Sampling Distributions of Sample Variances

Sample Variance

Let x₁, x_2,……..,x_n be a random sample of observations from a population. The quantity

 is called the sample variance, and its square root, s, is called the sample standard deviation. Given a specific random sample, we could compute the sample variance, and the sample variance would be different for each random sample because of differences in sample observations.

Chi-Square Distribution of Sample and Population Variances

Given a random sample of n observations from a normally distributed population whose population variance isand whose resulting sample variance is s2, it can be shown that

has a distribution known as the(chi-square) distribution with n – 1 degrees of freedom.

Sampling Distribution of the Sample Variance

Let s2 denote the sample variance for a random sample of n observations from a population with a variance. Then

1.    The sampling distribution of s2 has mean

2.    The variance of the sampling distribution of s2 depends on the underlying population distribution. If that distribution is normal, then

3.    If the population distribution is normal, thenis distributed as

Thus, if we have a random sample from a population with a normal distribution, we can make inferences about the sample variance by using s2 and the chi-square distribution.

Example: George Samson is responsible for quality assurance at Integrated Electronics. He has asked you to established a quality monitoring process for the manufacture of control device A. The variability of the electrical resistance, measured in ohms, is critical for this device. Manufacturing standards specify a standard deviation of 3.6, the population distribution of resistance measures is normal. The monitoring process requires that a random sample of n = 6 observations be obtained from the population of devices and the sample variance computed. Determine an upper limit for the sample variance such that the probability of exceeding this limit, given a population standard deviation of 3.6, is less than 0.05.

Solution: For this problem we have n = 6 and=12.96. Using the chi-square distribution, we can state that

where K is the desired upper limit and is the upper 0.05 critical value of the chi-square distribution with 5 degrees of freedom from row 5 of Table 7. The required upper limit for s2—labeled as K—can be obtained by solving

If the sample variance, s2, from a random sample of size n = 6 exceeds 28.69, there is strong evidence to suspect that the population variance exceeds 12.96 and that the manufacturing process should be halted and appropriate adjustments performed.