A local news station claims that 60% of teens in your town get their news online rather than from TV or print. You're skeptical — it feels higher than that to you. So you survey a random sample of 350 teens, and 230 of them say they get their news online. That's about 65.7%.
Here's the question that makes or breaks AP Statistics: Is 65.7% convincingly higher than 60%, or could a sample this far off easily happen by random chance even if the true proportion really is 60%?
In Lesson 21 you learned the logic of hypothesis testing — null vs. alternative, p-values, significance levels. Now we make it concrete for proportions. You'll learn to turn "I'm skeptical" into a number, the test statistic, and then into a p-value that measures how surprising your data would be if the claim were true.
One warning before we start, because it's the #1 mistake on this entire topic: the standard error for a test uses the claimed value p0, not your sample p̂. Keep that in your back pocket. We'll hammer it home.
We use a one-sample z-test for a proportion when we want to test a claim about a single population proportion p. The data is one categorical variable (success/failure), summarized by a count of successes x out of a sample size n, giving the sample proportion p̂ = x/n.
The whole procedure follows PANIC: Parameter, Assumptions, Name the procedure, Interval/test, Conclusion in context. Let's run the warm-up scenario through it.
Define the parameter in words: let p = the true proportion of all teens in the town who get their news online.
State the hypotheses:
H0: p = 0.60 — the claim, the "nothing new here" baseline.Ha: p > 0.60 — what we suspect (we thought it was higher).The claimed value, 0.60, is called the null value and we label it p0. The direction of Ha comes from the question, not from the data. Here we suspected "higher," so it's a one-sided (right-tailed) test.
Three conditions, every time:
n ≤ 0.10N. There are surely more than 350 × 10 = 3,500 teens in the town, so ✓.p0, the null value, to check that the expected numbers of successes and failures are both at least 10: - np0 = 350(0.60) = 210 ≥ 10 ✓
- n(1 − p0) = 350(0.40) = 140 ≥ 10 ✓
Because we assume H0 is true while testing, we use p0 here — not p̂.
One-sample z-test for a proportion,
Ha: p > 0.60,α = 0.05.
The test statistic measures how many standard errors p̂ sits from the claimed p0:
z = (p̂ − p0) / sqrt( p0(1 − p0) / n )
The denominator is the standard error under the null. Notice the formula uses p0 inside the square root — because we build the sampling distribution assuming the null is true. This is the crux of the lesson:
TEST (this lesson): SE = sqrt( p0(1 − p0) / n ) ← uses the CLAIMED p0
INTERVAL (Lesson 20): SE = sqrt( p̂(1 − p̂) / n ) ← uses the SAMPLE p̂Plug in p̂ = 230/350 = 0.65714, p0 = 0.60, n = 350:
SE = sqrt( 0.60 × 0.40 / 350 ) = sqrt( 0.24 / 350 ) = sqrt(0.00068571) = 0.026186
z = (0.65714 − 0.60) / 0.026186 = 0.05714 / 0.026186 = 2.182Now the p-value: the probability of getting a z this extreme or more so, in the direction of Ha, if H0 were true. Since Ha: p > 0.60 is right-tailed, we want the area to the right of z = 2.182:
TI-84: normalcdf(2.182, 1E99, 0, 1) = 0.0145So p-value ≈ 0.0146. (Reading off the standard normal, the area above z = 2.18 is about 0.0146 — a small probability.)
TI-84: STAT → TESTS → 1-PropZTest
Input: p0 = 0.60
x = 230
n = 350
prop: > p0 (right tail, matches Ha)
Calculate
Output: z = 2.1822
p = 0.0145
p̂ = 0.6571429The calculator's z = 2.182 and p = 0.0145 match our by-hand work exactly. (Note: x must be a whole number. If a problem gives you a percentage, compute x = p̂ × n and round to the nearest integer before entering it.)
Compare the p-value to α = 0.05:
Because the p-value ≈ 0.0146 < α = 0.05, we reject H0. There is convincing statistical evidence that the true proportion of teens in the town who get their news online is greater than 0.60.
A few rules for the conclusion that AP readers insist on:
That's the entire procedure. Every one-proportion test you'll ever do is this same five-step PANIC march. Change the numbers, change the tail, but the skeleton never moves.
Problem. A vaccine manufacturer claims that at most 45% of a certain age group will accept a new flu shot. A public-health researcher believes the acceptance rate is actually higher. In a random sample of 150 adults from this group, 78 say they would accept the shot. Test at α = 0.05.
P. Let p = the true proportion of adults in this age group who would accept the shot.
H0: p = 0.45 vs. Ha: p > 0.45. (One-sided, right tail — the researcher suspects higher.)
A. Random: random sample ✓. 10%: 150 is surely ≤ 10% of all adults in the age group ✓. Large Counts (use p0): np0 = 150(0.45) = 67.5 ≥ 10 ✓ and n(1 − p0) = 150(0.55) = 82.5 ≥ 10 ✓.
N. One-sample z-test for a proportion.
I. p̂ = 78/150 = 0.520.
SE = sqrt(0.45 × 0.55 / 150) = sqrt(0.001650) = 0.040620
z = (0.520 − 0.45) / 0.040620 = 0.070 / 0.040620 = 1.723
p-value = normalcdf(1.723, 1E99, 0, 1) = 0.0424
1-PropZTest (p0=.45, x=78, n=150, prop >p0): z = 1.7233, p = 0.0424C. Since p-value ≈ 0.0424 < 0.05, we reject H0. There is convincing evidence that more than 45% of adults in this age group would accept the flu shot.
Problem. A coin is suspected of being unfair — in either direction. It's flipped 80 times and lands heads 34 times. Is there convincing evidence at α = 0.05 that the coin is not fair?
P. Let p = the true probability the coin lands heads.
H0: p = 0.50 vs. Ha: p ≠ 0.50. ("Not fair, in either direction" → two-sided.)
A. Random: flips are independent and identically distributed ✓ (10% condition is unnecessary for a physical process — flips aren't drawn from a finite population, so we just need independence, which physical coin flips give us). Large Counts: np0 = 80(0.50) = 40 ≥ 10 ✓, n(1 − p0) = 40 ≥ 10 ✓.
N. One-sample z-test for a proportion.
I. p̂ = 34/80 = 0.425.
SE = sqrt(0.50 × 0.50 / 80) = sqrt(0.003125) = 0.055902
z = (0.425 − 0.50) / 0.055902 = −0.075 / 0.055902 = −1.342Because Ha is two-sided, the p-value is both tails — double the one-tail area:
p-value = 2 × normalcdf(−1E99, −1.342, 0, 1) = 2 × 0.0899 = 0.1797
1-PropZTest (p0=.50, x=34, n=80, prop ≠p0): z = −1.3416, p = 0.1797C. Since p-value ≈ 0.180 > 0.05, we fail to reject H0. There is not convincing evidence that the coin is unfair. (We do not conclude the coin is fair — absence of evidence is not evidence of absence.)
Problem. A nutritionist tests H0: p = 0.30 vs. Ha: p < 0.30 with a random sample of n = 40, where p0 = 0.30. (a) Are the Large Counts conditions met? (b) The test gives a p-value of 0.018. Interpret this p-value in context, where p is the proportion of customers who order dessert.
Solution. (a) np0 = 40(0.30) = 12 ≥ 10 ✓ and n(1 − p0) = 40(0.70) = 28 ≥ 10 ✓ — both met. (b) Interpretation: If the true proportion of customers who order dessert really were 0.30, there is only about a 0.018 probability of getting a sample proportion as low as — or lower than — the one observed, just by random sampling variation. Because that's small, it's evidence against H0.
A p-value is P(data this extreme or more | H0 true) — it is not the probability that H0 is true.
Problem. In a random sample of 200 voters, 112 support a ballot measure. (a) Test H0: p = 0.50 vs. Ha: p ≠ 0.50 at α = 0.05. (b) A 95% confidence interval from the same data is (0.491, 0.629). Explain how the interval is consistent with your test decision.
Solution.
(a) p̂ = 112/200 = 0.560. Test SE uses p0 = 0.50:
SE = sqrt(0.50 × 0.50 / 200) = 0.035355
z = (0.560 − 0.50) / 0.035355 = 1.697
p-value = 2 × normalcdf(1.697, 1E99, 0, 1) = 2 × 0.0448 = 0.0897
1-PropZTest (p0=.50, x=112, n=200, prop ≠p0): z = 1.6971, p = 0.0897Since 0.0897 > 0.05, fail to reject H0 — not convincing evidence that p ≠ 0.50.
(b) The 95% confidence interval (0.491, 0.629) contains 0.50. For a two-sided test at α = 0.05, "the C% interval contains p0" corresponds to "fail to reject H0 at α = 1 − C." Since 0.50 is a plausible value inside the interval, we don't reject it — perfectly consistent. (Heads-up: the interval's SE uses p̂ = 0.56, while the test's SE uses p0 = 0.50, so the agreement is close but not algebraically perfect. The directional logic always matches; borderline cases can occasionally differ.)
SE = sqrt(p0(1−p0)/n) — the claimed value. Students who plug in p̂ get the wrong z and a wrong p-value. (The interval is the one that uses p̂.) Memorize: test → p0, interval → p̂.Ha: p > p0, take the area to the right of z. If Ha: p < p0, area to the left. A negative z with Ha: p > should make you suspicious — your data went the wrong way and the p-value will be large.Ha: p ≠ p0, you must double the one-tail area: p-value = 2 × P(Z > |z|). Forgetting to double makes the result look twice as significant as it is. Conversely, don't double a one-sided test.MC 1. For a one-proportion z-test of H0: p = 0.40, which is the correct standard error?
sqrt(0.40 × 0.60 / n)sqrt(p̂(1 − p̂) / n)sqrt(0.40 × 0.60 × n)0.40 / sqrt(n)MC 2. A test has Ha: p > 0.25 and yields z = 1.88. The p-value is closest to:
MC 3. A two-sided test gives z = −2.05. The p-value is closest to:
MC 4. Which condition for a one-proportion z-test is checked using the null value p0 rather than the sample proportion?
np0 ≥ 10 and n(1 − p0) ≥ 10)MC 5. A researcher tests H0: p = 0.50 vs. Ha: p ≠ 0.50 and obtains a p-value of 0.12 at α = 0.05. The correct conclusion is:
p ≠ 0.50.MC 6. In a one-proportion z-test, decreasing the p-value (making it smaller) corresponds to:
MC 7. A 1-PropZTest with Ha: p < p0 returns z = 0.90. Without computing, the p-value is:
MC 8. A manufacturer claims 5% of its chips are defective. A quality inspector suspects the rate is higher and samples 400 chips, finding 28 defective. The standard error used in the test statistic is:
sqrt(0.07 × 0.93 / 400)sqrt(0.05 × 0.95 / 400)sqrt(0.05 × 0.95 / 28)sqrt(0.07 × 0.93 / 28)Short Answer 9 (in context). A polling firm claims that 70% of adults support a new park. A city council member suspects support is lower. In a random sample of 240 adults, 168 support the park. Carry out a full PANIC test at α = 0.05. (This is the FRQ scenario worked below — try it first, then check.)
Short Answer 10 (in context). Explain, in one or two sentences, why a one-proportion test uses p0 in its standard error while a one-proportion confidence interval uses p̂.
Short Answer 11. A two-sided test gives z = 1.40. Find the p-value and state the decision at α = 0.05.
Short Answer 12. A right-tailed test (Ha: p > 0.30) of a random sample with n = 200, x = 72 is run. Check Large Counts, compute z, and find the p-value.
Statistical Practices 3 (Analyze) + 4 (Interpret)
A coffee shop's loyalty program advertises that 75% of its customers return within a week of their first visit. A regional manager suspects the true return rate is lower than advertised. She takes a random sample of 240 first-time customers and finds that 168 of them returned within a week.
(a) State the hypotheses for an appropriate significance test, defining any parameter. (2 pts)
(b) Identify the appropriate test and verify that the conditions for inference are met. (3 pts)
(c) Calculate the test statistic and the p-value. (3 pts)
(d) At the
α = 0.05significance level, state your conclusion in the context of the problem. (2 pts)
(a) Parameter and hypotheses.
Let p = the true proportion of all first-time customers at this coffee shop who return within a week.
H0: p = 0.75vs.Ha: p < 0.75
The alternative is one-sided (left-tailed) because the manager suspects the rate is lower than the advertised 0.75.
(b) Procedure and conditions.
Procedure: a one-sample z-test for a proportion.
240 ≤ 0.10N means N ≥ 2400, which is reasonable). ✓np0 = 240(0.75) = 180 ≥ 10 and n(1 − p0) = 240(0.25) = 60 ≥ 10. Both ≥ 10. ✓(c) Test statistic and p-value.
p̂ = 168/240 = 0.700. Standard error uses the null value p0 = 0.75:
SE = sqrt( 0.75 × 0.25 / 240 ) = sqrt(0.00078125) = 0.027951
z = (0.700 − 0.75) / 0.027951 = −0.050 / 0.027951 = −1.789Because Ha: p < 0.75 is left-tailed, the p-value is the area to the left of z = −1.789:
p-value = normalcdf(−1E99, −1.789, 0, 1) = 0.0368
1-PropZTest (p0=.75, x=168, n=240, prop <p0): z = −1.7889, p = 0.0368, p̂ = 0.70(d) Conclusion in context.
Because the p-value ≈ 0.0368 < α = 0.05, we reject H0. There is convincing statistical evidence that the true proportion of first-time customers who return within a week is less than 0.75 (lower than the advertised 75%).
| Part | Criterion | Points |
|---|---|---|
| (a) | Correctly defines parameter p in context (the true proportion of first-time customers who return) | 1 |
| (a) | Correct hypotheses H0: p = 0.75, Ha: p < 0.75 (left-tailed) | 1 |
| (b) | Names the one-sample z-test for a proportion | 1 |
| (b) | States Random and 10% conditions and ties them to the context | 1 |
| (b) | Checks Large Counts using p0: np0 = 180 ≥ 10 and n(1−p0) = 60 ≥ 10 | 1 |
| (c) | Correct SE using p0 and correct test statistic z ≈ −1.79 | 1 |
| (c) | Correct tail / setup for the p-value (left tail) | 1 |
| (c) | Correct p-value ≈ 0.037 | 1 |
| (d) | Correct decision linked to comparing p-value with α (0.037 < 0.05 → reject H0) | 1 |
| (d) | Conclusion in context with direction (evidence that p < 0.75) | 1 |
| Total | 10 |
p̂ = 0.70 instead of p0 = 0.75 in the Large Counts check — costs the Large Counts point. (Here both still pass, but the method is wrong and readers penalize it.)p̂ in the SE instead of p0 — the single most common error; produces z ≈ −1.74 and the wrong p-value, and forfeits the SE/test-statistic point.z = −1.79, p = 0.037 with no formula or labeled conditions can cap part (c) — show the SE setup with p0 and label which calculator routine you used.MC 1 — (A). A test uses SE = sqrt(p0(1−p0)/n) = sqrt(0.40 × 0.60 / n). (B) is the interval's SE (uses p̂) — the classic trap. (C) multiplies by n instead of dividing. (D) is not a standard error at all.
MC 2 — (A). Right-tailed: p-value = normalcdf(1.88, 1E99) = 0.0301. (B) doubles it (that's the two-sided value). (C) and (D) give the area to the left of z — wrong tail.
MC 3 — (B). Two-sided: 2 × normalcdf(−1E99, −2.05) = 2 × 0.0202 = 0.0404. (A) is the one-tail area (forgot to double). (C)/(D) are left-over complements — wrong region entirely.
MC 4 — (C). Large Counts uses p0 because we test under the assumption H0 is true. (A) Random and (B) 10% don't involve any proportion value. (D) is not a condition for this procedure.
MC 5 — (C). 0.12 > 0.05 → fail to reject H0; not convincing evidence that p ≠ 0.50. (A)/(B) wrongly reject. (D) says "accept H0," which is never correct.
MC 6 — (B). A smaller p-value = stronger evidence against H0 (data more surprising under the null). (A) is backwards. (C) and (D) are unrelated to p-value size.
MC 7 — (C). Ha: p < p0 wants the left tail, but z = +0.90 is on the right side — data went against Ha — so the left-tail p-value is large: normalcdf(−1E99, 0.90) = 0.816. (A)/(B)/(D) all understate it.
MC 8 — (B). Test SE uses the claimed p0 = 0.05: sqrt(0.05 × 0.95 / 400). (A)/(D) use p̂ = 28/400 = 0.07 — the interval's value, wrong for a test. (C)/(D) wrongly use 28 instead of 400 in the denominator.
SA 9. This is the FRQ scenario. Full solution: p = true proportion of first-time customers who return within a week; H0: p = 0.75, Ha: p < 0.75. Conditions: Random ✓, 10% (N ≥ 2400) ✓, Large Counts np0 = 180 ≥ 10, n(1−p0) = 60 ≥ 10 ✓. p̂ = 0.700; SE = sqrt(0.75×0.25/240) = 0.027951; z = (0.700 − 0.75)/0.027951 = −1.789; left-tailed p-value = 0.0368. Since 0.0368 < 0.05, reject H0 — convincing evidence the return rate is below 75%. (1-PropZTest confirms z = −1.7889, p = 0.0368.)
SA 10. A test assumes H0 is true, so we build the null sampling distribution using the hypothesized value p0 — hence SE = sqrt(p0(1−p0)/n). A confidence interval makes no claim about p; it estimates p from the data, so it uses the best available estimate p̂ — hence SE = sqrt(p̂(1−p̂)/n).
SA 11. Two-sided: p-value = 2 × normalcdf(1.40, 1E99) = 2 × 0.0808 = 0.1615. Since 0.1615 > 0.05, fail to reject H0 — not convincing evidence that p ≠ p0.
SA 12. p̂ = 72/200 = 0.360. Large Counts: np0 = 200(0.30) = 60 ≥ 10 ✓, n(1−p0) = 200(0.70) = 140 ≥ 10 ✓. SE = sqrt(0.30×0.70/200) = sqrt(0.00105) = 0.032404; z = (0.360 − 0.30)/0.032404 = 1.852; right-tailed p-value = normalcdf(1.852, 1E99) = 0.0320. At α = 0.05, reject H0 — evidence that p > 0.30.
See the model response and rubric in section (f). Key recomputed values: p̂ = 0.700, SE = sqrt(0.75 × 0.25 / 240) = 0.027951 (uses p0 = 0.75), z = −1.789, left-tailed p-value = 0.0368. Verified against 1-PropZTest: z = −1.7889, p = 0.0368. Decision: 0.0368 < 0.05 → reject H0; convincing evidence the return rate is below 0.75.
StatsIQ · Lesson 22 of 30 · Unit 3: Inference for Categorical Data: Proportions · Phase 4
This lesson is study material aligned to the new 2026–27 AP Statistics Course and Exam Description (first exam May 2027). "AP" is a trademark of the College Board, which was not involved in the production of and does not endorse this product.
Accuracy review: All test statistics and p-values were recomputed independently and confirmed against TI-84 1-PropZTest output. The test standard error uses the null value p0 throughout, distinct from the confidence-interval SE (which uses p̂). Reviewed for statistical accuracy by a retired actuary.