Error propagation question

  • Thread starter davidp92
  • Start date
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+...)


Insights Author
2018 Award
If you take the arithmetic mean, then you should divide the sum under the root by three, otherwise you add up errors, which aren't additive. Another measure would be the arithmetic mean of the percentages.

Want to reply to this thread?

"Error propagation question" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads