# 4.1 - Sampling Distribution of the Sample Mean

4.1 - Sampling Distribution of the Sample MeanLet’s put some numbers into Ellie’s example.

**Note!**The sampling method is done without replacement.

## Sample Means with a Small Population: Runner’s MIleage

In this example, the population is the mileage of six runners. Ellie is going to try to guess the true average mileage of the six runners by taking a random sample without replacement from the population.

Mileage | A | B | C | D | E | F |
---|---|---|---|---|---|---|

19 | 14 | 15 | 9 | 10 | 17 |

Since we know the miles from the population, we can find the population mean.

\(\mu=\dfrac{19+14+15+9+10+17}{6}=14\) miles

To demonstrate the sampling distribution, let’s start with obtaining all of the possible samples of size \(n=2\) from the populations, sampling without replacement. The table below show all the possible samples, the weights for the chosen runners the sample mean and the probability of obtaining each sample. Since we are drawing at random, each sample will have the same probability of being chosen.

View Full Table

Sample | Mileage | \(\boldsymbol{\bar{x}}\) | Probability |

A, B | 19, 14 | 16.5 | \(\frac{1}{15}\) |

A, C | 19, 15 | 17.0 | \(\frac{1}{15}\) |

A, D | 19, 9 | 14.0 | \(\frac{1}{15}\) |

A, E | 19, 10 | 14.5 | \(\frac{1}{15}\) |

A, F | 19, 17 | 18.0 | \(\frac{1}{15}\) |

B, C | 14, 15 | 14.5 | \(\frac{1}{15}\) |

B, D | 14, 9 | 11.5 | \(\frac{1}{15}\) |

B, E | 14, 10 | 12.0 | \(\frac{1}{15}\) |

B, F | 14, 17 | 15.5 | \(\frac{1}{15}\) |

C, D | 15, 9 | 12.0 | \(\frac{1}{15}\) |

C, E | 15, 10 | 12.5 | \(\frac{1}{15}\) |

C, F | 15, 17 | 16.0 | \(\frac{1}{15}\) |

D, E | 9, 10 | 9.5 | \(\frac{1}{15}\) |

D, F | 9, 17 | 13.0 | \(\frac{1}{15}\) |

E, F | 10, 17 | 13.5 | \(\frac{1}{15}\) |

We can combine all of the values and create a table of the possible values and their respective probabilities.

\(\boldsymbol{\bar{x}}\) | 9.5 | 11.5 | 12.0 | 12.5 | 13.0 | 13.5 | 14.0 | 14.5 | 15.5 | 16.0 | 16.5 | 17.0 | 18.0 |

Probability | \(\frac{1}{15}\) | \(\frac{1}{15}\) | \(\frac{2}{15}\) | \(\frac{1}{15}\) | \(\frac{1}{15}\) | \(\frac{1}{15}\) | \(\frac{1}{15}\) | \(\frac{2}{15}\) | \(\frac{1}{15}\) | \(\frac{1}{15}\) | \(\frac{1}{15}\) | \(\frac{1}{15}\) | \(\frac{1}{15}\) |

The table is the probability table for the sample mean and it is the sampling distribution of the sample mean mileage of the runners when the sample size is 2. It is also worth noting that the sum of all the probabilities equals 1. It might be helpful to graph these values.

One can see that the chance that the sample mean is exactly the population mean is only 1 in 15, very small. (In some other examples, it may happen that the sample mean can never be the same value as the population mean.) When using the sample mean to estimate the population mean, some possible error will be involved since the sample mean is random.

Now that we have the sampling distribution of the sample mean, we can calculate the mean of all the sample means. In other words, we can find the mean (or expected value) of all the possible \(\bar{x}\)’s.

The mean of the sample means is

\(\mu_\bar{x}=\sum \bar{x}_{i}f(\bar{x}_i)=9.5\left(\frac{1}{15}\right)+11.5\left(\frac{1}{15}\right)+12\left(\frac{2}{15}\right)\\+12.5\left(\frac{1}{15}\right)+13\left(\frac{1}{15}\right)+13.5\left(\frac{1}{15}\right)+14\left(\frac{1}{15}\right)\\+14.5\left(\frac{2}{15}\right)+15.5\left(\frac{1}{15}\right)+16\left(\frac{1}{15}\right)+16.5\left(\frac{1}{15}\right)\\+17\left(\frac{1}{15}\right)+18\left(\frac{1}{15}\right)=14\)

Even though each sample may give you an answer involving some error, the expected value is right at the target: exactly the population mean. In other words, if one does the experiment over and over again, the overall average of the sample mean is exactly the population mean.

Now, let's do the same thing as above but with sample size \(n=5\)

Sample |
Mileage |
\(\boldsymbol{\bar{x}}\) |
Probability |

A, B, C, D, E |
19, 14, 15, 9, 10 |
13.4 |
1/6 |

A, B, C, D, F |
19, 14, 15, 9, 17 |
14.8 |
1/6 |

A, B, C, E, F |
19, 14, 15, 10, 17 |
15.0 |
1/6 |

A, B, D, E, F |
19, 14, 9, 10, 17 |
13.8 |
1/6 |

A, C, D, E, F |
19, 15, 9, 10, 17 |
14.0 |
1/6 |

B, C, D, E, F |
14, 15, 9, 10, 17 |
13.0 |
1/6 |

The sampling distribution is:

\(\boldsymbol{\bar{x}}\) |
13.0 |
13.4 |
13.8 |
14.0 |
14.8 |
15.0 |

Probability |
1/6 |
1/6 |
1/6 |
1/6 |
1/6 |
1/6 |

The mean of the sample means is...

\(\mu=(\dfrac{1}{6})(13+13.4+13.8+14.0+14.8+15.0)=14\) miles

The following dot plots show the distribution of the sample means corresponding to sample sizes of \(n=2\) and of \(n=5\).

Again, we see that using the sample mean to estimate population mean involves sampling error. However, the error with a sample of size \(n=5\) is on the average smaller than with a sample of size\(n= 2\).

## Sampling Error and Size

- Sampling Error
- The error resulting from using a sample characteristic to estimate a population characteristic.

**Sample size and sampling error:** As the dot plots above show, the possible sample means cluster more closely around the population mean as the sample size increases. Thus, the possible sampling error decreases as sample size increases.

What happens when the population is not small?

## Sample Means with Large Samples

An instructor of an introduction to statistics course has 200 students. The scores out of 100 points are shown in the histogram.

The population mean is \(\mu=69.77\) and the population standard deviation is \(\sigma=10.9\).

Let's demonstrate the samping distribution of the sample means using the StatKey website. The first video will demonstrate the sampling distribution of the sample mean when n = 10 for the exam scores data. The second video will show the same data but with samples of n = 30.

You should start to see some patterns. The mean of the sampling distribution is very close to the population mean. The standard deviation of the sampling distribution is smaller than the standard deviation of the population.

In the examples so far, we were given the population and sampled from that population.

What happens when we do not have the population to sample from? What happens when all that we are given is the sample? Fortunately, we can use some theory to help us. The mathematical details of the theory are beyond the scope of this course but the results are presented in this lesson.

In the next two sections, we will discuss the sampling distribution of the sample mean when the population is Normally distributed and when it is not.

# 4.1.2 - Population is Not Normal

4.1.2 - Population is Not NormalWhat happens when the sample comes from a population that is not normally distributed? This is where the Central Limit Theorem (CLT) comes in.

#### Central Limit Theorem

For a large sample size (we will explain this later), \(\bar{x}\) is approximately normally distributed, regardless of the distribution of the population one samples from. If the population has mean \(\mu\) and standard deviation \(\sigma\), then the distribution of \(\bar{x}\) has mean \(\mu\) and standard deviation \(\dfrac{\sigma}{\sqrt{n}}\).

We should stop here to break down what this theorem is saying because the Central Limit Theorem is very powerful!

The Central Limit Theorem applies to a sample mean from any distribution. We could have a left-skewed or a right-skewed distribution. As long as the sample size is large, the distribution of the sample means will follow an approximate Normal distribution.

For the purposes of this course, a sample size of \(n>30\) is considered a large sample.

For many people just learning statistics there is a "so what" thought about the CLT. Why is this important and why do I care? If you recall, when we introduced the idea of Z scores we did so with the caveat that the distribution was normal. We take the observed data, that is normally distributed, and convert the data to z scores creating a standard normal distribution. We then leveraged this distribution to find percentiles (and will in future units leverage this to find probabilities.

The CLT allows us to assume a distribution IS normal as long as the sample size is greater than 30 observations. With this, we can apply most of our inferential statistics without having to compensate for non-normal distributions. This will take on greater relevance as we move through the course.

## Sampling Distribution of the Sample Mean

With the Central Limit Theorem, we can finally define the sampling distribution of the sample mean.

**Sampling Distribution of the Sample Mean**

The sampling distribution of the sample mean will have:

- the same mean as the population mean, \(\mu\)
- Standard deviation [standard error] of \(\dfrac{\sigma}{\sqrt{n}}\)

It will be Normal (or approximately Normal) if either of these conditions is satisfied

- The population distribution is Normal
- The sample size is large (greater than 30).

# 4.1.1 - Population is Normal

4.1.1 - Population is NormalIf the population is normally distributed with mean \(\mu\) and standard deviation \(\sigma\), then the sampling distribution of the sample mean is also normally distributed no matter what the sample size is. When the sampling is done with replacement or if the population size is large compared to the sample size, then \(\bar{x}\) has mean \(\mu\) and standard deviation \(\dfrac{\sigma}{\sqrt{n}}\). We use the term standard error for the standard deviation of a statistic, and since sample average, \(\bar{x}\) is a statistic, standard deviation of \(\bar{x}\) is also called standard error of \(\bar{x}\). However, in some books you may find the term standard error for the estimated standard deviation of \(\bar{x}\). In this class we use the former definition, that is, standard error of \(\bar{x}\) is the same as standard deviation of \(\bar{x}\).

**Standard Deviation of \(\boldsymbol{\bar{x}}\) [Standard Error]**

\(SD(\bar{X})=SE(\bar{X})=\dfrac{\sigma}{\sqrt{n}}\)