Tests of the variance of a normal distribution

In addition to the need for tests based on the sample mean and sample proportion, there are a number of situations where we want to determine if the population variance is a particular value or set of values. The basis for developing particular tests lies in the fact that the random variable

follows a Chi-square distribution with degrees of freedom.

The value of the test statistic is calculated as

We are given a random sample of n observations from a normally distributed population with variance . If we observe the sample variance , then the following tests have significance level :

1. To test either null hypothesis

or against the alternative

the decision rule is

Reject if

2. To test either null hypothesis

or against the alternative

the decision rule is

Reject if

3. To test the null hypothesis

against the two sided alternative

the decision rule is

Reject if or

where is a Chi-square random variable and .

Example:

Variance of yearly earnings of all state employees for all 40 states is

$49000 square dollars. A sample of 29 employees selected from state A produced a variance of their earnings equal to $600 000 square dollars. Test at 5% significance level if the variance of yearly earnings of state employees in state A is different from $490 000 square dollars. Assume that the yearly earnings of all state employees in state A have an (approximate) normal distribution.

Solution:

From the given information,

; ;

The null and alternative hypotheses are

We use Chi square distribution to use. The decision rule is

Reject if or

; ;

Then from Table 3 of appendix we obtain

and

The value of the test statistic is

The value of the test statistic 34.286 is between the two critical values of , 15.308 and 44.461, and falls in the nonrejection region. Consequently we fail to reject and conclude that the population variance of yearly earnings of all employees in state A is not different from 490000 square dollars.

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