AP® Statistics review sheet from Aim for Five (aimforfive.com/stats/units/3/3-7)
Unit 3 · Topic 3.7
3.7 Carrying Out a Test for a Population Proportion
Carrying out a one-proportion z-test means computing the test statistic and p-value, comparing the p-value with α and writing a conclusion in context. The conclusion is where most points are won or lost.
Key terms
- z test statistic
- significance level (α)
- reject / fail to reject H₀
- conclusion in context
The test statistic
z = (p̂ − p₀) ÷ √(p₀(1 − p₀)/n). It tells you how many standard deviations p̂ is from the null value. This matches the general form on the formula sheet: (statistic − parameter) ÷ (standard deviation of the statistic).
Use p₀ in the denominator, not p̂. The test assumes H₀ is true, so you use the standard deviation p̂ would have if p really were p₀. When H₀ is true, z follows a standard normal distribution.
Significance level and the decision
The significance level α is a cutoff chosen before collecting data, usually 0.05 (sometimes 0.01 or 0.10). It's the probability of rejecting H₀ when H₀ is actually true.
If the p-value ≤ α, reject H₀: the result is statistically significant, and there's convincing evidence for Hₐ. If the p-value > α, fail to reject H₀: there isn't convincing evidence for Hₐ.
If no α is given, use 0.05 and say so.
You can also judge the strength of evidence without a fixed α: a p-value of 0.30 gives essentially no evidence against H₀, 0.04 gives moderate evidence and 0.001 gives very strong evidence.
Writing the conclusion
A full conclusion has three parts: a comparison ("because the p-value of 0.015 is less than α = 0.05"), a decision ("we reject H₀") and a statement in context about Hₐ ("there is convincing evidence that the true proportion of all the company's deliveries that arrive on time is less than 0.90").
Use careful language. Never say "accept H₀" or "prove." When you fail to reject, say "there is not convincing evidence that…" followed by Hₐ in context.
Tests and intervals agree
A two-sided test at α = 0.05 and a 95% confidence interval usually tell the same story. If p₀ is outside the interval, the test will usually reject; if it's inside, the test usually won't. The interval adds something a test can't: a range of plausible values for p.
The four steps
Use these four steps for every test. And remember: statistically significant doesn't mean large or important. It means the result would be unlikely if H₀ were true, so always look at the size of the difference, too.
- State: hypotheses, the parameter defined in context and α.
- Plan: name the test and check the random, 10% and normal conditions with numbers.
- Do: compute p̂, the test statistic and the p-value.
- Conclude: compare the p-value with α, make a decision and state it in context in terms of Hₐ.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Complete one-sided test
Continuing 3.5: in the random sample of 150 deliveries, 127 arrived on time. Test H₀: p = 0.90 vs. Hₐ: p < 0.90 at α = 0.05.
Show the solutionHide the solution
- Step 1: Conditions were checked in 3.5 (random, 10%, np₀ = 135 and n(1 − p₀) = 15).
- Step 2: p̂ = 127/150 ≈ 0.847.
- Step 3: SD under H₀: √(0.90 × 0.10/150) ≈ 0.0245.
- Step 4: z = (0.847 − 0.90)/0.0245 ≈ −2.18.
- Step 5: p-value = P(z ≤ −2.18) ≈ 0.015.
- Step 6: 0.015 < 0.05, so reject H₀.
Answer: z ≈ −2.18, p-value ≈ 0.015. Because 0.015 < α = 0.05, we reject H₀. There is convincing evidence that the true proportion of all the company's deliveries last month that arrived on time is less than 0.90.
- Example 2Calculator allowed
Two-sided test, fail to reject
A school's records show 18% of students were absent at least once last week in previous years. This year, in a random sample of 250 of its 2,700 students, 52 were absent at least once (p̂ = 0.208). Test at α = 0.05 whether the proportion has changed.
Show the solutionHide the solution
- Step 1: H₀: p = 0.18; Hₐ: p ≠ 0.18, where p = the true proportion of all the school's students absent at least once last week.
- Step 2: Random sample; 250 ≤ 270 (10% of 2,700); np₀ = 45 and n(1 − p₀) = 205, both ≥ 10.
- Step 3: SD = √(0.18 × 0.82/250) ≈ 0.0243. z = (0.208 − 0.18)/0.0243 ≈ 1.15.
- Step 4: p-value = 2 × P(z ≥ 1.15) ≈ 0.25.
- Step 5: 0.25 > 0.05, so fail to reject H₀.
Answer: z ≈ 1.15, p-value ≈ 0.25. Because 0.25 > 0.05, we fail to reject H₀. There is not convincing evidence that the true proportion of the school's students absent at least once in a week has changed from 0.18.
Common mistakes
- Using p̂ instead of p₀ in the test statistic's denominator.
- Concluding "we accept H₀" or "the proportion is 0.18." Say there isn't convincing evidence for Hₐ.
- Giving a decision without linking it to the p-value and α.
- Writing a conclusion about the sample instead of the population parameter.
On the exam
- Readers look for the comparison of the p-value with α, the decision and the context. A correct decision with no link to the p-value usually loses credit.
- Show the test statistic formula with numbers substituted, even if technology does the arithmetic.
Connected topics
Videos
Check yourself
3 questions on 3.7 Carrying Out a Test for a Population Proportion. Pick an answer to see if you got it, and why.
A city's mayor claims that 60% of the city's voters approve of the job she is doing. A reporter suspects the true percent is lower. In a random sample of 400 city voters, 222 (55.5%) say they approve.
Described scenario with invented results
What are the test statistic and p-value?
What conclusion should the reporter make at α = 0.05?
Would the conclusion change at α = 0.01?
0 of 3 answered