A power supply manufacturer is interested in the variability of the output voltage, a variable known to follow a normal law. He tried $12$ selected random blocks and obtained the following results, the standard deviation of which is $0.322$: $5.34$, $5.00$, $5.07$, $5.25$, $5.65$, $5.55$, $5.35$, $5.35$, $4.76$, $5.54$, $5.44$, $4.61$.
If the true value of $\sigma^2$ is $1.0$, what is the probability that the hypothesis to be rejected with $\alpha = 0.05$.
So the hypothesis is $H_0: \sigma^2 = 0.5$ and $H_1: \sigma^2 \not= 0.5$.
$$\beta = P(\chi_{1-\frac{\alpha}{2}; 11} \leq U_0 \leq \chi_{\frac{\alpha}{2};11} | \sigma_1^2 = 1)$$ $$=P(3.82 \leq U_0 \leq 21.82 | \sigma_1^2 = 1)$$ $$=P(3.82 \leq \frac{(n-1) S^2}{\sigma_0^2} \leq 21.82 | \sigma_1^2 = 1)$$ $$=P(\frac{3.82\sigma_0^2}{\sigma_1^2} \leq \frac{(n-1) S^2}{\sigma_1^2} \leq \frac{21.82 \sigma_0^2}{\sigma_1^2})$$ $$P(1.91 \leq U \leq 10.91)\text{, where } U \sim \chi^2_{11}$$ So $$1 - \beta = P(U > 10.91) + P(U < 1.91)$$
The answer is $0.45$, but I obtained $0.55$. What am I doing wrong?
Setting $U=\frac{(n-1)S^2}{\sigma^2}\sim \chi_{(11)}^2$, you have
$$1-\alpha=0.95=\mathbb{P}[5.124\leq U\leq 21.920]$$
thus the "Acceptance region" for $H_0$ is
$$0.2329\leq S^2\leq 0.9964$$
Given $\sigma^2=1$ you can calculate the same region by
$$\mathbb{P}[2.56\leq U\leq 10.96]=0.995-0.447\approx0.55$$
leading to the desired result of
$$1-0.55=0.45$$