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

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.

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Arya 1 week 2021-11-19T06:27:44+00:00 1 Answer 0 views 0

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    2021-11-19T06:29:26+00:00

    Answer:

     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

    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.

     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

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