I am a bit confused when looking at the confidence interval formula for one of my class and I am looking to see if anyone can clarify it.

When constructing, let's say, a 95% confidence interval of a sample mean the formula is:

$\overline{Y} \displaystyle \pm 1.96 * SE$ where the $SE = \frac{\sqrt{\sigma^2}}{\sqrt{N}}$. However, in some cases the $SE =\sqrt{\sigma^2} $. How can I differentiate between the two? Is it because $ \sqrt{\sigma^2_\overline{Y}} = \frac{\sqrt{\sigma^2_x}}{\sqrt{N}}$?

Sorry if that sounds confusing but it's been giving me a bit trouble to wrap my head around.


1 Answer 1


OK, this might be longer than I care for it to be. Go along with it because I think it will clarify some things".

Your estimate of the mean is actually a random variable. That is because $\bar{Y}$ is a function of random variables (more specifically a scaled sum $\bar{Y} = \sum_i x_i/n$), and a function of random quantities is itself a random quantity.

It is completely sensible to ask "if $\bar{Y}$ is a random variable, then what is its distribution". According to the central limit theorem, the asymptotic (that is, with enough data) sampling distribution of $\bar{Y}$ is

$$\bar{Y} \sim \mathcal{N} \left(\mu, \dfrac{\sigma^2_x}{N} \right)$$

Here $\mu$ is the true mean of the distribution that gave you the data $x_i$ and $\sigma^2_x$ is the variance of the distribution that gave you the data $x_i$. $N$ is the size of the sample used to compute $\bar{Y}$. We call the quantity $\sqrt{\sigma^2_x}/\sqrt{N} = \sigma_{\bar{Y}}$ the standard error of the mean. The standard error is the standard deviation of the sampling distribution.

So when you use the CI formula, you are essentially telling people where the middle 95% probability exists for that sampling distribution.

So, to answer your question

  • $\sigma_{\bar{Y}}$ is the standard deviation of the sampling distribution. It is computed using...

  • $\sigma_x$ which is the standard deviation of the data in your sample.


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