Why Check Conditions Before Testing?
A one-proportion \(z\)-test uses sample data to assess a claim about a population proportion. Before calculating a test statistic, you need to check whether the study’s sampling process and the expected counts support that procedure. This tutorial applies the Random, 10%, and Large Counts conditions to a customer-preference study, then shows how the checks fit into a complete test response.
In Observational Data and Limits on Generalization, you learned to connect a sampling method to the population a study can represent. The same idea matters here: a calculation cannot make a self-selected poll behave like a random sample. As in Random Condition for Proportion Inference and Independence and the 10 Percent Condition, explain what the study actually did and use the population from which the data were obtained.
The Large Counts check for a test differs from the check for a confidence interval. As explained in Checking the Success-Failure Condition for Tests, a test evaluates data against the null hypothesis. Therefore, check the counts expected under the null proportion \(p_0\), not the observed counts.
What Each Condition Checks
Random condition. Identify how the individuals entered the study. For a customer survey, a simple random sample from a suitable customer list can support inference to the population represented by that list. A voluntary online poll is not random simply because many people respond. If a study uses random assignment rather than random sampling, explain that process and identify the units to which the inference applies. Do not claim that random assignment makes participants a random sample of a broader population.
10% condition. When sampling without replacement from a finite population, check that the sample is no more than 10% of the population it came from. In symbols, \(n\leq0.10N\), where \(n\) is the sample size and \(N\) is the size of the source population. This check supports treating the observations as approximately independent. Use the actual source population, not a larger population that the study did not sample from. When the population is unknown, use clear evidence that it is at least ten times the sample size, as discussed in When Population Size Is Unknown or Large.
Large Counts condition. For a one-proportion \(z\)-test of \(H_0:p=p_0\), calculate the expected number of successes, \(np_0\), and expected number of failures, \(n(1-p_0)\). Both must be at least 10. These are expected counts under the null hypothesis, not the counts actually observed in the sample. If either expected count is below 10, the usual Normal approximation for the \(z\)-test is not supported by this condition.
A condition check is an argument, not just a label. State what the study did, show the relevant comparison or calculation, and say whether it meets the condition. If one condition fails, do not proceed as though the standard one-proportion \(z\)-test is justified.
Worked Examples
Worked Example: Testing a Preference for a New Drink
A beverage company wants to know whether more than half of its active loyalty-program customers prefer a new citrus drink to its current berry drink. The company randomly selects 200 customers from a list of 5,000 active loyalty customers. Each selected customer tastes both drinks and chooses one; 117 choose citrus. Carry out a one-proportion \(z\)-test at the 5% significance level, beginning with the conditions.
Let \(p\) be the proportion of active loyalty-program customers represented by the list who prefer the new citrus drink to the current berry drink. Define a success as a selected customer choosing citrus. The hypotheses are \(H_0:p=0.50\) and \(H_a:p>0.50\). The company is testing whether a majority prefer citrus.
Use a one-proportion \(z\)-test if the Random, 10%, and Large Counts conditions are met. The customers were randomly selected from the loyalty-customer list, so the sampling method supports inference to customers represented by that list. The sample is no more than 10% of the 5,000 listed customers because \(200\leq0.10(5{,}000)=500\). Under \(H_0\), the expected successes are \(np_0=200(0.50)=100\), and the expected failures are \(n(1-p_0)=200(0.50)=100\). Both expected counts are at least 10, so all three conditions are met.
The sample proportion choosing citrus is \(\hat{p}=117/200=0.585\). The test statistic uses the null proportion in its standard error:
For the upper-tail alternative, the p-value is \(P(Z\geq2.404)\approx0.0081\), rounded to four decimal places. This is the probability, assuming the null model and conditions are correct, of obtaining a sample proportion at least as high as 0.585.
Because the p-value, 0.0081, is less than the significance level of 0.05, reject \(H_0\). The sample provides convincing evidence that more than half of the active loyalty-program customers represented by the list prefer the new citrus drink. The conclusion applies to that represented customer population; it does not automatically apply to all beverage consumers.
Worked Example: A Random Sample With Too Few Expected Successes
A snack shop randomly selects 40 customers from its list of 1,200 customers to ask whether they prefer a new spicy cracker. The shop wants to test the claim that 20% of its customers prefer the cracker, using a one-proportion \(z\)-test. Assess the conditions before attempting the test.
Let \(p\) be the proportion of customers represented by the shop’s list who prefer the new cracker, and define a success as a customer preferring it. The proposed null hypothesis is \(H_0:p=0.20\). The sample is random, and the 10% condition is met because \(40\leq0.10(1{,}200)=120\).
The Large Counts check must use the null proportion. There are \(np_0=40(0.20)=8\) expected successes and \(n(1-p_0)=40(0.80)=32\) expected failures. Although the expected failures meet the threshold, 8 expected successes is less than 10. Therefore, the Large Counts condition fails, and the usual one-proportion \(z\)-test is not justified by the stated conditions. Do not proceed to calculate and interpret a standard \(z\)-test p-value as if the condition were met.
The failure does not mean the claim \(p=0.20\) is true or false. It means that this Normal-based test is not supported by the Large Counts check for this sample size and null proportion. The Random and 10% conditions being met cannot compensate for the failed Large Counts condition.
Worked Example: Many Responses Do Not Make a Poll Random
A home-goods store posts an optional poll on its website asking whether customers prefer a proposed blue mug to a green mug. Of the 600 customers who choose to respond, 282 select blue. The store wants to test whether fewer than 40% of all its customers prefer blue. It has 10,000 customers on its mailing list. Evaluate whether a one-proportion \(z\)-test is appropriate.
Let \(p\) be the proportion of customers on the store’s mailing list who prefer the blue mug; a success is a customer preferring blue. The hypotheses for the proposed test are \(H_0:p=0.40\) and \(H_a:p<0.40\). The 600 respondents chose whether to take part in the poll. They were not randomly selected, so the Random condition is not met. Customers with strong preferences or more interest in the proposed mug could be more likely to respond.
The sample size is less than 10% of the mailing-list population because \(600\leq0.10(10{,}000)=1{,}000\). This numerical comparison does not turn a voluntary-response poll into a random sample; it only addresses the 10% comparison if sampling without replacement were being considered.
Under the proposed null hypothesis, the expected counts would be \(np_0=600(0.40)=240\) successes and \(n(1-p_0)=600(0.60)=360\) failures. Both are at least 10, so the Large Counts condition is met. However, the Random condition fails. The store should not treat a one-proportion \(z\)-test based on these self-selected responses as a sound test about all customers on the list. It can accurately describe the responses: \(282/600=0.47\), or 47% of the people who chose to answer, selected blue. The large response count does not remove the selection problem.
Worked Example: A Sample That Is Too Large Relative to Its Source Population
A small outdoor store has 1,000 customers in its current membership database. It randomly selects 150 members without replacement to ask whether they prefer a new lightweight tent to the store’s standard tent. The owner proposes testing \(H_0:p=0.50\), where \(p\) is the proportion of database members who prefer the new tent. Check whether a one-proportion \(z\)-test is supported.
The Random condition is met because 150 members were randomly selected from the database. But the 10% condition fails: \(0.10(1{,}000)=100\), and \(150\not\leq100\). The sample is 15% of the source population. Thus, the usual approximate-independence justification from the 10% condition is not met.
For the proposed null proportion of 0.50, the expected counts are \(np_0=150(0.50)=75\) successes and \(n(1-p_0)=150(0.50)=75\) failures. The Large Counts condition is met, but that does not repair the failed 10% condition. As described, the ordinary one-proportion \(z\)-test is not supported by all three required checks. The issue is not that a random sample was absent; it is that the sample is too large relative to the finite population from which it was drawn.
Writing Condition Checks Clearly
For an AP Statistics response, make each check verifiable. A statement such as “the counts are large” is incomplete unless it shows which counts were checked and how they were obtained. In a test, write the null proportion into the Large Counts calculations. A concise but complete explanation for the main example would say: “The 200 customers were randomly selected from the active loyalty-customer list. Since \(200\leq0.10(5{,}000)=500\), the 10% condition is met. Under \(H_0:p=0.50\), there are \(200(0.50)=100\) expected successes and \(200(0.50)=100\) expected failures, both at least 10.”
Keep the target population precise, too. If the sample was drawn from a store’s loyalty list, name the customers represented by that list rather than claiming a result for all shoppers everywhere. This connects the conditions check to the scope of inference in Observational Data and Limits on Generalization.
Common Mistakes and AP Exam Tips
- Using the observed counts for a test’s Large Counts check. For a one-proportion \(z\)-test, calculate \(np_0\) and \(n(1-p_0)\). The observed successes and failures are used for the interval check, not this test condition.
- Using \(\hat{p}\) instead of \(p_0\) in the test standard error. The test statistic is evaluated under the null hypothesis, so its standard error is based on \(p_0\). Keep this separate from the one-proportion interval formula.
- Calling an optional poll random because it has many responses. Describe how respondents were selected. A large voluntary-response group may still differ systematically from the target population.
- Checking the 10% condition against the wrong population. Compare the sample size with the population or frame from which the sample was actually drawn. Do not use the size of a broad target population if the sample came from a smaller list.
- Continuing with a \(z\)-test after one condition fails. State which condition is not met and explain why that prevents the usual procedure from being justified. Do not hide the failure behind calculations.
- Writing only “Random, 10%, and Large Counts are satisfied.” A full-credit check names the method, shows the comparison or expected counts, and states the result of each check.
Key Takeaway
A one-proportion \(z\)-test depends on how the data were collected, how the sample relates to its source population, and whether the null model predicts enough successes and failures. The checks are related, but each answers a different question: random selection supports inference to a represented population, the 10% condition supports approximate independence for sampling without replacement, and Large Counts supports the Normal approximation under the null.
Check Your Understanding
For each situation, identify the relevant condition checks and decide whether the usual one-proportion \(z\)-test is supported.
- A store randomly samples 120 customers from 4,000 loyalty accounts. For \(H_0:p=0.30\), calculate the expected successes and failures and assess the three conditions.
- A random sample of 35 people is taken without replacement from a membership group of 250. The proposed null proportion is 0.50. Which conditions are met, and which condition needs careful attention?
- A website poll receives 900 voluntary responses. The proposed null proportion is 0.60, so both null expected counts exceed 10. Does the Large Counts check make the poll suitable for a test about all customers? Explain.
- A random sample of 70 customers is taken without replacement from a finite list of 500. What comparison determines whether the 10% condition is met?
- In a one-proportion test, the observed sample has 14 successes and 46 failures, with \(n=60\) and \(p_0=0.10\). What expected counts should be checked for Large Counts, and does the condition pass?