Another use of the F distribution is testing two variances. It is often desirable to compare two variances rather than two averages. For instance, college administrators would like two college professors grading exams to have the same variation in their grading. For a lid to fit a container, the variation in the lid and the container should be the same. A supermarket might be interested in the variability of check-out times for two checkers.
To perform a F test of two variances, it is important that the following are true:
- The populations from which the two samples are drawn are normally distributed.
- The two populations are independent of each other.
Unlike most other tests in this book, the F test for equality of two variances is very sensitive to deviations from normality. If the two distributions are not normal, the test can give higher p-values than it should, or lower ones, in ways that are unpredictable. Many texts suggest that students not use this test at all, but in the interest of completeness we include it here.
Suppose we sample randomly from two independent normal populations. Let and be the population variances and and be the sample variances. Let the sample sizes be n1 and n2. Since we are interested in comparing the two sample variances, we use the F ratio
F has the distribution F ~ F(n1 – 1, n2 – 1),
where n1 – 1 are the degrees of freedom for the numerator and n2 – 1 are the degrees of freedom for the denominator.
If the null hypothesis is , then the F ratio becomes .
The F ratio could also be . It depends on Ha and on which sample variance is larger.
If the two populations have equal variances, then and are close in value and is close to 1. But if the two population variances are very different, and tend to be very different, too. Choosing as the larger sample variance causes the ratio to be greater than 1. If and are far apart, then is a large number.
Therefore, if F is close to 1, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than 1, then the evidence is against the null hypothesis. A test of two variances may be left-tailed, right-tailed, or two-tailed.
Two college instructors are interested in whether or there is any variation in the way they grade math exams. They each grade the same set of 30 exams. The first instructor’s grades have a variance of 52.3. The second instructor’s grades have a variance of 89.9. Test the claim that the first instructor’s variance is smaller. In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors. The level of significance is 10 percent.
Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively.
n1 = n2 = 30.
H0: and Ha: .
Calculate the test statistic: By the null hypothesis , the F statistic is
Distribution for the test: F29,29 where n1 – 1 = 29 and n2 – 1 = 29.
Graph: This test is left-tailed.
Draw the graph, labeling and shading appropriately.
Probability statement: p-value = P(F < 0.5818) = 0.0753.
Compare α and the p-value: α = 0.10 α > p-value.
Make a decision: Since α > p-value, reject H0.
Conclusion: With a 10 percent level of significance from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller.
Using the TI-83, 83+, 84, 84+ Calculator
STAT and arrow over to
TESTS. Arrow down to
ENTER. Arrow to
Stats and press
ENTER after each. Arrow to
ENTER. Arrow down to
Calculate and press
ENTER. F = 0.5818 and p-value = 0.0753. Do the procedure again and try
Draw instead of
The New York Choral Society divides male singers into four categories from highest voices to lowest: Tenor1, Tenor2, Bass1, and Bass2. In the table are heights of the men in the Tenor1 and Bass2 groups. One suspects that taller men will have lower voices, and that the variance of height may go up with the lower voices as well. Do we have good evidence that the variance of the heights of singers in each of these two groups (Tenor1 and Bass2) are different?