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.

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)

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 of the area on the tails and 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

Step-by-step 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.

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

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:

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 of the area on the tails and 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