Test Statistic vs Standardized Test Statistic

A test statistic is one number computed from the sample whose behavior under $H_0$ is known in advance. That second half is what makes it worth computing: without a named reference distribution, and its degrees of freedom where it has them, a test statistic converts into nothing. The common ones in an introductory course are $z$, $t$, and $\chi^2$ (chi-square), and the first two share the single recipe in standardized test statistic.

A sample of $n = 16$ has $\bar{x}$ (x-bar) equal to 52.4 and $s = 6$, tested against $H_0: \mu = 50$. The standard error is $6/\sqrt{16} = 1.5$, so $t = \frac{52.4 - 50}{1.5} = 1.60$, meaning the sample mean sits 1.60 standard errors above the value the null claims. On 15 degrees of freedom the one-sided p-value is 0.0652. Read that same 1.60 against the standard normal curve and you get 0.0548, so the statistic on its own does not settle the answer.

The misreading: "$z = 4.1$, so the effect is huge." The magnitude counts standard errors, not units of the variable, and the standard error shrinks as $\sqrt{n}$ grows. A difference far too small to act on produces a large statistic once the sample is large enough. Read the size of an effect from the estimate and its confidence interval, and read surprise from the test statistic.

Not every test statistic is signed. $\chi^2$ squares its terms, so it is never negative, carries no direction, and is read in one tail only: a $\chi^2$ of 4.11 and a $z$ of 4.11 have nothing to do with each other. Whether one tail or two get counted also belongs to the alternative hypothesis rather than the statistic. Here $t = 1.60$ gives 0.0652 one-sided and 0.1304 two-sided.

Computing the statistic is the carrying-out half of each inference topic, for instance AP Statistics topic 3.7, Carrying Out a Test for a Population Proportion.

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.