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 *σ*^{2}.

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 *σ ^{2}*, 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:

H_{0}: the standard deviation of the pipe length is ≤ 1.2 cm

H_{1}: 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

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

H_{0}: 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.

Hello Dr. Zaiontz,

I found an online calculator that shows how to calculate a confidence interval for Chi square here: http://www.mathcelebrity.com/chiconf.php?n=+30&variance=+1.21&conf=+95&pl=Variance+Confidence+Interval . I put your Example 1 into it, and I don’t get the same answer for the confidence interval. I’m just wondering what is going on here, or what the mathcelebrity website calculator is doing differently? Thanks!

Jesse,

In Example 1 we are finding the confidence interval for s^2 given that we know the population variance. The calculation you are referring to finds the confidence interval for the population variance given that you know the sample variance.

Charles

Charles:

I am seeing an inconsistency in certain aspects of the literature concerning this test. I would like your input in this matter. In conducting a two tailed chi-squared test for a single population variance, it is clear that one computes the p value by doubling the smaller area on a side of the calculated value of the test statistic. Does one then compare that p-value to alpha or to alpha/2? I say alpha as with all other hypothesis tests. However, if one does not double the aforementioned smaller area, then I suspect that the proper comparison is to alpha/2. That said, then one should not call the non doubled area the p value, correct? You would actually be comparing p-value/2 to alpha/2 which would be a valid comparison, but you would not be using the p value per se. What do you think?

Alan,

If the p-value for the one-tail test is say .02, then for a symmetric distribution the p-value for the two-tailed test is .04, which you can compare with alpha (say .05).

Equivalently, you can use the one-tailed p-value of .02 and compare it with alpha/2 = .025.

This is true, for example, for the t test. You need to be careful with the chi-square distribution since it is not symmetric.

Charles

What would the nonparametric equivalent be?

Patty,

Wilcoxon Signed Ranks Test

Charles

Hi Charles,

Is it possible to construct a hypothesis test where H0: variance = 0 ; H1: variance > 0?

Lionel,

You are asking for a one-tailed test. Example 2 of the referenced website shows a one-tailed test (while Example 1 shows a two-tailed test).

Note: since variance is always a non-negative number, variance > 0 is equivalent to variance <> 0.

Charles

I have a same queation but can not find the result, I am looking for a test to test if the variability of my sample is close to zero, if i want to use chi-square test i would have zero in the denominator.!!!

Parisa,

This is a good question. I hadn’t thought about it before. Essentially zero variance means that all the data is the same (and so is equal to the mean). Thus, let w = abs(x-mean) where mean is the mean of your sample. Thus you could perform a one sample t test to see whether the w values have a mean of zero.

Charles

Hello Charles, how do I calculate p-value for test where H0: VAR = anything; H1: VAR anything? (two tailed test). Do I have two P values?

Even in a two tailed test, you only have one p-value. This single value needs to take into account both the right tail and the left tail.

Charles