# Confidence interval for mean of a normal population

Consider a normal population with unknown mean $$\mu$$ and variance $$\sigma^2=9$$. To test $$H_{0}:\mu=0$$,against $$H_{1}:\mu \ne 0$$. 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)

Now, if we construct the test statistic $$\tau=\frac{\bar{X_n}-\mu}{0.3} \sim N(0,1)$$ Therefore our test ,actually based on the observed value of $$\tau=\tau_{0}$$ is given by $$|\tau_{0}|> \tau_{\frac{\alpha}{2}}$$ where $$\alpha=0.05$$ Now,we get $$K=\frac{3}{10}\tau_{0.025}=0.588$$ Thus the 95% confidence interval for $$\mu$$ is coming to be $$(\bar{X_n}-0.588 ,\bar{X_n}+0.588)$$So, looking at the options it seems that both (A) , (C) are correct but only (C) is given to be correct. Please help!

• Consider the width of the CI. – GNUSupporter 8964民主女神 地下教會 Dec 2 '18 at 14:51
• sorry,i edited one of the options – Legend Killer Dec 2 '18 at 14:52
• @GNUSupporter8964民主女神地下教會they have the same width,I mean options A,C – Legend Killer Dec 2 '18 at 14:55