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Sigma / sqrt (n) — why is it used?

Statistics Blog > sigma / sqrt (n)

You’ll come across a couple of different formulas when calculating z-scores:
sigma / sqrt(n)
z value formula
When I first started learning statistics, the different formulas completely confused me. They still do (a little) and I have to think about what I’m doing every time I calculate a z-score. In order to figure out which formula to use, and why we use sigma / sqrt (n), I tried to really understand what I was doing by asking questions:

Q. When do I use sigma /sqrt(n)?

A. You always divide by sqrt(n). However, occasionally the square root of n sometimes equals 1 (making it just σ in the denominator. for example, if you are choosing one person and trying to figure out the probability their weight is under x pounds, then n=1. In other words, if you are calculating a z-score, you can always use √(n).

Q. Why do we have to use sigma / sqrt(n)?

When you are estimating the standard error, SE, for the mean (the SE is the standard deviation of the means of samples), the larger your sample size, the smaller the standard deviation. for example, if you took a sample of 200, you would be much more likely to get close to the true mean than if you took a sample of 2. In other words, the larger your “n”, the smaller the standard deviation.

To see a visualization of this fact, try out this applet, which shows you how the sampling distribution of the sample mean approaches a normal distribution (you can also visualize how the sample standard deviation gets smaller as n gets larger.

Need a mathematical proof? Dr.Math gave an excellent answer to the question of proof. You can read the response here.


If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track.

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Sigma / sqrt (n) — why is it used? was last modified: January 17th, 2018 by Stephanie

2 thoughts on “Sigma / sqrt (n) — why is it used?

  1. Andrew

    I think it’s misleading to say σ / √(n) estimates σ. That’s silly since you would need to know σ to compute σ / √(n)! However, σ / √(n) does estimate the standard deviation of means taken from samples of size n (see Central Limit Theorem).

    SE is NOT just another name for the standard deviation of your sample (usually called s). Usually people define SE = σ / √(n) when σ is known, and SE = s / √(n) when σ is unknown. So SE estimates the standard deviation of MEANS OF samples.

  2. Andale Post author

    I went back and reread my post and can’t figure out for the life of me why I wrote that σ/√(n) estimates σ. Not enough coffee at time of writing? Anyway, the errors are fixed.
    Thanks for taking the time to point those errors out,