Question

If two independent samples of sizes m=26 and n=8 are taken from normal population, what is the probability that the variance of the first sample will be at least as large as the variance of the second sample?

Select one:a. 0.05b. 0.10 c.0.20 d. 0.01

Answer 1

Let $S^2_1$ be the sample variance for sample 1 and $S^2_2$ be the sample variance for sample 2. This question requires us to find,

$p(S^2_1\geqslant s^2_2)$. By expressing the two sample variances as a ratio, this probability can be written as,

$p(S_1^2/S_2^2\geqslant1)$. Now the ratio, $S_1^2/S_2^2$ follows an $F$ distribution with $n_1-1$ and $n_2-1$ degrees of freedom.

Therefore, we can also write this probability as,

$p(F\geqslant F_0)$ where, $F_0=1$ is a point in the $F$ distribution tables that leaves an area equal to $\alpha$ to the right with $n_1-1=26-1=25$ and $n_2-1=8-1=7$ degrees of freedom. This can be written as,

$F_{\alpha,25,7}=1$

What we need to determine is the value of $\alpha$.

From the Tables, this value is approximately equal to 0.20.

Therefore, the probability that the variance of the first sample will be at least as large as the variance of the second sample is 0.2.