One sample $t$ test: sampling distribution of the sample mean and its standard error

Definition of the sampling distribution of the sample mean $\bar{y}$ and its standard error

Sampling distribution of the sample mean $ \bar{y}$:

When we draw a sample of size $ N$ from the population, we can compute the sample mean of a variable $ y$: $ \bar{y}$. Now suppose that we would draw many more samples. Specifically, suppose that we would draw an infinite number of samples, each of size $ N$. In each sample, we could compute the sample mean $ \bar{y}$. Different samples will give different sample means. The distribution of all these sample means is the sampling distribution of $ \bar{y}$. Note that this sampling distribution is purely hypothetical. We will never really draw an infinite number of samples, but hypothetically, we could.

Standard error:

Suppose that the assumptions of the one sample $ t$ test hold:
  • The variable $ y$ is normally distributed in the population, with mean $\mu$ and standard deviation $\sigma$
  • The sample is a simple random sample from the population. That is, observations are independent of one another
Then the sampling distribution of $ \bar{y}$ is normal with mean $\mu$ and standard deviation $\sigma / \sqrt{N}$. Since the one sample $ t$ test does not make the assumption that the value of $\sigma$ is known (like the $ z$ test does), we need to:
  • estimate $\sigma$ with $ s$: the standard deviation of $ y$ in the sample
  • estimate $\sigma / \sqrt{N}$ with $ s / \sqrt{N}$
That is, we estimate the standard deviation of the sampling distribution of $\bar{y}$, $\sigma / \sqrt{N}$, with $ s / \sqrt{N}$. We call this estimated standard deviation of the sampling distribution of $\bar{y}$ the standard error of $\bar{y}$.

Note that the $ t$ statistic $ t = \frac{\bar{y} - \mu_0}{s / \sqrt{N}}$ thus indicates how many standard errors the observed sample mean $\bar{y}$ is removed from $\mu_0$: the population mean according to H0.