It looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results.

# Statistics

## Sample Size

We can calculate the required sample size to reach a certain confidence level.

For the confidence level, the value we add/subtract to the mean is called the sampling error, E. Thus, the confidence intervals could be expressed as

$\mu \approx \bar{x} \pm E$

where

$E=z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right)$

We can rearrange E to solve for the sample size required to reach a specific confidence level.

 Determining the Sample Size for a Confidence Interval for $$\mu$$. $n=\frac{(z_{\alpha/2})^2\sigma^2}{E^2}$ when you do not have $$\sigma$$ and the sample size is expected to be small. $n=\frac{(t_{\alpha/2})^2\sigma^2}{E^2}$
 Determining the Sample Size for a Confidence Interval for Proportions, $$p$$. $E=z_{\alpha/2}\sqrt{\frac{pq}{n}}$ $n=\frac{(z_{\alpha/2})^2 pq}{E^2}$

Example: Suppose you wish to estimate a population mean correct to within 0.15 with a confidence level of 0.90. $$\sigma^2$$ is approximately equal to 5.4. Find the sample size required.

Solution:

The population variance is 5.4, so we know we can use the z-statistic a 90% confidence level. $$z_{0.10/2}=1.645, E=0.15$$

$n=\frac{(1.645)^2(5.4)}{(0.15)^2}=649.446$

You need more than 649 data points, so we round up to a sample size 650 to obtain a 90% confidence interval with a sampling error within 0.15. 