I am currently learning about confidence intervals for the population mean. Assume we do not know the variance of the population. Let $\bar{x}$ be the sample mean, $s$ be the sample variance and $n$ the sample size. I learnt the following:
- Use $\bar{x}\pm t_{\alpha,n-1}\frac{s}{\sqrt{n}}$ if the data is normally distributed, where $t_{\alpha,n-1}$ is the $\alpha$ t-score from the $T$ distribution with $n-1$ degrees of freedom.
- If the data is not normally distributed, then with a large enough sample we can use the z-interval $\bar{x}\pm z_{\alpha}\frac{s}{\sqrt{n}}$ where $z_{\alpha}$ is a suitable z-score. The reason is that the T-ratio $T=\bar{x}-\mu/(s/\sqrt{n})$ becomes approximately normal with a large enough sample.
The lecture notes mention that we can consider $n=30$ a large enough sample size but without providing any explanation. So my question is simple: Why $n=30$?
