Calculating 95% confidence interval for mean for a normal population Consider a normal population with unknown $\mu$ and variance $\sigma^2=9$. To test $H_0:\mu=0$ against $H_1:\mu\neq 0$, a random sample of size 100 is taken. Based on this sample, the test of the form $|\bar{X_n}|>K$ rejects the null hypothesis at 5% level of significance. Then, which of the following is a possible 95% confidence interval for $\mu$ ?
(A)$(-0.488,0.688)$ (B)$(-1.96,1.96)$ (C)$(0.422,1.598)$ (D)$(0.588,1.96)$
$\begin{aligned} 0.95 & = 1-\alpha=P(-z \le Z \le z)=P \left(-1.96 \le \frac {\bar X-\mu}{\sigma/\sqrt{n}} \le 1.96 \right) \\[6pt] & = P \left( \bar X - 1.96 \frac{\sigma}{\sqrt{n}} \le \mu \le \bar X + 1.96 \frac{\sigma}{\sqrt{n}}\right)\\[6pt] & = P \left( \bar X - 1.96 \frac{3}{10} \le \mu \le \bar X + 1.96 \frac{3}{10}\right) \\ & = P \left( \bar X - 0.588 \le \mu \le \bar X + 0.588\right) \end{aligned}$
Both (A) and (C) seem to work here with sample mean being 0.1 and 1.01 respectively. But if the sample mean were 1.01, we would be reject $H_0$ using hypothesis testing at 5% significance level ($z=3.366>1.65$). 
So, should (A) be the answer ?
 A: Option A)\begin{align} \bar{x}-0.588&=-0.488\implies \bar{x}=0.1 \\ \bar{x}+0.588&=0.688\implies \bar{x}=0.1 \end{align}
So if the sample mean $\bar{x}$ was $0.1$ the 95% confidence interval  for the population mean $\mu$  would be $(-0.488\le \mu \le0.688)$.  $0$ is contained within the confidence interval so this is correct.
Option B)\begin{align} \bar{x}-0.588&=-1.96\implies \bar{x}=-1.372 \\ \bar{x}+0.588&=1.96\implies \bar{x}=1.372 \end{align}
So the sample mean $\bar{x}$ is not consistent for both sides of the confidence interval as it should be. This option is incorrect.
Option C)\begin{align} \bar{x}-0.588&=0.422\implies \bar{x}=1.01 \\ \bar{x}+0.588&=1.598\implies \bar{x}=1.01\end{align}So if the sample mean $\bar{x}$ was $1.01$ the 95% confidence interval  for the population mean $\mu$  would be $(0.422\le \mu \le1.598)$.  $0$ is not contained within the confidence interval so this is rejected.
Option D)\begin{align} \bar{x}-0.588&=0.588\implies \bar{x}=1.176 \\ \bar{x}+0.588&=1.96\implies \bar{x}=1.372 \end{align}
So the sample mean $\bar{x}$ is not consistent for both sides of the confidence interval as it should be. This option is incorrect.
A: we are given that $P(|\bar{X_n}|>K|H_0 \text{ is true})=0.05$
Hence, $P(\sqrt{n}\frac{|\bar{X_n}-0|}{\sigma}>\frac{\sqrt{n}}{\sigma}K)=0.05$
Hence, $\frac{\sqrt{n}}{\sigma}K=q_{0.025}=1.96$
Hence, $K=0.588$
Now, since the test $|\bar{X_n}|>K$ rejects the null hypothesis at $5\%$ level of significance, hence we are given that the realized value of  $|\bar{X_n}|$
i.e. $\bar{x_n}$ is such that $|\bar{x_n}|>0.588$
Also, as you have already derived, the CI for $\mu$ is of the form
$ \bar{X_n} - 0.588 \le \mu \le \bar{X_n} + 0.588$
Now, you have to figure out which one of the four options can be written in the form $ [\bar{x_n} - 0.588 , \bar{x_n} + 0.588]$ such that $|\bar{x_n}|>0.588$
It turns out, option C is the only option that can be written in this form.
