Question

Suppose F has an F-distribution with parameters r1 = 5 and r2 = 10. Using only 95th percentiles of F-distributions, 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. Hint: Write P(F ≤ a) = P(1/F ≥ 1/a) = 1 − P(1/F ≤ 1/a), and use the result of Exercise 3.6.9 and R.

Answer 1

`P(F \leq a) =0.05`

We can use the following excel code: "=F.INV(0.05,5,10)" and we got a = 0.21119

With R the code is:

> qf(0.05,5,10,TRUE)

`P(F \leq b) =0.95`

We can use the following excel code: "=F.INV(0.95,5,10)" and we got b=3.326

The R code is:

> qf(0.95,5,10,TRUE)

Now for the last case we want to find two values a and b who accumulates 0.90 of the area on the middle so then we need
`\alpha=1-0.9=0.1` of the area on the tails and
`\alpha/2 =0.05` of the area on each tail.

And the two values on this case are a=0.21119 and b =3.326

And we can check this using the following code: "=F.DIST(3.326,5,10,TRUE)-F.DIST(0.21116,5,10,TRUE)"

And we satisfy that
`P(a=0.21116<F<b=3.326) =0.9`

Explanation:

For this case we know that F follows a F distribution with parameters r1= 5 degrees of freddom for the numerator and r2= 10 degrees of freedom for the denominator.

`F \sim F (r_1 =5, r_2 =10)`

First we want to calculate the value of a who satisfy:

`P(F \leq a) =0.05`

We can use the following excel code: "=F.INV(0.05,5,10)" and we got a = 0.21119

With R the code is:

> qf(0.05,5,10,TRUE)

Then we want to calculate a value of b who satisfy:

`P(F \leq b) =0.95`

We can use the following excel code: "=F.INV(0.95,5,10)" and we got b=3.326

The R code is:

> qf(0.95,5,10,TRUE)

Now for the last case we want to find two values a and b who accumulates 0.90 of the area on the middle so then we need
`\alpha=1-0.9=0.1` of the area on the tails and
`\alpha/2 =0.05` of the area on each tail.

And the two values on this case are a=0.21119 and b =3.326

And we can check this using the following code: "=F.DIST(3.326,5,10,TRUE)-F.DIST(0.21116,5,10,TRUE)"

And we satisfy that
`P(a=0.21116<F<b=3.326) =0.9`