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Averaging errors

  1. Feb 10, 2009 #1
    This isn't a set problem itself, rather it is just a small part of my chemisty coursework.

    I have two values with known errors

    a) -0.000379272 ± 1.75895E-05
    b) -0.000576206 ± 4.13448E-05

    What happens to the errors when i average the two values?
  2. jcsd
  3. Feb 10, 2009 #2
    Good question and one which too few people ask. The first question to ask is what kind of errors are they. Are they Gaussian distributed or evenly distributed? If Gaussian, how many standard deviations does the given error value represent?

    You might want to start by just adding and subtracting the error values to the data and averaging and see what happens. Using a random number generator or a Gaussian random number generator might give you a better picture what you might encounter with real data. Maybe after doing it a few times you can come up with a rule for what happens.
  4. Feb 10, 2009 #3
    Thanks for your reply. I was hoping it would be something straight-forward. It's not hugely important, so I won't put much more effort into now.
  5. Feb 10, 2009 #4
    I know you were, but lots of times you learn a lot more by investigating little things like these and you're able to use your findings your whole career. Since very few other people investigate these things, it puts you slightly above the rest.
  6. Feb 10, 2009 #5

    D H

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    Staff Emeritus
    Science Advisor

    Before you even start to do something like averaging these values, you should ask whether it makes sense to do so. In this case it does not. The two values, along with their errors, are incommensurate. It would be better in this case to throw one out than to average them. Which one? Better make another measurement.

    If you insist in proceeding, you need to ask whether the errors are statistically independent and how they are distributed. You also should be careful how you do the averaging. Value (a) has a much smaller error than value (b), so the "average" should be closer to (a) than (b). What you want is a weighted average, with the value with the smaller error receiving the greater weight. One widely used weight is the inverse of the variance.

    If you compute the average as the variance-weighted mean and if the errors are uncorrelated from each other, the error in the weighted mean is given by

    [tex]\frac 1 {\sigma^2} = \sum_i \frac 1 {{\sigma_i}^2}[/tex]
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