# Showing $F(1,v) = t^2(v)$ Using Charts

As the title states, I don't see how this is the case from the charts. I don't see how the F-ratio (with $1$ and $v$ d.f.) is equal to the square of the t-ratio (with $v$ d.f.).

Example:

Lets say we have $v=8$ degrees of freedom and $\alpha = 0.05$.

Using the t-chart, we see that the corresponding t-value is $1.86$.

Using the F-chart, we see that the corresponding F-value (using $1$ and $8$ degrees of freedom respectively) is $5.32$

However, $1.86^2 = 3.4596 \ne 5.32$

So how can this be the case? Am I doing something wrong in the thought process? I'm aware that it can be proven via formula, but my book tells me I can validate this through the charts, and that is what I am looking to do for this.

Any help would be greatly appreciated.

The t-distribution can take on negative values as well whereas F-distribution can only take positive values. Thus, you should look at the $0.025$ tail for t-distribution and not $0.05$. The t-distribution's critical value for $0.025$ with $8$ degrees of freedom is $2.306$ and $2.306^2 \approx 5.32$.