In AP Statistics, the large counts condition for chi-square tests requires that all expected counts in the two-way table be at least 5, which ensures the chi-square distribution is an accurate model for the test statistic (Topic 8.5, learning objective 8.5.C).
Large counts is one of the conditions you have to verify before running a chi-square test for homogeneity or independence. The conservative check is simple. Compute the expected count for every cell in your two-way table and confirm that each one is at least 5. Not the observed counts, the expected counts. That distinction trips up a lot of people on the exam.
Here's the intuition. The chi-square statistic doesn't actually follow a chi-square distribution perfectly. It only approximates one, and that approximation gets better as the counts grow. If a cell has an expected count of 1 or 2, that tiny number sits in the denominator of the chi-square calculation and can blow up the statistic, making your p-value unreliable. Checking large counts is how you justify the shape of the sampling distribution, the same way Normality checks work in earlier inference units.
This condition lives in Unit 8: Inference for Categorical Data: Chi-Square, specifically Topic 8.5, and it directly supports learning objective AP Stats 8.5.C (verify the conditions for chi-square inference). The CED lists three checks for a chi-square test on a two-way table: random data collection (simple random sample for independence, stratified random sample or randomized experiment for homogeneity), the 10% condition when sampling without replacement, and large counts for shape. On the AP exam, the conditions step is graded explicitly in inference FRQs. If you skip the large counts check or check observed counts instead of expected counts, you lose credit even if your calculations are perfect. Writing 'all expected counts ≥ 5' with the actual expected counts shown is the move that earns the point.
Expected Frequencies (Unit 8)
Large counts is checked against expected frequencies, not observed ones. You compute each expected count as (row total × column total) ÷ table total, then verify every one is at least 5 before trusting the chi-square approximation.
Large Counts for Proportions (Unit 6)
You first met a 'large counts' condition in Unit 6, where it was np ≥ 10 and n(1−p) ≥ 10 for inference about proportions. Same idea, different threshold. Both exist because an approximation (Normal there, chi-square here) only works when counts are big enough.
Chi-Square Test (Unit 8)
Large counts is a gatekeeper for every chi-square procedure in Unit 8, including goodness of fit in Topics 8.2-8.3. No matter which chi-square test you pick, you can't report a p-value until this condition checks out.
Sampling Without Replacement (Unit 8)
Large counts is verified alongside the 10% condition (n ≤ 10% of N). Don't merge them. The 10% condition justifies independence of observations, while large counts justifies the shape of the sampling distribution.
Multiple-choice questions ask you why the condition exists and what the conservative check is. The answer they want is that all expected counts must be at least 5 so the chi-square distribution accurately approximates the sampling distribution of the test statistic. One common stem describes a statistician verifying that all expected counts are at least 5 and asks for the primary reason, and the correct answer points to the validity of the chi-square approximation, not 'making the sample representative' or other distractors. On the free-response side, no released FRQ uses the phrase 'large counts' as a question topic, but every chi-square inference FRQ expects you to verify conditions. The scoring guidelines reward you for showing the expected counts (or a table of them) and explicitly stating that all are at least 5. State it, show it, then move on to the mechanics.
These share a name but use different rules. For one- and two-proportion z-procedures in Unit 6, large counts means at least 10 expected successes and 10 expected failures (np ≥ 10 and n(1−p) ≥ 10). For chi-square tests in Unit 8, large counts means every expected cell count in the table is at least 5. If you write 'np ≥ 10' on a chi-square FRQ, you've checked the wrong condition. Match the threshold to the procedure.
The large counts condition for chi-square tests is satisfied when every expected count in the two-way table is at least 5.
Check expected counts, not observed counts; an observed cell of 3 is fine as long as its expected count is 5 or more.
The condition exists because the chi-square distribution is only an approximation, and it becomes accurate when counts are large.
Large counts is the 'shape' check; random sampling and the 10% condition (n ≤ 10% of N) handle the other inference conditions in 8.5.C.
The threshold is different from Unit 6, where large counts for proportions means at least 10 expected successes and 10 expected failures.
On FRQs, show the expected counts and explicitly state that all are at least 5 to earn the conditions credit.
It's the requirement that all expected counts in a chi-square test be at least 5. It guarantees the chi-square distribution is an accurate model for the test statistic, which makes your p-value trustworthy.
Expected counts. This is the classic mistake. A cell can have an observed count of 2 and still pass the condition, as long as its expected count (row total × column total ÷ table total) is at least 5.
No. For z-procedures with proportions (Unit 6), the rule is at least 10 expected successes and 10 failures. For chi-square tests (Unit 8), the conservative check is that every expected cell count is at least 5.
The chi-square approximation may be inaccurate, so the p-value can't be trusted. On the exam, you should state that the condition is not met rather than plowing ahead, since the conditions check is graded.
It's a conservative rule of thumb. Chi-square tests become more accurate with more observations, and requiring every expected count to reach 5 keeps small cells from distorting the test statistic and inflating it artificially.