Checking the Success-Failure Condition for Tests

Which Counts Matter for a Test?

A one-proportion $z$-test asks whether sample data provide evidence against a stated population proportion. Before relying on the Normal-based test, check whether the test’s Large Counts condition is met. Unlike the confidence-interval check in Large Counts Condition for Confidence Intervals, this check uses the proportion specified by the null hypothesis.

That difference follows from the purpose of each procedure. A confidence interval estimates an unknown population proportion using the observed sample proportion $\hat{p}$. A test evaluates the sample result under the assumption that the null hypothesis is true. If the null hypothesis says $p=p_0$, the test’s expected success and failure counts are calculated using $p_0$.

Condition: For a one-proportion $z$-test of $H_0:p=p_0$, check the Large Counts condition using the null proportion: $$ np_0\geq10 \qquad\text{and}\qquad n(1-p_0)\geq10. $$ These are the expected numbers of successes and failures in a sample of size $n$ if the null hypothesis is true. Both must be at least 10.

Here, “expected” means the average counts predicted by the null model over repeated samples. The values $np_0$ and $n(1-p_0)$ do not have to be whole numbers, and they are not the counts actually observed in the sample. The observed counts are $x$ successes and $n-x$ failures.

Why a Test Uses the Null Proportion

The test’s $z$-statistic measures how far the observed sample proportion $\hat{p}$ is from the null value $p_0$, in standard-error units. The standard error for the test is calculated as though $H_0$ is true:

SE_{\hat{p},\,H_0} =\sqrt{\frac{p_0(1-p_0)}{n}}

The Large Counts check addresses whether a Normal approximation for the sampling distribution of $\hat{p}$ under that null model is reasonable. If the null model predicts fewer than 10 successes or fewer than 10 failures, the usual one-proportion $z$-test is not supported by this condition. A large observed count in one category cannot make up for a small expected count in the other.

For an interval, by contrast, there is no null hypothesis supplying a value $p_0$. As discussed in the previous tutorial, its Large Counts condition uses the observed counts, $n\hat{p}=x$ and $n(1-\hat{p})=n-x$. These checks can give different results for the same sample.

\begin{array}{ll} \text{One-proportion }z\text{-test:} & np_0,\quad n(1-p_0)\\[4pt] \text{One-proportion }z\text{-interval:} & n\hat{p}=x,\quad n(1-\hat{p})=n-x \end{array}

Keep this distinction clear: the test condition describes counts expected if the null claim is true; the interval condition describes counts observed in the sample. One does not replace the other, even when a test and interval use the same data.

A Reliable Test Condition Check

The Large Counts condition is one part of the full check, not a substitute for the sampling conditions. As in Why Inference Procedures Need Conditions, consider whether the data come from a random sample or a suitable random process. If the sample was drawn without replacement from a finite population, check the 10% condition as covered in When Population Size Is Unknown or Large and Independence and the 10 Percent Condition.

Identify $n$ and $p_0$.
Use the full sample size and the specific population proportion in the null hypothesis.

Check the sampling process.
Verify that the sample is random or otherwise supports treating observations as random. For sampling without replacement, check that $n\leq0.10N$, where $N$ is the source population size.

Calculate both null expected counts.
Find $np_0$ and $n(1-p_0)$. Do not use $\hat{p}$ for this test condition.

State whether the condition is met.
Both expected counts must be at least 10 to meet the Large Counts condition for the usual one-proportion $z$-test.

Worked Examples

Worked Example: Testing a Claimed Proportion of 0.30

A fictional city randomly selects 120 households without replacement from 4,000 eligible households. Forty-two report using a particular water-saving device. The city wants to test whether the population proportion differs from 0.30. Check the conditions and carry out the test at the 0.05 significance level.

State. Let $p$ be the proportion of all 4,000 eligible households in this fictional city that use the device. The hypotheses are $$ H_0:p=0.30 \qquad\text{and}\qquad H_a:p\ne0.30. $$ The sample size is $n=120$, and the observed sample proportion is $$ \hat{p}=\frac{x}{n}=\frac{42}{120}=0.35. $$

Plan: check the conditions. The households were randomly selected, so the Random condition is met. The sample was drawn without replacement from a finite population, so check the 10% condition: $$ 120\leq0.10(4{,}000)=400. $$ This is true, so the 10% condition is met. For the test’s Large Counts condition, use the null value $p_0=0.30$: $$ np_0=120(0.30)=36, \qquad n(1-p_0)=120(1-0.30)=120(0.70)=84. $$ Both expected counts are at least 10, so the Large Counts condition is met. The conditions support using a one-proportion $z$-test.

Do. Under $H_0$, the test standard error is $$ SE_{\hat{p},\,H_0} =\sqrt{\frac{p_0(1-p_0)}{n}} =\sqrt{\frac{(0.30)(0.70)}{120}} =\sqrt{0.00175} \approx0.0418. $$ The test statistic is $$ z=\frac{\hat{p}-p_0}{SE_{\hat{p},\,H_0}} =\frac{0.35-0.30}{0.041833} \approx1.195. $$ For the two-sided alternative, results at least this far from $p_0$ in either direction have a p-value of approximately $0.232$ (rounded). A calculator’s 1-PropZTest command with $x=42$, $n=120$, $p_0=0.30$, and the two-sided alternative gives the same result, rounded.

Conclude. Since the p-value $0.232$ is greater than $\alpha=0.05$, fail to reject $H_0$. The sample does not provide convincing evidence that the proportion of eligible households in this fictional city using the device differs from 0.30.

Worked Example: The Test Check Fails Even Though the Sample Has 12 Successes

A fictional wildlife team randomly samples 100 sites from a very large set of eligible sites. Twelve sites have a specified habitat feature. The team plans to test $H_0:p=0.04$ against a two-sided alternative. Check the Large Counts condition for the test.

The null hypothesis gives $p_0=0.04$, so calculate the expected counts under that claim: $$ np_0=100(0.04)=4, \qquad n(1-p_0)=100(0.96)=96. $$ The expected success count is only 4, which is less than 10. Therefore, the Large Counts condition for the one-proportion $z$-test is not met. The fact that the sample actually contains 12 successes does not change this test-condition check: it uses the null value $p_0$, not the observed proportion $\hat{p}=12/100=0.12$. The usual Normal-based $z$-test is not supported by this condition for these data.

Worked Example: Test Condition Passes While the Interval Check Fails

A fictional technology club randomly surveys 30 members, and 1 member reports using a particular accessibility feature. Consider a test of $H_0:p=0.50$. Compare the Large Counts checks for the test and for a confidence interval based on these same observations.

For the test, use $p_0=0.50$: $$ np_0=30(0.50)=15, \qquad n(1-p_0)=30(0.50)=15. $$ Both null expected counts are at least 10, so the test’s Large Counts condition is met.

For the confidence interval, use the observed counts instead. There is 1 observed success and $30-1=29$ observed failures. The success count is below 10, so the interval’s Large Counts condition is not met. Thus the same sample meets the test’s Large Counts condition but fails the interval’s Large Counts condition. This comparison is only about the Large Counts checks; each procedure still requires its other conditions as well.

Common Mistakes and AP Exam Tip

Using $\hat{p}$ for a test’s Large Counts condition. The test evaluates the sample under $H_0$, so calculate $np_0$ and $n(1-p_0)$. Save $n\hat{p}$ and $n(1-\hat{p})$ for the interval check.
Using observed counts as if they were expected under the null. The sample may have $x$ observed successes, but the test’s expected success count is $np_0$. Identify which quantity the procedure requires.
Checking only the larger expected count. Both null expected counts must reach 10. In the wildlife example, 96 expected failures do not offset only 4 expected successes.
Saying only that the sample size is large enough. The condition is not simply $n\geq10$. Its result depends on both $n$ and the null proportion $p_0$.
Treating a failed check as a conclusion about $p$. An unmet condition means the usual $z$-test’s Normal approximation is not supported by this check; it does not prove that the null hypothesis is false.
Forgetting the other conditions. Passing the Large Counts condition alone does not establish that the sample is random or that the 10% condition is met when it applies.

AP Exam Tip: Show the null value and both calculations: “Under $H_0:p=p_0$, $np_0=\_\_$ and $n(1-p_0)=\_\_$. Both expected counts are at least 10, so the Large Counts condition for the one-proportion $z$-test is met.” If one is below 10, name it and state that the condition is not met.

Key Takeaway

A test and an interval use different Large Counts checks because they answer different questions. The test checks expected counts under the null hypothesis; the interval checks the counts observed in the sample. In both cases, both the success and failure counts must be at least 10, and the Random and applicable 10% conditions must also be considered.

Key takeaway: For a one-proportion $z$-test of $H_0:p=p_0$, check $np_0\geq10$ and $n(1-p_0)\geq10$. For a one-proportion $z$-interval, check $n\hat{p}\geq10$ and $n(1-\hat{p})\geq10$. Use the null proportion for the test and the observed sample proportion for the interval.

Check Your Understanding

For each situation, identify the appropriate counts and state what the Large Counts check tells you.

A test uses $n=80$ and $H_0:p=0.25$. Calculate both expected counts and decide whether the test’s Large Counts condition is met.
A test uses $n=150$ and $H_0:p=0.05$. Which expected count should you check carefully, and is the condition met?
A sample has 18 successes out of 40, and a test uses $H_0:p=0.50$. What counts are used for the test’s Large Counts condition?
For a confidence interval based on 18 successes out of 40, which counts are used instead? Does that interval Large Counts condition pass?
In a test, one null expected count is 8 and the other is 42. Explain why the condition is not met and what that does—and does not—tell you.