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:

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


  • 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 $.

  • $\begingroup$ $\alpha$ 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 $\alpha$. $\endgroup$ Commented Apr 3, 2020 at 9:35
  • $\begingroup$ 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) $ ? $\endgroup$
    – Question
    Commented Apr 3, 2020 at 10:04
  • $\begingroup$ Why the condition on $\sigma^2=\sigma_1^2$? It is the null hypothesis so condition on $\sigma^2=1$. $\endgroup$ Commented Apr 3, 2020 at 14:37
  • $\begingroup$ Thank you for your comment, I have edited the post. However, could you answer my questions? $\endgroup$
    – Question
    Commented Apr 3, 2020 at 15:44

1 Answer 1


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 $E_{\sigma_1^2}[\varphi]>0.95$.

  • $\begingroup$ Thank you for such a comprehensive answer. Could you explain what does $\ E_{\sigma_1^2}[\phi]$ mean? $\endgroup$
    – Question
    Commented Apr 3, 2020 at 20:11
  • $\begingroup$ 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 $\sigma^2=\sigma_1^2$. $\endgroup$ Commented Apr 3, 2020 at 20:35

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