable number of answers added 273 characters in body |
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Nov 30 |
comment |
Sample size for a variable number of answers @whuber Oh, I know. At the very least the above approach will get the OP started with something. One can compute margin of errors associated with the probability estimates and use that to compute required sample sizes. For $K=2$ the above collapses to the well known polling problem for which sample sizes are computed using margin of error. I am a bit lazy to actually look at MLE estimates and associated margin of errors! |
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Nov 30 |
answered | Sample size for a variable number of answers |
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Nov 29 |
answered | Tobit model explanation |
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Nov 29 |
comment |
We're trying to analyze which monsters drop what in an MMO Logic stays the same. If you observe that two monsters $n$ and $n'$ do not drop the $m^\text{th}$ object then the probability of that happening is: $(1-p_{nm}) (1-p_{n'm})$ Include this probability as well in the likelihood function and maximize the likelihood function to estimate $p_{nm}, p_{n'm}$. You can also get confidence intervals for your estimates via maximum likelihood theory. |
