search
Search
Login
Unlock 100+ guides
menu
menu
web
search toc
close
Comments
Log in or sign up
Cancel
Post
account_circle
Profile
exit_to_app
Sign out
What does this mean?
Why is this true?
Give me some examples!
search
keyboard_voice
close
Searching Tips
Search for a recipe:
"Creating a table in MySQL"
Search for an API documentation: "@append"
Search for code: "!dataframe"
Apply a tag filter: "#python"
Useful Shortcuts
/ to open search panel
Esc to close search panel
to navigate between search results
d to clear all current filters
Enter to expand content preview
icon_star
Doc Search
icon_star
Code Search Beta
SORRY NOTHING FOUND!
mic
Start speaking...
Voice search is only supported in Safari and Chrome.
Navigate to

Guide on standard error in statistics

schedule Aug 12, 2023
Last updated
local_offer
Tags
mode_heat
Master the mathematics behind data science with 100+ top-tier guides
Start your free 7-days trial now!
Theorem.

The expected value and variance of the sample mean

Suppose that a random sample of $n$ observations is taken from some population with mean $\mu_X$ and variance $\sigma_X^2$. Each random variable $X_1,X_2,…,X_n$ of the random sample will have the same distribution (iid). If such is the case, then the sample mean, $\bar{X}$, will have mean and variance as given below:

$$\begin{align*} \mu_\bar{X}&=\mathbb{E}(\bar{X})=\mu\\ \sigma_\bar{X}^2&=\mathbb{V}(\bar{X})=\frac{\sigma_X^2}{n} \end{align*} $$

Note the following points:

  • these results hold true regardless of the underlying probability distribution of $X$.

  • only the expected value and variance of the sample mean $\bar{X}$ can be calculated - not the distribution of $\bar{X}$.

The expected value of the sample mean is:

$$\begin{align*} \mathbb{E}(\bar{X})&=\mathbb{E} \Big(\frac{1}{n}\sum^{n}_{i=1}X_i\Big)\\ &=\frac{1}{n}\sum^{n}_{i=1}\mathbb{E}(X_i)\\ &=\frac{1}{n}\sum^{n}_{i=1}\mu_X\\ &=\frac{1}{n}(n\mu_X)\\ &=\mu_X\\ \end{align*}$$

The variance of the sample mean is:

$$\begin{align*} \mathbb{V}(\bar{X})&=\mathbb{V}\Big(\frac{1}{n}\sum^{n}_{i=1}X_i\Big)\\ &=\frac{1}{n^2}\sum^{n}_{i=1}\mathbb{V}(X_i)\\ &=\frac{1}{n^2}\sum^{n}_{i=1}\sigma^2_X\\ &=\frac{1}{n^2}(n\sigma^2_X)\\ &=\frac{\sigma^2_X}{n}\ \end{align*}$$
Definition.

Standard error

The standard error is defined as the standard deviation of an estimator or statistic.

Example.

Standard error of the sample mean

What is the standard error of the sample mean?

When we try to estimate the population mean using the sample mean, we know that the sample mean won't exactly equal the population mean. In order to know how good the estimate is, we use a statistic called the standard error of the sample mean.

A high value for the standard error means that there is high uncertainty in our estimate, whereas a low standard error means that we can be certain our estimate isn't off by much.

Recall from theorem 5 that the variance of the sample mean $\bar{X}$ is:

$$\begin{align*} \sigma_\bar{X}^2&=\mathbb{V}(\bar{X})=\frac{\sigma_X^2}{n} \end{align*} $$

The standard error of the sample mean $\bar{X}$ is the standard deviation of $\bar{X}$:

$$\begin{align*} \sigma_\bar{X}&=\sqrt{\mathbb{V}(\bar{X})}=\frac{\sigma_X}{\sqrt{n}} \end{align*} $$

As we can see, the standard error decreases as the sample size $n$ increases. This should be intuitive because the more data we use to compute the sample mean, the more precise the sample mean is in estimating the population mean. As the sample size tends to infinity, the standard error approaches zero.

robocat
Published by Isshin Inada
Edited by 0 others
Did you find this page useful?
thumb_up
thumb_down
Comment
Citation
Ask a question or leave a feedback...
thumb_up
0
thumb_down
0
chat_bubble_outline
0
settings
Enjoy our search
Hit / to insta-search docs and recipes!