Like sample mean, if we calculate the sample variance for each samples drawn from a population then it shows also a random variable. We have the following result: If a random sample of size *n *with sample variance S^{2 }is taken from a normal population having the variance σ _{2}, then

is a random variable having the chi-square distribution with the degrees of freedom v = n - 1.

(In chi-square distribution table X_{a}^{2 }represents the area under the chi-square distribution to its right is equal to a).

If S_{1}^{2} and S_{2}^{2} are the variances of independent random sample of size n1 and n2 respectively, taken from two normal populations having the same variance, then

F = S_{1}^{2} / S_{2}^{2}

is a random variable having the F distribution with the degrees of freedoms v1 = n1 - 1 and v2 = -1.

**Example 4.** If two independent random samples of size n1 = 9 and n2 = 16 are taken from the normal population, what is the probability that the variance of the first sample will be at least four times as large as that of the second sample ?

**Solution.**** **Here v 1 = 9 - 1 = 8, v2 16- 1 = 15

S_{1}^{2} = 4 S_{2}^{2}

From F distribution table we find that

F_{0.01 }= 4.00 for v1= 8 and v2 = 15

Thus, the desired probability is 0.01.

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