# 13.4 Test of two variances

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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, collegeadministrators 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 lidand 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:

1. The populations from which the two samples are drawn are normally distributed.
2. The two populations are independent of each other.

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 ${n}_{1}$ and ${n}_{2}$ . Since we are interested in comparing the two sample variances, we use the F ratio

$F=\frac{\left[\frac{\left({s}_{1}{\right)}^{2}}{{\left({\sigma }_{1}\right)}^{2}}\right]}{\left[\frac{\left({s}_{2}{\right)}^{2}}{{\left({\sigma }_{2}\right)}^{2}}\right]}$

$F$ has the distribution $F$ ~ $F\left({n}_{1}-1,{n}_{2}-1\right)$

where ${n}_{1}-1$ are the degrees of freedom for the numerator and ${n}_{2}-1$ are the degrees of freedom for the denominator.

If the null hypothesis is ${\sigma }_{1}^{2}={\sigma }_{2}^{2}$ , then the F-Ratio becomes $F=\frac{\left[\frac{\left({s}_{1}{\right)}^{2}}{{\left({\sigma }_{1}\right)}^{2}}\right]}{\left[\frac{\left({s}_{2}{\right)}^{2}}{{\left({\sigma }_{2}\right)}^{2}}\right]}=\frac{\left({s}_{1}{\right)}^{2}}{{\left({s}_{2}\right)}^{2}}$ .

The $F$ ratio could also be $\frac{\left({s}_{2}{\right)}^{2}}{{\left({s}_{1}\right)}^{2}}$ . 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 $F=$ $\frac{\left({s}_{1}{\right)}^{2}}{{\left({s}_{2}\right)}^{2}}$ 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 $\frac{\left({s}_{1}{\right)}^{2}}{{\left({s}_{2}\right)}^{2}}$ to be greater than $1$ . If ${s}_{1}^{2}$ and ${s}_{2}^{2}$ are far apart, then $F=$ $\frac{\left({s}_{1}{\right)}^{2}}{{\left({s}_{2}\right)}^{2}}$ 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, 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 30exams. 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 nearlythe same among instructors.) The level of significance is 10%.

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

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

${H}_{o}$ : ${\sigma }_{1}^{2}={\sigma }_{2}^{2}$ and ${H}_{a}$ : ${\sigma }_{1}^{2}$ $()$ ${\sigma }_{2}^{2}$

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

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

Distribution for the test: ${F}_{29,29}\phantom{\rule{20pt}{0ex}}$ where ${n}_{1}-1=29$ and ${n}_{2}-1=29$ .

Graph: $\phantom{\rule{20pt}{0ex}}$ This test is left tailed.

Draw the graph labeling and shading appropriately.

Probability statement: $\text{p-value}=P$ ( $F$ $()$ $0.5818$ ) $=0.0753$

Compare $\alpha$ and the p-value: $\alpha =0.10$ $\phantom{\rule{20pt}{0ex}}\alpha >\text{p-value}$ .

Make a decision: Since $\alpha >\text{p-value}$ , reject ${H}_{o}$ .

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 $\sqrt{\left(52.3\right)}$ , 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 $\text{p-value}=0.0753$ . Do the procedure again and try Draw instead of Calculate .

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