Trying to calcualte the uncertainty

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    Uncertainty
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Discussion Overview

The discussion revolves around calculating the uncertainty of the mean from two repeated measurements, A and B. Participants explore different approaches to express uncertainty, particularly in terms of the difference between the measurements, δ(A-B).

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes a method to calculate the uncertainty of the mean using the relationship between the uncertainties of A and B, specifically δ(A-B) and δA.
  • Another participant agrees with the approach but suggests avoiding the substitution of δB with δA, even if they are equal, to prevent confusion.
  • A later post questions the relevance of differentiation in the context of the uncertainty calculations, leading to a disagreement about the application of differentiation in this scenario.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculating uncertainty but disagree on the clarity of substituting δB with δA and the relevance of differentiation in this context.

Contextual Notes

There are unresolved assumptions regarding the definitions of δA and δB, as well as the implications of using differentiation in the calculations.

lavster
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if i have two repeat measuremetns, A and B, is the following correct? I am trying to calculate the uncertainty... I know there are easier ways to do this but i need to have the uncertainty of the mean in terms of δ(A-B).

from pairs of measurements A and B at each setting we have:
δ(A-B)=〖(〖δA〗^2+〖δB〗^2)〗^(1/2)
=√2 δA
(since δA=δB).

Rearrangement of this equation gives the uncertainty of a single measurement:

δA=(δ(A-B))/√2

Now considering the mean of each pair of measurements:

δ((A+B)/2)=δA/√2=(δ(A-B))/2

In particular, is is correct to have the √2 in the denominator in the last line?

Thanks
 
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It is correct, but I would not replace δB with δA even if they are equal, that is just confusing.
You can get the same result without the detour of δ(A-B), as δ(A+B)=δ(A-B) and δ(X/2)= (δX)/2 for all X.
 
Thanks :) is this just differentiation? So delta F(A) = da times dF/da ?
 
Where do you see differentiation?
So delta F(A) = da times dF/da ?
No.
 

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