### Chapter 5. Sampling: Sampling Distributions

### Sampling Distribution of the Sample Mean

When using a sample mean to estimate a population mean, the *sampling **distribution of the sample mean *can be used to determine how much estimation error is reasonable to expect.

#\phantom{0}#

Sampling Distribution of the Sample Mean

The **sampling distribution of the sample mean** is the probability distribution of the sample means of every possible sample of a particular size #n# that can be drawn from a population.

The *mean* of the distribution of sample means is called the **expected value of the sample mean **and is denoted #\mu_{\bar{X}}#.

The *standard deviation* of the distribution of sample means is called the **standard error of the mean **and is denoted #\sigma_{\bar{X}}#. The standard error is a measure of how much discrepancy to expect between a sample mean #\bar{X}# and the population mean #\mu#.

Normally Distributed Population

If a random sample of size #n# is drawn from a population which is *normally* distributed with parameters #\mu# and #\sigma#, then the sampling distribution of #\bar{X}# will have the following characteristics:

**Shape***.*The sampling distribution of the sample mean will be*normally*distributed.**Central tendency***.*The mean of the sampling distribution, #\mu_{\bar{X}}#, will be equal to the population mean #\mu#.

\[\mu_{\bar{X}} = \mu\]**Variability**. The standard deviation of the sampling distribution, #\sigma_{\bar{X}}#, is calculated as follows:

\[\sigma_{\bar{X}} = \cfrac{\sigma}{\sqrt{n}}\]

\[\bar{X} \sim N(\mu, \cfrac{\sigma}{\sqrt{n}})\]

#\mu_{\bar{X}} = 100#

#\sigma_{\bar{X}} = 2.3717#

Since the population from which the sample is drawn is *normally *distributed, we know that the *sampling distribution of the sample mean* is the normally distributed as well.

The *expected value of the sample mean*, #\mu_{\bar{X}}#, is equal to the population mean #\mu#:

\[\mu_{\bar{X}} = \mu = 100\]

The *standard error of the sample mean*, #\sigma_{\bar{X}}#, is calculated as follows:

\[\sigma_{\bar{X}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{15}{\sqrt{40}}=2.3717\]

So #\bar{X} \sim N(100, 2.3717)#.

#\phantom{0}#

Suppose, however, that a random sample is drawn from a population that is not normally distributed. In these cases, the

*Central Limit Theorem*can be invoked to determine the characteristics of the

*sampling distribution of the sample mean*.

#\phantom{0}#

Central Limit Theorem

The **Central Limit Theorem** states that, as the sample size increases, the *sampling distribution of the sample mean* approaches a normal distribution with a mean of #\mu_{\bar{X}}=\mu# and a standard deviation of #\sigma_{\bar{X}}=\sigma/\sqrt{n}#, regardless of the shape of the population from which the sample is drawn.

A sample size of #30# or more is large enough to consider the *sampling distribution of the sample mean *to be approximately normal.

*simple random sample*of size #80# is drawn from from a population having a

*right-skewed*

*distribution with a mean of #\mu = 130# and a standard deviation of #\sigma = 5#.*

Determine the

*expected value of the sample mean*, #\mu_{\bar{X}}#, and the

*standard error of the sample mean*, #\sigma_{\bar{X}}#.

#\mu_{\bar{X}} = 130#

#\sigma_{\bar{X}} = 0.5590#

A sample size of #n=80# is considered large enough for the *Central Limit Theorem *to apply.

This means that, although the sample in question comes from a population having a *right-skewed** *distribution, the *sampling distribution* *of the sample* *mean* is approximately normal.

Now that we know that the *sampling distribution of the sample mean *is approximately normal, we can determine its characteristics:

- #\mu_{\bar{X}} = \mu = 130#
- #\sigma_{\bar{X}} = \cfrac{\sigma}{\sqrt{n}} = \cfrac{5}{\sqrt{80}} = 0.5590#

So #\bar{X} \sim N(130, 0.5590)#.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.