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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 $V$

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  • $\begingroup$ 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

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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.

Figure

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 $7.76.$

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