Let $X_1,X_2,\ldots,X_n$ a random sample of a normal population with mean $0$ and variance $σ ^2$. Consider the random variable define as $T=\frac{\bar{X}\sqrt{n}}{S}$. Which is the distribution of $T^2$?
Standardizing, $Z=\frac{\bar{X}-0}{\sqrt{\sigma^2/n}}=\frac{\bar{X}\sqrt{n}}{\sigma}$, because $\bar{X}\sim N(0,\sigma^2/n)$. Also, $Z^2\sim \chi^2_{(1)}$.
On the other hand, we have that $$\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{(n-1)}$$
How can I get to this distributions from $T^2=\left(\frac{\bar{X}\sqrt{n}}{S}\right)^2$, in order to say that it distributes $F_{(1,n-1)}$?
Note that $\bar{X}$ and $S^2$ are estimators of $\mu$ and $\sigma^2$, respectively.
Best Answer
If $U\sim \chi^2_p$ and $V\sim\chi^2_q$ and $U$ and $V$ are independent, then the ratio $$ \frac{U/p}{V/q} $$ has $F_{p,q}$ distribution. This is either the definition of the $F$ distribution, or can be proved as a theorem. Either way, your conclusion follows immediately from the additional fact that $\bar X$ and $S^2$ are independent.