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I have run an experiment 3 times and measured the results by gamma spectrometry.

For example I get values like this (1 $\sigma$):

$100 (10)$

$90 (8)$

$110 (12)$

The above 1 sigma error is based on counting statistics Poisson distribution

The mean is $100$

I also have the overall 1 $\sigma$ error of $100, 90, 110$ as being $8.16$

So I have the counting error for each run and an error for the measurements.

  • How can I combine these errors to get a single value of 100 with a single total error combing all the errors shown?

UPDATE

Experiment |  Ra-226   |   2sigma |     N    |    SUM   |   
           |  Bq/g     |          |          |          |
-----------+-----------+----------+----------+----------+---------
   9x -    |   126.8   |  46.7 -  |   7.37   |   934.5  |
   9y -    |   157.5 - |  51.5 -  |   9.32   |  1467.9  |
   9z -    |   170.9 - |  52.2 -  |  10.72   |  1832.0  |

Grand mean = $4234.4/27.41$ = $154.5$

overall 2 $\sigma$ = $(\sqrt27.41)$ = $5.24$

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    $\begingroup$ Would Cross Validated be a better home for this question? $\endgroup$ – Qmechanic Jan 4 '16 at 14:05
  • $\begingroup$ I have a hard time understanding where the "overall 1 sigma error" of 8.16 comes from. Is that unrelated to the counting error? $\endgroup$ – Floris Jan 4 '16 at 14:09
  • $\begingroup$ @Floris Thanks for your promt response. the 8.16 1 sig error is unrelated to the counting error it comes from: 1)Work out the Mean (the simple average of the numbers) 2)Then for each number: subtract the Mean and square the result. 3)Then work out the mean of those squared differences. 4)Then take the square root of that and that gives 8.16 i want to include the 8.16 in my total error $\endgroup$ – JoelGarner1990 Jan 4 '16 at 14:23
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The "1 sigma" error, when based on Poisson statistics, is $\frac{\mu}{\sqrt{N}}$ so you can get the value of $N$ from the individual errors. The final error is then the mean divided by $\sqrt{\sum{N}}$.

Now you have clarified where the 8.16 comes from, I want to make a few more comments.

Using the above equation, I can estimate the number of events that went into each of the three measurements, and the sum (N*mean):

exp  mean   std    N     sum 
 1    100    10   100  10000
 2     90     8   127  11430
 3    110    12    84   9240

If I wanted to combine these three, I would come up with a total sum of 30670, a total N of 311, and a grand mean of 9240/311 = 98.6 with a 1 sigma limit of 5.6 ($\sqrt{311}$).

You might have the actual values of N, in which case you can do this calculation more accurately.

Note that if you were to use the "standard error" method for estimating the error in the mean of three measurements that have different values, then you could use the standard deviation of the measurements (3 measurements of 100, 90, 110 have a standard deviation of 10 ... remember to use (n-1) in the denominator as this is a sample, not the whole population) and divide by the square root of the number of measurements ($\sqrt{3}$) - this would get you a value of 100 ± 5.8

I consider the two answers to be equivalent although they will obviously not give exactly the same values.

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  • $\begingroup$ ok thanks again: is the μ the mean? so 100 in this case. $\endgroup$ – JoelGarner1990 Jan 4 '16 at 14:29
  • $\begingroup$ Really you should use the mean of each measurement - 100, 90, 110 for the three. But I assume you actually know what $N$ is for each measurement? $\endgroup$ – Floris Jan 4 '16 at 14:31
  • $\begingroup$ Also, now that you have clarified where the 8.16 comes from, I believe you would be counting your error twice if you tried using the fact that you got three different measurements (which will of course have different values) and then want to use the fact they are different to estimate the error in the mean. Give me a second and I will try to update... $\endgroup$ – Floris Jan 4 '16 at 14:34
  • $\begingroup$ 100 90 and 110 are the results of 3 separate experiments of the same conditions. So if by N you mean the number of repeats then N = 3 (100, 90, 110) $\endgroup$ – JoelGarner1990 Jan 4 '16 at 14:43
  • $\begingroup$ @JoelGarner1990 - I have updated my answer... $\endgroup$ – Floris Jan 4 '16 at 15:01

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