Determining the uncertainity in Geiger Counter data

AI Thread Summary
The discussion centers on using Poisson statistics to determine uncertainty in Geiger Counter data from a single trial with multiple counts. It is confirmed that the uncertainty can be calculated as the square root of the counts, which results in small error bars due to the high count values. Participants agree that using Poisson for this purpose is reasonable, despite the resulting small uncertainty appearing insignificant. The conversation highlights the necessity of making assumptions about the distribution when calculating error bars. Ultimately, the method applied is deemed appropriate for the data presented.
learningastronomy
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Homework Statement
I have a list of counts that are taken from a Geiger Counter based off only one trial. There were 20 runs taken with different voltages. The voltage ranged from 500 to 1000 and the counts ranged from 20,000 to 100,000.
Relevant Equations
Let c be counts therefore uncertainty is ##\sqrt{c}##.
From what I understand thus far is the counting involves Poisson therefore the uncertainty is just the square root of the counts, correct? But when I take the square root of the counts it produces a very small number compared to the count which makes it insignificant therefore the error bars that I want to produce aren't visible because the uncertainty is so small. Am I correct in using Poisson to measure the uncertainty? There was only one trial done.
 
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learningastronomy said:
Homework Statement:: I have a list of counts that are taken from a Geiger Counter based off only one trial. There were 20 runs taken with different voltages. The voltage ranged from 500 to 1000 and the counts ranged from 20,000 to 100,000.
Relevant Equations:: Let c be counts therefore uncertainty is ##\sqrt{c}##.

From what I understand thus far is the counting involves Poisson therefore the uncertainty is just the square root of the counts, correct? But when I take the square root of the counts it produces a very small number compared to the count which makes it insignificant therefore the error bars that I want to produce aren't visible because the uncertainty is so small. Am I correct in using Poisson to measure the uncertainty? There was only one trial done.
The description is not clear to me.

Is it that, on the basis that the process is Poisson, you want the uncertainty in the Poisson rate as deduced from a single count? My understanding is that if you estimate the parameter from a single Poisson count the standard error in that estimate is indeed the square root of the count. Yes, with such large counts this will be a relatively small number.
 
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haruspex said:
The description is not clear to me.

Is it that, on the basis that the process is Poisson, you want the uncertainty in the Poisson rate as deduced from a single count? My understanding is that if you estimate the parameter from a single Poisson count the standard error in that estimate is indeed the square root of the count. Yes, with such large counts this will be a relatively small number.

The process does not have to be Poisson, I just assumed to use Poisson to compute the uncertainty. If there was only one trial in the experiment how do I solve for the uncertainty/error bars? Will Poisson suffice even though the uncertainty is so small it becomes insignificant?
 
learningastronomy said:
The process does not have to be Poisson, I just assumed to use Poisson to compute the uncertainty. If there was only one trial in the experiment how do I solve for the uncertainty/error bars? Will Poisson suffice even though the uncertainty is so small it becomes insignificant?
To get an answer you have to make an assumption about the nature of the distribution, and Poisson seems entirely reasonable.
I see no objection to what you have done. The error bars will really be that small.
 
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haruspex said:
To get an answer you have to make an assumption about the nature of the distribution, and Poisson seems entirely reasonable.
I see no objection to what you have done. The error bars will really be that small.
Oh I see, thank you!
 
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