Problem source

\textit{In this question, $\Phi(z)$ is the cumulative distribution function of a standard normal random variable.}
A random variable is known to have a Normal distribution with mean $\mu$ and standard deviation either $\sigma_0$ or $\sigma_1$, where $\sigma_0 < \sigma_1\,$. 
The mean, $\overline{X}$, of a random sample of $n$ values of $X$ is to be used to test the hypothesis $\mathrm{H}_0: \sigma = \sigma_0$ against the alternative $\mathrm{H}_1: \sigma = \sigma_1\,$.
Explain carefully why it is appropriate to use a two sided test of the form: accept $\mathrm{H}_0$ if  $\mu - c < \overline{X} < \mu+c\,$, otherwise accept $\mathrm{H}_1$.
Given that the probability of accepting $\mathrm{H}_1$ when $\mathrm{H}_0$ is true is $\alpha$, determine $c$ in terms of $n$,  $\sigma_0$ and $z_{\alpha}$, where $z_\alpha $ is defined by $\displaystyle\Phi(z_{\alpha}) = 1 - \tfrac{1}{2}\alpha$.
The probability of accepting $\mathrm{H}_0$ when $\mathrm{H}_1$ is true is denoted by $\beta$. Show that $\beta$ is independent of $n$.
Given that $\Phi(1.960)\approx 0.975$ and that $\Phi(0.063) \approx 0.525\,$, determine, approximately, the minimum value of $\displaystyle \frac{\sigma_1}{\sigma_0}$ if $\alpha$ and $\beta$ are both to be less than $0.05\,$.

Solution source

If $\sigma$ is smaller we should expect our sample to have a mean closer to the true mean. Therefore we should use a two sided test which accepts $\mathrm{H}_0$ if the mean is very close to the true mean.

Suppose $\textrm{H}_0$ is true, ie $\sigma = \sigma_0$, then note that $X \sim N(\mu, \frac{\sigma_0^2}{n})$

\begin{align*}
&& 1-\alpha &= \mathbb{P}(\mu - c < X < \mu + c) \\
&&&= \mathbb{P}(\mu - c < \frac{\sigma_0}{\sqrt{n}} Z + \mu < \mu + c) \\
&&&= \mathbb{P}(- \frac{c\sqrt{n}}{\sigma_0} < Z<\frac{\sqrt{n}c}{\sigma_0}) \\
&&&= \mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0}) -\mathbb{P}( Z<-\frac{\sqrt{n}c}{\sigma_0}) \\
&&&= \mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0}) -(1-\mathbb{P}( Z<\frac{\sqrt{n}c}{\sigma_0})) \\
&&&= 2\mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0})-1 \\
\Rightarrow && \Phi(\frac{\sqrt{n}c}{\sigma_0})&=1 - \tfrac12 \alpha \\
\Rightarrow && \frac{\sqrt{n}c}{\sigma_0} &= z_{\alpha} \\
&& c &= \frac{\sigma_0 z_{\alpha}}{\sqrt{n}}
\end{align*}

Under $\mathrm{H}_1$, $\sigma = \sigma_1$ so
\begin{align*}
&& \beta &= \mathbb{P}(\mu - c < X < \mu + c) \\
&&&= \mathbb{P}(-\frac{c\sqrt{n}}{\sigma_1} < Z < \frac{\sqrt{n}c}{\sigma_1}) \\
&&&= \mathbb{P}(-\frac{\sigma_0}{\sigma_1} z_{\alpha}< Z < \frac{\sigma_0}{\sigma_1} z_{\alpha}) \\
&&&= 2\Phi(\frac{\sigma_0}{\sigma_1} z_{\alpha})-1
\end{align*}
which does not depend on $n$.

Suppose both $\alpha<0.05$ and $\beta<0.05$, then $z_{\alpha} > 1.96$ and $\Phi(\frac{\sigma_0}{\sigma_1}1.96)<0.525 \Rightarrow \frac{\sigma_0}{\sigma_1}1.96 < 0.063 \Rightarrow \frac{\sigma_1}{\sigma_0} > \frac{1.96}{0.063} = 31.1 $ so the ratio of variances needs to be larger than $31.1$.