# The most powerful test for variance of normal distribution

Let $$X_1, \ldots, X_{10}$$ be simple samples from normal distribution with expected value$${}= 0$$ and variance$${}= \sigma^2$$. Write the most powerful test for $$\alpha = 0.05$$ to verify the hypothesis that $$\sigma^2 =1$$ versus $$\sigma^2 = \sigma^2_1$$ for $$\sigma^2_1 >1$$. For which $$\sigma^2_1$$ the power would be greater than 0.95?

I have estimated likelihood ratio as:

$$\left(\frac{\sigma^2}{\sigma^2_1}\right)^{10}\exp\left(\left(\frac{1}{2\sigma^2}-\frac{1}{2\sigma^2_1}\right)\sum_{i=1}^{10}X_i^2\right) \geq k$$

Given $$\ \chi^2(0.95, 10) = 18.3$$ and $$\ \chi^2(0.05, 10) = 3.94$$, I write that:

$$\alpha = P\left(\sum_{i=1}^{10}\frac{X_i^2}{\sigma} \geq \frac{k}{\sigma^2} \,\Big|\, \sigma^2 = 1 \right)$$

Questions:

• is it correct?

• is it correct to set $$k=\sigma^2\chi^2(0.05, 10)$$? Not to use (0.95, 10)?

• how to move on with this?

• For which $$\sigma^2_1$$ the power would be greater than 0.95? I am really confused with this question. I have thought that the power is independent regarding $$\sigma^2_1$$.

• $\alpha$ equals the probability of rejecting null hypothesis when the null is true, i.e. when $\sigma^2=1$. This is not reflected in your equation for $\alpha$. Commented Apr 3, 2020 at 9:35
• Sorry, I have discovered stats after posting on math. @StubbornAtom, do you mean: $\alpha = P\left(\sum_{i=1}^{10}X_i^2\geq k \,\Big|\, \sigma^2 = \sigma^2_1 \right)$ ? Commented Apr 3, 2020 at 10:04
• Why the condition on $\sigma^2=\sigma_1^2$? It is the null hypothesis so condition on $\sigma^2=1$. Commented Apr 3, 2020 at 14:37
• Thank you for your comment, I have edited the post. However, could you answer my questions? Commented Apr 3, 2020 at 15:44

The rejection region for testing the null $$H_0:\sigma^2=1$$ against the alternative $$H_1:\sigma^2>1$$ is of the form $$T>k$$, where $$T=\sum\limits_{k=1}^{10} X_i^2$$ and $$k$$ is so chosen that size of the test is $$0.05$$.

You also have $$\frac T{\sigma^2}\sim \chi^2_{10}$$

This implies $$P_{H_0}\left[T>k\right]=P_{H_0}\left[\chi^2_{10}>k\right]=0.05\,,$$

so that in terms of the upper $$5\%$$ point of a chi-square distribution you have $$k=\chi^2_{10,0.05}\,,$$

whose value can be found from a chi-square table or software.

So the test function is

$$\varphi=\begin{cases}1&,\text{ if }T>\chi^2_{10,0.05} \\ 0 &,\text{ else }\end{cases}$$

Now the power at $$\sigma^2=\sigma_1^2(>1)$$ is

\begin{align} E_{\sigma_1^2}[\varphi]&=P_{\sigma_1^2}\left[\frac T{\sigma_1^2}>\frac1{\sigma_1^2}\chi^2_{10,0.05}\right] \\&=P\left[\chi^2_{10}>\frac1{\sigma_1^2}\chi^2_{10,0.05}\right] \end{align}

This would give you the value of $$\sigma_1^2$$ for which this power equals $$0.95$$:

$$\frac1{\sigma_1^2}\chi^2_{10,0.05}=\chi^2_{10,0.95} \implies \sigma_1^2 = \frac{\chi^2_{10,0.05}}{\chi^2_{10,0.95}} = k' \,(\text{say})$$

The nature of the power function $$E_{\sigma^2}[\varphi]=P_{\sigma^2}\left[T>\chi^2_{10,0.05}\right]$$ (i.e. increasing/decreasing in $$\sigma^2$$) would then suggest the possible values of $$\sigma^2_1$$ for which $$E_{\sigma_1^2}[\varphi]>0.95$$.

• Thank you for such a comprehensive answer. Could you explain what does $\ E_{\sigma_1^2}[\phi]$ mean? Commented Apr 3, 2020 at 20:11
• It means the expectation of $\phi$ where the subscript $\sigma_1^2$ emphasizes that the quantity depends on $\sigma_1^2$; specifically it is the power function evaluated at $\sigma^2=\sigma_1^2$. Commented Apr 3, 2020 at 20:35