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Error Propogation

  • #1
This isn't a homework problem. I'm just having ALOT of trouble understanding how to do error propogation.
[EDIT] These are actually lab results. But its not a "homework" question. Its how to do uncertainty. Just trying to be clear.

Consider we take different measurements of an objects acceleration with uncertainties:
-0.2590 ± 0.0065 m/s^2
-0.2760 ± 0.0019 m/s^2
-0.2800 ± 0.0057 m/s^2
-0.2510 ± 0.0230 m/s^2
-0.2640 ± 0.0073 m/s^2
I want to take the average acceleration with the uncertainty of the average.
So the average is fine and easy however;
There are two ways to take the uncertainty:
1. Standard deviation of the acceleration divided by the square root of the number of accelerations.
2. Since to get the mean accelerate we summed accelerations, we use standard error propogation rules for addition and sum the square of the uncertainties and take their square root.

Notice in both of the above, we ignore the other method. In the first, we ignore the fact that each value has its own uncertainty and just look at the standard deviation.

In the second, we ignore the distribution of our data and only look at their individual uncertainties.

So what do I do...? :(.
 
Last edited:

Answers and Replies

  • #2
D H
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So the average is fine and easy however.
Is it? It's not. One thing you don't want to do is to let that noisy measurement of -0.2510 have near as much influence as those other measurements, and that very clean measurement of -0.2760 should have a lot more influence than any of the other measurements. There are lots of ways to do this, but one widely used approach is to weight the measurements by the inverse of the variance. It's a lot easier if the weights sum to one:
$$w_i = \frac{1/\sigma_i^2}{\sum_j 1/\sigma_j^2}$$
With this, the weighted mean is
$$\bar x = \sum w_i x_i$$

What about the variance? An estimate of the variance that follows the same paradigm is given by
$$s^2 = \sum w_i^2 \sigma_i^2$$
Note: This is a biased estimate. It underestimates the variance, with the bias most marked with a small sample size.
 
  • #3
Ugh... I'm just a first year. My statistics class is next year and I only have to follow standard rules for now, so even if I wanted to, I probably shouldn't look up formulas, especially when there are apparently more than one way to do something (as per yourself).
Also, why should one thing have extra weight and the other less? That seems like we are biasing the results and doesn't make a lot of sense to me. So I will refrain.

Back to my original problem. How do I go about calculating the final uncertainty of the unweighted average?

I really really appreciate your response though.
 

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