One Sample, Two Different Checks
A one-proportion \(z\)-test and a one-proportion \(z\)-interval can use the same sample but have different Large Counts checks. The reason is that the procedures use the data in different ways: a test evaluates the sample under a null hypothesis, while an interval estimates an unknown population proportion from the observed sample.
In Large Counts Condition for Confidence Intervals, you learned to check the observed success and failure counts for an interval. In Checking the Success-Failure Condition for Tests, you learned to check the expected success and failure counts under the null hypothesis for a test. This tutorial puts those checks side by side so you can select the right proportion without mixing the procedures.
For a test, the expected success count is \(np_0\), and the expected failure count is \(n(1-p_0)\). These describe what the null model predicts on average if \(H_0\) is true. For an interval, \(n\hat{p}=x\) is the number of observed successes and \(n(1-\hat{p})=n-x\) is the number of observed failures.
In either check, both counts must be at least 10. The calculations are different, but the threshold is the same. A useful habit is to write the procedure name beside each pair of calculations before deciding whether its Large Counts condition is met.
A Side-by-Side Checking Routine
When a question asks you to compare an interval and a test based on the same observations, organize your work into two separate columns. First record the sample size \(n\), the observed number of successes \(x\), and \(\hat{p}=x/n\). Then record the test’s null value \(p_0\). Use \(p_0\) only for the test check and \(x\), or equivalently \(\hat{p}\), for the interval check.
Write \(n\), \(x\), and \(\hat{p}=x/n\). If there is a test, also write the null value \(p_0\).
Consider whether the data come from a random sample or suitable random process. If sampling without replacement from a finite population, check the 10% condition.
For the test, calculate \(np_0\) and \(n(1-p_0)\). For the interval, calculate \(x\) and \(n-x\).
State whether both counts for that procedure are at least 10. Passing one check does not determine the result of the other.
The Random condition and, when applicable, the 10% condition are considered for both procedures. Passing the Large Counts check does not replace either of them. Also, these conditions concern whether a Normal-based inference procedure is supported; they do not say whether the null hypothesis is true or whether an interval contains the population proportion.
Worked Examples
Worked Example: The Test Check Passes but the Interval Check Fails
A fictional community center randomly selects 80 members without replacement from 2,000 eligible members. Eight selected members say they use a particular exercise room each week. Compare the Large Counts checks for a test of \(H_0:p=0.20\) and for a one-proportion \(z\)-interval using the sample.
The sample size is \(n=80\), the observed success count is \(x=8\), and $$ \hat{p}=\frac{x}{n}=\frac{8}{80}=0.10. $$ Here, \(p\) is the proportion of all eligible community center members who use the room each week.
Check the sampling conditions. The members were randomly selected, so the Random condition is met. Since selection was without replacement, check the 10% condition: $$ 80\leq0.10(2{,}000)=200. $$ This is true, so the 10% condition is met.
Check the test. The null value is \(p_0=0.20\), so calculate expected counts under \(H_0\): $$ np_0=80(0.20)=16, \qquad n(1-p_0)=80(0.80)=64. $$ Both expected counts are at least 10. The test’s Large Counts condition is met.
Check the interval. Use the observed counts: there are \(x=8\) successes and \(n-x=80-8=72\) failures. Because the observed success count \(8\) is less than 10, the interval’s Large Counts condition is not met.
For this sample, then, the test check passes and the interval check fails. The reason is that the test’s null model predicts 16 successes, while the sample actually contains only 8. Meeting the test’s Large Counts condition does not make the interval’s observed-count condition pass.
Worked Example: The Interval Check Passes but the Test Check Fails
A fictional environmental group randomly selects 120 monitoring locations without replacement from 5,000 eligible locations. Twenty locations show a specified sign of erosion. Compare the interval check with the check for a test of \(H_0:p=0.05\).
The observed sample proportion is $$ \hat{p}=\frac{x}{n}=\frac{20}{120}\approx0.1667. $$ The parameter \(p\) is the proportion of eligible monitoring locations that show the specified sign of erosion.
Check the sampling conditions. The locations were randomly selected, so the Random condition is met. The 10% condition is also met: $$ 120\leq0.10(5{,}000)=500. $$
Check the test. Under the null hypothesis, the expected success and failure counts are $$ np_0=120(0.05)=6, \qquad n(1-p_0)=120(0.95)=114. $$ The expected success count is below 10. Therefore, the test’s Large Counts condition is not met, even though the expected failure count is well above 10.
Check the interval. The observed counts are 20 successes and \(120-20=100\) failures. Both are at least 10, so the interval’s Large Counts condition is met.
This is the reverse of the community center example: the interval check passes, but the test check fails. The test uses \(p_0=0.05\), which predicts only 6 successes; the interval uses the 20 successes actually observed. A failed test check means the usual Normal-based \(z\)-test is not supported by this condition. It is not, by itself, evidence that \(H_0\) is false.
Worked Example: Both Checks Pass, but the Counts Still Come from Different Sources
A fictional school district randomly selects 100 students without replacement from 1,200 students. Forty students report bringing a reusable water bottle to school. Compare the Large Counts checks for a one-proportion \(z\)-interval and a test of \(H_0:p=0.35\).
Here, \(n=100\), \(x=40\), and $$ \hat{p}=\frac{40}{100}=0.40. $$ The population proportion \(p\) refers to all 1,200 students in the district.
Check the sampling conditions. The students were randomly selected, meeting the Random condition. For the 10% condition, $$ 100\leq0.10(1{,}200)=120, $$ so this condition is met as well.
Check the interval. There are 40 observed successes and \(100-40=60\) observed failures. Both counts are at least 10, so the interval’s Large Counts condition is met.
Check the test. Under the null value \(p_0=0.35\), the expected counts are $$ np_0=100(0.35)=35, \qquad n(1-p_0)=100(0.65)=65. $$ Both expected counts are at least 10, so the test’s Large Counts condition is met.
Both checks pass, but their counts are not interchangeable: the interval uses the observed counts 40 and 60, while the test uses the null expected counts 35 and 65. A shared result—both conditions are met—does not mean the calculations were the same.
Compare Before You Conclude
The examples show three possible patterns: the test check can pass while the interval check fails; the interval check can pass while the test check fails; or both can pass. The correct result depends on the sample and, for the test, the specific null value. Neither procedure’s check can be inferred from the other.
| Question to ask | For the test | For the interval |
|---|---|---|
| Which proportion determines the success count? | The null proportion \(p_0\) | The observed proportion \(\hat{p}\) |
| What success count is checked? | Expected: \(np_0\) | Observed: \(n\hat{p}=x\) |
| What failure count is checked? | Expected: \(n(1-p_0)\) | Observed: \(n(1-\hat{p})=n-x\) |
| What must be true? | Both counts are at least 10 | Both counts are at least 10 |
If a question asks for both checks, write out both pairs and label them. This makes it clear which values are expected under a null claim and which are observed in the sample. It also prevents a common error: using \(p_0\) for an interval just because a related test has a null hypothesis, or using \(\hat{p}\) for a test because it is easy to calculate.
Common Mistakes and AP Exam Tip
- Using one pair of counts for both procedures. Even when the interval and test use the same sample, the test uses \(np_0\) and \(n(1-p_0)\); the interval uses \(x\) and \(n-x\).
- Using the sample proportion for the test check. The test’s Large Counts check is based on the model under \(H_0\). State \(p_0\) and calculate both null expected counts.
- Using the null proportion for the interval check. The interval check is based on the sample’s observed successes and failures. Calculate \(x\) and \(n-x\), or show the equivalent calculations with \(\hat{p}\).
- Checking only one category. Both success and failure counts must be at least 10. A count well above 10 in one category cannot compensate for a count below 10 in the other.
- Calling expected counts observed counts. For a test, \(np_0\) and \(n(1-p_0)\) are predictions under the null model, not a report of what happened in the sample.
- Treating a failed check as a decision about the claim. A failed Large Counts condition says that this Normal-based procedure is not supported by the check. It does not prove or disprove the population proportion’s value.
Key Takeaway
The Large Counts threshold is the same for a one-proportion \(z\)-test and a one-proportion \(z\)-interval, but the counts being checked come from different sources. A test checks what the null model expects; an interval checks what the sample actually contains.
Check Your Understanding
For each situation, identify the counts required for the stated procedure and decide whether its Large Counts condition is met.
- A one-proportion \(z\)-test has \(n=90\) and \(H_0:p=0.20\). Find both expected counts and state whether the test check passes.
- A sample has 7 successes out of 70. For a one-proportion \(z\)-interval, find the observed success and failure counts. Does the interval check pass?
- A sample has 24 successes out of 100. Compare the interval check with the test check for \(H_0:p=0.10\). Which check, if either, fails?
- In a test, the null expected counts are 9 successes and 41 failures. Explain whether the Large Counts condition is met and why the larger count does not change the result.
- Why can the same sample meet the test’s Large Counts condition but fail the interval’s Large Counts condition?