In the discussion on measures of spread, we showed that the sample variance of a set of values can be computed as:

thus

Now if the population from which this sample data has been drawn is Normal, the sum of squared deviations in this expression will be distributed as a chi-square variate. A simple test that a Normal population has a variance equal to σ02 is to evaluate the expression:

Example: Variation of a chemical process

The following is a set of 13 burning times in seconds for a chemical compound (after Crow et al., 1960, P.71, [CRO1]):

0.516,0.508,0.517,0.529,0.501,0.521,0.539,0.509,0.521,0.532,0.547,0.504,0.525

Assuming the burning times are Normally distributed, do these sample measurements indicate whether the population standard deviation of burning times is 0.01 (the null hypothesis)?

The sample variance of this data is 0.000187 seconds (i.e. a standard deviation of 0.0137), so using the formula above we find that the chi-square value to be used for testing is 22.427, with 12 degrees of freedom (df). The critical upper 2.5% tail of the chi-square for 12df is 23.34, and the lower 2.5% level is 4.404 so the sample variance is close to, but less than the upper value of the two-tailed 95% confidence interval. We therefore cannot reject the null hypothesis at this level. On the other hand, a one-tailed test at the upper end has a critical chi-square value of 21.03, so if the null hypothesis was that the standard deviation was less than or equal to 0.01 we would reject this null hypothesis, since our sample suggest the value is actually greater than 0.01.

Confidence intervals

Using the above computations, with minor re-arrangement, 100(1-α)% confidence intervals for the standard deviation of a Normal population can be obtained. These are:

Note that because these limits involve the chi-square distribution, which is asymmetric, the limits are not symmetric.

References

[CRO1] Crow E L, Davis F A, Maxfield M W (1960) Statistics Manual. Dover Publications, New York