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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?

John
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2 Answers2

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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.

Ross Millikan
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  • For context, this relates to data in chemistry. I was explicitly told that mean values had to be rounded to the "least significant decimal figure" of the data points. I guess my main question is just can a standard deviation be more precise than a mean i.e. a mean rounded to tenths place and then a SD to hundreths place? – John May 30 '20 at 19:48
  • If you have a lot of samples then you can know the mean to within $\pm 0.01$ even if every sample is rounded to $\pm 0.1$. The usual rule is to round a sum to the same number of decimals. If you have at least $10$ items in the sum, the count of items is exact, so the uncertainty in the mean is smaller by a factor $10$ than the uncertainty in the sum. I suspect that whoever says you have to round to the same number of decimals is not thinking about how the division influences the error. – Ross Millikan May 30 '20 at 20:02
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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.

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.

BruceET
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