In my textbook it says that if two experimental results vary less than 3$\sigma$ then they can be considered to have arrived at the same result. My question is how do you determine this "x$\sigma$".

For example if i did an experiment to calculate $g$ and my result was $g$=9.79$\pm$0.07 and i want to compare it to $g$$^´$=9.80$\pm$0.22 , should i just use $$\frac{g^´-g}{error}\,\,\,\,?$$ But then what error do i use?If i use o.o7 i get a difference of 1.4$\sigma$ but if i use 0.22 i get 0,45$\sigma$. Or should i use a combination of both?

  • $\begingroup$ The right answer depends sensitively on the meaning intended for the two errors you have to start with. This can be an issue if you are talking about comparing measurements form papers. If they both represent 1 std. dev. random error intervals then there is a good answer (that involves a little more computation than the ones you suggest). Could you clarify? $\endgroup$ – dmckee Sep 10 '17 at 21:46
  • $\begingroup$ Yes, they represent 1 std. deviation. $\endgroup$ – Danbur Sep 10 '17 at 21:52

Given that both measurement have a 1 std. dev. random error estimate (all bets are off when systemantics and model dependencies rear their ugly heads!), then you are effectively comparing the difference of the measurements with zero.

So the error you use is the error of the difference.

Which means propagating the error in the usual way (i.e. adding the errors in quadrature) $$ \left[\frac{g' - g}{\sqrt{(\Delta g')^2 + (\Delta g)^2}}\right] \stackrel{?}{\le} 1 \;.$$

If the computed value is

  • less than one the measurements are in agreement.

  • more than a few the measurements clearly disagree.

  • bewtween 1 and—say—3 is more ambiguous.

    But your instructor may want you to treat them as disagreeing so that you can have a yes/no answer.


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