The F-test provides a means to compare paired data variances. It is a variance hypothesis test.

If we are exploring the precision of one measuring device or another, or we are comparing assembly processes, we often want to know if the variance is different or not.

Working with data from normal distributions from two different processes or devices, we know from statistical theory that the ratio (s1)2 / (s2)2 is described by the F distribution.

There are three hypothesis test possible, basically to test if the population variances are different, or one is less than or greater than the other. The following details the three test’s null and alternative hypotheses.

Assuming the population variances are equal

$$ \large\displaystyle \begin{array}{l}{{H}_{0}}:\sigma _{1}^{2}=\sigma _{2}^{2}\\{{H}_{1}}:\sigma _{1}^{2}\ne \sigma _{2}^{2}\end{array}$$

Assuming population variance 1 is less then or equal to population variance 2

$$ \large\displaystyle \begin{array}{l}{{H}_{0}}:\sigma _{1}^{2}\le \sigma _{2}^{2}\\{{H}_{1}}:\sigma _{1}^{2}\gt\sigma _{2}^{2}\end{array}$$

Assuming population variance 1 is greater than or equal to population variance 2

$$ \large\displaystyle \begin{array}{l}{{H}_{0}}:\sigma _{1}^{2}\ge \sigma _{2}^{2}\\{{H}_{1}}:\sigma _{1}^{2}\lt\sigma _{2}^{2}\end{array}$$

Recall that the shape of the F distribution is non-symmetrical and depends on the degrees, DF, of freedom of the two sample variances, s12 and S22. We will use v1 and v2 respectively, and v1 is the DF in the numerator. By convention we use the larger sample variance as s12 and place it in the numerator.

## Example Problem

For a medical delivery device, the ability for a specific part to remain strong (measured in psi) is important even after a 2 year period of storage. Assuming a 95% confidence level, use the data to determine if the variation is equal or better after aging.

At the start we tested 9 units and found a standard deviation of 900 psi. After two years of aging, we tested 7 units and determined a 300 psi variation.

## Solution

The null and alternative hypothesis are

$$ \large\displaystyle \begin{array}{l}{{H}_{0}}:\sigma _{1}^{2}\le \sigma _{2}^{2}\\{{H}_{1}}:\sigma _{1}^{2}>\sigma _{2}^{2}\end{array}$$

and, DF1 = 8 and DF2 = 6

We are using a one-tailed test to explore if after aging the units have a smaller variance. The entire α risk will be in the right tail. From the F distribution, the critical value of F is 4.15. Thus we will reject the null hypothesis if the ratio of the two sample variances is greater than 4.15.

$$ \large\displaystyle F=\frac{s_{1}^{2}}{s_{2}^{2}}=\frac{{{900}^{2}}}{{{300}^{2}}}=9$$

Since, 9 is greater than 4.15 we have sufficient evidence, 95% confidence, to reject the null hypothesis and conclude strength variation is reduced after aging for two years.

Related:

Hypothesis Tests for Variance Case I (article)

Hypothesis unequal variance (article)

Levene’s Test (article)

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