Test of Two Variances

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 ${\sigma }_1^2$ and ${\sigma }_2^2$ be the population variances and $s_1^2$ and $s_2^2$ 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

$$F=\frac{\left[\frac{{(s_1)}^2}{{({\sigma }_1)}^2}\right]}{\left[\frac{{(s_2)}^2}{{({\sigma }_2)}^2}\right]}.$$

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 ${\sigma }_1^2={\sigma }_2^2$, then the F ratio becomes $F=\frac{\left[\frac{{(s_1)}^2}{{({\sigma }_1)}^2}\right]}{\left[\frac{{(s_2)}^2}{{({\sigma }_2)}^2}\right]}=\frac{{(s_1)}^2}{{(s_2)}^2}$.

If the two populations have equal variances, then $s_1^2$ and $s_2^2$ are close in value and $F=\frac{{(s_1)}^2}{{(s_2)}^2}$ is close to 1. But if the two population variances are very different, $s_1^2$ and $s_2^2$ tend to be very different, too. Choosing $s_1^2$ as the larger sample variance causes the ratio $\frac{{(s_1)}^2}{{(s_2)}^2}$ to be greater than 1. If $s_1^2$ and $s_2^2$ are far apart, then $F=\frac{{(s_1)}^2}{{(s_2)}^2}$ 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.

Example 13.5

Problem

Two college instructors are interested in whethe 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.

Show solution

Solution

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

n₁ = n₂ = 30.

H₀: ${\sigma }_1^2={\sigma }_2^2$ and H_a: ${\sigma }_1^2\text{ < }{\sigma }_2^2$.

Calculate the test statistic: By the null hypothesis $\text{(}{\sigma }_{\text{1}}^{\text{2}}\text{ = }{\sigma }_{\text{2}}^{\text{2}}\text{)}$, the F statistic is

$F=\frac{\left[\frac{{(s_1)}^2}{{({\sigma }_1)}^2}\right]}{\left[\frac{{(s_2)}^2}{{({\sigma }_2)}^2}\right]}=\frac{{(s_1)}^2}{{(s_2)}^2}=\frac{52.3}{89.9}=\mathrm{0.5818.}$

Distribution for the test: F_29,29 where n₁ – 1 = 29 and n₂ – 1 = 29.

Graph: This test is left-tailed.

Draw the graph, labeling and shading appropriately.

Figure 13.7

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

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 $\sqrt{\left(52.3\right)}$, 30, $\sqrt{\left(89.9\right)}$, 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.