One-Proportion Z-Test, Step by Step (Worked Example)

The four steps of a one-proportion z-test

A one-proportion z-test asks whether a single population proportion equals some claimed value. Every version of the test runs the same four steps, and AP scoring awards points one step at a time.

1. State the parameter and the two hypotheses, both about the population proportion.
2. Check three conditions: random, 10%, and large counts, using the null value.
3. Compute the standardized test statistic and the p-value.
4. Conclude by comparing the p-value to $\alpha$ and writing a sentence in context.

One example runs through this entire page. A game designer says a redesigned puzzle is a coin flip: half of first-time players solve it. In a random sample of 100 first-time players, 62 solve it. So $n = 100$, the count of successes is 62, and the hypothesized value is $p_0 = 0.5$ (read "p-naught").

Two symbols do most of the work. $p$ is the population proportion, the unknown number the claim is about. $\hat{p}$ (read "p-hat") is the sample proportion, the fraction of successes you actually observed. Keeping those two apart is what the rest of this page is about.

Step 1: write hypotheses about p, never about p-hat

Define the parameter in words first, then write the two hypotheses.

Let $p$ be the true proportion of all first-time players who solve the redesigned puzzle. Then

The null hypothesis $H_0$ (read "H-naught") sets $p$ equal to the claimed value $p_0$. The alternative $H_a$ states what you are collecting evidence for. Use $\neq$ when the question asks whether the proportion differs, and $>$ or $<$ when the question names a direction.

Never write $H_0: \hat{p} = 0.5$. You already know $\hat{p}$ exactly: it is 0.62 in this sample. A hypothesis is a claim about something unknown, and the only unknown here is $p$. Guidance on wording both hypotheses is in null vs alternative hypothesis.

Pick the direction from the question before you look at the data. Choosing $H_a: p > 0.5$ because $\hat{p}$ landed above 0.5 is circular reasoning, and it halves your p-value for free. The AP course covers this under setting up a test for a population proportion.

Step 2: check the three conditions with the null value

The one-sample z-test for a population proportion requires three conditions, and the AP course lists them explicitly.

• Random. The data come from a random sample or a randomized experiment.
• 10%. When sampling without replacement, the population is at least 10 times the sample size, so $n \leq 0.10N$.
• Large counts (normality). The expected counts under the null hypothesis are at least 10: $np_0 \geq 10$ and $n(1 - p_0) \geq 10$.

For the puzzle example, the 100 players are a random sample, which settles the first condition. They are fewer than 10% of all first-time players as long as the game has at least 1,000 of them. For large counts, $np_0 = 100(0.5) = 50 \geq 10$ and $n(1 - p_0) = 100(0.5) = 50 \geq 10$.

Both counts use $p_0$, not $\hat{p}$. A test assumes the null hypothesis is true while it runs, so every quantity inside the test is computed under that assumption. A confidence interval assumes no null value, which is why its version of this condition uses the observed counts $n\hat{p}$ and $n(1-\hat{p})$ instead. Why 10 successes and 10 failures explains where the threshold comes from.

Show the arithmetic. Writing "large counts met" without the two products is the most common way to lose the condition point.

Step 3: compute the test statistic

The standardized test statistic counts how many standard deviations $\hat{p}$ sits from $p_0$:

Work the denominator first. It is the standard deviation of the sampling distribution of $\hat{p}$ when $H_0$ is true, and it is the piece students get wrong most often.

For the puzzle example, $\hat{p} = \frac{62}{100} = 0.62$ and

Then

A z of 2.4 says the observed sample proportion sits 2.4 standard deviations above the value the null hypothesis predicts. The proportion z-test calculator reproduces this value if you want to check your work.

Step 4: find the p-value and compare it to alpha

The p-value is the probability, computed assuming $H_0$ is true, of a test statistic at least as extreme as the one you got. The tail you use comes from $H_a$.

• $H_a: p > p_0$ gives $P(Z \geq z)$.
• $H_a: p < p_0$ gives $P(Z \leq z)$.
• $H_a: p \neq p_0$ gives $P(Z \leq -|z|) + P(Z \geq |z|)$, which is twice one tail.

The puzzle test is two-sided, so use both tails. From the z-table, $P(Z < 2.4) = 0.9918$, so one tail is $1 - 0.9918 = 0.0082$ and the p-value is $2(0.0082) = 0.0164$.

Now compare to the significance level $\alpha$ (the Greek letter alpha), the false alarm rate you fixed in advance. With $\alpha = 0.05$ you have $0.0164 < 0.05$, so reject $H_0$. For what that number does and does not say, see what does a p-value mean.

Writing the conclusion in context

A conclusion needs three pieces: the comparison, the decision, and the context. Dropping the context is what turns a full-credit answer into a partial one.

Template. Because the p-value of 0.0164 is less than $\alpha = 0.05$, we reject $H_0$. There is convincing statistical evidence that the true proportion of first-time players who solve the redesigned puzzle differs from 0.5.

If the p-value had come out above $\alpha$, the wording changes to "we fail to reject $H_0$" and "there is not convincing statistical evidence that ...". You never accept $H_0$, and you never say the proportion equals $p_0$. A large p-value means the data are consistent with $p_0$, not that $p_0$ is correct.

You can also interpret the p-value itself. If the true solve rate really were 0.5, about 1.64% of random samples of 100 players would produce a sample proportion at least as far from 0.5 as 0.62 is, in either direction.

The mistake that costs the most points: p-hat in the standard error

The single most common error on this test is using $\hat{p}$ in place of $p_0$ inside the square root. The whole procedure runs under the assumption that $H_0$ is true, so the standard deviation of $\hat{p}$ has to be built from the null value: $\sqrt{p_0(1 - p_0)/n}$.

Watch what the swap does here. The correct denominator is $\sqrt{0.5(0.5)/100} = 0.05$, giving $z = 2.4$ and a two-sided p-value of 0.0164. The wrong denominator is $\sqrt{0.62(0.38)/100} = \sqrt{0.002356} \approx 0.04854$, giving $z \approx 2.47$ and a two-sided p-value of about 0.0135.

Both versions reject at $\alpha = 0.05$, so this time the mistake hides. It does not always, and it does not even push in a fixed direction. Take $n = 200$, $p_0 = 0.30$, and $\hat{p} = 0.37$: the correct statistic is $z \approx 2.16$ with a one-sided p-value of 0.0154, while the $\hat{p}$ version gives $z \approx 2.05$ and 0.0202. Near a cutoff, a gap that size flips the decision.

The rule worth memorizing is short: **tests use $p_0$, intervals use $\hat{p}$.** An interval has no null value to borrow, so it must estimate the standard error from the sample. That difference is also why a confidence interval and a p-value can disagree in borderline cases. When the claim involves two groups rather than one, switch procedures entirely (one-proportion vs two-proportion z-test).