10.3: Difference between Two Variances - the F Distributions

Here we have to assume that the two populations (as opposed to sample mean distributions) have a distribution that is almost normal as shown in Figure 10.2.

Figure 10.2: Two normal populations lead to two \(\chi^{2}\) distributions that represent distributions of sample variances. The \(F\) distribution results when you build up a distribution of the ratio of the two \(\chi^{2}\) sample values.

The ratio \(\frac{s_{1}^{2}}{s_{2}^{2}}\) follows an \(F\)-distribution if \(\sigma_{1} = \sigma_{2}\). That \(F\) distribution has two degrees of freedom: one for the numerator (d.f.N. or \(\nu_{1}\)) and one for the denominator (d.f.D. or \(\nu_{2}\)). So we denote the distribution more specifically as \(F_{\nu_{1}, \nu_{2}}\). For the case of Figure 10.2, \(\nu_{1} = n_{1} - 1\) and \(\nu_{2} = n_{2} - 1\). The \(F\) ratio, in general is the result of the following stochastic process. Let \(X_{1}\) be random variable produced by a stochastic process with a \(\chi^{2}_{\nu_{1}}\) distribution and let \(X_{2}\) be random variable produced by a stochastic process with a \(\chi^{2}_{\nu_{2}}\) distribution. Then the random variable will, by definition, have a \(F_{\nu_{1}, \nu_{2}}\) distribution.

The exact shape of the \(F_{\nu_1, \nu_2}\) distribution depends on the choice of \(\nu_1\) and \(\nu_2\), But it roughly looks like a \(\chi^2\) distribution as shown in Figure 10.3.

Figure 10.3: A generic \(F\) distribution.

\(F\) and \(t\) are related :

\[F_{1,\nu} = t^{2}_{\nu}\]

so the \(t\) statistic can be viewed as a special case of the \(F\) statistic.

For comparing variances, we are interested in the follow hypotheses pairs :

Right-tailed	Left-tailed	Two-tailed
\(H_0: \sigma^2_1 \leq \sigma^2_2\)	\(H_0: \sigma^2_1 \geq \sigma^2_2\)	\(H_0: \sigma^2_1 = \sigma^2_2\)
\(H_1: \sigma^2_1 > \sigma^2_2\)	\(H_1: \sigma^2_1 < \sigma^2_2\)	\(H_1: \sigma^2_1 \neq \sigma^2_2\)

We’ll always compare variances (\(\sigma^2\)) and not standard deviations (\(\sigma\)) to keep life simple.

The test statistic is

\[ F_{\rm test} = F_{\nu_1, \nu_2} = \frac{s^{2}_{1}}{s^{2}_{2}} \]

where (for finding the critical statistic), \(\mu_{1} = n_{1} - 1\) and \(\mu_{2} = n_{2} - 1\).

Note that \(F_{\nu_1, \nu_2} = 1\) when \(s_{1}^{2}=s_{2}^{2}\), a fact you can use to get a feel for the meaning of this test statistic.

Values for the various \(F\) critical values are given in the F Distribution Table in the Appendix. We will denote a critical value of \(F\) with the notation :

\[F_{\rm crit} = F_{\alpha, \hspace{.1in}\nu_{1}, \hspace{.1in} \nu_2}\]

Where:

\(\alpha\) = Type I error rate
\(\nu_{1}\) = d.f.N.
\(\nu_{2}\) = d.f.D.

The F Distribution Table gives critical values for small right tail areas only. This means that they are useless for a left-tailed test. But that does not mean we cannot do a left-tail test. A left-tail test is easily converted into a right tail test by switching the assignments of populations 1 and 2. To get the assignments correct in the first place then, always define populations 1 and 2 so that \(\sigma^{2}_{1} > \sigma^{2}_{2}\). Assign population 1 so that it has the largest sample variance. Do this even for a two-tail test because we will have no idea what \(F_{\rm crit}\) on the left side of the distribution is.

Example 10.3 : Given the following data for smokers and non-smokers (maybe its about some sort of disease occurrence, who cares, let’s focus on dealing with the numbers), test if the population variances are equal or not at \(\alpha = 0.05\).

Smokers	Nonsmokers
\(n_{1} = 26\)	\(n_{2} = 18\)
\(s_{1}^{2}=36\)	\(s_{2}^{2}=10\)

Note that \(s_{1}^{2} > s_{2}^{2}\) so we’re good to go.

Solution :

1. Hypothesis.

\[\begin{equation*} $H_0: \sigma^2_1 = \sigma^2_2$ \\ $H_1: \sigma^2_1 \neq \sigma^2_2$ \end{equation*}\]

2. Critical statistic.

Use the F Distribution Table; it is a bunch of tables labeled by “\(\alpha\)” that we will designate at \(\alpha_{T}\), the table values that signify right tail areas. Since this is a two-tail test, we need . Next we need the degrees of freedom:

\[\mbox{d.f.N.} = \nu_{1} = n_{1} - 1 = 26-1 = 25\] \[\mbox{d.f.D.} = \nu_{2} = n_{2} - 1 = 18-1 = 17\]

So the critical statistic is

3. Test statistic.

\[F_{\nu_1, \nu_2} = \frac{s^2_1}{s^2_2}\] \[F_{\rm test} = F_{25, 17} = \frac{36}{10} = 3.6\]

With this test statistic, we can estimate the \(p\)-value using the F Distribution Table. To find \(p\), look up all the numbers with d.f.N = 25 and d.f.N = 17 (24 \(\&\) 17 are the closest in the tables so use those) in all the the F Distribution Table and form your own table. For each column in your table record \(\alpha_{T}\) and the \(F\) value corresponding to the degrees of freedom of interest. Again, \(\alpha_{T}\) corresponds to for a two-tailed test. So make a row above the \(\alpha_{T}\) row with \(p = 2 \alpha_{T}\). (For a one-tailed test, we would put \(p=\alpha_{T}\).)

\(p\) \(\alpha_{T}\)	0.20 0.10 0.05 0.02 0.01 0.10 0.05 0.025 0.01 0.005
\(F\)	1.84 2.19 2.56 3.08 3.51 3.6 is over here somewhere so \(p<0.01\)

Notice how we put an upper limit on \(p\) because \(F_{\rm test}\) was larger than all the \(F\) values in our little table.

Let’s take a graphical look at why we use \(p=2\alpha\) in the little table and for finding \(F_{\rm crit}\) for two tailed tests :

But in a two-tailed test we want \(\alpha\) split on both sides:

4. Decision.

Reject \(H_{0}\). The \(p\)-value estimate supports this :

\[ ( p < 0.01) < (\alpha = 0.05) \]

5. Interpretation.

There is enough evidence to conclude, at \(\alpha = 0.05\) with an \(F\)-test, that the variance of the smoker population is different from the non-smoker population.

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