# How many sample do we need for normality of t-ratio? [duplicate]

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 $$n=30$$?

• It is generally not true that $n=30$ is sufficient.
– Tim
Commented Apr 5, 2023 at 19:19
• If $\alpha=0.05$ you get $z_\alpha=1.96$ in your two-tailed example, which people sometimes round to $2$. The equivalent $t_{\alpha,30}\approx 2.04$ which also rounds to $2$ Commented Apr 5, 2023 at 19:34