Answer to Question #270143 in Statistics and Probability for NIHAL

Question

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

Answer 1

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

$$F=\dfrac{s_1^2}{s_2^2}$$

is a random variable having the $F$-distribution with parameters $n_1-1$ and $n_2-1$

$$\dfrac{s_1^2/\sigma_1^2}{s_2^2/\sigma_2^2}\sim F_{n_1-1, n_2-1}$$

It is assumed that sample variances are calculated using the formula

$$\dfrac{1}{n-1}\displaystyle\sum_{i=1}^n(x_1-\bar{x})^2$$

$$P(s_1^2>2s_2^2)=P(\dfrac{s_1^2/\sigma_1^2}{s_2^2/\sigma_2^2}>2\dfrac{\sigma_2^2}{\sigma_1^2})$$

$$=P( F_{8-1, 10-1}>2\cdot\dfrac{36}{20})=P( F_{7, 9}>3.6)$$

From the distribution tables required probability lies between $0.01$ and $0.05.$