5.2.2 Two Samples: Estimating 𝜇1 - 𝜇2

What do we know

Case 1: Confidence Interval for $\mu_1-\mu_2$, with known $\sigma^2_1$ and $\sigma^2_2$

If $\bar x_1$ and $\bar x_2$ are means of independent random samples of sizes $n_1, n_2 \ge 30$ from populations with variances $\sigma^2_1, \sigma^2_2$ respectively, a $100(1-\alpha)\%$ C.I. for $\mu_1 - \mu_2$ is given by

$$ (\bar x_1-\bar x_2)-z_{\alpha / 2}\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}\lt \mu_1-\mu_2\lt (\bar x_1-\bar x_2)+z_{\alpha / 2}\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}} $$

Note: Derivation

Case 2: Confidence Interval for $\mu_1-\mu_2$, with $\sigma^2_1 = \sigma^2_2$ but both unknown

Using $t$-distribution, $t =\frac{Z}{\sqrt{V / v}}$

If $\bar x_1$ and $\bar x_2$ are respectively the means of small independent random samples of sizes $n_1$$n_2$ (where $n_1,n_2 \lt 30$), from normal populations with unknown by equal variances($\sigma^2_1 = \sigma^2_2$), a $100(1-\alpha)\%$ C.I. for $\mu_1-\mu_2$ is given by

$$ (\bar x_1 - \bar x_2) - t_{\alpha / 2}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \lt \mu_1 - \mu_2 \lt (\bar x_1 - \bar x_2) + t_{\alpha / 2}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} $$

This derivation is a little bit hard, but useful.

Note: Derivation

Case 3: Confidence Interval for $\mu_1-\mu_2$, with $\sigma^2_1\neq \sigma^2_2$ and both unknown

Using $t$-distribution, $T = \frac{\bar X - \mu}{\sqrt{S^2 / n}}$ For 2 samples, $T'=\frac{(\bar X_1 - \bar X_2) - (\mu_1 - \mu_2)}{\sqrt{S^2_1 / n_1 + S^2_2 / n_2}}$ The equation is similar to Case 2, but $z \rightarrow t$, $\sigma \rightarrow s$, the thing next to $t$ is different and $v$ is very complicated.

If $\bar x_1, s^2_1$ and $\bar x_2, s^2_2$ are respectively the means and variances of small independent random samples of sizes $n_1$ and $n_2$ (where $n_1,n_2\lt 30$) from normal populations with unknown and unequal variances, a $100(1-\alpha)\%$ C.I. for $\mu_1 - \mu_2$ is

$$ (\bar x_1 - \bar x_2) - t_{\alpha / 2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} \lt \mu_1 - \mu_2 \lt (\bar x_1 - \bar x_2) + t_{\alpha / 2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} $$

where $t_{\alpha / 2}$ is the $t$-value with

$$ v = \frac{(s^2_1 / n_1 + s^2_2 / n_2)}{[(s^2_1 / n_1)^2 / (n_1 - 1)]+[(s^2_2 / n_2)^2 / (n_2 - 1)]} $$

degrees of freedom (rounded down), leaving an area of $\alpha / 2$ to the right.