Standard deviation sig figs

Question

Let's say I calculated a mean to be 2.475, but the data values had the least significant figure in the tenths place (i.e. 2.6, 2.8 etc.) so I round the mean value to 2.5 for correct sig figs. However, my standard deviation is 0.0835, and choosing my 1 sig fig for SD, I round to 0.08 So my SD is uncertain in the hundredths place, but my mean is in the tenths place....should I round my SD to 0.1 to match the mean? Or is it okay to have uncertainty at a decimal place further right than your mean?

Answer 1

You can know the mean more accurately than the data is known. If your data is rounded to one decimal, each item is uncertain by $\pm 0.05$. The variance of the uniform distribution of total width $0.1$ is $\frac 1{12}\cdot 0.1^2$. The variance of the sum of $N$ items is then $\frac N{1200}$. The standard deviation of the mean is about $\frac 1{35\sqrt N}$. You can use this as a guide on how many decimals to quote the mean to.

Answer 2

There are many rules about how many decimal places to show, some formulated by statisticians (seeking useful answers), others by editors (seeking pretty typeset pages).

You're on the right track, seeking to show only as many decimal places as convey useful information.

As a guide to the precision of parameter estimates, you can look at the widths of 95% confidence intervals. Crucially missing from the information about your example, as given in your question, are the sample size $n,$ the sample mean $\bar X$ and the sample SD $S.$ So I can only speculate.

Suppose $n = 100, \bar X = 50.0, S = 4.0,$ then a 95% CI for $\mu$ would be $50 \pm 2(4)/10$ or $50 \pm 0.8.$ Also a 95% CI for $\sigma$ would use the relationship $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = n-1).$ This leads to the 95% CI $(3.51,4.65)$ for $\sigma.$

sqrt((99*(16)/qchisq(c(.975,.025), 99)))
[1] 3.512027 4.646701

Both CIs seem to suggest that one-place accuracy is enough. However, if original data were rounded to two or three places, I might show the mean and SD (and if appropriate, their CIs) to two places. This would be more a matter of tact than of fact, but civility is a worthwhile goal.

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Note: Without regret, I gave two-place accuracy in my 95% CI $(3.51,4.65)$ for $\sigma.$ But if anyone thinks the second decimal place is crucial, then ponder the following 96% and 94% CIs. (As a matter of practical decision making, does anyone really consider the three intervals to be crucially different?)

sqrt((99*(16)/qchisq(c(.98,.02), 99)))
[1] 3.490703 4.681064
sqrt((99*(16)/qchisq(c(.97,.03), 99)))
[1] 3.530201 4.618012

But $(3.512, 4.647)$ for the 95% CI would pretend a degree of precision that is only a fantasy.