Hypothesis Test vs Chi-Square Test

A hypothesis test, also called a significance test, assumes a claim about a population parameter and asks how unusual the observed sample would be if that claim held. The machinery is fixed: state $H_0$ and $H_a$, check the conditions, compute a test statistic, turn it into a p-value using the null distribution, and compare that p-value with a significance level $\alpha$ (alpha) fixed beforehand. The output is one of two verdicts, reject $H_0$ or fail to reject $H_0$, and never a probability that either hypothesis is true.

Test $H_0: \mu = 100$ against $H_a: \mu > 100$ with $\sigma$ (sigma) known to be 15 and $n = 25$, so the standard error is $15/\sqrt{25} = 3$. A sample mean of 106.2 gives $z = \frac{106.2 - 100}{3} = 2.07$ and a p-value of 0.0194. At $\alpha = 0.05$ that is a rejection: samples this far above 100 turn up under 2 percent of the time when 100 is the truth.

The misreading that costs points: "the p-value was 0.16, so we accept the null hypothesis and conclude the mean is 100." A test never accepts $H_0$. A sample mean of 103 in that same setup gives $z = 1.00$ and a p-value of 0.1587, and a 95 percent interval of $103 \pm 1.96(3)$, or 97.12 to 108.88. Every value in that range is as consistent with the sample as 100 is, so failing to reject rules nothing in. The asymmetry is deliberate: evidence can contradict one specific value, and no sample can confirm one.

A test also answers only the question its hypotheses posed and does not audit its own scope. Whether a conclusion reaches a population depends on random sampling, and whether it supports a cause depends on random assignment: a rejection from a convenience sample is still a rejection and still worth very little. Significant is not the same as large, the subject of statistical vs practical significance.

In the Fall 2026 AP Statistics course, one-proportion tests are Unit 3 topics 3.5 through 3.7, and tests for means are Unit 4 topics 4.4 and 4.5.

A chi-square test asks whether a table of counts sits further from a model than sampling variability explains. Every version computes $\chi^2 = \sum \frac{(O - E)^2}{E}$ across the cells, where $O$ is an observed count and $E$ the expected count under the null. Squaring throws away the direction of each gap, so every discrepancy pushes the statistic up instead of cancelling another, and all the evidence against the null lives in the upper tail of the chi-square curve.

Two schools are surveyed about biking to school: School A has 45 bikers out of 100, School B 45 out of 140. The expected counts from the margins are 37.5 and 62.5 in the first row, 52.5 and 87.5 in the second, so the four terms are 1.5, 0.9, 1.0714, and 0.6429. That gives $\chi^2 = 4.1143$ on $(2 - 1)(2 - 1) = 1$ degree of freedom, and a p-value of 0.0425. At $\alpha = 0.05$ that is convincing evidence the two schools differ in their biking rates.

The error that ruins the test: "School A bikes at 45 percent and School B at 32.1 percent, so those go in the formula." Percentages are not counts, and $\chi^2$ scales with how much data stands behind it. Double both schools while holding those two percentages fixed and the statistic doubles to 8.2286 with a p-value of 0.0041. Raw counts carry the sample size; percentages discard it.

Three tests share the formula and differ in where $E$ comes from and how the degrees of freedom are counted. Homogeneity compares separate samples on one categorical variable, which is the table above. Independence takes one sample and cross-classifies it two ways. Both read $E$ off the margins and use $(r - 1)(c - 1)$ degrees of freedom. Goodness of fit checks one variable against claimed proportions using $E = np_0$.

Homogeneity and independence are AP Statistics topics 3.14 and 3.15. Goodness of fit was removed from the course for Fall 2026, so it is not examinable, though it stays standard in a college introductory course.