Two samples of sizes 8 and 10 are drawn from two normally distributed populations having variances 20 and 36, respectively. Find the probability that the variance of the first sample is more than twice the variance of the second
If "s_1^2" and "s_2^2" are the variances of two indepenent random samples of sizes "n_1" and "n_2" respectively from normal populations with the same variance, then
is a random variable having the "F"-distribution with parameters "n_1-1" and "n_2-1"
It is assumed that sample variances are calculated using the formula
"P(s_1^2>2s_2^2)=P(\\dfrac{s_1^2\/\\sigma_1^2}{s_2^2\/\\sigma_2^2}>2\\dfrac{\\sigma_2^2}{\\sigma_1^2})"
"=P( F_{8-1, 10-1}>2\\cdot\\dfrac{36}{20})=P( F_{7, 9}>3.6)"
From the distribution tables required probability lies between "0.01" and "0.05."
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