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Question:For example, the degrees of freedom (numerator) is equal to 3, the degrees of freedom (denomenator) is equal to 21, and the given F-statistic is equal to 5.25. Using a p-value calculater like this one (http://www.danielsoper.com/statcalc/calc07.aspx) I came up with the answer of .0073, but how does one calculate this number manually? Thanks.

Answers:1. F-statistic [ratio] is computed 2. P-value is area under distribution to 'right' of calculated F ratio ["look-up" value] 3. Large value of F ratio leads to rejecting Null Hypothesis; small value does not reject.

Question:For a random variable F: Find the value of the point "c" such that P(F

Question:If Y is distributed according to F distribution with (7,4) degrees of freedom, find probability of Y being more than 4.12 ? How to solve this question when I do not even know the significance level? Please help me to solve this question. Thank you! please tell me which statistical table to refer to as well. thanks

Answers:No need for significance levels. Any such question, irregardless of the actual distribution, is answered by determining the cumulative distribution function,F(x), representing the probability that the variable X is greater than x. For the F distribution, the cumulative distribution is: F(x)=BI[d1* x/(d1* x + d2)](d1 / 2, d2 / 2) BI := Incompelete beta function with d1 and d2 degrees of freedom. In the sources, I've pasted a link that claims to calculate this function. Otherwise, from the CRC Standard Mathematical Tables and Formulae source (p. 645-650 in 30th edition), you can see that for d1=4 and d2=7: F(3.98) = 0.90 and F(6.09) = 0.95 One can then linearly interpolate these results for a quick estimate.

Question:Recall that f(x) is a density function for a random variable X if and only if f(x) 0 for all x and f(x)dx = 1 (i.e., the area under the graph of f(x) is 1). Consider the following function: f(x) = .25, 1 < x < 5; f(x) = 0, otherwise. 1. Verify f(x) is a density function: a. Give a geometric argument based on the graph of f(x). b. Use calculus. 2. Recall that the expected value (long run mean value) for a continuous random variable is defined to be: = E(X) x f(x)dx Based on the graph of f(x), what value do you think is? Explain. 3. Use the definition for E(X) to verify the value in (2).

Answers:1. Pretty much trivial a. Rectangle of height 0.25 and base 4=> 4*0.25 + lim(x->infinity) x*0 = 1 b. We integrate piecewise: integral(-infinity,1)0dx = lim(x->-infinity) 0-0*x = 0 integral(1,5) 0.25dx = 4*0.25 = 1 integral(5,infinity) 0dx = lim(x->infinity)0*x-0=0 Total integral = 0+1+0=1 2. The expected value is, by symmetry considerations, 3. 3. Integral (1 to 5) 0.25xdx +0 = x^2/8|(1 to 5) = 25/8-1/8 = 3 Really, just apply definitions (the calculations are not at all difficult).

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