How does Standard Error of Mean = Standard Deviation of Sampling Distribution?


As we discussed in class, the standard error of the mean is defined as the standard deviation of the sampling distribution of the mean, that is, the average difference between the population mean and a sample mean of size n. Formally:

\sigma _{\bar{x} } = \sigma/\sqrt{N}     [1]


Note that equation [1] refers to the standard error of the mean for an entire population of sample means. In practice, you often have to estimate the standard error of the mean, because the population standard deviation is not known:

\hat{s} _{\bar{x} } = \hat{s}/\sqrt{N}     [2]

For this discussion, we'll be discussing the standard error of the mean for a population of sample means and be using equation [1].


Say that we have a population with N = 4 scores {1, 2, 2, 4}. This population has a mean of μ = 2 and a standard deviation of σ = 0.707. If we randomly sampled n = 2 scores from this population, then the standard error of the mean should be: 

\sigma _{\bar{x} } = \sigma/\sqrt{N} \\ \sigma _{\bar{x} } = 0.707/\sqrt{2} \\ \sigma _{\bar{x} } = 0.707 / 1.414 \\ \sigma _{\bar{x} } = 0.500

Say that we create a sampling distribution of all possible samples of n = 2 scores from this population; that is, we sample, rith replacement, all possible combinations of n = 2 scores. We would have the following distribution of means:

X1X2\bar{X}
\left( \bar{X} - \mu_{\bar{X}} \right)\left( \bar{X} - \mu_{\bar{X}} \right)^{2}
111.0-1.01.00
121.5-0.50.25
121.5-0.50.25
132.000
211.5-0.50.25
222.000
222.000
232.50.50.25
211.5-0.50.25
222.000
222.000
232.50.50.25
312.000
322.50.50.25
322.50.50.25
333.011







\sum{\bar{X} = 32

SS = 4


\mu_{\bar{X} = 2



Notice that the mean of the sampling distribution of the means is 2, which is equal to the population mean. Also, the sum of squares is SS = 4, which means that the standard deviation of this sampling distribution of the means is:

\sigma _{\bar{X} =\sqrt{\frac{SS}{N} }} \\ \sigma_{\bar{X} =\sqrt{\frac{4}{16} } } \\ \sigma_{\bar{X} =\sqrt{0.250 } } \\ \sigma_{\bar{X} = 0.500



Importantly, this standard deviation of the sampling distribution of the mean is equal to the standard error of the mean that we calculated earlier.