**Instructions:** Use this calculator to estimate the effect of a finite population on the calculation of the standard error. Please provide the standard deviation \((\sigma)\), the sample size (\(n\)), and the population size (\(N\)), in the form below:

#### Finite Population Correction

The standard error for the sampling distribution of sample means is computed as:

\[\sigma (\bar X) = \displaystyle \frac{\sigma}{\sqrt n}\]where \(\sigma\) is the population standard deviation of the underlying distribution. This expression holds in the case that the population size is infinite (in which case the sampling processes can be considered as sampling with replacement). But the above expression won't be accurate if the population size is finite, equal to \(N\). In such case, there is a correction factor:

\[ cf = \sqrt{\frac{N-n}{N-1}} \]and the standard error is computed instead as:

\[\sigma (\bar X) = \displaystyle \frac{\sigma}{\sqrt n} \sqrt{\frac{N-n}{N-1}} \]Notice that the correction factor converges to 1 as \(N\) approaches to infinity. If you are dealing with sampling with an infinite population size, use instead this standard error calculator.

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