A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance or population standard deviation. The test statistic is



  • n = the total number of data,
  • s2 = sample variance, and
  • σ2 = population variance.

You may think of s as the random variable in this test. The number of degrees of freedom is df = n – 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. Example 11.10 will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.

Example 11.10

Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance, or standard deviation, may be more important than the average.

Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?

Solution 11.10

Even though we are given the population standard deviation, we can set up the test using the population variance as follows:

  • H0: σ2 = 52
  • Ha: σ2 > 52
Try It 11.10

A scuba instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be?

Example 11.11

With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.

With a significance level of 5 percent, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers.

Solution 11.11

Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ2, or the population standard deviation, σ.

Random variable: The sample standard deviation, s, is the random variable. Let s = standard deviation for the waiting times.

  • H0: σ2 = 7.22
  • Ha: σ2 < 7.22

The word less tells you this is a left-tailed test.

Distribution for the test

 χ242χ242, where

  • n = the number of customers sampled, and
  • df = n – 1 = 25 – 1 = 24.

Calculate the test statistic

χ2=(n  1)s2σ2=(25  1)(3.5)27.22=5.67χ2=(n  1)s2σ2=(25  1)(3.5)27.22=5.67

where n = 25, s = 3.5, and σ = 7.2.


This is a nonsymmetrical chi-square curve with values of 0 and 5.67 labeled on the horizontal axis. The point 5.67 lies to the left of the peak of the curve. A vertical upward line extends from 5.67 to the curve and the region to the left of this line is shaded. The shaded area is equal to the p-value.
Figure 11.8

Probability statement: p-value = P ( χ2 < 5.67) = 0.000042

Compare α and the p-value: α = 0.05; p-value = 0.000042; α > p-value

Make a decision: Since α > p-value, reject H0. This means that you reject σ2 = 7.22. In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less.

Conclusion: At a 5 percent level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes.

Using the TI-83, 83+, 84, 84+ Calculator

In 2nd DISTR, use 7:χ2cdf. The syntax is (lower, upper, df) for the parameter list. For Example 11.11, χ2cdf(-1E99,5.67,24). The p-value = 0.000042.

Try It 11.11

The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August 2012, the standard deviation of internet speeds across internet service providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, calculate the test statistic, sketch the graph of the p-value, and draw a conclusion. Test at the 1 percent significance level.