Standardized test statistic

One line generates every $z$ and $t$ statistic in the course: take the statistic your sample produced, subtract the value $H_0$ claims for the matching parameter, and divide by the standard error of that statistic. That is the AP formula sheet line, statistic minus parameter over standard error, with the parameter pinned to its null value. What comes out is a count of standard errors, never a probability.

Test $H_0: p = 0.50$ with 60 successes in 100 trials. Then $\hat{p}$ (p-hat) is 0.60, the standard error under the null is $\sqrt{0.50(0.50)/100} = 0.05$, and $z = (0.60 - 0.50)/0.05 = 2.00$. Said aloud: the sample landed two standard errors above where $H_0$ said it would. For a mean the letter changes, because the spread has to be estimated too: $\bar{x} = 52$ (x-bar) against $H_0: \mu = 50$ with $s = 8$ and $n = 25$ gives $8/\sqrt{25} = 1.6$ and $t = 2/1.6 = 1.25$ on 24 degrees of freedom.

The wrong version is "divide by the standard deviation." That turns $t = 1.25$ into $2/8 = 0.25$, and its two-sided p-value from 0.223 into 0.805. $s$ describes how spread out the data are; $s/\sqrt{n}$ describes how spread out $\bar{x}$ is from sample to sample, and only the second is the right ruler for a statistic. The slip shrinks the statistic by a factor of $\sqrt{n}$, so it buries effects rather than inventing them.

The denominator is also where a test and an interval part company. A one-proportion test builds its standard error from the null value, $\sqrt{0.50(0.50)/100} = 0.05$ above, because everything in a test is computed as though $H_0$ were true. The matching interval has no null hypothesis in the room and uses $\hat{p}$: $\sqrt{0.60(0.40)/100} = 0.049$. Means have no such split, since a null about $\mu$ (mu) says nothing about $\sigma$ (sigma).

A standardized value is unreadable until you name the curve: 2.00 has a two-sided p-value of 0.0455 on the standard normal and 0.0569 on a $t$ curve with 24 degrees of freedom. Report the degrees of freedom, or the reader cannot finish.