F Distribution and ANOVA: Test of Two Variances

Module by: Susan Dean, Barbara Illowsky, Ph.D.

Summary: 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.

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Example 1:	$\frac{\sqrt{\frac{1}{2}}}{\sqrt{{f(x+y)}^2}}\frac{\sqrt{1}}{2}\langle ℝ\rangle$	Example 2:		(Hide examples)

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:

Suppose we sample randomly from two independent normal populations. Let σ12σ12 and σ22σ22 be the population variances and s12s12 and s22s22 be the sample variances. Let the sample sizes be n1n1 and n2n2. 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= [ ( s1 )2 ( σ1 ) 2 ] [ ( s2 )2 ( σ2 ) 2 ]

FF has the distribution FF ~ F(n1-1,n2-1)F(n1-1,n2-1)

where n1-1n1-1 are the degrees of freedom for the numerator and n2-1n2-1 are the degrees of freedom for the denominator.

If the null hypothesis is σ12=σ22σ12=σ22, then the F-Ratio becomes F= [ ( s1 )2 ( σ1 ) 2 ] [ ( s2 )2 ( σ2 ) 2 ] = ( s1 )2 ( s2 ) 2 F= [ ( s1 )2 ( σ1 ) 2 ] [ ( s2 )2 ( σ2 ) 2 ] = ( s1 )2 ( s2 ) 2 .

If the two populations have equal variances, then s12s12 and s22s22 are close in value and F=F= ( s1 )2 ( s2 ) 2 ( s1 )2 ( s2 ) 2 is close to 11. But if the two population variances are very different, s12s12 and s22s22 tend to be very different, too. Choosing s12s12 as the larger sample variance causes the ratio ( s1 )2 ( s2 ) 2 ( s1 )2 ( s2 ) 2 to be greater than 11. If s12s12 and s22s22 are far apart, then F=F= ( s1 )2 ( s2 ) 2 ( s1 )2 ( s2 ) 2 is a large number.

Therefore, if FF is close to 11, the evidence favors the null hypothesis (the two population variances are equal). But if FF is much larger than 11, then the evidence is against the null hypothesis.

A test of two variances may be left, right, or two-tailed.

Example 1

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.

Problem 1

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%.

Solution

Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively.

n1=n2=30n1=n2=30.

HoHo: σ12=σ22σ12=σ22 and HaHa: σ12σ12<σ22σ22

Calculate the test statistic: By the null hypothesis (σ12=σ22)(σ12=σ22), the F statistic is

F= [ ( s1 )2 ( σ1 ) 2 ] [ ( s2 )2 ( σ2 ) 2 ] = ( s1 )2 ( s2 ) 2 =52.389.9=0.6 F= [ ( s1 )2 ( σ1 ) 2 ] [ ( s2 )2 ( σ2 ) 2 ] = ( s1 )2 ( s2 ) 2 =52.389.9=0.6

Distribution for the test: F29,29F29,29where n1-1=29n1-1=29 and n2-1=29n2-1=29.

Graph:This test is left tailed.

Draw the graph labeling and shading appropriately.

Figure 1

Probability statement: p-value=Pp-value=P (FF < 0.5820.582) =0.0755=0.0755

Compare αα and the p-value: α=0.10α=0.10α>p-valueα>p-value.

Make a decision: Since α>p-valueα>p-value, reject HoHo.

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)(52.3), 30, (89.9)(89.9), and 30. Press ENTER after each. Arrow to σ1: and <σ2. Press ENTER. Arrow down to Calculate and press ENTER. F=0.5818F=0.5818 and p-value=0.0753p-value=0.0753. Do the procedure again and try Draw instead of Calculate.

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Last edited by Susan Dean on Feb 5, 2009 9:31 pm US/Central.