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If I conducted 6 trials for a lab, all with the uncertainty of (+/- 0.1 mL)... After averaging the 6 trials to plot the results, what is the new uncertainty?

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closed as off-topic by bon, M.A.R., Jan, Todd Minehardt, Martin - マーチン Oct 25 '15 at 12:18

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    $\begingroup$ This is a homework question. We ‎have a policy which states that you should show your thoughts and/or efforts into solving the ‎problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this ‎question may get closed.‎ $\endgroup$ – M.A.R. Oct 24 '15 at 21:47
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In any case, you should calculate the mean $\overline X$, the standard deviation $\sigma$, and the standard error of the mean $\sigma_{\overline X}$ for the six trials.

However, the estimated uncertainty $u(y)$ of the final result $y$ depends on what is meant by “all with the uncertainty of $\pm0.1\ \mathrm{ml}$”, i.e. $u(x)=0.1\ \mathrm{ml}$.

For example, if this uncertainty $u(x)=0.1\ \mathrm{ml}$ is caused by the unknown error of a pipette and the same pipette was used for all six trials, each of the six individual results includes the same systematic error. You can neither eliminate this error nor reduce the caused uncertainty by making additional trials. In this case, you should combine (by using the rules for error propagation) the calculated standard error $\sigma_{\overline X}$ and the assumed uncertainty $u(x)=0.1\ \mathrm{ml}$. Therefore, the estimated uncertainty $u(y)$ of the final result cannot be smaller than $u(x)=0.1\ \mathrm{ml}$.

If the uncertainty $u(x)=0.1\ \mathrm{ml}$ is caused by a random error of each individual trial, it is already included in the observed standard deviation $\sigma$ and thus already taken into account for the calculated standard error $\sigma_{\overline X}$. Therefore, the observed standard deviation $\sigma$ should be greater than (or at least equal to) the assumed uncertainty $u(x)=0.1\ \mathrm{ml}$. In this case, you may assume that the estimated uncertainty of the final result $u(y)$ is approximately equal to the calculated standard error $\sigma_{\overline X}$.

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