0
$\begingroup$

Suppose to take three measurements of the same thing. Say for example of the area of the surface of a table. In this situation we have the bases and heights values and errors like

$b\pm \sigma_b$

$h\pm \sigma_h$

Then we calculate the area $A$ values as

$A=b h \pm \sigma_A$

Where $\sigma_A=\sqrt{(\frac{\sigma_b}{b})^2+(\frac{\sigma_h}{h})^2}$

Therefore we have three values of the area with three different errors.

Say $A_1\pm \sigma_{A_1},A_2\pm \sigma_{A_2},A_3\pm \sigma_{A_3}$

Now we want to find the average value of the area with its error.

I read that if I have a small amount of measurement ($n\leq 3$) of the same thing, the best error to give to the average value is the semidispersion, i.e.

$s.d.=\frac{A_{max}-A_{min}}{2}$

Therefore we should get

$A_{average} \pm s.d.$

My doubt is how are $\sigma_{A_1},\sigma_{A_2},\sigma_{A_3}$ important then? I mean, we could have $\sigma_{A_1}=\sigma_{A_2}=\sigma_{A_3}=10^{-6} m^2$ and still have a quite big error if the semidispersion is big.

In other words is there any difference between two sets of measurements with equal values for the area but one with $\sigma_{A_1},\sigma_{A_2},\sigma_{A_3}$ very big and the other with $\sigma_{A_1},\sigma_{A_2},\sigma_{A_3}$ very small? Or will the average area error be exactly the same?

Or maybe there is a better way to get the error on the average area in a situation like this?

$\endgroup$
  • 1
    $\begingroup$ Maybe I'm not reading carefully enough here, but I don't think that $\sigma_A$ has the right units. It looks unitless where it should have dimensions of area. $\endgroup$ – march Feb 26 '16 at 18:52
  • $\begingroup$ There is no such thing as "one best error estimate". How you estimate your errors depends on the actual statistical and systematic error. You may, for instance, know that your measurement device has an average offset or an integral non-linearity error (yes, most measurement devices have that). That's a totally different problem from a Gaussian distribution for which (and only for which!) the usual square root error formulas work. This is independent of the number of measurements. $\endgroup$ – CuriousOne Feb 26 '16 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.