Is chi-square goodness of fit on the AP exam?

By Jude Wallis · Published

No. The chi-square goodness-of-fit test was removed when AP Statistics was redesigned for Fall 2026, first examined May 2027. It was old topics 8.2 and 8.3. The chi-square tests for homogeneity and independence stayed, as Unit 3 topics 3.14 and 3.15.

AP Statistics: Unit 3 (topics 3.14 Setting Up a Chi-Square Test for Homogeneity or Independence, 3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence). Chi-square tests for homogeneity and independence are Unit 3 topics 3.14 and 3.15 in the Fall 2026 AP Statistics course. The goodness-of-fit test was old topics 8.2 and 8.3 and is listed as removed on College Board's AP Statistics future revisions page, so it is not assessed from May 2027 onward, though it remains standard college content.

No, goodness of fit was removed

The chi-square goodness-of-fit test is not on the redesigned AP Statistics exam. College Board lists it among the removed topics on its AP Statistics future revisions page, naming old topics 8.2 and 8.3, and the change applies from the May 2027 exam onward.

Chi-square itself did not go anywhere. What survived is the two-way table version:

  • 3.14 Setting Up a Chi-Square Test for Homogeneity or Independence
  • 3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence

Both sit in Unit 3, Inference for Categorical Data: Proportions, and the chi-square statistic is on the formula sheet you get during the exam.

So the answer has two halves that are easy to run together. Chi-square: yes. Goodness of fit: no.

What the goodness-of-fit test was

A goodness-of-fit test took one categorical variable from one sample and asked whether its distribution matched a set of claimed proportions. Is this die fair? Do these blood types appear in the proportions the textbook claims?

It used the same statistic as every other chi-square test:

χ2=∑(Observed−Expected)2Expected\chi^2 = \sum \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}}

The two things that made it its own procedure were how you built the expected counts and how you counted degrees of freedom. Expected counts came from the claimed proportions rather than from a table's margins, E=np0E = n p_0, and the degrees of freedom were the number of categories minus 1.

That is the whole of what was cut: one way of generating expected counts, and one df rule.

What stayed, and how to tell it apart

The tests for homogeneity and independence both start from a two-way table, build expected counts from the row and column totals, and use df=(rows−1)(columns−1)df = (\text{rows} - 1)(\text{columns} - 1).

Two questions separate the AP-tested tests from the removed one:

  1. Is your data a two-way table? If the counts sit in a grid with row and column totals, you are in 3.14 and 3.15 territory. If they are a single row of categories compared against claimed proportions, that is goodness of fit, which is no longer assessed.
  2. If it is a two-way table, how were the data collected? One sample measured on two variables is a test for independence. Two or more separate samples measured on one variable is a test for homogeneity. The arithmetic is identical; the label comes from the design.

Chi-square tests explained works all three through side by side, including a full two-way worked example, and which statistical test to use covers the wider choice if chi-square may not be the right family at all.

What this changes about your prep

Less than you might hope, and that is the useful part.

The statistic is unchanged, so every hour spent on ∑(O−E)2/E\sum (O - E)^2 / E still counts. The expected count idea is unchanged; only one of the two ways of producing expected counts is gone. Conditions are unchanged: random data, independent observations, and every expected count at least 5.

What changes is which df rule you need. Categories minus 1 belonged to goodness of fit. For 3.14 and 3.15 it is always (rows−1)(columns−1)(\text{rows} - 1)(\text{columns} - 1), and if you are reading a critical value, the chi-square table is indexed by that df.

To drill the version that is examined, use chi-square two-way practice or run the numbers with the chi-square calculator.

Why it is still worth learning

Goodness of fit remains standard content in a one-semester college introductory statistics course, which is the course AP Statistics stands in for. If you use your AP credit and skip ahead, you may still meet it.

It is also the natural test for a real class of question the two-way version cannot answer. Any time a claimed distribution exists in advance, a fair die, a published population breakdown, a genetics ratio, goodness of fit is the test that compares data to it. There is no two-way table in those situations, because there is only one variable.

The practical position: know what it is and when it applies, spend your exam revision on 3.14 and 3.15.

Why review books still cover it

Most AP Statistics material describes the nine-unit course, where goodness of fit was topics 8.2 and 8.3 inside a chi-square unit. Those books are not wrong about the test. They are aimed at an exam that no longer includes it.

A quick check on any resource: if it lists nine units, if it puts chi-square in Unit 8, or if it teaches inference for a regression slope, it predates the Fall 2026 redesign. In the current course, chi-square is Unit 3 and there is no unit on slope inference at all.

The current exam format and what is on the AP Statistics exam both describe the version you will actually sit.

A goodness-of-fit test, shown as the thing that was cut

A six-sided die is rolled 60 times, giving counts of 8, 12, 9, 14, 7, and 10 for faces 1 through 6. Test whether the die is fair at the 0.05 level. This procedure is no longer assessed on the AP exam; it is worked here so you can recognize the setup.

  1. State the hypotheses. H0H_0: each face has probability 1/61/6. HaH_a: at least one face has a probability different from 1/61/6.

  2. Build the expected counts from the claimed proportions, not from a table's margins: E=np0=60×16=10E = n p_0 = 60 \times \frac{1}{6} = 10 for every face.

  3. Check the random and independence conditions. The 60 rolls are independent repetitions of the same chance process, which is what those two conditions ask for here.

  4. Check the large-counts condition. Every expected count is 10, which is at least 5, so the condition holds.

  5. Compute each term. Face 1: (8−10)2/10=4/10=0.4(8 - 10)^2 / 10 = 4/10 = 0.4. Face 2: (12−10)2/10=4/10=0.4(12 - 10)^2 / 10 = 4/10 = 0.4. Face 3: (9−10)2/10=1/10=0.1(9 - 10)^2 / 10 = 1/10 = 0.1. Face 4: (14−10)2/10=16/10=1.6(14 - 10)^2 / 10 = 16/10 = 1.6. Face 5: (7−10)2/10=9/10=0.9(7 - 10)^2 / 10 = 9/10 = 0.9. Face 6: (10−10)2/10=0(10 - 10)^2 / 10 = 0.

  6. Add the terms: χ2=0.4+0.4+0.1+1.6+0.9+0=3.4\chi^2 = 0.4 + 0.4 + 0.1 + 1.6 + 0.9 + 0 = 3.4.

  7. Find the degrees of freedom with the goodness-of-fit rule, categories minus 1: df=6−1=5df = 6 - 1 = 5.

  8. Compare with the critical value. At α=0.05\alpha = 0.05 with df=5df = 5, the chi-square table gives a critical value of 11.070, and 3.4<11.0703.4 < 11.070.

χ2=3.4\chi^2 = 3.4 on df=5df = 5, well below the 11.070 critical value at α=0.05\alpha = 0.05, so you fail to reject H0H_0: there is not convincing evidence that the die is unfair. Note the two features that made this a goodness-of-fit test, expected counts from claimed proportions and df=df = categories −1- 1. Both are what the redesign removed.

Which of these chi-square settings the exam can still ask

Classify each setting, and say whether the redesigned AP Statistics exam can assess it. (a) 120 randomly chosen students are each recorded by year group and by whether they cycle to school. (b) A bag of sweets is claimed to be 30% red, 30% green, and 40% yellow; a sample of 200 is counted by color. (c) Separate random samples of 80 students are taken at each of three schools and each student is recorded as cycling or not.

  1. Setting (a): one sample, two categorical variables recorded on each person, so the counts form a two-way table with row and column totals. That is a test for independence.

  2. Setting (b): one categorical variable compared against claimed proportions, with no second variable and no two-way table. That is a goodness-of-fit test.

  3. Setting (c): one categorical variable, cycling or not, measured on three separate samples. That is a test for homogeneity.

  4. Apply the course scope. Independence and homogeneity are topics 3.14 and 3.15, so (a) and (c) are examinable. Goodness of fit was removed, so (b) is not.

  5. Check the df you would use for the two examinable ones. (a) has 2 columns and as many rows as year groups, so df=(rows−1)(2−1)df = (\text{rows} - 1)(2 - 1). (c) is a 3 by 2 table, so df=(3−1)(2−1)=2df = (3 - 1)(2 - 1) = 2.

(a) independence and (c) homogeneity are both examinable, as topics 3.14 and 3.15. (b) is a goodness-of-fit test and is not on the redesigned exam. The tell is whether the data form a two-way table or a single row of categories held against claimed proportions.

Frequently asked questions

Is chi-square still on the AP Statistics exam?

Yes. The chi-square tests for homogeneity and independence are Unit 3 topics 3.14 and 3.15 in the Fall 2026 course, and the chi-square statistic is on the formula sheet. Only the goodness-of-fit test was removed.

Which topics were the goodness-of-fit test?

Old topics 8.2 and 8.3, in the nine-unit course the Fall 2026 redesign replaced. College Board's future revisions page names them in its list of removed content; what was removed from AP Statistics has the other four entries and what each one covered.

Does the chi-square formula change now that goodness of fit is gone?

No. Every chi-square test uses the same statistic, the sum of observed minus expected squared over expected. What differed about goodness of fit was how the expected counts were built and that its degrees of freedom were categories minus 1.

Do I still need the chi-square table?

Yes, for topics 3.14 and 3.15. You look up a critical value by degrees of freedom and significance level exactly as before, using the two-way rule for df, rows minus 1 times columns minus 1.

Should I still learn goodness of fit?

Not for the AP exam. It is worth knowing for a college introductory statistics course, and it is the right test whenever a claimed distribution exists in advance and there is only one variable, such as testing whether a die is fair.