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.

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

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

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 \(n=5\)

The sampling distribution is:

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\(n= 2\).