Conditions for inference: the complete checklist

Q: Do I check the 10% condition on data from an experiment?

No. The AP course notes that the 10% condition is unnecessary when the data come from a randomized experiment, because random assignment rather than random sampling supplies the independence. Check randomization by describing the random assignment, then move to the shape condition.

Q: Is it enough to write "np is at least 10"?

No. A condition is verified only when the numbers appear. Write np₀ = 250(0.40) = 100 ≥ 10, not np₀ ≥ 10. The same applies to the 10% condition, where both 0.10N and n have to be on the page, and to expected counts, where each cell has to be computed.

Q: What do I do when a condition fails?

Say which condition failed, give the number that failed it, and state that the procedure may not be valid. Then follow the prompt: if it asks you to carry out the test anyway, do so and flag that the result may not be trustworthy. Silently computing is the error that costs the most.

Q: Why does a sample of 40 pass the condition for a test but fail it for an interval?

Because the two procedures assume different things about p. A test computes everything under H₀: p = p₀, so it checks np₀ and n(1-p₀). An interval assumes nothing, so it checks the observed np-hat and n(1-p-hat). When p-hat sits well below p₀, the interval's success count is the smaller of the two and can fall under 10.

Q: Are there conditions to check in Unit 5 regression?

Not in the Fall 2026 AP course. Inference for regression slopes was removed, so Unit 5 has no inference procedure and no condition checklist. What Unit 5 does ask is whether a linear model fits at all, which you argue from the scatterplot and residual plot rather than from a condition list. College courses still teach regression inference with its own conditions.

By Jude Wallis · Updated July 31, 2026

Every AP inference procedure needs Random and, when sampling without replacement, the 10% condition. Proportion procedures add Large Counts (use $p_0$ for a test, $\hat{p}$ for an interval), mean procedures add the sample data condition, and chi-square needs every expected count above 5.

AP Statistics: Unit 3 (topics 3.3 Constructing a Confidence Interval for a Population Proportion, 3.5 Setting Up a Test for a Population Proportion, 3.14 Setting Up a Chi-Square Test for Homogeneity or Independence, 4.2 Constructing a Confidence Interval for a Population Mean or Population Mean Difference, 4.4 Setting Up a Test for a Population Mean or Population Mean Difference). Condition checking is AP skill 4.E, justifying the use of a chosen statistical inference method by verifying conditions, and it runs through every inference topic in Units 3 and 4 of the Fall 2026 course; Unit 5 has no inference procedure now that inference for regression slopes has been removed.

Every condition check has the same three parts

Whichever procedure you run, the conditions answer three questions in the same order.

Randomization. How were the data collected? Inference about a population needs a random sample; inference about a treatment effect needs random assignment.
Independence. Are the observations independent enough? When you sample without replacement, the 10% condition covers this.
Shape. Is the sampling distribution close enough to the model you are about to use? This is Large Counts for proportions, the sample data condition for means, and expected counts for chi-square.

The AP course names this skill directly: justify the use of a chosen statistical inference method by verifying conditions. The word is verifying, not listing, so every condition below has to be written out with the actual numbers from the problem. If you are not yet sure which procedure you need, start with which statistical test to use.

The whole checklist in one table

Procedure	Randomization condition	10% condition	Third condition
One-sample z-interval for $p$	Random sample	$n \leq 0.10N$	Normality: $n\hat{p} \geq 10$ and $n(1-\hat{p}) \geq 10$
One-sample z-test for $p$	Random sample	$n \leq 0.10N$	Normality: $np_0 \geq 10$ and $n(1-p_0) \geq 10$
Two-sample z-interval for $p_1 - p_2$	Two independent random samples or a randomized experiment	$n_1 \leq 0.10N_1$ and $n_2 \leq 0.10N_2$ , not needed for an experiment	Normality: all four observed counts $n_1\hat{p}_1$ , $n_1(1-\hat{p}_1)$ , $n_2\hat{p}_2$ , and $n_2(1-\hat{p}_2)$ at least 10
Two-sample z-test for $p_1 - p_2$	Two independent random samples or a randomized experiment	Same as above, not needed for an experiment	Normality: all four counts built from the pooled $\hat{p}_c$ at least 10
Chi-square test for homogeneity or independence	Independence: one random sample. Homogeneity: independent random samples or a randomized experiment	$n \leq 0.10N$ , not needed for an experiment	Expected counts: every expected count greater than 5
One-sample t-interval or t-test for $\mu$ or a mean difference	Random sample or randomized experiment	$n \leq 0.10N$	Sample data: population stated approximately normal, or $n \geq 30$ , or the sample is free from strong skewness and outliers
Two-sample t-interval or t-test for $\mu_1 - \mu_2$	Two independent random samples or a randomized experiment	$n_1 \leq 0.10N_1$ and $n_2 \leq 0.10N_2$ , not needed for an experiment	Sample data: both $n \geq 30$ or both populations stated approximately normal; if either $n < 30$ , both samples free from strong skewness and outliers

Name the symbols once. $p$ is the population proportion, $\hat{p}$ ("p-hat") is the sample proportion, $p_0$ ("p-naught") is the value the null hypothesis names, $\hat{p}_c$ is the combined or pooled proportion, $\mu$ ("mu") is the population mean, $n$ is the sample size, and $N$ is the population size.

Randomization: say how the data were collected

The randomization condition asks where the data came from, and it is checked by quoting the problem, not by computing anything. Two versions exist.

For inference about a population, the data should come from a random sample. Write the sentence in context: "The 250 households were selected by a simple random sample of the city's households." If the problem describes volunteers or an unspecified group, the condition is not met and you say so.

For two-sample procedures, the requirement is two independent random samples, or a randomized experiment in which treatments were randomly assigned to the experimental units.

One wrinkle worth memorizing: the chi-square test for independence expects a single random sample from one population, while the chi-square test for homogeneity expects independent random samples or a randomized experiment. Same statistic, different data-collection story.

The 10% condition: only when sampling without replacement

When you sample without replacement, each selection changes what is left, so the observations are not strictly independent. The 10% condition keeps that dependence small enough to ignore: the population must be at least 10 times larger than the sample, written $n \leq 0.10N$ .

Check it with both numbers. "The city has 90,000 households and $0.10(90{,}000) = 9{,}000$ , and $250 \leq 9{,}000$ , so the 10% condition is met." Writing only "the population is large" is not a check.

Two cases skip it. Sampling with replacement already gives independence, so the condition does not apply; see with replacement vs without replacement. And for data from a randomized experiment, the AP course notes the condition is unnecessary in the two-sample and chi-square procedures, because random assignment supplies the independence.

The error the AP course flags here is mislabeling: calling the 10% condition the randomization condition, or folding them into one line. Keep them separate and label each.

Large Counts for proportions: p-naught for a test, p-hat for an interval

Proportion procedures use a normal model, and that model is only reasonable when both outcomes are common enough. The AP course calls this the normality condition; most textbooks call it Large Counts. The bar is 10 successes and 10 failures.

Which proportion goes into the check depends on what the procedure assumes, and this is where points quietly disappear.

Significance test. Everything in a test is computed as though $H_0: p = p_0$ were true, so use the null value: $np_0 \geq 10$ and $n(1 - p_0) \geq 10$ . These are the expected counts.
Confidence interval. An interval assumes nothing about $p$ , so use the data: $n\hat{p} \geq 10$ and $n(1 - \hat{p}) \geq 10$ . These are the observed counts, which is to say the successes and failures you actually counted.
**Sampling distribution with $p$ given.** When a problem hands you the true $p$ , use $np \geq 10$ and $n(1-p) \geq 10$ .

The same data can pass one check and fail the other, and that is not a contradiction. For the reasoning behind the number 10, see why 10 successes and 10 failures.

The sample data condition for means: three ways to satisfy it

Mean procedures use a t-distribution, which needs the sampling distribution of $\bar{x}$ ("x-bar") to be roughly normal. The AP course calls this the sample data condition and accepts three arguments. Any one is enough, and you should name which one you are using.

The problem states it. If the population distribution is described as approximately normal, you are done, regardless of sample size.
The sample is large. If $n \geq 30$ , the sampling distribution of the sample mean can be modeled approximately by a normal distribution. The AP caveat: for an extremely skewed population, a sample size much larger than 30 may be needed.
The sample looks well behaved. If $n < 30$ and normality is not stated, look at a graph of the sample data and argue it is free from strong skewness and outliers. A dotplot, boxplot, or stemplot works, and you cite what it shows. The 1.5 IQR rule backs up the outlier half of that claim.

Matched pairs follow the same rule applied to the differences: at least 30 differences, or a sample of differences free from strong skewness and outliers. Never check $n \geq 30$ on a proportion problem, and never check Large Counts on a means problem.

Expected counts for chi-square

A chi-square test for homogeneity or independence replaces the shape condition with one about expected counts: all expected counts should be greater than 5. The AP course's own exam-preparation note states the same check as all expected counts (rather than actual counts) being at least 5, so either phrasing is acceptable on the exam. What the note is really guarding against is the word expected: a check built from the counts in your table earns nothing.

Expected counts are not the numbers in your table. Compute each one from the margins under the null hypothesis:

\text{expected count} = \frac{\text{row total} \times \text{column total}}{\text{table total}}

So the check has real arithmetic in it. Compute every cell, then state that the smallest one clears 5. Writing "all expected counts are greater than 5" without producing them is the kind of unverified claim that loses the point, and the smallest expected count often sits in a corner of the table you would not have guessed. The worked example below finds one at 4.8.

One scope note: the chi-square goodness-of-fit test was removed from the AP exam in the Fall 2026 redesign, so the conditions above cover homogeneity and independence only. Goodness-of-fit is still standard college content, where the same expected-count rule applies. See chi-square tests explained; the symbol $\chi^2$ is read "chi-square".

Named versus verified: the same answer, two scores

A lab takes a random sample of 12 lithium cells from a production run of 5,000 and wants a 95% confidence interval for the mean capacity. Compare two condition checks.

Named only, and worth little:

"Random: yes. 10%: yes. Normal: yes, it's fine."

Verified, and worth full credit:

"Randomization: the problem states the 12 cells were randomly selected from the production run. 10% condition: the run contained 5,000 cells and $0.10(5{,}000) = 500$ , and $12 \leq 500$ , so the condition is met. Sample data condition: $n = 12 < 30$ and the population is not stated to be normal, so I examined a dotplot of the 12 capacities. It is roughly symmetric with no outliers, so the sample is free from strong skewness and outliers and a one-sample t-interval is appropriate."

The second version names each condition, supplies the numbers, and says which of the three sample data arguments it is using. Inference is its own free-response question on the exam (Question 3, worth 12.5 percent of the score), and condition checking is a scored part of it. The AP Statistics FRQ guide covers how the rest of that response is structured.

When a condition fails, name it and give the number that failed it, then say the procedure may not be valid. A silent calculation earns nothing.

Check all three conditions for a one-proportion procedure, both ways

A city with 90,000 households claims that 40% of them have a compost bin. A simple random sample of 250 households finds 87 with a bin. Verify the conditions for (a) a one-sample z-test of $H_0: p = 0.40$ and (b) a one-sample z-interval for $p$ .

Randomization, same for both. The problem states a simple random sample of the city's households, so the randomization condition is met.
10% condition, same for both. $0.10(90{,}000) = 9{,}000$ , and $250 \leq 9{,}000$ , so the sample is no more than 10% of the population and the observations are close enough to independent.
(a) The test assumes $H_0$ is true, so the normality check uses the null value $p_0 = 0.40$ . Expected successes: $np_0 = 250(0.40) = 100$ .
Expected failures: $n(1 - p_0) = 250(0.60) = 150$ . Both 100 and 150 are at least 10, so the condition is met and a one-sample z-test is appropriate.
(b) The interval assumes nothing about $p$ , so the check uses the data. $\hat{p} = \frac{87}{250} = 0.348$ .
Observed successes: $n\hat{p} = 250(0.348) = 87$ . Observed failures: $n(1 - \hat{p}) = 250(0.652) = 163$ . As a check, $87 + 163 = 250$ , the sample size.
Both 87 and 163 are at least 10, so the condition is met and a one-sample z-interval is appropriate.
Notice the two checks used different numbers (100 and 150 versus 87 and 163) from identical data. That is the point: the test uses expected counts under $H_0$ , the interval uses observed counts.

Both procedures pass. Test: random sample, $250 \leq 9{,}000$ , and $np_0 = 100$ with $n(1-p_0) = 150$ , both at least 10. Interval: random sample, $250 \leq 9{,}000$ , and $n\hat{p} = 87$ with $n(1-\hat{p}) = 163$ , both at least 10. The test uses $p_0 = 0.40$ ; the interval uses $\hat{p} = 0.348$ .

A chi-square check that fails on one cell

A ski resort takes a random sample of 90 riders from the 4,000 on the mountain that day and records which of three lifts they rode first and whether they rented gear. Renters: 18 on Lift A, 12 on Lift B, 6 on Lift C. Riders with their own gear: 27 on Lift A, 21 on Lift B, 6 on Lift C. Verify the conditions for a chi-square test for independence.

Randomization. The problem states a random sample of riders from a single population, which is what a test for independence requires. Met.
10% condition. $0.10(4{,}000) = 400$ , and $90 \leq 400$ . Met.
Find the margins. Renter row total: $18 + 12 + 6 = 36$ . Own-gear row total: $27 + 21 + 6 = 54$ . Column totals: Lift A $18 + 27 = 45$ , Lift B $12 + 21 = 33$ , Lift C $6 + 6 = 12$ . Table total: $36 + 54 = 90$ , and $45 + 33 + 12 = 90$ agrees.
Compute the expected counts for the renter row using row total times column total divided by table total. Lift A: $\frac{36 \times 45}{90} = \frac{1620}{90} = 18$ . Lift B: $\frac{36 \times 33}{90} = \frac{1188}{90} = 13.2$ . Lift C: $\frac{36 \times 12}{90} = \frac{432}{90} = 4.8$ .
Compute the expected counts for the own-gear row. Lift A: $\frac{54 \times 45}{90} = \frac{2430}{90} = 27$ . Lift B: $\frac{54 \times 33}{90} = \frac{1782}{90} = 19.8$ . Lift C: $\frac{54 \times 12}{90} = \frac{648}{90} = 7.2$ .
Check the expected counts against their own margins. Renter row: $18 + 13.2 + 4.8 = 36$ . Own-gear row: $27 + 19.8 + 7.2 = 54$ . Both match, so the expected counts are right.
Compare every expected count to 5. Five of the six clear it, but the renter and Lift C cell has an expected count of 4.8, which is not greater than 5.
State the verdict in context. The expected counts condition is not met, because the expected number of renters who rode Lift C first is 4.8, so a chi-square test for independence may not be valid for these data. A larger sample is the fix.

Randomization is met and $90 \leq 400$ satisfies the 10% condition, but the expected counts condition fails: the smallest expected count is $\frac{36 \times 12}{90} = 4.8$ , which is not greater than 5. Report the failure by name and number rather than running the test as if nothing were wrong.

Frequently asked questions

Do I check the 10% condition on data from an experiment?

No. The AP course notes that the 10% condition is unnecessary when the data come from a randomized experiment, because random assignment rather than random sampling supplies the independence. Check randomization by describing the random assignment, then move to the shape condition.

Is it enough to write "np is at least 10"?

No. A condition is verified only when the numbers appear. Write $np_0 = 250(0.40) = 100 \geq 10$ , not $np_0 \geq 10$ . The same applies to the 10% condition, where both $0.10N$ and $n$ have to be on the page, and to expected counts, where each cell has to be computed.

What do I do when a condition fails?

Say which condition failed, give the number that failed it, and state that the procedure may not be valid. Then follow the prompt: if it asks you to carry out the test anyway, do so and flag that the result may not be trustworthy. Silently computing is the error that costs the most.

Why does a sample of 40 pass the condition for a test but fail it for an interval?

Because the two procedures assume different things about $p$ . A test computes everything under $H_0: p = p_0$ , so it checks $np_0$ and $n(1-p_0)$ . An interval assumes nothing, so it checks the observed $n\hat{p}$ and $n(1-\hat{p})$ . When $\hat{p}$ sits well below $p_0$ , the interval's success count is the smaller of the two and can fall under 10.

Are there conditions to check in Unit 5 regression?

Not in the Fall 2026 AP course. Inference for regression slopes was removed, so Unit 5 has no inference procedure and no condition checklist. What Unit 5 does ask is whether a linear model fits at all, which you argue from the scatterplot and residual plot rather than from a condition list. College courses still teach regression inference with its own conditions.