Timeline for How to calculate a confidence level for a Poisson distribution?
Current License: CC BY-SA 3.0
13 events
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| Jun 13, 2023 at 8:23 | comment | added | asmaier | If $n=1$ and $\lambda=2$ the left bound of this 95%-interval would be negative: $$ 2 - 1.96 \sqrt{2} = -0.7719 < \lambda < 4.7719 = 2 + 1.96 \sqrt{2} $$ Using symmetric confidence intervals for a right-skewed distribution like the poisson distribution is not a good idea. | |
| Mar 13, 2019 at 17:40 | comment | added | Alex |
Here they are using the standard deviation of the mean right? That is, SE = sig/sqrt(N) = sqrt(lam/N) ? This would make sense since the standard deviation of single values sig tells us about the likelihood of drawing random samples from the Poisson distribution, whereas the SE as defined above tells us about our confidence in lam, given the number of samples we've used to estimate it.
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| May 6, 2014 at 20:14 | history | edited | Nick Stauner | CC BY-SA 3.0 |
capitalization
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| S May 6, 2014 at 20:14 | history | suggested | TheBigAmbiguous | CC BY-SA 3.0 |
Made formula more mathemtically precise.
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| May 6, 2014 at 20:12 | review | Suggested edits | |||
| S May 6, 2014 at 20:14 | |||||
| Jul 25, 2012 at 17:59 | comment | added | user12849 | How did you calculate the exact interval for this problem given the information on that website given by whuber? I couldn't follow because that site seems to only indicate how to proceed when you have one sample. Maybe I'm just not understanding something simple but my distribution has a much smaller value of lambda(n) so I can't use the normal approximation and I don't know how to compute the exact value. Any help would be greatly appreciated. Thanks! | |
| Apr 17, 2012 at 19:26 | comment | added | maddyblue | For others confused like I was: here is a description of where the 1.96 comes from. | |
| Sep 9, 2011 at 15:36 | history | edited | whuber♦ | CC BY-SA 3.0 |
TeX
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| Sep 9, 2011 at 15:34 | comment | added | whuber♦ | This is fine when $n \lambda$ is large, for then the Poisson is adequately approximated by a Normal distribution. For smallish values or higher confidence, better intervals are available. See math.mcmaster.ca/peter/s743/poissonalpha.html for two of them along with an analysis of their actual coverage. (Here, the "exact" interval is (45.7575, 48.6392), the "Pearson" interval is (45.7683, 48.639), and the Normal approximation gives (45.7467, 48.617): it's a little too low, but close enough, because $n \lambda = 4152$.) | |
| Sep 9, 2011 at 15:31 | history | edited | whuber♦ | CC BY-SA 3.0 |
Merged with previous reply.
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| Sep 9, 2011 at 14:57 | history | migrated | from stackoverflow.com (revisions) | ||
| Sep 9, 2011 at 12:47 | vote | accept | CommunityBot | ||
| Sep 9, 2011 at 12:40 | history | answered | Dan | CC BY-SA 3.0 |