Understanding Error Propagation in Averaging Measurements

davidp92
Messages
12
Reaction score
0
Not sure if this is the right section to post this..
I have 3 measurements and was trying to take the average of the measurements and calculate the error of the average:
replicate 1 = 8.9 (+/-) 0.71mg
replicate 2 = 9.3 (+/-) 0.69mg
replicate 3 = 8.8 (+/-) 0.70mg

I get an average of 8.9333 (+/-) e where e=sqrt((rep 1 error)^2 + (rep 2 error)^2 + (rep 3 error)^2) which gives me a value of 1.21. But why is the error value so much higher in the average?
What step am I missing? I don't know the derivation behind the error propagation formula - so I just use it as it is: e=sqrt((e1)^2+(e2)^2+...)
 
Mathematics news on Phys.org
If you take the arithmetic mean, then you should divide the sum under the root by three, otherwise you add up errors, which aren't additive. Another measure would be the arithmetic mean of the percentages.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top