Assumed distribution of the $y$ scores in population 1 and population 2:

Sampling distribution of the sample mean $\bar{y}_1$, and sampling distribution of the sample mean $\bar{y}_2$:
What is the sampling distribution of the sample mean? Suppose that we would draw an infinite number of samples of size $n$ from a population (e.g., from population 1), and in each sample we computed the sample mean of the scores on $y$. Different samples would result in different sample means, due to the randomness of the data. The distribution of all these sample means is the sampling distribution of the sample mean.

Sampling distribution of the difference between the two sample means $\bar{y}_1 - \bar{y}_2$:
What is the sampling distribution of the difference between the two sample means? Suppose that we would draw a sample of size $n_1$ from population 1 and a sample of size $n_2$ from population 2, computed the mean of the scores on $y$ in each sample, and then computed the difference between the two sample means $\bar{y}_1 - \bar{y}_2$. Then suppose that we would repeat these steps an infinite number of times, resulting in many differences $\bar{y}_1 - \bar{y}_2$. The distribution of all these sample mean differences $\bar{y}_1 - \bar{y}_2$ is the sampling distribution of $\bar{y}_1 - \bar{y}_2$.

Sampling distribution of the difference between the two sample means $\bar{y}_1 - \bar{y}_2$, if $\mu_1 - \mu_2 = 0$ (i.e., if the null hypothesis were true):

Sampling distribution of $z = \frac{(\bar{y}_1 - \bar{y}_2) - 0}{\sigma \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} = \frac{\bar{y}_1 - \bar{y}_2}{\sigma \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, if $\mu_1 - \mu_2 = 0$:

Sampling distribution of $t = \frac{\bar{y}_1 - \bar{y}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, if $\mu_1 - \mu_2 = 0$:

Probability of finding the observed $t$ value (red dot) or a more extreme value, if $\mu_1 - \mu_2 = 0$ (two sided $p$ value):

Rejection region for two sided test, with significance level $\alpha$: