Question

If F has an F-distribution with parameters r1 = 5 and r2 = 10, find a and
b so that P(F ≤ a) = 0.05 and P(F ≤ b) = 0.95, and, accordingly, P(a < F < b) =
0.90.

Answer 1

Usually you can find F tables for \(\alpha\) values with right-tail probabilities 0.05 and 0.10.
Essentially, the table will give you the quantiles for an F-distribution for a specified \(\alpha\) value.

Here, If you want \(P(F \le b)\), then you want to left side of the probability density function (left of the F-quantile) to be 0.95. Thus, your right-tail probability is 0.05, so you look at the value in a \(\alpha=0.05\) table and look at the intersection where you have degree of freedom 5 and 10 (usually r1 are the column values, and r2 are the row values).

For \(P(F\le a)=0.05\), what you can do is:
1) on the computer, calculate the quantile using a statistical program such as R (because usually you don't find tables with \(\alpha\) values of 0.95.

2) By hand: You can use the property that \(1/F\sim~F(r_2,r_1)\), that is 1/F has an F distribution with the degrees of freedom swapped. Thus,
\[P\left(F \le a\right) =P\left( \frac{1}{F} \ge \frac{1}{a} \right)=1-P\left( \frac{1}{F}\le \frac{1}{a}\right)=0.05\]

The 2nd equality just comes from the fact that you are using the complement probability (it doesn't matter if you use \(\le\) or \(<\) since the distribution is continuous.

Now, You should deduce from the 2nd equality that \( P\left( \frac{1}{F}\le \frac{1}{a}\right)\)=0.95. Thus, you have acumulative probability of 0.95, and thus a tail probability \(\alpha=0.05\). You can now you the table with \(\alpha=0.05\) values, but now taking into account that you are looking at a \(1/F\) distribution. So the column degree of free is 10, and the row degree of freedom is 5. And thus you find the \(1/a\) value, and just reciprocate it to get \(a\).

Answer 2

You could also use this site to compute F values: http://stattrek.com/online-calculator/f-distribution.aspx