# Confidence in a range estimate, and +- 2sigma rule of thumb

It's common to see people report variability in their realized sample through the $$\bar{x} \pm 2s$$ range, where $$\bar{x}$$ is the realized mean and $$s$$ the realized standard deviation estimate in their sample.

Beyond informing about the realized sample, I am guessing $$\bar{x} \pm 2s$$ is often reported in the hope that it provides information about $$\mu \pm 2\sigma$$.

But unless the sample is very large, $$\bar{x}$$ and $$s$$ will be imperfect estimates of $$\mu$$ and $$\sigma$$, and $$\bar{x} \pm 2s$$ will therefore be an inaccurate estimate of $$\mu \pm 2\sigma$$.

Whether or not it's a good idea to try and estimate $$\mu \pm 2\sigma$$, this leads me to wonder about standard practices to do so, and to assess the accuracy of the estimated range (including the relationship between accuracy and sample size $$n$$).

I guess you could construct standard CIs around both $$\mu$$ and $$\sigma$$ and somehow "combine" these two CI looking at worse case scenarios to get a "maximum" CI for $$\mu \pm 2\sigma$$ you're equally confident in (say by taking the highest value in the CI for $$\sigma$$ plus/minus the highest/lowest value in the CI for $$\mu$$).

But that seems super hacky and not well-founded theoretically (among other things, I suspect there are issues with $$s$$ appearing in the CI formula for both $$\mu$$ and $$\sigma$$?).

If that's indeed not a good idea, what would be a better approach to estimating $$\mu \pm 2\sigma$$ with confidence? In the classical CI sense, and under some reasonable distributional assumptions, is there an observable random interval $$[\underline{F}(X_1, \dots, X_n; \alpha), \bar{F}(X_1, \dots, X_n; \alpha)]$$ that is guaranteed to include the whole $$\mu \pm 2\sigma$$ range $$(1-\alpha)$$ percent of the time (in a similar way that $$(1-\alpha)$$ percent of the time, the range $$\bar{X} \pm z_{\alpha/2} (s/\sqrt{n})$$ includes $$\mu$$ itself)?

That is, I am looking for a family of statistics $$\underline{F}(X_1, \dots, X_n; \alpha)$$ and $$\bar{F}(X_1, \dots, X_n; \alpha)$$ such that:

$$P\big(\underline{F}(X_1, \dots, X_n; \alpha) \leq \mu - 2\sigma < \mu + 2\sigma \leq \bar{F}(X_1, \dots, X_n; \alpha) \big) = 1-\alpha.$$

• You are right that the sample mean and standard deviation $\bar{x}$ and $s$ are estimates of the population mean $\mu$ and standard deviation $\sigma$. But you seem to confuse the standard deviation $s$ with the standard error of the mean $s/\sqrt{n}$. These are not the most intuitive of concepts; see here. Apr 24 at 11:02
• @dipetkov: When I wrote "is there an observable random interval similar to 𝑋¯±𝑧𝛼/2(𝑠/𝑛√) that includes the whole 𝜇±2𝜎 range (1−𝛼) percent of the time?", what I meant was: "is there an observable random interval that includes the whole 𝜇±2𝜎 range (1−𝛼) percent of the time ? (in the same way that similar to 𝑋¯±𝑧𝛼/2(𝑠/𝑛√) includes 𝜇 (1−𝛼) percent of the time)". Maybe that's where the impression of confusion came from? I'll make a correction to the question anyways.
– FZS
Apr 24 at 23:20
• Re "I am guessing x¯±2s is often reported in the hope that it provides information about μ±2σ." Not usually. You describe a tolerance interval. This has important applications, especially in quality control. Those are computed using special formulas that account for the uncertainty in $\mu\pm2\sigma.$ Please search our site for more information. stats.stackexchange.com/questions/26702 would be a good starting point.
– whuber
Apr 25 at 13:11
• @whuber: Thanks a lot for mentioning the notion of "tolerance interval" which I wasn't familiar with. Most likely the keyword I was missing in my search. Answers are still welcome, but I will look into it myself and hopefully be able to answer my own answer on that basis soon.
– FZS
Apr 25 at 14:12

If I understand things correctly so far, one connection with the $$\bar{x} + 2s$$ rule of thumb is that, with normal data, as the sample-size $$n$$ tends to infinity, $$\bar{x} + 1.96s$$ obviously gets closer and closer to including 95% of the population (as $$n$$ tends to infinity, this is true, I believe, for any level of confidence, be it 95%, 99%, 99.9%,...).
However, for finite $$n$$, intervals including 95% of the population with, say, 95% confidence are larger than $$\bar{x} + 1.96s$$ (e.g., the interval is around $$\bar{x} + 2.75s$$ when $$n=20$$, where calculations are from https://statpages.info/tolintvl.html with underlying statistical assumption --- in particular, normal data --- described in https://www.itl.nist.gov/div898/handbook/prc/section2/prc263.htm).