My question is what is the equation to find n if you know standard deviation, and the CI of 95%. Like what is the equation for n?
2026-03-31 01:08:45.1774919325
How to find n=# of samples
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We know $\bar x\pm \frac{z^*\sigma}{\sqrt n}$ is the confidence interval. Call the margin of error $m$. The value $z^*$ is the corresponding z-score to a 95% CI (value that has area 0.025 to the right of it).
$$m=\frac{z^*\sigma}{\sqrt n}\implies n=(\frac{z^*\sigma}{m})^2$$