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.