sample size determination using a population range


I have a problem about sample size determination.

The population of interest is somewhat vague. What I know about the population is only the population range R, i.e, R = max value - min value in the population. I don't know the population distribution, also.

Is it possible to determine a sample size estimating the population mean, so to speak, for a given Δ, to make (sample mean - Δ, sample mean + Δ) be a (1-α)% confidence level?
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TS Contributor
I just find this in Wiki:'s_inequality_on_variances

The population variance is bounded by \( \frac {1} {4} R^2 \)

So the SE of the sample mean is bounded by \( \frac {1} {4} \frac {R^2} {\sqrt{n}} \)

Then by CLT, you solve the inequality

\( \frac {1} {4} \frac {R^2} {\sqrt{n}} z_{\alpha} < \Delta
\iff n > \frac {1} {16} \frac {R^4} {\Delta^2} z_{\alpha}^2 \)

Since you state that only the population range is available, this is the most conservative estimate we have.
Thanks to BGM.

I summarized the result to be useful to others.

This is Popoviciu's inequality, which implies the standard error of the sample mean is bounded as following.

(BGM missed to square-root the upper bound, R^2 / 4, when he calculated the standard error of the sample mean.)

Anyway, the folloiwng inequalities can be obtained when we estimate the population mean using the sample mean with a (1-α) confidence level.

So for a given limit of error Δ, the sample size should satisfy the following inequities.

Thanks again, BGM.
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