# How to construct a confidence interval of the mean, when the distribution is unknown and the sample is small?

I'm looking for the preferred approach to construct a finite sized confidence interval for the population mean, assuming:

• The distribution of the population is unknown
• The sample size is low
• The population standard deviation is unknown

The usual approaches do not work in this setting:

• Using the usual t-distribution to construct the confidence interval is not possible because we do not assume normality
• We can't use the central limit theorem because of the low sample size
• The standard deviation is unknown, so we can't use Chebyshev's inequality

I found out that if we assume the distribution is unimodal & symmetric, we can construct a confidence interval for the population mean from a single value. However, it is unclear to me how to generalize this to higher sample sizes (say, 10 or 15 observations), and I wonder if the unimodal & symmetric assumptions are necessary.

• Maybe something as described by Zhou & Dinh (2005)? Mar 28, 2021 at 9:21
• @COOLSerdash: I was read to write bootstrap t-procedure but your suggestion is even better. When/if you write it as an answer, let me know to upvote properly. Mar 28, 2021 at 11:05

In general this is impossible. Suppose you have a function $$f$$ for which it is claimed that for any population with a finite mean $$\mu$$, $$f$$ applied to a sample of size $$n$$ from that population will return a finite length interval which is a $$100\alpha\%$$ confidence interval for $$\mu$$. Let $$I = f(0, ..., 0)$$ be the confidence interval $$f$$ produces when every value in the sample is zero. Pick $$c \notin I$$. Now suppose the population distribution is as follows:
$$P(X = 0) = 1 - P(X = \frac{c}{1 - \alpha^{\frac {1}{2n}}}) = \alpha^{\frac 1 {2n}}$$
Then the mean of this distribution is $$c$$ which does not belong to $$I$$. But with probability $$> \alpha$$, a sample of size $$n$$ from this distribution will consist of $$n$$ zeroes, so will give you an interval which does not contain the true population mean.
So $$f$$ does not work on this distribution.
(This argument assumes $$f$$ is deterministic. It can be tweaked to work even if $$f$$ is random.)