Answer :
Answer:
[tex] P(F \leq a) =0.05[/tex]
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)
[tex] P(F \leq b) =0.95[/tex]
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 [tex]\alpha=1-0.9=0.1[/tex] of the area on the tails and [tex]\alpha/2 =0.05[/tex] 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 [tex] P(a=0.21116<F<b=3.326) =0.9[/tex]
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.
[tex] F \sim F (r_1 =5, r_2 =10)[/tex]
First we want to calculate the value of a who satisfy:
[tex] P(F \leq a) =0.05[/tex]
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:
[tex] P(F \leq b) =0.95[/tex]
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 [tex]\alpha=1-0.9=0.1[/tex] of the area on the tails and [tex]\alpha/2 =0.05[/tex] 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 [tex] P(a=0.21116<F<b=3.326) =0.9[/tex]