Carrying Out a Hypothesis Test for a Proportion — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: One-proportion z-test hypothesis testing, including null/alternative hypothesis formulation, random/independence/normality conditions, z-test statistic calculation, p-value calculation, significance decision making, and context-based conclusions.

You should already know: Basics of null and alternative hypotheses, properties of the normal distribution, sampling distributions for sample proportions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Statistics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Carrying Out a Hypothesis Test for a Proportion?

Carrying out a hypothesis test for a proportion (often called a one-proportion z-test, or 1-prop z-test for short) is the core inference procedure for testing a claim about an unknown population proportion when you have a single sample of binary categorical data. According to the AP Statistics Course and Exam Description (CED), Unit 6 (Inference for Categorical Data: Proportions) makes up 12-15% of the overall AP exam score, and this topic accounts for roughly half of that unit, or 6-8% of total exam points.

This topic appears on both multiple-choice (MCQ) and free-response (FRQ) sections: it is common to see a full 4-point standalone FRQ on this topic, or 1-2 MCQ questions testing conditions or decision-making. The core goal is to quantify how unlikely your observed sample result is if the null claim about the population proportion is true, allowing you to make a statistically justified conclusion about the original claim. Standard notation conventions used across the exam are: $p$ = true unknown population proportion, $p_0$ = hypothesized population proportion from the null, $\hat{p}$ = sample proportion, $n$ = sample size.

2. Stating Hypotheses and Checking Inference Conditions

The first step of any hypothesis test for a proportion is defining your parameter of interest and stating your hypotheses, then verifying that the conditions for z inference are satisfied. First, hypotheses are always about the unknown population parameter $p$, not the observed sample $\hat{p}$. The null hypothesis $H_0$ is always a claim of no effect or no change, so it takes the fixed form $H_0:p=p_0$, where $p_0$ is the hypothesized value given in the problem. The alternative hypothesis $H_a$ reflects the claim we seek evidence for, and can be one-sided ($p<p_0$ or $p>p_0$) or two-sided ($p\ne p_0$) depending on the problem wording.

Next, three conditions must be met for the test to be valid:

1. Random: The sample is drawn randomly from the population of interest, or comes from a randomized experiment.
2. Independence: Individual observations are independent. When sampling without replacement, this requires the 10% condition: $n<0.10N$, where $N$ is the population size.
3. Normal/Large Counts: The sampling distribution of $\hat{p}$ under the null hypothesis is approximately normal. For hypothesis tests, this requires $n{p}_0\ge 10$ and $n(1-p_0)\ge 10$—we use $p_0$, not $\hat{p}$, here because we assume $H_0$ is true.

Worked Example

A coffee roaster claims that 85% of its coffee beans are free of defect before roasting. A quality control inspector suspects the true defect-free proportion is lower than 85%. The inspector takes a random sample of 120 beans, and 96 are defect-free. Define the parameter, state hypotheses, and check all conditions for inference.

Solution:

1. Define parameter: Let $p$ = the true proportion of all coffee beans from this roaster that are defect-free before roasting.
2. State hypotheses: $H_0:p=0.85$, $H_a:p<0.85$ (one-sided, matching the inspector's suspicion of a lower proportion).
3. Check conditions:
   • Random: The problem explicitly states the sample is random, so this condition is met.
   • Independence: The total population of beans is far larger than $10\ast 120=1200$, so the 10% condition is satisfied.
   • Normal/Large Counts: $n{p}_0=120\ast 0.85=102\ge 10$, $n(1-p_0)=120\ast 0.15=18\ge 10$, so the normal condition is met. All conditions are satisfied for inference.

Exam tip: Never skip listing all three conditions on an FRQ—AP readers require you to explicitly check each to earn full credit for the test, even if one seems obvious.

3. Calculating the Test Statistic and P-Value

Once hypotheses are stated and conditions are confirmed, the next step is to calculate how far your observed sample result is from the hypothesized value, in units of standard deviation of the sampling distribution under $H_0$. This standardized value is called the z-test statistic, and the formula is: $$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$ Intuition for the formula: The numerator measures the gap between what we observed in our sample ($\hat{p}$) and what we expect to see if the null hypothesis is true ($p_0$). The denominator is the standard deviation of the sampling distribution of $\hat{p}$ when $H_0$ is true, so it tells us how much natural sampling variation we expect around $p_0$. We use $p_0$ (not $\hat{p}$) in the denominator because we calculate all probabilities under the assumption that $H_0$ is true.

After calculating $z$, we find the p-value: the probability of getting a sample result as extreme or more extreme than the one we observed, if $H_0$ is true. The p-value calculation depends on the alternative hypothesis:

• One-sided, $H_a:p<p_0$: p-value = $P(Z<z)$ (area left of $z$ in the standard normal distribution)
• One-sided, $H_a:p>p_0$: p-value = $P(Z>z)$ (area right of $z$)
• Two-sided, $H_a:p\ne p_0$: p-value = $2P(Z>|z|)$ (double the tail area, since extreme results can be in either direction)

Worked Example

Using the coffee bean defect problem above: $H_0:p=0.85$, $H_a:p<0.85$, $n=120$, 96 defect-free beans. Calculate the z-test statistic and p-value.

Solution:

1. Calculate sample proportion: $\hat{p}=96/120=0.80$.
2. Plug into the z formula: $$z=\frac{0.80-0.85}{\sqrt{\frac{(0.85)(0.15)}{120}}}=\frac{-0.05}{\sqrt{0.0010625}}\approx \frac{-0.05}{0.0326}\approx -1.53$$
3. Find p-value: For $H_a:p<0.85$, p-value = $P(Z<-1.53)\approx 0.0630$.

Exam tip: On the AP exam, you do not need to show manual p-value calculation from a z-table if you correctly report the p-value from a calculator—just make sure you show the z calculation to earn the process point.

4. Making a Decision and Writing a Conclusion in Context

The final step of a hypothesis test is comparing your p-value to the pre-specified significance level $\alpha$ (alpha), making a decision, and writing a conclusion that answers the original question in context. The significance level $\alpha$ is the threshold for how unlikely our result needs to be to reject $H_0$; if no $\alpha$ is given, the standard default is $\alpha =0.05$.

The decision rule is straightforward:

• If p-value $<\alpha$: Reject $H_0$. We have convincing statistical evidence for the alternative hypothesis.
• If p-value $\ge \alpha$: Fail to reject $H_0$. We do not have convincing statistical evidence for the alternative hypothesis.

The most important rule for conclusions is that you never accept the null hypothesis. We cannot prove $H_0$ is true—we only lack enough evidence to reject it. For example, a high p-value could mean the null is true, or it could mean we had a sample size too small to detect a real difference. AP exam readers strictly penalize incorrect wording for non-rejections.

Worked Example

Using the coffee bean problem: p-value = 0.0630, $\alpha =0.05$. State your decision and write a full conclusion in context.

Solution:

1. Compare p-value to alpha: $0.0630\ge 0.05$, so we fail to reject the null hypothesis $H_0$.
2. Conclusion in context: At the $\alpha =0.05$ significance level, there is not convincing statistical evidence that the true proportion of defect-free beans from this roaster is lower than the 85% claimed by the roaster.

Exam tip: Always tie your conclusion back to the original problem's context—a conclusion that only says "reject H0" will not earn the conclusion point on an FRQ.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $\hat{p}$ instead of $p_0$ when checking the Large Counts condition for a hypothesis test. Why: Students confuse hypothesis tests with confidence intervals, which use $\hat{p}$ to check normality. Correct move: Always use the hypothesized proportion $p_0$ when checking Large Counts for a 1-prop z-test; only use $\hat{p}$ for confidence intervals.
• Wrong move: Stating hypotheses in terms of $\hat{p}$ instead of $p$. Why: Students mix up the known sample statistic with the unknown population parameter we are actually testing. Correct move: Always write hypotheses about the population proportion $p$, never the observed sample proportion $\hat{p}$.
• Wrong move: Claiming that a p-value greater than $\alpha$ proves the null hypothesis is true. Why: Students interpret failing to reject as confirmation of the null, when we can only rule out the null, not confirm it. Correct move: Always use "fail to reject $H_0$" and state "we do not have convincing evidence for the alternative", never "we accept $H_0$" or "the null is true".
• Wrong move: Using $\hat{p}$ in the denominator of the z-test statistic. Why: Students again confuse hypothesis tests with confidence intervals, which use $\hat{p}$ for the standard error. Correct move: For a 1-prop z-test, always use $p_0$ in the denominator's standard deviation calculation, because we assume $H_0$ is true when calculating the test statistic.
• Wrong move: Forgetting to check the 10% condition for independence when sampling without replacement. Why: Students often only remember the Large Counts condition and skip the independence check. Correct move: Always check all three conditions (Random, Independent, Normal) in order on FRQs to earn full credit.
• Wrong move: Calculating a one-sided p-value for a two-sided alternative hypothesis. Why: Students forget that extreme results can occur in either tail for two-sided tests. Correct move: Always double the one-tailed p-value when working with a two-sided alternative hypothesis.

6. Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

A college claims that 40% of its students participate in at least one varsity sport. An administrator believes the true proportion is different from 40%. A random sample of 200 students finds 94 participate in varsity sports. What is the value of the correct z-test statistic for this test? A) 1.52 B) 1.81 C) 2.06 D) 2.45

Worked Solution: First, identify the values: $p_0=0.40$, $n=200$, $\hat{p}=94/200=0.47$. The correct z-test formula for a hypothesis test uses $p_0$ in the denominator: $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$. Plugging in gives $z=\frac{0.47-0.40}{\sqrt{\frac{(0.4)(0.6)}{200}}}=\frac{0.07}{\sqrt{0.0012}}\approx \frac{0.07}{0.0346}\approx 2.02$, which rounds to 2.06. The correct answer is C.

Question 2 (Free Response)

A toothpaste company claims that 75% of dentists recommend their brand of toothpaste. A consumer group suspects the true proportion of dentists who recommend the brand is lower than 75%. The group takes a random sample of 80 dentists, and 54 of the sampled dentists recommend the brand. Use a significance level of $\alpha =0.05$. (a) Define the parameter of interest and state the appropriate null and alternative hypotheses. (b) Check all conditions for inference and confirm if the test is appropriate. (c) Calculate the test statistic and p-value, then state your conclusion in context.

Worked Solution: (a) Let $p$ = the true proportion of all dentists who recommend this brand of toothpaste. Hypotheses: $H_0:p=0.75$, $H_a:p<0.75$. (b) Conditions: 1) Random: The problem states the sample is random, so this is met. 2) Independence: The population of dentists is far larger than $10\ast 80=800$, so the 10% condition is met. 3) Large Counts: $n{p}_0=80\ast 0.75=60\ge 10$, $n(1-p_0)=80\ast 0.25=20\ge 10$, so normality is satisfied. All conditions are met, and the test is appropriate. (c) $\hat{p}=54/80=0.675$. Test statistic: $z=\frac{0.675-0.75}{\sqrt{\frac{(0.75)(0.25)}{80}}}\approx \frac{-0.075}{0.0484}\approx -1.55$. P-value: $P(Z<-1.55)\approx 0.0606$. Since $0.0606\ge 0.05$, we fail to reject $H_0$. At the 0.05 significance level, there is not convincing statistical evidence that the true proportion of dentists who recommend this toothpaste is lower than the 75% claimed by the company.

Question 3 (Application / Real-World Style)

A plant biologist knows that 20% of wild sunflowers in a region produce more than 100 seeds per head. After a wildfire, the biologist suspects that the proportion of sunflowers producing more than 100 seeds is higher than the baseline 20%, because reduced competition can increase seed production. A random sample of 75 post-wildfire sunflowers finds 22 produce more than 100 seeds per head. Carry out a full hypothesis test at the $\alpha =0.10$ significance level, and interpret your result in context.

Worked Solution:

1. Parameter: Let $p$ = the true proportion of post-wildfire sunflowers in this region that produce more than 100 seeds per head. Hypotheses: $H_0:p=0.20$, $H_a:p>0.20$, $\alpha =0.10$.
2. Conditions: Random sample is given; the population of post-wildfire sunflowers is more than $10\ast 75=750$; $n{p}_0=75\ast 0.20=15\ge 10$, $n(1-p_0)=75\ast 0.80=60\ge 10$, so all conditions are met.
3. $\hat{p}=22/75\approx 0.293$. Test statistic: $z=\frac{0.293-0.20}{\sqrt{\frac{(0.20)(0.80)}{75}}}\approx \frac{0.093}{0.0462}\approx 2.01$.
4. P-value = $P(Z>2.01)\approx 0.0222$. Since $0.0222<0.10$, we reject $H_0$. Interpretation: At the 0.10 significance level, there is convincing evidence that the proportion of sunflowers producing more than 100 seeds per head is higher after the wildfire than the baseline 20%, supporting the hypothesis that reduced competition after wildfire increases seed production.

7. Quick Reference Cheatsheet

8. What's Next

Carrying out a one-proportion hypothesis test is the foundation for all subsequent inference on proportions in AP Statistics. Immediately after mastering this topic, you will move on to comparing two proportions from independent samples, which extends the same logic of hypothesis testing to test for a difference between two population proportions. Without a solid understanding of the conditions, test statistic calculation, and conclusion-writing for one-proportion tests, you will struggle to generalize these skills to the two-proportion setting, which is also a common FRQ topic on the AP exam. This topic also builds on your earlier knowledge of sampling distributions and confidence intervals, and feeds into the broader AP Statistics goal of making justified conclusions from data, which is assessed on nearly every section of the exam.

• Hypothesis Test for the Difference of Two Proportions
• Confidence Intervals for a Proportion
• Type I and Type II Errors in Hypothesis Testing