Confidence Intervals for One Means: Finite Population Correction

When we sample without replacement from a finite population of size $N$, the standard error of the sample mean must be reduced by the finite population correction (FPC) factor

$$\sqrt{\dfrac{N-n}{N-1}}.$$

Applying this correction to the usual $t$-interval for a population mean $\mu$ yields

$$\bar{x} \;\pm\; t^*_{n-1} \cdot \dfrac{s}{\sqrt{n}} \cdot \sqrt{\dfrac{N-n}{N-1}},$$

where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, and $t^*_{n-1}$ is the critical value from the $t$-distribution with $n-1$ degrees of freedom at the desired confidence level.

For example, suppose we draw $n=16$ items without replacement from a population of $N=80$, obtaining $\bar{x}=12$ and $s=4$. To build a 95% CI we use df $=15$, so $t^*\approx 2.131$:

$$\dfrac{s}{\sqrt{n}} = \dfrac{4}{4} = 1,$$ $$\sqrt{\dfrac{N-n}{N-1}} = \sqrt{\dfrac{64}{79}} \approx 0.9001,$$ $$\text{ME} = 2.131 \cdot 1 \cdot 0.9001 \approx 1.918.$$

The 95% CI is therefore $12 \pm 1.918 \approx (10.08,\,13.92).$