I am learning about the Central Limit Theorem (CLT) and confidence intervals from this online course.

I have learnt that that the CLT states that the theoretical sampling distribution of a sample statistic (such as the sample mean) will be approximately normal, have an average of the true, population-level value being estimated and have variability that is a function of the variation of individual values in the population (standard deviation) and the size of the sample the statistic is based upon.

Further, since the distribution of estimates around their truth is normal, 95% of the time, you'll get an estimate that falls ± 2 standard errors from the truth.

However, there is something that I do not understand with regards to 95% confidence intervals. I have learnt from this course that if you take your estimate and add ± 2 standard errors, you will get an interval that contains the unknown truth 95% of the time (i.e. the 95% confidence interval).

I understand that in a theoretical sampling distribution, the true value being estimated is at the center so this is why 95% of the estimates you'll get fall ± 2 standard errors from the truth (since this is a property of normal distributions).

However, the estimate is not likely to be at the center of the distribution in contrast to the truth. Why is it the case that when you add ± 2 standard errors to your estimate, you get an interval that contains the truth 95% of the time?

  • 1
    $\begingroup$ To clarify your problem: are you aware of the properties of the normal distribution (en.wikipedia.org/wiki/Normal_distribution) or are you struggling with the overall approach of defining confidence intervals? $\endgroup$
    – Pitouille
    Sep 4, 2021 at 5:05
  • $\begingroup$ Is this a question from a course or textbook? If so, please add the self-study tag & read its wiki. $\endgroup$ Sep 4, 2021 at 18:14
  • $\begingroup$ @Pitouille: Please don't add the tag self-study to posts, it is better to comment as in my comment, and let the OP tag such themselves. This is discussed at stats.meta.stackexchange.com/questions/5611/… $\endgroup$ Sep 4, 2021 at 18:20
  • $\begingroup$ Ooops, I did not know this rule… I was trying to be helpful. I like when things are properly tagged. Well noted @kjetil b halvorsen! $\endgroup$
    – Pitouille
    Sep 4, 2021 at 18:32
  • $\begingroup$ @Pitouille: Just keep up the work! The reason for this rule is that users must show they are aware of the special rules for self-help questions. $\endgroup$ Sep 4, 2021 at 18:55

1 Answer 1


The reason why the confidence interval has this coverage property is due to some probability manipulations. It might be instructive to do some math.

The Central Limit Theorem states that as $n \to \infty$,

$$ z=\dfrac{\bar{X}- \mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0, 1) $$

Let $z_\alpha$ be the $\alpha^{th}$ quantile of a standard normal distirbution. If $\alpha>0.5$, then

$$ P( \vert z \vert \leq z_{1-\alpha/2} ) = 1-\alpha $$


$$ P \left( -z_{1-\alpha/2}\dfrac{\sigma}{\sqrt{n}} \leq \bar{X} - \mu \leq z_{1-\alpha/2} \dfrac{\sigma}{\sqrt{n}}\right) = 1-\alpha $$

Rearranging to put $\mu$ in the middle of the inequality

$$ P\left( \bar{X} - z_{1-\alpha/2}\dfrac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{X} + z_{1-\alpha/2}\dfrac{\sigma}{\sqrt{n}} \right) = 1-\alpha$$

The argument in $P$ is familliar, right? These are the end points of a confidence interval.


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