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?

  • $\begingroup$ You mention the central limit theorem, but if we assume that $X_1,...,X_n$ are i.i.d. $N(\mu,\sigma^2)$, then $W$ has exactly a $N(0,1)$ distribution (no limit theorems involved). $\endgroup$ Dec 3, 2019 at 21:19
  • $\begingroup$ @LeanderTilstedKristensen that makes sense of cause, thank you for pointing that out. $\endgroup$
    – NewDev90
    Dec 3, 2019 at 21:50

1 Answer 1


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.

  • $\begingroup$ Hmm I follow most of it I believe. I am a bit unsure on the part where you define $q_1$ and $q_2$ in terms of $P(W < q_1) = \alpha/2 = P(W > q_2)$: Where does $\alpha/2$ come from? In terms of the symmetry of the normal distribution, I suppose you calculate $q_1 = -1.96$ from a standard z-table using $\Phi^{-1} \alpha/2$ = 0.0250, which reads exactly -1.96? And similar for $\Phi^{-1} 1-\alpha/2 = 0.975$, which has the value 1-96. I have not seen the $\Phi$ notation before, but is this how I am supposed to use it? $\endgroup$
    – NewDev90
    Dec 3, 2019 at 21:43
  • $\begingroup$ If the confidence level is $(1-\alpha)$ then intuitively we would like to create the interval in such a way, that the probability of guessing too high or too low is the same, so we $\alpha/2$ such that the total probability of "missing" is $\alpha$. $\endgroup$ Dec 3, 2019 at 22:30
  • $\begingroup$ Obviously notation might vary, but $\Phi$ usually refers to the CDF of the standard normal distribution (at least in the context of statistics). $\Phi^{-1}$ is simply the inverse, which means that $\Phi^{-1}(\Phi(x)) = x$ for all x, and $\Phi(\Phi^{-1}(y))=y$ for $y\in (0,1)$. Computing $\Phi^{-1}$ can be done using a table, but most statistical software should have it implemented. (for example qnorm() in R or norminv() in matlab) $\endgroup$ Dec 3, 2019 at 22:32
  • $\begingroup$ The symmetry comes from the fact that $\Phi(-x) = 1 - \Phi(x)$, hence $\Phi(-q_2)=1-\Phi(q_2) = 1- (1-\alpha/2)= \alpha/2$. $\endgroup$ Dec 3, 2019 at 22:42
  • $\begingroup$ Alright great, thanks for explaining, that really clarified things for me, in terms of engaging problems of this sort. $\endgroup$
    – NewDev90
    Dec 4, 2019 at 8:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.