1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Error propagation question

  1. Oct 20, 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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Error propagation question
  1. Error propagation (Replies: 1)

  2. Error Propagation (Replies: 3)

Loading...