Combining errors. Gamma spectrometry, Poisson distribution

Question

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

Answer 1

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.