###### The Standard Normal Distribution

The standard normal distribution refers to a normal distribution that has been standardized... **Read More**

The standard error (SE) of the sample mean refers to the standard deviation of the distribution of the sample means. It gives analysts an estimate of the variability they would expect if they were to draw multiple samples from the same population. While the standard deviation measures the variability obtained within one sample, the standard error gives an estimate of the variability among many samples.

Provided the population standard deviation, σ, is known, analysts use the following formula to estimate the standard error of the sample mean, denoted as σ_{x}:

$$ \sigma_x=\cfrac {\sigma}{\sqrt n} $$

Where *n* is the sample size.

It is noteworthy that the population standard deviation, σ, is usually unknown. When the population standard deviation, σ, is unknown, the following formula is used to estimate the standard error of the sample mean, also denoted as S_{x}:

$$ S_x =\cfrac {S}{\sqrt n} $$

Where:

\(S\) = The sample standard deviation; and

\(S^2 =\cfrac {\sum \left(X_i- X \right)^2}{n- 1} \).

The standard error of the sample mean gives analysts an idea of how **precisely **the sample mean estimates the population mean. A lower value of the standard error indicates a more precise estimation of the population mean. On the other hand, a larger value of the standard error indicates a less precise estimate of the population mean.

It is also important to note that the standard error becomes smaller as the sample size increases. This happens because increasing the sample size ultimately brings the sample mean closer to the true value of the population mean.

In a certain property investment company with an international presence, workers have a mean hourly wage of $12 with a population standard deviation of $3. Given a sample size of 30, the standard error of the sample mean is *closest* to:

**Solution**

$$ \begin{align*} \sigma_x & =\cfrac {\sigma}{\sqrt n} \\ & =\cfrac {3}{\sqrt {30}} \\ & = $0.55 \\ \end{align*} $$

* Interpretation: *If we were to draw several samples of size 30 from the employee population and construct a sampling distribution of the sample means, we would end up with a mean of $12 and a standard error of $0.55.

A sample of 30 latest returns on XYZ stock reveals a mean return of $4 with a sample standard deviation of $0.13. The standard error of the sample mean is *closest* to:

**Solution**

$$ \begin{align*} S_x & =\cfrac {S}{\sqrt n} \\ & =\cfrac {0.13}{\sqrt {30}} \\ & = $0.02 \\ \end{align*} $$

* Interpretation: *If we were to draw more samples from the population of yearly returns on XYZ stock and construct a sample mean distribution, we would end up with a mean of $4 and a standard error of $0.02.

QuestionSuppose we increased the sample size to 80 in the example above and derived similar values for the mean and standard deviation of returns, then the standard error of the sample mean would be

closestto:

- 0.01.
- 0.02.
- 0.08.

SolutionThe correct answer is

A.$$ \begin{align*} S_x & =\cfrac {S}{\sqrt n} \\ & = \cfrac {0.13}{\sqrt {80}} \\ & = $0.01 \\ \end{align*} $$

This clearly proves that increasing the sample size reduces the SE of the sample mean.