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# T distribution and F distribution

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Please see attached problem, and please explain all aspects of both problems involving T and F distributions.

a) Z is a standard normal random variable, and U is distributed as X(v)^2. Z and U are independent. State the distribution of T = Z/sqrt(U/v) . What is the distribution of T^2?
b) Y is a random variable distributed as F(1,5) . Use your answer to part (a), together with the tables of the t (not the F) distribution, to find P(Y > 4) . Also find the value of c such that P(Y > c) = 0.01.

https://brainmass.com/statistics/f-test/distribution-distribution-465245

## SOLUTION This solution is FREE courtesy of BrainMass!

a) Z is a standard normal random variable, and U is distributed as X(v)^2. Z and U are independent. State the distribution of T = Z/sqrt(U/v) . What is the distribution of T^2?

T, by definition, is called the student t distribution of degrees of freedom v (where v is the d.f. of the chi square). Furthermore, if T is a student t distribution of d.f. = v, then T^2 follows an F(1,v), where F is the F distribution.

b) Y is a random variable distributed as F(1,5) . Use your answer to part (a), together with the tables of the t (not the F) distribution, to find P(Y > 4) . Also find the value of c such that P(Y > c) = 0.01.

We note that if Y is distributed as F(1,5), and t is distributed as t(5), then t^2 = F.

Thus, P(Y > 4) = P(t^2 > 4) = P(t > 2) + P(t < 2)

From a t table (for example here http://en.wikipedia.org/wiki/Student%27s_t-distribution), we can see that

P(t > 2) = 0.05 (roughly), and since t is symmetric with respect to the origin, P(t^2 > 4) = P(t > 2) + P(t < 2) = 2 X 0.05 = 0.1

If we wish to find P(Y > c) = 0.01, we are essentially looking for P(t > sqrt c) + P(t < sqrt c) = 0.01.

Since t is symmetric, each part should equal to 0.005.

Using the same t table, this number is 4.032. Thus c = 4.032^2 = 16.26.

You may check the answers by computing the probabilities using the F distribution. Here is a calculator http://stattrek.com/online-calculator/f-distribution.aspx.

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