### Learning Outcomes

- Conduct and interpret chi-square single variance hypothesis tests

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:

[latex]\displaystyle\frac{\left(n-1\right)^2}{\sigma^2}[/latex]

where:

*n*= the total number of data*s*^{2}= sample variance*σ*^{2}= 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 will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.

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?

### try it

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?

### try it

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%, test the claim that **a single line causes lower variation among waiting times (shorter waiting times) for customers**.

### try it

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 of 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, the test statistic, sketch the graph of the *p*-value, and draw a conclusion. Test at the 1% significance level.

# References

“AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013).

Data from the World Bank, June 5, 2012.

# Concept Review

To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation).

# Formula Review

χ2= (n−1)⋅s2σ2 Test of a single variance statistic where:

*n*: sample size

*s*: sample standard deviation

*σ*: population standard deviation*df* = *n* – 1 Degrees of freedom

- Use the test to determine variation.
- The degrees of freedom is the number of samples – 1.
- The test statistic is (n–1)⋅s2σ2, where
*n*= the total number of data,*s*^{2}= sample variance, and*σ*^{2}= population variance. - The test may be left-, right-, or two-tailed.