You might recall from chapter 6 that when we take samples from a normally distributed population and calculate the standard deviation or variance, these sample standard deviations are not normally distributed, but are distributed according to the chi-square distribution. When testing a claim about a standard deviation or variance, you also compare against the chi-square distribution.

The process of doing the hypothesis test is exactly the same as in the previous sections, however we have a new test statistic:

Test statistic for testing a hypothesis about :

Where n is the sample size, is the sample variance and is the population variance - here are some worked examples

Example 1: Tests in Mr. Wildmans past statistics classes have scores with a standard deviation equal to 14.1. One of his current classes now has 27 test scores with a standard deviation of 9.3. Use a 0.01 level of significance to test the claim that this current class has less variation than past classes.

Claim:

Recall to get the chi-square distribution you need the degrees of freedom (n-1), so in this case 26. The test is also one tailed and in fact it is left tailed, so to get the critical value from the table, you need the point on the chi squared distribution with 0.01 to left of it. To use table A-4, you need the area to the right, so you need the point (with 26 degrees of freedom) with .99 area to the right, from the table this is 12.198

Now for the critical region

So we reject the null hypothesis since our test statistic is in the critical region. Our conclusion is:

The sample data supports the claim that the variation of the current class is less than 14.1

Example 2: For randomly selected adults IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A sample of 24 randomly selected college professors resulted in IQ scores having a standard deviation of 10. Test the claim that the IQ scores for college professors is the same as the general population, that is 15. Use a 0.05 level of significance.

Claim:

There are 23 degrees of freedom. This is a two tailed test - so you need both critical values - the left critical value has .975 area to its right, so from table A-4, this is 11.689 and the right critical value has .025 area to its right, again from the table this is 38.076. Here are the critical regions:

So we reject the null hypothesis. Our conclusion:

There is sufficient evidence to warrant rejection of the claim that the standard deviation of college professors on the IQ test is 15.