We have two independent (i.e., disjoint) populations. We want to determine whether there is a statistically significant difference between the means of each population. We donâ€™t know with certainty the standard deviation of one or both populations, \(\sigma_1\) and \(\sigma_2\), so we estimate them using the sample standard deviations, \(s_1\) and \(s_2\). We have a sample of size \(n_1\) from population 1, and a sample of size \(n_2\) from popluation 2. Population 1 yields sample mean \(\bar{x}_1\), and population 2 yields sample mean \(\bar{x}_2\).

To apply a two-sample t-test for means, one should verify the following conditions:

- If \(n_1 + n_2 < 15\), it is critical that the populations are normally distributed.
- If \(n_1 + n_2 \ge 15\), one may proceed in the absence of outliers or strong skewness.
- If \(n_1 + n_2 \ge 40\), procedures are generally robust.

One can enhance robustness by planning \(n_1 \approx n_2\).

**Null hypothesis, \(H_0\):** \(\mu_1 - \mu_2 = 0\) (i.e., there is no difference between the two populations.)

**Alternative hypothesis, \(H_a\):** \(\mu_1 - \mu_2 \ne/>/< 0\) (i.e., there is a difference between the two populations.)

\[ t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

Where T is distributed according to the \(t(\text{min}(n_1-1, n_2-1))\) distribution *(conservative approach)*,

- For \(H_a : \mu_1 - \mu_2 < 0\): \(p = P(T \le t)\)
- For \(H_a : \mu_1 - \mu_2 > 0\): \(p = P(T \ge t)\)
- For \(H_a : \mu_1 - \mu_2 \ne 0\): \(p = 2P(T \le -|t|)\)

By choosing the smaller of \(n_1\) and \(n_2\) as the basis for our degrees of freedom, we take a *conservative approach*. As a result, our p-values are slightly larger than the true p-value.

- For \(H_a : \mu_1 - \mu_2 < 0\):
`p = stats.t.cdf(t, df=min(n1-1,n2-1))`

- For \(H_a : \mu_1 - \mu_2 > 0\):
`p = 1 - stats.t.cdf(t, df=min(n1-1,n2-1))`

- For \(H_a : \mu_1 - \mu_2 \ne 0\):
`p = 2 * stats.t.cdf(-abs(t), df=min(n1-1,n2-1))`

\[ \text{C% confidence interval} = (\bar{x}_1 - \bar{x}_2) \pm t^\star \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\\ \text{choose } t^\star \text{ s.t. area on }t(\text{min}(n_1-1,n_2-1))\text{ distribution from }(-t^\star,t^\star)\text{ = C} \]

Because we employ a *conservative approach*, this confidence interval will be slightly wider than the true confidence interval.

If you want to perform inference on the mean of a single population, a one-sample t-test for means is more appropriate.

If you have access to both of the population standard deviation values, a two-sample z-test for means is more appropriate.

A two-sample t-test conducted at confidence level \(\alpha\) will reject the null hypothesis if and only if the value corresponding to the null hypothesis, \(0\), is completely outside of the \(C = 1-\alpha\) t-confidence interval for the true difference between means.