Based on Theorem 2 of Chi-square Distribution and its corollaries, we can use the chi-square distribution to test the variance of a distribution.

Example 1: A company produces metal pipes of a standard length. Twenty years ago it tested its production quality and found that the lengths of the pipes produced were normally distributed with a standard deviation of 1.1 cm. They want to test whether they are still meeting this level of quality by testing a random sample of 30 pipes, and finding the 95% confidence interval around σ².

By Theorem 2 of of Chi-square Distribution

Note that since the chi-square distribution is not symmetric, the confidence interval is not symmetric around σ², and so the approach used in Confidence Intervals for Sampling Distributions and Confidence Interval for t-test needs to be modified somewhat, in that we need to calculate the lower and upper values of the confidence interval based on different critical values of the distribution:

Upper limit = 0.042*CHIINV(.025, 29) = 0.042 ∙ 45.72 = 1.91

Lower limit = 0.042*CHIINV(.975, 29) = 0.042 ∙ 16.05 = 0.67

And so the confidence interval is (0.67, 1.91). We see that the variance of  (1.1)² = 1.21 is  in this range, but the sample is too small to get much precision.

Example 2: A company produces metal pipes of a standard length, and claims that the standard deviation of the length is at most 1.2 cm. One of its clients decides to test this claim by taking a sample of 25 pipes and checking their lengths. They found that the standard deviation of the sample is 1.5 cm. Does this undermine the company’s claim?

We perform a one tail test based on the following hypotheses:

H₀: the standard deviation of the pipe length is ≤ 1.2 cm
H₁: the standard deviation of the pipe length is > 1.2 cm

If we assume that the population has a normal distribution then by Corollary 3 of Chi-square Distribution, we know that

Since

it follows that

p-value = CHIDIST(x, df) = CHIDIST(37.5, 24) = 0.039 < .05 = α

and so the null hypothesis is rejected, leading the client to conclude with 95% confidence that the company is no longer meeting their claimed quality standard.

Alternatively, x-crit = CHIINV(α, df) = CHIINV(.05, 24) = 36.42 < 37.5 = x and so once again the null hypothesis is rejected.

Observation: In Example 2 we used a one-tail test. For a two-tail test, proceed as follows using the null hypothesis

H₀: the standard deviation of the pipe length is = 1.2 cm

We reject the null hypothesis if either CHIDIST(x, df) = .039 < .025 = α/2 or CHIDIST(x, df) = .039 > 1 – α/2 = .975. As result, in this case, we can’t reject the null hypothesis. Alternatively, we reject the null hypothesis if either 37.5 = x > CHIINV(α/2, df) = CHIINV(.025, 24) = 39.4 or 37.5 = x < CHIINV(1–α/2, df) = 12.4, and so once again we cannot reject the null hypothesis in the two-tail test.