Chapter 11: The Chi-Square Distribution
11.6 Test of a Single Variance
Learning Objectives
By the end of this section, the student should be able to:
- calculate the test of a single variance
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:
where:
- n = the total number of data
- s2 = sample variance
- 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. [link] will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.
Example
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
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
Your Turn!
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?
Solution
H0: σ2 = 32
Ha: σ2 < 32
Example
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.
Solution
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:[latex]{\chi }_{24}^{2}[/latex], where: