Timeline for When combining p-values, why not just averaging?
Current License: CC BY-SA 3.0
22 events
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| Aug 12, 2015 at 9:30 | history | edited | amoeba |
adding new tag
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| Jul 13, 2015 at 20:36 | comment | added | DWin | I would ask that every one read Duncan Murdoch's piece "P-values are Random Variables" in "The American Statistician". I find a copy online at: hypergeometric.files.wordpress.com/2013/09/… | |
| S Jul 13, 2015 at 16:53 | history | bounty ended | Glen_b | ||
| S Jul 13, 2015 at 16:53 | history | notice removed | Glen_b | ||
| Jul 13, 2015 at 1:39 | history | tweeted | twitter.com/#!/StackStats/status/620407148653248512 | ||
| S Jul 12, 2015 at 16:31 | history | bounty started | Glen_b | ||
| S Jul 12, 2015 at 16:31 | history | notice added | Glen_b | Reward existing answer | |
| Jul 10, 2015 at 8:10 | answer | added | amoeba | timeline score: 33 | |
| Jul 10, 2015 at 7:55 | history | edited | amoeba | CC BY-SA 3.0 |
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| Jul 9, 2015 at 23:51 | vote | accept | Alby | ||
| Jun 23, 2015 at 0:38 | comment | added | Glen_b | Note that Fisher's method is based on the product, which is what I'm describing as natural -- because you multiply independent probabilities to find their joint probability. Considering GM is not really different from product other than there's then an additional step in figuring out what the corresponding combined p-value is because having worked out the GM ($g$, say) by taking the product, you'd then need to look at $-2n \log g=-2 \log (g^n)$ get the combined p-value. Which is to say you'd converting the GM back to the product before taking logs to find the combined p-value. | |
| Jun 22, 2015 at 20:22 | comment | added | Alby | @Glen_b if product is the obvious way to proceed, what about geometric mean? | |
| Dec 7, 2013 at 22:37 | answer | added | Elvis | timeline score: 46 | |
| Dec 7, 2013 at 20:14 | vote | accept | Alby | ||
| Jul 9, 2015 at 23:51 | |||||
| Dec 6, 2013 at 5:41 | comment | added | Glen_b | There are other ways to combine p-values, of course, though the product is the most natural way to do it. One could add the p-values for example; under the joint null the sum of them should have a triangular distribution. Or one could convert the p-values to z-values and add those (and if you were combining results from similar size not-too-small samples from a normal population, this would make a lot of sense). But the product is the obvious way to proceed; it makes logical sense every time. | |
| Dec 6, 2013 at 5:28 | comment | added | Glen_b | Say I have 2 independent samples from the same population (let's say we have a one sample t-test). Imagine the sample mean and standard deviations are just about the same. So the p-value for the first sample is 0.0666 and for the second sample is 0.0668. What should the overall p-value be? Well, should it be 0.0667? Actually, it's quite obvious it must be smaller. In this case the "right" thing to do is combine the samples, if we have them. We'd have about the same mean and standard deviation, but twice the sample size. The std. error of the mean is smaller, and the p-value must be smaller. | |
| Dec 5, 2013 at 23:41 | comment | added | whuber♦ | It comes down to a basic axiom of probability: p-values are probabilities and probabilities for the outcomes of independent experiments don't add, they multiply. Where multiplication is concerned, logarithms simplify a product to a sum: that's where $\sum\log(X_i)$ comes from. (That it has a chi-squared distribution is then an ineluctable mathematical consequence.) Far from begin "convoluted," this is perhaps the simplest and most natural (legitimate) procedure conceivable. | |
| Dec 5, 2013 at 3:47 | answer | added | DWin | timeline score: 11 | |
| S Dec 5, 2013 at 1:48 | history | edited | Nick Cox | CC BY-SA 3.0 |
corrected mathematical symbols; corrected text in math fomulas
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| S Dec 5, 2013 at 1:48 | history | suggested | yanfei kang | CC BY-SA 3.0 |
corrected mathematical symbols; corrected text in math fomulars
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| Dec 5, 2013 at 1:26 | review | Suggested edits | |||
| Dec 5, 2013 at 1:48 | |||||
| Dec 4, 2013 at 23:11 | history | asked | Alby | CC BY-SA 3.0 |