2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $2.306^2 \approx 5.32$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.