Although I am aware of the relationship "square of the t distribution is F distribution", I'm having a little trouble applying it to the following problem:
Use the t table to find the 80th percentile of the F$_{1,30}$ distribution.
Even if I use a simple function call in r to find the quartile of t$_{30}$ at 80th percentile, and square it, it is not equal to F$_{1,30}$ . Am I misinterpreting the relationship?
If $X\sim t(\nu)$, then $Y:=X^2\sim F(1,\nu)$. The $\alpha$ quantile of $Y$ is the positive number $q_{1,\nu,\alpha}$ such that $$\mathbb P(Y\leqslant q_{1,\nu,\alpha})=\alpha. $$ Since $x^2 \leqslant a$ if and only if $|x|\leqslant a^{\frac12}$, we have \begin{align} \frac45&=\mathbb P\left(Y\leqslant q_{1,30,\frac45}\right)\\ &= \mathbb P\left(|X|\leqslant q_{1,30,\frac45}^{\frac12}\right)\\ &= 1-2\mathbb P\left(X\leqslant -q_{1,30,\frac45}^{\frac12}\right), \end{align} where the last inequality follows from symmetry of the $t$ distribution. Hence $$P\left(X\leqslant -q_{1,30,\frac45}^{\frac12}\right)=\frac1{10}, $$ and the $t$-table yields $-q_{1,30,\frac45}^{\frac12} \approx -1.310415$. It follows that $q_{1,30,\frac45}\approx 1.717188$.