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Homework Help: Error propagation question

  1. Oct 23, 2011 #1
    Not sure if this is the right section to post this..
    I have 3 measurements and was trying to take the average of the measurements and calculate the error of the average:
    replicate 1 = 8.9 (+/-) 0.71mg
    replicate 2 = 9.3 (+/-) 0.69mg
    replicate 3 = 8.8 (+/-) 0.70mg

    I get an average of 8.9333 (+/-) e where e=sqrt((rep 1 error)^2 + (rep 2 error)^2 + (rep 3 error)^2) which gives me a value of 1.21. But why is the error value so much higher in the average?
    What step am I missing? I don't know the derivation behind the error propagation formula - so I just use it as it is: e=sqrt((e1)^2+(e2)^2+...)
  2. jcsd
  3. Oct 23, 2011 #2


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    Homework Helper

    When I was in first year physics, they told me to divide the error by the square root of the number of measurements I had. This is contrary to the usual rules of error propagation. I found it in one place
    Look at section 3.10 near the bottom of the page.
    Unfortunately it is very difficult to read due to typos and anyway, I think they are just saying there is a rule somewhere else that you divide by root n. It certainly makes sense that your average should be more accurate than each measurement is.

    EDIT: more on how it is derived here: http://en.wikipedia.org/wiki/Standard_error_(statistics)
    Last edited: Oct 24, 2011
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