Accuracy of average measurements of a group

[Lord Vader]

If you can think of a better tag please let me know - I have no clue

You have n number of devices measuring something, each with 'a' amount of accuracy.
Average the measurement values.
The question is, what is the accuracy of the average measurement (A)

hmm something like A = F(n,a)

Another way to think about this might be
Some number of overlapping circles. Each center has some amount of inaccuracy.
Find the centre point, how accurate is that point?

or perhaps line of best fit through points with some level of inaccuracy.
What are the error bars of the best fit line?

It would seem obvious that increasing n increases A.

I'm trying to model something. I could go for just one measurement that is 99.999999....% accurate...or I got to thinking I could, effectively sum the accuracy of less accurate models.
aaah but how many models at say 60% accuracy to get overall accuracy of 99.9%?

I'm not mathematically inclined but I'm decent with a spreadsheet so don't go overboard with a solution...and ideally it's something I could plug into google sheets. I'm just trying to wrap my head around a solution.

Oh, some limits.
whatever formula...the value of A can't exceed 100%, nor 0.
..and if it's only 1 measurement (n) with accuracy 'a', then A has to equal 'a'

Group Accuracy - Google Drive

docs.google.com

 

[blamocur]

If the errors of your devices are independent random variables, then the average error for [imath]n[/imath] devices should be [imath]\frac{a}{\sqrt{n}}[/imath]. Here you can find a more detailed discussion.

 

[Dr.Peterson]

For your information, your first suggested formula,

​

is what you would use if you knew the probability of a given individual being right on a yes/no question, and wanted to know the probability that at least one of n people would be right. So if each person were right 60% of the time, then among 3 people, at least one would be right 94% of the time. (That's because the probability of a given one being wrong is 40%, and the probability that all 3 are wrong is 0.40^3 = 0.064 = 6.4%.)

This is, of course, nothing like what you are looking for. That's why you don't use random formulas to solve a problem.

 

[NotJeffM]

Let’s say you have three different devices, each with validated accuracy of 60%.

You have the following readings: 9, 10, and 12. Notice that each reading is within 60% of any other (if this were not the case, the hypothesis that each measuring device is 60% accurate would be contradicted).

Based on that you can conclude that the true value lies between 4.8 and 14.4

The mean of 9, 10, and 12 is about 10.3. So the accuracy of the mean reading is about 47%.

You do get improved accuracy with multiple measuring devices, but it is hardly impressive. With just a spread sheet, you could explore numerically how increasing the number of devices increases accuracy.