Say I have a given $y \in \mathbb{R}$ that is fixed, and $\mathbf{X}_1,...,\mathbf{X}_n$ be i.i.d with distribution $N(\mu, \sigma^2)$.
I can further estimate $\hat{\mu} = \frac{1}{n}\sum_{i=1}^n \mathbf{X}_i$ (is needed below).
Given: $$(1)~~~~~~~~~~\sqrt{n}\frac{\hat{\mu}-\mu}{\sigma} \approx N(0,1)$$
Say (1) holds (for this I do not need to show otherwise), how can I construct a $(1-y)$-confidence interval for $\hat{\mu}$?
I am really unsure how to engage this problem at all as I have no confidence level (at least not defined, to find a critical value $z$), nor any values specified for any of the data to actually start computing sample mean, standard deviation etc. to finally construct a confidence interval.
Can anyone point me in the right direction?
The usual idea for creating a confidence interval is to find a test statistic, which has a distribution which does not depend on the parameter itself. Here the test statistic $$W=\sqrt{n}\frac{\bar{X}-\mu}{\sigma}$$ has a distribution $N(0,1)$ which does not depend on $\mu$. Therefore if we choose $q_1,q_2$, such that $$P(W<q_1)= \frac{\alpha}{2} = P(W> q_2)$$ then $$1-\alpha =P(q_1 \leq W \leq q_2) = P(q_1 \leq \sqrt{n}\frac{\bar{X}-\mu}{\sigma} \leq q_2).$$ Can you see how the above can be used to create a confidense interval for $\mu$? The values $q_1$ and $q_2$ are quantiles of the normal distribution, that is $q_1=\Phi^{-1}(\alpha/2)$ and $q_2 = \Phi^{-1}(1-\alpha/2)$. The symmetry of the normal distribution actually gives that $q_2 = -q_1$. For $\alpha = 0.05$ (the usual case) we would have $q_1=-1.96$ and $q_2=1.96$.
Remark: In the above computations i assumed that $\sigma^2$ is a known value, but if this is not the case we would have to estimate using the sample variance, and that would give a t-distribution with n-1 degrees of freedom for W.