# sample size determination using a population range

#### rolji

##### New Member
Hi.

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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#### BGM

##### TS Contributor
I just find this in Wiki:

http://en.wikipedia.org/wiki/Popoviciu'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.

#### rolji

##### New Member
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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