$\frac{s_1}{s_2}$

$p_2-p_1$

$\mu_1-\mu_2$

 $\sigma^2$  

 $\left(\overline{x}_1-\overline{x}_2\right)\pm t_{0.025,\ v\ }\sqrt{...}$  

 $\left(\overline{x}_1-\overline{x}_2\right)\pm z_{0.025\ }\sqrt{...}$  

 $\left(\overline{x}_1-\overline{x}_2\right)\pm t_{0.025,\ v\ }\times s_p\sqrt{...}$  

$\left(p_1-p_2\right)\pm z_{0.01}\sqrt{...}$

$\left(p_2-p_1\right)\pm t_{0.01,\nu}\sqrt{...}$

$\left(p_1-p_2\right)\pm z_{0.02}\sqrt{...}$

$\left(\pi_1-\pi_2\right)\pm z_{0.01}\sqrt{...}$

$t$ score

$z$ score

$\chi^2$ score

$f$ score

We are 95% confident that there is no difference between the population means.

We are 95% confident that there is a difference between the population means.

We are 95% confident that  $\mu_2$  is larger than  $\mu_1$  .

We are 95% confident that  $\mu_1$  is larger than  $\mu_2$  .

No difference because the 99% CI contains 1.

Yes, there is a difference because the 99% CI does not contain 0.

No difference because the upper limit of 99% CI is greater than the lower limit.

YES, because negative lower limit implies  $\pi_1$  is smaller than  $\pi_2$  .

NO, because any proportion must always be positive.

 $z_{0.025}$  

 $z_{0.05}$  

 $t_{0.025,\ \nu}$  

 $t_{0.05,\nu}$  

$f_{0.05,4,8}=\frac{1}{f_{0.05,\ 8,\ 4}}$

$f_{0.95,\ 4,\ 8}=\frac{1}{f_{0.05,\ 4,\ 8}}$

$f_{0.95,\ 4,\ 8}=\frac{1}{f_{0.05,\ 8,\ 4}}$

$f_{0.05,\ 4,\ 8}=1-f_{0.95,\ 4,\ 8}$