I have given following information:
$n=15$ - sample size
$S_n=0.013$ - sample standard deviation

I have to test null hypothesis $H_0(\sigma=0.01)$ agaist alternative hypothesis $H_1(\sigma>0.01)$ on a significance level $\alpha=0.05$.
Then I have to compute type II error $\beta$ and test power $M=1-\beta$ if I know that exact value of $\sigma$ is $0.02$.
I will write my work and I would like you to tell me where I am making a mistake. Thanks!

For the first part I've done the following:

$H_0(\sigma=0.01)$, $H_1(\sigma>0.01)$
$\alpha=0.05=P_{H_0}\{H_0 \ will\ be\ rejected\}=P_{H_0}\{S_n^2>c\}=P_{H_0}\{\frac{nS_n^2}{\sigma_0^2}>\frac{nc}{\sigma_0^2}\}=P_{H_0}\{\chi^2_{14}>\frac{nc}{\sigma_0^2}\}$

Using probability calculator I compute that $\frac{nc}{\sigma_0^2}=23.685$, which gives me $c=0.0001579$ for $n=15$ and $\sigma=0.01$. Hence, critical area is $C=[0.0001579,\infty)$
Since $S_n^2=0.000169\in C$, $H_0$ will be rejected and $H_0(\sigma>0.01)$ will be accepted.

For the second part I tried the following but I didn't get the result we are told we should get:
$H_0(\sigma=0.01)$, $H_1(\sigma=0.02)$, $n=15$ and $\sigma_1=0.02$

$\beta=P_{H_1}\{H_1 \ will\ be\ rejected\}=P_{H_1}\{H_0\ will\ be\ accepted\}=P_{H_1}\{S_n^2<c\}=P_{H_1}\{S_n^2<0.0001579\}=P_{H_1}\{\frac{nS_n^2}{\sigma_1^2}<\frac{0.0001579n}{\sigma_1^2}\}=P_{H_1}\{\frac{nS_n^2}{\sigma_1^2}<5.92125\}=P_{H_1}\{\chi_{14}^2<5.92125\}$

Using probability calculator I compute that $\beta=0.031563$ and $M=1-\beta=0.968437$

  • $\begingroup$ Do you have your $\beta$ and your $M$ switched at the end there? You are certainly missing the $1-$ part of "$P(H_0\text{ will be rejected})=1-P(H_0\text{ will be accepted})$". $\endgroup$ – tabstop Feb 26 '14 at 22:13
  • $\begingroup$ I made a typing mistake. $H_0$ is to be accepted in the second part. Thanks. Can you help me further? $\endgroup$ – gov Feb 26 '14 at 22:16

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