# Significant figures in a practical experimental problem [duplicate]

Here is a question that combines uncertainties in measurements and significant figures.

Consider the following results to measure the value of g:

$$g={9.7,9.8,9.7,10.0,10.1,10.3}$$

Then to four decimal places $\bar{g} = 9.9333$, while the uncertainty of the mean is $$0.2422/\sqrt{6} = 0.0989$$

But clearly we would not quote these values as they stand.

So, what would be the statement of the mean value of g and its uncertainty to the correct number of significant figures? Would it be $\bar{g}=9.9 \pm 0.1$ or $\bar{g} =9.93 \pm 0.10$? The original measurements had two significant figures, but it would seem using the mean and its uncertainty would permit an additional significant figure. And what is the general rule for the significant figures for mean values and their uncertainties?

• Possible duplicate of Multiple measurements of the same quantity - combining uncertainties – Kyle Kanos Aug 6 '18 at 16:49
• shouldn't it be dividing by sqrt(5), i.e. sqrt(n-1)? – Walter Aug 6 '18 at 17:18
• Isn't uncertainty built into significant figures in the last written digit? To explicitly show the uncertainty in a measurement, I would completely avoid sig figs and instead write the most precise estimate $\pm$ the error. – Quantumness Aug 6 '18 at 17:21
• – sammy gerbil Aug 6 '18 at 18:00
• Walter: the Standard Error of the Mean (SEM) is given by $$\sigma_\bar{x} = \cfrac{\sigma}{\sqrt{n}}$$, where $\sigma$ is the standard deviation of the population (that's where $n-1$ comes in) and $n$ is the number of samples to determine the mean. – MSulkanen Aug 6 '18 at 19:20

Here's a significant-figures-only way to think of it. When you add your six values, you get $\sum g = 59.6$, where the third significant figure is trustworthy because it came from the trustworthy digits in each of the six individual values. So your average, $\overline g = \sum g / N$, should also have three significant figures.
• Well put. OK, so this also translates to the uncertainty as well, consistent with the significant figures of the mean? $$\sigma_\bar{g} = \cfrac{0.242}{\sqrt{6}} \rightarrow \sigma_\bar{g} = 0.10$$ That seems to hang together with the significant figure of the mean. – MSulkanen Aug 7 '18 at 13:56
• I think reporting an uncertainty of $\pm0.10$ is more appropriate than an uncertainty of $\pm0.1$, because the former implies that small differences in the data might have shifted the result to $\pm0.09$ or $\pm0.11$, while the latter suggests a small difference in the data might have shifted the result to $\pm0.02$. Keeping an extra significant figure if the leading digit is a one (or sometimes a two) is a good housekeeping idea. – rob Aug 7 '18 at 15:30