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 $V$

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

1 Answer 1


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 $7.76.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.