# Hypothesis Testing for the equality of 2 variances

Suppose I have 2 distributions: $$X\sim N(\mu_1,\sigma_1^2)$$ and $$Y\sim N(\mu_2,\sigma_2^2)$$
such that $$\sigma_1^2$$ and $$\sigma_2^2$$ are unknown.

How do I test $$H_0: \sigma_1^2 = 10\sigma_2^2$$ and $$H_1: \sigma_1^2 \ne 10\sigma_2^2$$ ?

Do I transform the $$Y\sim N(\mu_2,\sigma_2^2)$$ distribution to $$V=\sqrt{10}Y\sim N(\sqrt{10}\mu_2,10\sigma_2^2)$$
Then proceed to test $$X$$ with $$V$$

• Hello Emily, is this a homework question for a school assignment? If not I am happy to provide an answer. Nov 12, 2021 at 20:30
• Yes, the problem is a bit more complicated though. By the way, is my attempt correct or flawed? Thanks. Nov 12, 2021 at 22:38

Your strategy is a good one. Let's justify it.

In a sample $$x_1, x_2, \ldots, x_m$$ of $$m$$ independent draws of $$X,$$ the statistic

$$(x_1-\mu_1)^2 + \cdots + (x_m-\mu_1)^2$$

behaves like $$\sigma_1^2$$ times a $$\chi^2(m)$$ distribution. A similar result holds for an independent sample of $$n$$ draws from $$Y.$$ Therefore, under the null hypothesis $$\sigma_1^2=10\sigma_2^2,$$ the ratio

$$F=\frac{10\left[x_1-\mu_1)^2 + \cdots + (x_m-\mu_1)^2\right]/m}{\left[(y_1-\mu_2)^2 + \cdots + (y_n-\mu_2)^2\right]/n}$$

behaves like $$10\sigma_1^2 / (10\sigma_1^2) = 1$$ times an $$F(m,n)$$ distribution. For a symmetric two-tailed test with confidence $$1-\alpha,$$ compare this ratio to the $$\alpha/2$$ and $$1-\alpha/2$$ quantiles of this $$F$$ distribution.

The principal noteworthy element of this result is that the parameters of $$F$$ (its "degrees of freedom") are $$m$$ and $$n$$ rather than the conventional $$m-1$$ and $$n-1$$ used for distributions with unknown variances.

For example, here is a histogram of $$F$$ for 100,000 simulated datasets with $$m=3$$ and $$n=5$$ where the null hypothesis holds. (Because values of $$F$$ can be extremely large for small sample sizes, I have drawn it on a logarithmic scale.) On it is superimposed in red the density of the $$F(3,5)$$ distribution: it's a close fit, supporting the accuracy of the previous calculations and confirming that $$(m,n)$$ are the correct values for the degrees of freedom. The vertical dotted lines are the $$0.025$$ and $$1-0.025$$ limits for a test of $$100(1-0.05)=95\%$$ confidence. These limits are approximately $$0.067$$ and $$7.76.$$ Thus, with these sample sizes of $$m$$ and $$n,$$ you would reject the null when the sample ratio $$F$$ is less than $$0.067$$ or greater than $$7.76.$$