Conditions When the Large Counts Check Fails

When a Count Is Too Small

The Large Counts condition is a gatekeeper for one-proportion \(z\) procedures. It asks whether the relevant success and failure counts are both at least 10. If a count is below 10, the condition fails, and the Normal approximation that supports the procedure may be unreliable.

For a confidence interval, the check uses the counts observed in the sample. For a test of \(H_0:p=p_0\), it uses the counts expected under the null hypothesis. This distinction, covered in Interval Versus Test Condition Checks Compared, matters: six observed successes do not automatically mean that every one-proportion test condition fails. The test’s expected counts depend on \(p_0\).

Key idea: If either relevant count is less than 10, the Large Counts condition fails. Do not treat the usual one-proportion \(z\)-interval or \(z\)-test as a trustworthy inference just because a calculator returns numbers.

Why does the count matter? A Normal model is continuous and roughly symmetric, while counts of successes are discrete and can have a strongly skewed distribution when the expected number of successes or failures is small. With only a few expected successes, the shape may not be well approximated by the Normal curve used to calculate a \(z\)-statistic or interval.

What “Fails” Means for Each Procedure

For a one-proportion \(z\)-interval, the relevant quantities are the observed number of successes \(x=n\hat{p}\) and observed number of failures \(n-x=n(1-\hat{p})\). Both must be at least 10. If, for example, the sample contains six successes, then \(x=6<10\), so the interval’s Large Counts condition fails, regardless of how many failures there are.

For a one-proportion \(z\)-test, the question is different. Under \(H_0:p=p_0\), the expected counts are \(np_0\) successes and \(n(1-p_0)\) failures. Both must be at least 10. The observed number \(x\) is used to calculate the sample proportion and, if the test is appropriate, the test statistic. But the test’s Large Counts check is based on the null model.

\begin{array}{lll} \text{Procedure} & \text{Success count checked} & \text{Failure count checked}\\[4pt] \text{One-proportion }z\text{-interval} & x=n\hat{p} & n-x\\[4pt] \text{One-proportion }z\text{-test} & np_0 & n(1-p_0) \end{array}

A failed condition is not evidence that the null hypothesis is true or false. It means the usual Normal-based \(z\) method is not adequately supported by the conditions, so its interval or \(p\)-value should not be presented as a reliable result.

Worked Examples

Worked Example: A Confidence Interval with Six Successes

A fictional random sample of 40 water filters is inspected for a particular defect. Six filters have the defect. Can a one-proportion \(z\)-interval be used to estimate the proportion of all filters of this type with the defect? Assume the sample is drawn without replacement from a production lot of 2,000 filters.

Define success as “the filter has the defect.” The observed success count is \(x=6\), and the failure count is \(40-6=34\). The sample proportion is

\hat{p}=\frac{x}{n}=\frac{6}{40}=0.15

The sample is stated to be random. The 10% condition is met because \(40\leq0.10(2000)=200\). But the observed success count is \(6<10\), while the failure count is \(34\geq10\). Therefore, the Large Counts condition fails.

A calculator can still produce an answer if the one-proportion \(z\)-interval formula is entered. For illustration only, using a 95% confidence level gives \(z^*=1.96\), and the formula would produce

0.15\pm1.96\sqrt{\frac{0.15(0.85)}{40}} =0.15\pm1.96(0.05646) \approx(0.0393,\ 0.2607)

The arithmetic is possible, but the failed condition means this Normal-based interval is not justified as a reliable 95% interval. Do not report it as a valid confidence interval merely because the calculator supplied endpoints. The appropriate conclusion is that the usual one-proportion \(z\)-interval cannot be used here; a method suited to small counts or additional data is needed.

Worked Example: A Test That Fails Its Large Counts Check

A fictional random sample of 80 seedlings is checked for signs of a fungal infection. Six show signs. A researcher wants to test whether the population proportion with signs is greater than 10%. Assume the seedlings were sampled without replacement from a greenhouse population of 3,000.

State.
Let \(p\) be the proportion of seedlings in this greenhouse population that show signs of the infection. The hypotheses are \(H_0:p=0.10\) and \(H_a:p>0.10\).

Plan.
A one-proportion \(z\)-test would be considered if the conditions hold. The sample is random. The 10% condition holds because \(80\leq0.10(3000)=300\). For the Large Counts condition, use \(p_0=0.10\), not the observed sample proportion.

Do.
The expected success count under \(H_0\) is \(np_0=80(0.10)=8\). The expected failure count is \(n(1-p_0)=80(0.90)=72\). Since \(8<10\), the Large Counts condition fails. Do not calculate or report a one-proportion \(z\)-test \(p\)-value as though the Normal approximation were justified.

Conclude.
The Large Counts condition for the one-proportion \(z\)-test is not met because only 8 successes are expected under the null hypothesis. Therefore, this \(z\)-test does not provide a reliable basis for deciding whether there is convincing evidence that more than 10% of the greenhouse seedlings show signs of infection.

Notice that the observed sample proportion is \(6/80=0.075\), but it is not used to check the test’s Large Counts condition. The failed expected count comes from \(p_0=0.10\). This is the test-versus-interval distinction from Checking the Success-Failure Condition for Tests.

Worked Example: Six Observed Successes Do Not Always Fail a Test Check

A fictional random sample of 200 home composting systems finds six with a particular type of odor problem. Consider a one-proportion \(z\)-test of \(H_0:p=0.05\), where \(p\) is the proportion of systems in the target population with the problem. Assume a suitable random sample was taken without replacement from a population of 5,000 systems.

There are six observed successes and \(200-6=194\) observed failures. The sample proportion is \(6/200=0.03\). For this test, however, check the expected counts under \(H_0\):

np_0=200(0.05)=10 \qquad\text{and}\qquad n(1-p_0)=200(0.95)=190

Both expected counts are at least 10, so the Large Counts condition for the test is met. The random condition is met by the stated sampling method, and the 10% condition holds because \(200\leq0.10(5000)=500\). Thus, the count of six observed successes alone does not rule out the test’s Large Counts condition in this example.

This does not prove that every other aspect of the study is sound, nor does it determine the test’s result. It demonstrates only which counts belong in the Large Counts check: expected counts for a test, observed counts for an interval.

What to Do When the Check Fails

First, identify which count is below 10 and which procedure you were considering. Then state plainly that the Large Counts condition fails. Do not write “the condition is close enough,” round a count up to 10, or proceed with a usual \(z\)-procedure as if the check had passed.

If planning is still possible, collect more data. A larger sample can increase the expected counts. The sample must still be selected appropriately, and the other conditions must also be checked.
If the data have already been collected, use a method designed for small counts. In an appropriate setting, an exact binomial procedure may be available. Use it when it is part of the task or when suitable statistical guidance is available; do not present it as the one-proportion \(z\)-procedure.
If no suitable alternative is available, stop short of unsupported inference. You can describe the sample count and sample proportion, but do not claim that an unreliable \(z\)-interval or \(z\)-test establishes a population result.

When planning a test, a simple count calculation can show how large a sample would be needed just to meet the Large Counts threshold for a chosen null proportion. If \(p_0=0.08\), for instance, the expected success count reaches 10 when \(n(0.08)\geq10\), or \(n\geq125\). At \(n=125\), the expected failure count is \(125(0.92)=115\), also at least 10. This is only a condition check: meeting it does not guarantee adequate power or satisfy the random and 10% conditions.

Common Mistakes and AP Exam Tip

Using observed counts for a test. A test checks \(np_0\) and \(n(1-p_0)\), not \(x\) and \(n-x\). State the null proportion before calculating expected counts.
Using expected counts for an interval. An interval checks the observed counts \(x\) and \(n-x\). Do not substitute a claimed population proportion for the sample results.
Thinking a calculator output validates the procedure. A calculator can return a number even when a condition fails. A numerical answer does not repair an unsupported Normal approximation.
Calling a failed condition a result. Failure of the Large Counts check does not mean “reject \(H_0\)” or “fail to reject \(H_0\).” It means the usual \(z\)-procedure is not justified.
Claiming that six observed successes always invalidate a test. For a test, compare the expected counts under \(H_0\) with 10. The observed count of six is not the test’s Large Counts check.

AP Exam Tip: Name the relevant counts and show the comparison. For an interval, write “There are 6 observed successes, and \(6<10\), so the Large Counts condition fails.” For a test, show \(np_0\) and \(n(1-p_0)\). Then state that the usual \(z\)-procedure is not supported; do not report its interval or \(p\)-value as a reliable inference.

Key Takeaway

A small count is a warning about the Normal approximation, not a prompt to ignore the conditions or force a \(z\)-procedure. Check observed counts for an interval and null expected counts for a test. If either relevant count is below 10, explain that the Large Counts condition fails and seek more data or an appropriate small-count method.

Key takeaway: Six observed successes make the Large Counts condition fail for a one-proportion \(z\)-interval. For a one-proportion \(z\)-test, check expected counts under \(H_0\); six observed successes alone do not determine whether that test condition passes.

Check Your Understanding

For each situation, identify the relevant counts and decide whether the Large Counts condition is met.

A random sample of 50 products contains 6 with a defect. Does the Large Counts condition for a one-proportion \(z\)-interval pass? Explain.
A test of \(H_0:p=0.12\) uses a sample of 70. Calculate both expected counts and decide whether the Large Counts condition for the test passes.
A sample contains 6 successes out of 200. A test uses \(H_0:p=0.05\). What counts should be checked for the test, and does the Large Counts condition pass?
In a test, one expected count is 7 and the other is 93. What should you say about using the usual one-proportion \(z\)-test?
Give two reasonable next steps when a one-proportion \(z\)-procedure fails its Large Counts condition.