How can I determine sample size in interval estimation? The oridinary way of determining sample size $n$ would be like the following:

  • Let $n$ be big enough to the extent of Central Limit Theorem.
  • take a permitted error $\epsilon$ arbitrarily which corresponds to the extent of Confidence Interval(CI).
  • Let confidence coefficient "95" (1.96) , then $1.96 * \sigma / \sqrt{n} = \epsilon$, where $\sigma^2$ is the population variance.
  • Solve this for $n$ , $n = (1.96 * \sigma / \epsilon )^2$
  • This would be the estimate of sample size which gives 95% CI with error bar $\epsilon$.

What I want to know is how to calculate the population variance $\sigma^2$ which appears in $n = (1.96 * \sigma / \epsilon)^2$ in case that the population distribution is not given( not necessarily Gaussian) and the population variance $\sigma^2$ is unknown.

Since the population distribution is not given, we cannot use t-distribution. So, we have to calculate unbiased variance $s^2$ from a sample and use it as the estimate of population variance. However, in the calculation of unbiased variance from a sample, the sample size $n$ appears. It is circulating.

How can I estimate sample size in case that the population distribution is not given( not necessarily Gaussian) and the population variance $\sigma^2$ is unknown?

  • $\begingroup$ So basically you are asking: how many samples do I need when I do not have any assumption on my distribution? $\endgroup$
    – md5
    Jun 2, 2021 at 15:49
  • $\begingroup$ Related: stats.stackexchange.com/questions/517043/… $\endgroup$
    – Peter O.
    Jun 2, 2021 at 15:50
  • $\begingroup$ @md5 yes. Both population distribution and population variance are unknown. $\endgroup$
    – somia
    Jun 2, 2021 at 16:06
  • $\begingroup$ Actually, I am in a project of computer simulation of sampling some physical quantity which appears with some randomness. If we could confirm that the randomness obeys Gaussian distribution, the story would be quite simple, just using t-distribution. However, it is not clear it obeys Gaussian because the appearance of the randomness is quite complicated. $\endgroup$
    – somia
    Jun 2, 2021 at 16:14
  • $\begingroup$ @PeterO. I checked it. I don't fully understand but it seems the article is not only about the case that population distribution is unknown but also about the case the n is small to the extent central limit theorem cannot be used. In my case, n can be set as big as required, and the sample mean would obey Gaussian distribution. $\endgroup$
    – somia
    Jun 2, 2021 at 16:24

1 Answer 1


The central limit theorem applies only if the random variables involved are independent and identically distributed with finite, not infinite, variance.

In that case, if all you know is that the population variance is finite, but you don't know an upper bound on that variance, then it's impossible in general to determine a sample size to achieve any given confidence interval. The same applies if the population random variable has finite moments but their bounds are unknown.

See Kunsch et al. 2018, Theorem 3.4. The proof boils down to the problem of distinguishing two Bernoulli random variables (which both have finite variance and moments):

  • One takes on the value 0 with probability 1.
  • The other takes on an arbitrarily large value with probability arbitrarily close to 0, and 0 otherwise.

See also the following question, whose answer shows a similar impossibility result:

For Bernoulli and other bounded variables, you can determine a sample size using Chebyshev's inequality or Hoeffding's inequality; see the following:

If you can accept a weaker guarantee on confidence intervals (for example, "within a reasonably tight interval of the true mean with probability at least $1-\delta$" rather than the stronger "within $\epsilon$ of the true mean with probability at least $1-\delta$"), then recent work has offered algorithms that estimate the mean with the sole assumption that the mean is finite (or any moment in between the first and second is finite) (Cherapanamjeri et al. 2020). However, these algorithms are far from trivial.



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