Four-Step Process for a One-Proportion z-Test

From a Test Question to a Complete Response

A one-proportion z-test solution is more than a test statistic and a p-value. A complete response identifies the population proportion being tested, explains why the procedure is appropriate, shows the calculation, and answers the original question. The four-step structure—State, Plan, Do, Conclude—makes that reasoning easy to follow.

In Common Mistakes in Setting Up Hypothesis Tests, you practiced defining \(p\), writing the hypotheses, and choosing the alternative from the research question rather than from the sample result. This tutorial brings those decisions together with the condition checks and calculations in a full solution. The condition details are familiar from earlier tutorials; the new skill is organizing every piece into one connected response.

Key idea: A complete test response connects the research question to the hypotheses, justifies the one-proportion \(z\)-test, calculates the p-value using the direction of \(H_a\), and concludes in context.

The Four Steps

Use the four steps in order. The State step defines the target and test; the Plan step justifies the method; the Do step reports the evidence; and the Conclude step interprets that evidence for the question. If the problem gives a significance level \(\alpha\), include it in the State step.

State.
Define \(p\) in context. Write \(H_0:p=p_0\) and the alternative \(H_a\) that matches the research question. Identify \(\alpha\), if it is given.

Plan.
Name a one-proportion \(z\)-test and check the Random condition, the 10% condition when appropriate, and the test’s Large Counts condition using \(p_0\).

Do.
Calculate \(\hat{p}=x/n\), the null standard error, the \(z\) statistic, and the p-value in the tail or tails specified by \(H_a\).

Conclude.
Compare the p-value with \(\alpha\), then state whether to reject or fail to reject \(H_0\) and answer the research question in context.

For the Plan step, explicitly connect each condition to the study. Random selection or an appropriate random process supports inference to the population represented by the process. If sampling without replacement from a finite population of size \(N\), check that \(n\leq0.10N\). For a test of \(H_0:p=p_0\), the Large Counts condition is \(np_0\geq10\) and \(n(1-p_0)\geq10\). As explained in Checking the Success-Failure Condition for Tests, these are expected counts under the null, not the observed success and failure counts.

Formula: For a one-proportion \(z\)-test, calculate the null standard error and test statistic as follows:

SE_0=\sqrt{\frac{p_0(1-p_0)}{n}} \qquad z=\frac{\hat{p}-p_0}{SE_0}

The alternative hypothesis determines the p-value: use a lower tail for \(H_a:p<p_0\), an upper tail for \(H_a:p>p_0\), and both tails for \(H_a:p\ne p_0\).

The p-value is calculated assuming \(H_0\) is true. At significance level \(\alpha\), reject \(H_0\) when the p-value is less than or equal to \(\alpha\); otherwise, fail to reject \(H_0\). A decision to fail to reject does not establish that \(H_0\) is true.

Worked Examples

Worked Example: A Left-Tailed Test About Transit Use

A transit planning team wants to know whether fewer than 25% of registered riders use a new monthly route. In an invented scenario, a random sample of 400 of the 8,000 registered riders includes 88 who use the route. Test the claim at \(\alpha=0.05\).

State: Let \(p\) be the true proportion of the 8,000 registered riders who use the new monthly route. The question asks whether this proportion is below 0.25, so \(H_0:p=0.25\) and \(H_a:p<0.25\). The significance level is \(\alpha=0.05\).

Plan: Use a one-proportion \(z\)-test. The riders were randomly sampled, so the Random condition is met for the population represented by the list of registered riders. The sample is taken without replacement from 8,000 riders, and \(400\leq0.10(8{,}000)=800\), so the 10% condition is met. Under the null, the expected success count is \(np_0=400(0.25)=100\), and the expected failure count is \(n(1-p_0)=400(0.75)=300\). Both are at least 10, so the Large Counts condition is met.

Do: There are \(x=88\) successes in \(n=400\) observations. Calculate the sample proportion and null standard error:

\hat{p}=\frac{x}{n}=\frac{88}{400}=0.22 \qquad SE_0=\sqrt{\frac{0.25(0.75)}{400}} =\sqrt{0.00046875} \approx0.02165

Using the unrounded standard error, the test statistic is:

z=\frac{0.22-0.25}{\sqrt{0.25(0.75)/400}} \approx-1.3856

Because the alternative is left-tailed, the p-value is \(\text{normalcdf}(-1\text{E}99,-1.3856,0,1)\approx0.0829\), rounded. Since \(0.0829>0.05\), fail to reject \(H_0\).

Conclude: The sample does not provide convincing evidence that fewer than 25% of registered riders use the new monthly route. This conclusion does not prove that the true proportion is 25%; it says that this sample does not give convincing evidence for the proposed lower proportion at the 0.05 significance level.

Worked Example: A Right-Tailed Test About a Device Setting

A fictional technology support team asks whether more than 40% of customers with a certain device have enabled an optional accessibility setting. A random sample of 250 customers from a population of 4,000 includes 115 who have enabled it. Test the question at \(\alpha=0.05\).

State: Let \(p\) be the true proportion of customers in the population of 4,000 who have enabled the accessibility setting. The question asks whether this proportion exceeds 0.40. Thus, \(H_0:p=0.40\), \(H_a:p>0.40\), and \(\alpha=0.05\).

Plan: Use a one-proportion \(z\)-test. The sample was selected randomly, meeting the Random condition for this customer population. The sample is selected without replacement, and \(250\leq0.10(4{,}000)=400\), so the 10% condition is met. Under \(H_0\), the expected number who enabled the setting is \(np_0=250(0.40)=100\); the expected number who did not is \(n(1-p_0)=250(0.60)=150\). Both counts are at least 10, so the Large Counts condition is met.

Do: Here \(x=115\) and \(n=250\), so:

\hat{p}=\frac{115}{250}=0.46 \qquad SE_0=\sqrt{\frac{0.40(0.60)}{250}} =\sqrt{0.00096} \approx0.03098

Using the unrounded standard error:

z=\frac{0.46-0.40}{\sqrt{0.40(0.60)/250}} \approx1.9365

The alternative is right-tailed, so the p-value is \(\text{normalcdf}(1.9365,1\text{E}99,0,1)\approx0.0264\), rounded. Since \(0.0264<0.05\), reject \(H_0\).

Conclude: The sample provides convincing evidence that more than 40% of customers in this population have enabled the accessibility setting. The conclusion is about the population proportion, not just the 46% observed in this sample.

Worked Example: A Two-Sided Test About a Community Proposal

A fictional community group wants to know whether the proportion of registered residents who support a proposed park renovation differs from 50%. A random sample of 160 residents from a population of 3,000 includes 94 supporters. Use \(\alpha=0.01\).

State: Let \(p\) be the true proportion of the 3,000 registered residents who support the proposed renovation. Because the question asks whether the proportion differs from 0.50 in either direction, use \(H_0:p=0.50\) and \(H_a:p\ne0.50\). The significance level is \(\alpha=0.01\).

Plan: Use a one-proportion \(z\)-test. The residents were randomly sampled, meeting the Random condition for this population. Sampling is without replacement, and \(160\leq0.10(3{,}000)=300\), so the 10% condition is met. Under the null, the expected success count is \(np_0=160(0.50)=80\), and the expected failure count is \(n(1-p_0)=160(0.50)=80\). Both are at least 10, so the Large Counts condition is met.

Do: The sample proportion and null standard error are:

\hat{p}=\frac{94}{160}=0.5875 \qquad SE_0=\sqrt{\frac{0.50(0.50)}{160}} =\sqrt{0.0015625} \approx0.03953

Using unrounded values, calculate the test statistic:

z=\frac{0.5875-0.50}{\sqrt{0.50(0.50)/160}} \approx2.2136

For a two-sided alternative, results at least as extreme as \(2.2136\) in either direction count toward the p-value. Thus, \(2\text{normalcdf}(2.2136,1\text{E}99,0,1)\approx0.0269\), rounded. Since \(0.0269>0.01\), fail to reject \(H_0\).

Conclude: At the 0.01 significance level, the sample does not provide convincing evidence that the proportion of registered residents who support the renovation differs from 50%. The p-value is greater than the specified significance level; this does not prove that exactly 50% support the proposal.

Common Mistakes and AP Exam Tips

A good four-step response does not hide its reasoning behind a calculator result. Name the procedure, show the evidence for its conditions, and make the conclusion answer the question that was asked.

Skipping or abbreviating the conditions. “Conditions met” does not show why. Identify the random selection, calculate the 10% comparison when applicable, and show both expected counts under \(H_0\).
Using \(\hat{p}\) for the test’s Large Counts check. For a one-proportion \(z\)-test, check \(np_0\) and \(n(1-p_0)\), not the observed counts \(x\) and \(n-x\). The null model determines the expected counts used by the test.
Using the wrong p-value tail. The direction comes from \(H_a\), not from the sign of the observed result. A two-sided alternative requires both tails; a one-sided alternative does not.
Reporting only “significant” or “not significant.” State the decision using “reject” or “fail to reject” \(H_0\), then describe the evidence in context. If rejecting, say there is convincing evidence for the alternative; if failing to reject, say the data do not provide convincing evidence for it.
Claiming that failing to reject proves the null. A large p-value is not proof that \(H_0\) is true. It means the sample did not provide sufficiently convincing evidence against the null at the stated \(\alpha\).
Leaving out the population in the conclusion. A conclusion should say which population proportion the evidence concerns and what characteristic is being measured.

AP Exam Tip: Make the four steps easy to identify. A full-credit response defines \(p\) and states both hypotheses, names the test and justifies each relevant condition, shows the test statistic and correctly tailed p-value, then compares that p-value with \(\alpha\) and concludes in context.

Key Takeaway

The four-step structure turns a one-proportion test into a connected argument. State the population question, plan with justified conditions, do the calculation under the null model, and conclude by interpreting the evidence at the given significance level.

Key takeaway: A complete one-proportion \(z\)-test response includes the parameter and hypotheses, explicit condition checks, the null-based \(z\) statistic and correctly directed p-value, and a cautious conclusion in context.

Check Your Understanding

For each question, identify what belongs in the relevant step of a one-proportion \(z\)-test response.

A random sample of 300 people is taken without replacement from a population of 2,500. Is the 10% condition met? Show the comparison.
For \(H_0:p=0.30\), \(n=200\), and \(H_a:p<0.30\), state the two expected counts to check for Large Counts and determine whether the condition is met.
A test has \(z=1.72\) and \(H_a:p>p_0\). Which tail gives the p-value, and what calculator command could find it?
A test at \(\alpha=0.05\) has p-value 0.08. Write the decision and a careful general form of the conclusion.
Why is “the null hypothesis is true” not an appropriate conclusion after failing to reject \(H_0\)?