When testing whether the means of two normal distributions are equal, we often assume that their variances are the same. We now describe a hypothesis test to assess this assumption.
H0 : \(\sigma_1^2 = \sigma_2^2\)
HA : \(\sigma_1^2 \ne \sigma_2^2\)
We showed earlier that the two sample variances, \(S_1^1\) and \(S_2^2\), have distributions proportional to chi-squared distributions,
\[ \frac{n_1-1}{\sigma_1^2} S_1^2 \;\;\sim\;\; \ChiSqrDistn(n_1 - 1 \text{ df}) \]and similarly for \(S_2^2\).
Hypothesis test
The ratio of the two sample variances can be used as a test statistic for this test.
Test statistic
If \(\overline{X}_1\) and \(S_1^2\) are the mean and variance of a sample of \(n_1\) values from a \(\NormalDistn(\mu_1, \sigma_1^2)\) distribution and \(\overline{X}_2\) and \(S_2^2\) are the mean and variance of an independent sample of \(n_2\) values from a \(\NormalDistn(\mu_2, \sigma_2^2)\) distribution,
\[ F \;\;=\;\; \frac{S_1^2}{S_2^2} \;\;\sim\;\; \FDistn(n_1 - 1,\; n_2 - 1 \text{ df}) \]provided \(\sigma_1^2 = \sigma_2^2\).
(Proved in full version)
The p-value for the test can be found from the tail probabilities of this distribution.
Example
When analysing the data set about the effect of the hormone IAA on the growth of dwarf pea stems on the previous page, an assumption was made that the underlying normal distribution's variance was the same for both hormone levels. Test whether the data are consistent with this assumption.
(Solved in full version)