Turn the Claim Into a Question About a Population Proportion
A company may advertise that a large percentage of its products meet a standard. An agency may report that a stated percentage of residents use a service. To test either claim, first identify the population and the characteristic being counted. Then decide what the data are supposed to reveal: evidence that the proportion is below the claimed minimum, above a benchmark, or simply different from a reported value.
As in Defining the Parameter in Hypothesis Statements and Choosing One-Sided or Two-Sided Alternatives, hypotheses describe a population parameter, not the sample result. Let \(p\) be the true proportion of a clearly specified population with the characteristic of interest. The sample provides evidence about \(p\), but it does not change the hypotheses.
Pay close attention to the claim’s wording. “At least 90%” sets a minimum; a test asking whether the company falls short uses a left-tailed alternative. “More than 60%” sets a benchmark for evidence of a higher proportion; that question uses a right-tailed alternative. “The proportion differs from 25%” asks about departures in either direction and uses a two-sided alternative.
A claim may use an inequality, even though a standard one-proportion \(z\)-test writes the null with equality. For a claim such as “at least 90%,” test the boundary value \(p_0=0.90\) against the possibility that \(p\) is below it. The boundary is the benchmark used in the test calculation. In context, rejecting the boundary null in the lower-tail direction provides evidence that the true proportion falls short of the claimed minimum.
| Question about a benchmark | Hypotheses | What the alternative asks |
|---|---|---|
| Is the proportion below a stated minimum? | \(H_0:p=p_0\); \(H_a:p<p_0\) | Is there evidence the population proportion falls short? |
| Is the proportion above a benchmark? | \(H_0:p=p_0\); \(H_a:p>p_0\) | Is there evidence the population proportion exceeds it? |
| Does the proportion differ from a reported value? | \(H_0:p=p_0\); \(H_a:p\ne p_0\) | Is there evidence of a departure in either direction? |
The wording of a claim does not by itself determine the alternative. Ask what the investigation is trying to find evidence for. A test designed to check whether a “more than 60%” marketing claim is supported uses \(H_a:p>0.60\). A test designed to investigate whether the proportion falls short of a “at least 90%” promise uses \(H_a:p<0.90\). These tests can use similar data but answer different questions.
A Claim-Test Plan
The calculation follows the one-proportion \(z\)-test procedure from earlier in this course. The new work is translating the claim into a precise question and making the conclusion match that question. Use the four steps below to keep the reasoning visible.
Define \(p\) in context and write \(H_0\) and \(H_a\). Preserve the claim’s benchmark and choose the alternative from the question being investigated.
Name a one-proportion \(z\)-test. Check the Random condition, the 10% condition when sampling without replacement from a finite population, and the Large Counts condition using \(p_0\).
Calculate \(\hat p=x/n\), the test statistic and the p-value. Use the tail or tails specified by \(H_a\), not the direction that happens to favor the claim.
Compare the p-value with the significance level \(\alpha\). State whether to reject or fail to reject \(H_0\), then explain what the evidence says about the claim in context.
For the Large Counts condition, use the null proportion: check \(np_0\geq10\) and \(n(1-p_0)\geq10\). As covered in Large Counts Condition for Confidence Intervals and Checking the Success-Failure Condition for Tests, a test uses expected counts under \(H_0\), not the observed counts used for a one-proportion \(z\)-interval.
A small p-value is evidence against the null model in the direction named by the alternative. It is not the probability that the company’s claim is true, and it does not prove a claim false. Likewise, failing to reject \(H_0\) does not establish that the claim is true. The conclusion should say whether the sample provides convincing evidence for the particular alternative being tested.
Worked Examples
Worked Example: Test Whether a Product Falls Short of a Minimum
A fictional packaging company advertises that at least 90% of its reusable containers meet a specified leak-resistance standard. A quality team randomly selects 200 containers from a production lot of 12,000; 169 meet the standard. Test whether the true proportion of containers in this lot that meet the standard is below 0.90, using \(\alpha=0.05\).
State: Let \(p\) be the true proportion of containers in this production lot that meet the leak-resistance standard. The claim sets a minimum of 0.90, and the question is whether the proportion falls below it. The hypotheses are \(H_0:p=0.90\) and \(H_a:p<0.90\).
Plan: Use a one-proportion \(z\)-test. The containers were randomly selected from the lot, so the Random condition is met. The sample was taken without replacement from a lot of 12,000, and \(0.10(12{,}000)=1{,}200\); since \(200\leq1{,}200\), the 10% condition is met. Under the null, the expected number meeting the standard is \(np_0=200(0.90)=180\), and the expected number not meeting it is \(n(1-p_0)=200(0.10)=20\). Both are at least 10, so the Large Counts condition is met.
Do: There are \(x=169\) successes in \(n=200\) containers, so the sample proportion is:
The null standard error and test statistic are:
The alternative is left-tailed, so the p-value is the area to the left of \(-2.59\): \(\text{normalcdf}(-1\text{E}99,-2.59,0,1)\approx0.0048\), rounded. Since \(0.0048<0.05\), reject \(H_0\).
Conclude: The sample provides convincing evidence that less than 90% of the containers in this production lot meet the leak-resistance standard. This is evidence that the lot falls short of the company’s advertised minimum; it is not proof that every production lot falls short.
Worked Example: Test Whether a Marketing Claim Is Supported
A fictional internet provider says that more than 60% of its current customers know how to activate a particular emergency setting. An independent team randomly selects 150 customers from a list of 18,000 current customers; 99 know how to activate the setting. Test whether the true proportion exceeds 0.60, using \(\alpha=0.05\).
State: Let \(p\) be the true proportion of current customers on the provider’s list who know how to activate the emergency setting. The hypotheses are \(H_0:p=0.60\) and \(H_a:p>0.60\). This right-tailed test asks whether the data provide evidence in support of the “more than 60%” part of the claim.
Plan: Use a one-proportion \(z\)-test. The team randomly selected customers from the list, so the Random condition is met. Sampling was without replacement, and \(0.10(18{,}000)=1{,}800\); \(150\leq1{,}800\), so the 10% condition is met. Under \(H_0\), the expected success count is \(150(0.60)=90\), and the expected failure count is \(150(0.40)=60\). Both counts are at least 10, so the Large Counts condition is met.
Do: The sample proportion is:
Calculate the null standard error and test statistic:
Because \(H_a:p>0.60\), use the upper-tail area: \(\text{normalcdf}(1.50,1\text{E}99,0,1)\approx0.0668\), rounded. Since \(0.0668>0.05\), fail to reject \(H_0\).
Conclude: The sample does not provide convincing evidence that more than 60% of the provider’s current customers know how to activate the setting. The result does not establish that the claim is false or that the proportion equals 60%; it means this sample does not give strong enough evidence for the stated alternative at the 0.05 significance level.
Worked Example: Test Whether an Agency’s Reported Value Differs
A fictional regional agency reports that 25% of households in its service area use a home energy-monitoring device. An auditor randomly selects 160 households from a list of 5,000; 52 use such a device. Test whether the true proportion differs from 0.25, using \(\alpha=0.05\).
State: Let \(p\) be the true proportion of households in the agency’s service area that use a home energy-monitoring device. The question asks whether the proportion differs from the reported value, so the hypotheses are \(H_0:p=0.25\) and \(H_a:p\ne0.25\).
Plan: Use a one-proportion \(z\)-test. The auditor randomly selected households, meeting the Random condition. The sample was drawn without replacement from 5,000 households, and \(0.10(5{,}000)=500\); since \(160\leq500\), the 10% condition is met. Under the null, the expected number of households using a device is \(160(0.25)=40\), and the expected number not using one is \(160(0.75)=120\). Both are at least 10, so the Large Counts condition is met.
Do: The sample proportion is:
The null standard error and test statistic are:
The alternative is two-sided, so results at least as far from zero as \(z=2.19\) count in either tail. The p-value is \(2\text{normalcdf}(2.19,1\text{E}99,0,1)\approx0.0285\), rounded. Since \(0.0285<0.05\), reject \(H_0\).
Conclude: The sample provides convincing evidence that the true proportion of households in the agency’s service area that use a home energy-monitoring device differs from 25%. The observed sample proportion is higher, but the two-sided test establishes evidence of a difference in either direction, not a guarantee that the population proportion is above 25%.
Common Mistakes and AP Exam Tips
Claim wording can make a test feel like a judgment about whether a company or agency is trustworthy. A statistical test has a narrower job: it assesses whether the sample provides convincing evidence against a stated null hypothesis and in the direction named by the alternative.
- Putting the claim in the alternative automatically. The question may be whether a claim is supported or whether it is contradicted. Translate the actual question first. Testing whether a “more than 60%” claim is supported uses \(H_a:p>0.60\); investigating whether the proportion falls short of a 90% minimum uses \(H_a:p<0.90\).
- Writing an inequality as the null hypothesis. For the standard one-proportion \(z\)-test, state the boundary benchmark with equality, such as \(H_0:p=0.90\). Explain that the test examines evidence in the direction of falling below that boundary.
- Choosing the tail from the sample result. A sample proportion below the benchmark does not, by itself, make a test left-tailed. The research question determines \(H_a\), and \(H_a\) determines which results count as extreme.
- Saying “the claim is true” after failing to reject. A large p-value means the sample does not provide convincing evidence for the alternative at the chosen significance level. It does not prove the null or the company’s claim.
- Using observed counts for the test’s Large Counts condition. Check \(np_0\) and \(n(1-p_0)\), not \(x\) and \(n-x\), for a one-proportion \(z\)-test.
- Making a broader conclusion than the sample represents. Name the population used to define \(p\). A random sample from one product lot supports a conclusion about that lot, not automatically every product the company makes.
Key Takeaway
A company’s or agency’s percentage claim becomes a testable question when its population, characteristic, and benchmark are clear. “Falls below,” “exceeds,” and “differs from” lead to different alternatives, even when they refer to the same benchmark. Let the question—not the sample’s direction or the organization’s interest—determine the hypotheses and the p-value tail.
Check Your Understanding
For each scenario, identify the population proportion and the alternative that matches the question. Then consider the conditions and how a test conclusion should be worded.
- A fictional snack company says at least 85% of its packages contain the advertised net weight. An inspector asks whether the proportion falls short. What are the null and alternative hypotheses?
- A transit agency claims that more than 70% of riders can find its route information online. An auditor wants evidence supporting the claim. Which direction should the alternative use?
- An agency reports that 40% of eligible residents used a new service. A researcher asks whether the true proportion is different. Write the hypotheses.
- For a one-proportion \(z\)-test with \(n=120\) and \(p_0=0.30\), calculate the two expected counts for the Large Counts condition.
- In a test of whether a company’s proportion is below its advertised minimum, the p-value is 0.12 and \(\alpha=0.05\). What decision should be made, and what should the conclusion avoid claiming?