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I have a radiation source and every particle emitted is registered. And the mean number of produced ionizations in a time interval T is recorded.

Suppose N number of particles were emitted during the time interval T and the number of ionizations recorded is I. Then the mean number of ionization per emitted particle is I/N. What is the uncertainty of the mean number of ionization per emitted particle?

Example: 853 emitted particles and 2182 registered ionizations in 100 seconds. Hence 2182/853 = 2.56 ionizations per emitted particle. But what is the uncertainty (or margin of error?)?

Fabi
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2 Answers2

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Strictly speaking, you just don't have enough information. As far as we know, there could be anywhere from exactly $2.56$ ionisations per particle (zero variation) to an arbitrarily large variance. Do you have any more information, such as the full dataset of exactly how many ionisations occur for each?

That said, if you can assume some more information about the distribution, it may be possible. If you assume it is an exponential distribution (which seems pretty feasible for these sorts of radiation questions), it happens that knowing the mean is equivalent to knowing the variance; the variance is just the mean squared! So from that you could probably work out all you want to know. But this is not at all robust unless you already know the exact distribution of the data happens to be exponentially distributed, for example.

Jack Crawford
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  • Thank you Jack, No, unfortunately I don't have the full data set. The data is gaussian distributed. – Fabi Jul 04 '19 at 14:42
  • @Fabi Unfortunately, I think you might be out of luck. There's no way you can infer the variance of a gaussian distribution just from the mean, and I don't think you can from the other given data either. Perhaps you're just going to have to take a larger number of distinct samples or collect finer data, sorry. (Hopefully I'm wrong about this and someone better at statistics can swoop in and save us, but it does not look feasible to me.) – Jack Crawford Jul 04 '19 at 14:56
  • I think the answer is super simple. I just posted it above. Let me know if something is not clear. – Fabi Jul 04 '19 at 16:22
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From a gaussian sampling distribution the standard deviation is approximately 1 / (2 * sqrt(I)) for the ionizations. Hence the 95% confidence interval which is 2*standard deviation gives us the easy solution to the problem: mean +/- 1/sqrt(I). For the example above it would be (2182+1/sqrt(2182))/853 and (2182-1/sqrt(2182))/853. The result is then 2.55803 +/- 1/sqrt(2182)/853 = 2.55803 +/- 2.50971e-05

Fabi
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  • Where did you find this formula for the sample standard deviation? I’ve never seen it before but it looks incredibly useful if true. – Jack Crawford Jul 04 '19 at 22:37
  • http://faculty.ycp.edu/~dhovemey/fall2006/mat111/lecture/lecture35.html – Fabi Jul 05 '19 at 07:17
  • I’m afraid I don’t think this source is applicable to you. There is a popular CLT result stating that sample standard deviation of the mean $\sigma_{\bar{X}}$ is given by $\frac{\sigma}{/sqrt{n}}$ where $\sigma$ is the actual true (unknown) population standard deviation but that’s both not what you would want (you’re trying to find the actual variance, not the variance of the mean in the sampling distribution), but also assumes $\sigma=1$ (we’re taking samples out of a distribution with known variance $1$?). I can’t imagine where else they pulled this formula from, otherwise seems incorrect – Jack Crawford Jul 05 '19 at 12:56
  • Maybe I’m just insane, but I’ll see if I can come back with a clearer way of demonstrating my point – Jack Crawford Jul 05 '19 at 13:10
  • Basically, you're going to get a total of roundabouts 2182 total ionisations regardless of whether you draw 853 particles from a gaussian distribution with parameters (2.56, 1) or whether you draw them from a gaussian distribution with parameters (2.56, 1000). This is all the information you have and it is consistent with any (positive) choice of variance you could possibly consider. You really don't have any information here that could tell you anything about the actual variance of the data, and I think the source you're using already assumes an actual population variance or something. – Jack Crawford Jul 05 '19 at 13:25