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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?

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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