One Sample Hypothesis Testing of the Variance

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

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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

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Since
image802

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.

8 Responses to One Sample Hypothesis Testing of the Variance

  1. Alan Chesen says:

    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?

    • Charles says:

      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

  2. Patty says:

    What would the nonparametric equivalent be?

  3. Lionel says:

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

    • Charles says:

      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

  4. Julo says:

    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?

    • Charles says:

      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

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