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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.

\[n=\frac{(z_{\alpha/2})^2\sigma^2}{E^2}\] when you do not have \(\sigma\) and the sample size is expected to be small. \[n=\frac{(t_{\alpha/2})^2\sigma^2}{E^2}\] |

\[E=z_{\alpha/2}\sqrt{\frac{pq}{n}}\] \[n=\frac{(z_{\alpha/2})^2 pq}{E^2}\] |

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.

Statistics by Matthew Cheung. This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.

- Last Updated: Apr 20, 2023 12:47 PM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=717168
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