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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.

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    $\begingroup$ Maybe something as described by Zhou & Dinh (2005)? $\endgroup$ Mar 28, 2021 at 9:21
  • $\begingroup$ @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. $\endgroup$
    – usεr11852
    Mar 28, 2021 at 11:05

1 Answer 1

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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.)

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