Test of Two Variances

Learning Outcomes

Conduct a hypothesis test of two variances and interpret the conclusion in context

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. 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 an 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 do 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 ${{\sigma}_{1}^{2}},{{\sigma}_{2}^{2}}$ 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:

$\displaystyle{F}=\dfrac{{{\left[\dfrac{{({s}{1})}^{{2}}}{{(\sigma_{1})}^{{2}}}\right]}}}{{{\left[\dfrac{{({s}{2})}^{{2}}}{{(\sigma_{2})}^{{2}}}\right]}}}$

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 $\displaystyle{\sigma_{{1}}^{{2}}}={\sigma_{{2}}^{{2}}}$ then the F Ratio becomes $\displaystyle{F}=\dfrac{{{\left[\dfrac{{({s}{1})}^{{2}}}{{(\sigma{1})}^{{2}}}\right]}}}{{{\left[\dfrac{{({s}{2})}^{{2}}}{{(\sigma{2})}^{{2}}}\right]}}}$ = $\displaystyle\dfrac{{({s}_{1})}^{{2}}}{{({s}_{2})}^{{2}}}$ .

Recall: Fraction Division

Find the reciprocal of the fraction that follows the division symbol (ex. The reciprocal of $\frac{1}{2}$ is $\frac{2}{1}$ ).
Multiply the first fraction (the one before the division symbol) by the reciprocal of the second fraction (the one after the division symbol).

$\frac{3}{4} \div \frac{1}{2}$

$= \frac{3}{4} \times \frac{2}{1}$

$= \frac{3 \cdot 2}{4 \cdot 1}$

$= \frac{6}{4}$

$= 1 \frac{1}{2}$

Note

The F ratio could also be $\displaystyle\dfrac{{({s}_{1})}^{{2}}}{{({s}_{2})}^{{2}}}$ . It depends on H_a and on which sample variance is larger.

If the two populations have equal variances, then $s_{{1}}^{{2}}$ and $s_{{2}}^{{2}}$ are close in value and F = $\displaystyle\dfrac{{({s}_{1})}^{{2}}}{{({s}_{2})}^{{2}}}$ is close to one. 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 $\displaystyle\dfrac{{({s}_{1})}^{{2}}}{{({s}_{2})}^{{2}}}$ to be greater than one. If $s_{{1}}^{{2}}$ and $s_{{2}}^{{2}}$ are far apart, then F = $\displaystyle\dfrac{{({s}_{1})}^{{2}}}{{({s}_{2})}^{{2}}}$ is a large number.

Therefore, if F is close to one, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than one, then the evidence is against the null hypothesis. A test of two variances may be left, right, or two-tailed.

Example

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%.

Show Answer

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

$\displaystyle{n}_{{1}}={n}_{{2}}={30}$ .

H₀: $\displaystyle{\sigma}_{{1}}^{{2}}={\sigma}_{{2}}^{{2}}$ and H_a: ${\sigma}_{{1}}^{{2}}<{\sigma}_{{2}}^{{2}}$ Calculate the test statistic: By the null hypothesis (σ21 = σ22), the F statistic is:

$\displaystyle{F}=\dfrac{{{\left[\dfrac{{({s}_{1})}^{{2}}}{{(\sigma_{1})}^{{2}}}\right]}}}{{{\left[\dfrac{{({s}_{2})}^{{2}}}{{(\sigma_{2})}^{{2}}}\right]}}}$ = $\displaystyle\dfrac{{({s}_{1})}^{{2}}}{{({s}_{2})}^{{2}}}$ = $\dfrac{{52.3}}{{89.9}}={0.5818}$

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.

This graph shows a nonsymmetrical F distribution curve. The curve is slightly skewed to the right, but is approximately normal. The value 0.5818 is marked on the vertical axis to the right of the curve's peak. A vertical upward line extends from 0.5818 to the curve and the area to the left of this line is shaded to represent the p-value.

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% 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 ${\sqrt{(52.3)}}$ , 30, ${\sqrt{(89.9)}}$ , and 30. Press ENTER after each.
Arrow toσ1: 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.

try it

The New York Choral Society divides male singers into four categories from highest voices to lowest: Tenor1, Tenor2, Bass1, Bass2. In the table are the 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?

Tenor1	Bass2	Tenor 1	Bass 2	Tenor 1	Bass 2
69	72	67	72	68	67
72	75	70	74	67	70
71	67	65	70	64	70
66	75	72	66		69
76	74	70	68		72
74	72	68	75		71
71	72	64	68		74
66	74	73	70		75
68	72	66	72

Show Answer

Subscripts: T1= tenor1 and B2 = bass 2

The histograms are not as normal as one might like. Plot them to verify. However, we proceed with the test in any case.

The standard deviations of the samples are s_T1 = 3.3302 and s_B2 = 2.7208.

The hypotheses are:

$\displaystyle{H}_{{o}}:{\sigma}_{{T1}}^{{2}}={\sigma}_{{B2}}^{{2}}$ and $\displaystyle{H}_{{o}}:{\sigma}_{{T1}}^{{2}}\neq{\sigma}_{{B2}}^{{2}}$ (two-tailed test)

The F statistic is 1.4894 with 20 and 25 degrees of freedom.

The p-value is 0.3430. If we assume alpha is 0.05, then we cannot reject the null hypothesis.

We have no good evidence from the data that the heights of Tenor1 and Bass2 singers have different variances (despite there being a significant difference in mean heights of about 2.5 inches.)

Module 13: F-Distribution and One-Way ANOVA