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06/09/2005 00:52:40 »
I am doing a relative error problem that asks what is the possible relative error when 7.5 mL are measured using a 10mL measuring pipet. The % Tolerance for a 10mL measuring pipet is 0.6%.
So the equation is Error=(.6%*10mL)/7.5 mL which gives you a relative error of .8% How many sig figs? Do we take into account the 10mL and only give one sig fig or do we use two based on the 7.5mL?
Any help is greatly appreciated.
Re: Sig Figs
Reply #1 on:
06/09/2005 01:30:17 »
If the pipet is not graduated the error could be enormas. So I will assume graduation: you would then assume linarity in the error, and work 7.5 * 0.6%. Next ask what terms the error is expressed in in terms of statistics. The confidence levels are generally 67%, 95 or 99.7% confidence levels (+/- 1 sig, +/- 2 sig or +/- 3 sigma).
As a retired quality control manager I find that most undereducated people instinctively give +/- 2 sigma numbers when asked about variation. So off the top of my head, I would say you have been given insufficent information to give an accurate answer in terms of probability or confidence level. If the 0.6% is 1 sigma, you would say that at the 95% confidence level the error would be less than or equal to +/- (7.5 * 0.006 * 2 sigma) or 0.09 ml which is 1.2%. That seems awlful small to me for the pipets I have used, so let me go back to a non-graduated pipet solution and say that the error cannot be determined, and you should use the correct size pipet for the experiment.
Last Edit: 06/09/2005 01:32:05 by David Sparkman