F Distribution and ANOVA: Test of Two Variances 1.4 2008/06/24 12:34:21 GMT-5 2008/07/15 10:59:05.734 GMT-5 Barbara Illowsky illowskybarbara@deanza.edu Susan Dean deansusan@deanza.edu Connexions cnx@cnx.org ANOVA degrees of freedom F Distribution normal distribution populations sample statistics Test of Two Variances variance This module provides the assumptions to be considered in order to calculate a Test of Two Variances and how to execute the Test of Two Variances. An example is provided to help clarify the concept. Another of the uses 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. In order 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.In order 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. Suppose we sample randomly from two independent normal populations. Let σ12 and σ22 be the population variances and s12 and s22 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= [ ( s1 )2 ( σ1 ) 2 ] [ ( s2 )2 ( σ2 ) 2 ] 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.A test of two variances may be left, right, or two-tailed.Two college instructors are interested in whether or not 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%. Let 1 and 2 be the subscripts that indicate the first and second instructor.n1=n2=30.Ho: σ12=σ22Ha: σ12σ22Calculate the test statistic: By the null hypothesis (σ12=σ22), the F statistic is F= [ ( s1 )2 ( σ1 ) 2 ] [ ( s2 )2 ( σ2 ) 2 ] = ( s1 )2 ( s2 ) 2 =52.389.9=0.6 Distribution for the test: F29,29where 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(F0.582)=0.0755Compare α and the p-value: α=0.10α>p-value.Make a decision: Since α>p-value, reject Ho.Conclusion: With a 10% level of significance, from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller.TI-83+ and TI-84: Press STAT and arrow over to TESTS. Arrow down to D:2-SampFTest. Press ENTER. Arrow to Stats and press ENTER. For Sx1, n1, Sx2, and n2, enter (52.3), 30, (89.9), and 30. Press ENTER after each. Arrow to σ1: and σ2. Press 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 Calculate.