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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 $n=30$?

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    $\begingroup$ It is generally not true that $n=30$ is sufficient. $\endgroup$
    – Tim
    Commented Apr 5, 2023 at 19:19
  • $\begingroup$ 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 $2$ $\endgroup$
    – Henry
    Commented Apr 5, 2023 at 19:34
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    $\begingroup$ See this site search for many posts about this subject. $\endgroup$
    – whuber
    Commented Apr 5, 2023 at 21:32

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