19
$\begingroup$

While doing my homework and checking my answers with the book's answers I noticed that sometimes the standard deviation is divided by $\sqrt n$ where $n$ is the sample size. I'm a little confused. For my current problem I am trying to find the estimated standard error of the estimator. I had found in a previous part of the problem that $\hat \sigma=.33853$ and the sample consists of $16$ measurements. Now the standard error is $.084633$ which is indeed $\frac{\hat \sigma}{\sqrt{16}}$. When I found the standard deviation I didn't divide by $4$, so whats different this time?

$\endgroup$
1
  • $\begingroup$ If you write \sqrt n you see $\sqrt n$ and if you write \sqrt16 you see $\sqrt16,$ where the radical extends only over the $1$ and not the $6;$ so you do not see $\sqrt{16}.$ The radical covers the first object that follows it, so \sqrt123456 gives you $\sqrt123456$ and not $\sqrt{123456}.$ The latter is coded as \sqrt{1234567}. The $\{\text{curly braces}\}$ have the effect that {123456} is treated as a single object. $\qquad$ $\endgroup$ – Michael Hardy Sep 2 '18 at 20:04
4
$\begingroup$

In the normal distribution, if the expectation of the average of a sample size n is the same as the expectation, however, the standard deviation of your sample is to be divided by the square root of your sample size. You may read about Square Root n Law or Central Limit theorem, which should be in your stats book somewhere.

$\endgroup$
3
20
$\begingroup$

This formula may be derived from what we know about the variance of a sum of independent random variables.[4]

If $X_1, X_2 , \ldots, X_n$ are $n$ independent observations from a population that has a mean $\mu$ and standard deviation $\sigma$, then the variance of the total $T = (X_1 + X_2 + \cdots + X_n)$ is $n\sigma^2$.

The variance of $T/n$ must be $\frac{1}{n^2}n\sigma^2=\frac{\sigma^2}{n}$. And the standard deviation of $T/n$ must be $\sigma/{\sqrt{n}}$. Of course, $T/n$ is the sample mean $\bar{x}$.

More explanation @http://en.wikipedia.org/wiki/Standard_error

$\endgroup$
4
$\begingroup$

The answers here don't mention a couple key points. So just wanted to clarify based on my understanding.

This formula is used to calculate the standard deviation of a sample distribution of the mean (of a large number of samples from a population). In other words, it's only applicable when you are looking for the standard deviation of means each calculated from a sample of size $n$, taken from a population.

So lets say i take a population and sample it $10\,000$ times with a sample size of $n=2$. I then take the mean of each one of those samples (so my data contains $10\,000$ calculated means). This equation provides that with a large enough number of samples, the standard deviation of the sample means can be approximated using this formula:

$\frac{\sigma}{\sqrt{n}}$ or in this case $\frac{\sigma}{\sqrt{2}}$

It should be intuitive that as $n \to p$ (population size), your standard deviation of your sample means will become very small, or in other words the means from each sample will have very little variance (if every sample out of $10\,000$ you sampled the entire population you would get no variance from the population mean).

With certain inference conditions (our sample is random, normal, independent) we can actually use this standard deviation calculation to estimate the standard deviation of our population. Since this is just an estimate, its called the Standard Error. The condition for using this as an estimate is that your sample size n is greater than 30 (given by the central limit theorem) and meets the independence condition n <= 10% of population size.

$\endgroup$
1
  • $\begingroup$ +1 Wonderful! Really drilled this concept in my head! Finally someone who can say this is such simple terms. $\endgroup$ – user585380 Sep 19 '18 at 12:08
3
$\begingroup$

I found this paper that could be helpful to some of you: why do we divide by square root of n

It provides both proof and intuition behind the reason there is a difference between the standard deviation of a sample point and the population mean.

$\endgroup$
1
$\begingroup$

$$\sigma_{xbar} = \sigma/n^{1/2}$$

This is used to find the standard deviation of a an xbar distribution.

$\endgroup$

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.