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I understand that $$\sqrt{n}(\frac{\bar{X} - \mu}{\sigma})$$

has a standard normal distribution, as per definition.

How do I approximate this distribution to the t distribution? According to my tutor, this approximates to the t distribution with (n-1) degrees of freedom. And the equation will look like this.

$$\frac{\sqrt{n}(\bar{X} - \mu)}{s}$$

The only theorem I can think of that fits into this situation is that if Z~N(0,1) and U~$X_n^2$ and Z and U are independent, then the distribution of $\frac{Z}{\sqrt{U/n}}$ will be the t distribution with n degrees of freedom.

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  • $\begingroup$ For $X_i$ iid $N(\mu,\sigma^2)$, we know that $\bar X\sim N(\mu,\sigma^2/n)$ and $(n-1)s^2/\sigma^2\sim \chi^2_{n-1}$ independently, where $s^2$ is the sample variance with divisor $n-1$. Using these two we can form the $t_{n-1}$ variate your tutor is talking about. You know the definition of 't' distribution. $\endgroup$ – StubbornAtom Sep 18 '18 at 9:57
  • $\begingroup$ Sorry, I am not too familiar with what you mean. Would you be able to show me in greater detail? $\endgroup$ – statsguy21 Sep 18 '18 at 11:08

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