AP Statistics Topic 3.7: Carrying Out a Test for a Population Proportion

01 · From a sample to an evidence statement

The concept in 60 seconds

A school wants to know whether more than $60\%$ of its students prefer an earlier lunch. It takes a simple random sample of $500$ students from its $10{,}000$ students, without replacement. Of those sampled, $320$ say yes, so $\hat{p} =\frac{320}{500}= 0.64$ .

The observed $64\%$ is above $60\%$ , but a sample can land above a population value just through random variation. We need to ask: is the result far enough above $60\%$ to provide convincing evidence for the school’s claim?

A hypothesis test compares the observed sample with a null model. A complete answer connects the setup, the conditions, the calculation and the conclusion.

For this example, the one-proportion $z$ -test gives $z \approx 1.8257$ and a right-tailed $\text{p-value} \approx 0.0339$ . The school chose $\alpha = 0.05$ before collecting the data. Since $0.0339 < 0.05$ , we reject $H_{0}$ and find convincing statistical evidence that more than $60\%$ of all students at this school prefer an earlier lunch.

That conclusion is evidence, not certainty. It does not say that every student prefers the change or that a policy decision has been settled.

All contexts on this page are fictional teaching examples. Reported $p$ -values use the approximate standard normal model for a one-proportion $z$ -test.

Quick check: is “

64\%

is bigger than

60\%

” a complete reason to reject

H_{0}

No. We need to account for sampling variability, justify the model, find the correct $p$ -value, and compare it with the chosen significance level. A sample proportion can be above $p_{0}$ without providing convincing evidence that the population proportion is above $p_{0}$ .

02 · Keep the argument connected

The complete test workflow

The familiar State, Plan, Do, Conclude organization is useful because it makes your reasoning easy to follow. Each part answers a different question.

Visual guide 1: the complete test in three connected phases

State + PlanWhat are we testing?

Define $p$ , $H_{0}$ , $H_{a}$ and $\alpha$ . Identify the method and justify its conditions.

DoHow unusual is the sample?

Calculate $\hat{p}$ , the null SD, $z$ and the correct tail probability.

ConcludeWhat does the evidence support?

Compare $p$ -value with $\alpha$ . Give the decision and a cautious population statement.

A calculator’s $z$ and $p$ -value supply only the calculation phase. A complete response includes all three phases.

State the population parameter. Let $p$ be the proportion of all students at this school who prefer an earlier lunch.
State the hypotheses and $\alpha$ . $H_{0}: p = 0.60$ ; $H_{a}: p > 0.60$ . The predetermined significance level is $\alpha = 0.05$ .
Plan and justify the method. Use a one-sample $z$ -test for a population proportion, after checking random sampling, the $10\%$ condition when needed, and the expected counts under $H_{0}$ .
Do the calculation. Find $\hat{p}$ , the null standard deviation, $z$ and the $p$ -value for the alternative’s tail.
Make the formal decision. Explicitly compare the $p$ -value with $\alpha$ , then reject or fail to reject $H_{0}$ .
Conclude in context. Say whether the data provide convincing statistical evidence for $H_{a}$ , naming the parameter and population.

Write the hypotheses about $p$ , the unknown population proportion. The observed $\hat{p}$ is evidence used to test those hypotheses; it is not the parameter being tested.

Symbols in the school’s complete test.
Symbol	Role	School example
$p$	Unknown population proportion	Proportion of all school students who prefer earlier lunch
$p_{0}$	Null benchmark	$0.60$
$x$ and $n$	Observed success count and sample size	$320$ yes responses out of $500$
$\hat{p}$	Observed sample proportion	$0.64$
$\operatorname{SD}_0$	Standard deviation assumed under $H_{0}$	Approximately $0.0219089$
$z$	Observed standardized departure	Approximately $+1.8257$
$p$ -value	Null probability of the specified extreme region	Approximately $0.0339$
$\alpha$	Predetermined rejection threshold	$0.05$

Choose before looking: the research question determines $H_{a}$ , and the significance level is selected before the data are examined. Do not change either one to obtain a preferred decision.

Quick check: which hypothesis matches “Has the proportion changed from

60\%

?”

$H_{a}: p \ne 0.60$ , a two-sided alternative. “Changed” includes both an increase and a decrease. Use $H_{0}: p = 0.60$ .

03 · Justify the normal model

Check the conditions before calculating

For a categorical response with $2$ outcomes, record a success count $x$ out of $n$ observations. “Success” is simply the outcome you are counting; a defective item can be a success for the calculation even though a defect is undesirable.

Visual guide 2: the school’s three condition checks

SamplingSRS of $500$ students

The stated simple random sample supports the random-sampling condition.

Approximate independence $500 \le 1{,}000$

The sample is no more than $10\%$ of the $10{,}000$ -student population.

Normal null approximation $300$ successes; $200$ failures

Under $p_{0} = 0.60$ , both expected counts are at least $10$ .

Check these before using the normal $z$ -test. The expected-count check uses the hypothesized $p_{0}$ , not the observed $\hat{p}$ .

1. Random sampling

The data should come from a random sample of the population being studied. In the school example, the stated SRS supplies this condition. A voluntary online poll is not an SRS, even if it has thousands of responses.

2. The $10\%$ condition, when sampling without replacement

Check $n\le0.10N$ , where $N$ is the population size. This supports treating sampled outcomes as approximately independent when calculating the usual standard deviation.

School: $500\le0.10(10{,}000)=1{,}000$ . The condition is met.

If $N$ is not supplied, give a reasonable contextual justification where appropriate. Do not silently invent a population size.

3. Large expected counts under the null

Check $np_{0} \ge 10$ and $n(1 – p_{0}) \ge 10$ . These expected counts justify the approximate normal null distribution used by the proportion $z$ -test.

School: $500(0.60)=300$ expected successes and $500(0.40)=200$ expected failures. Both are at least $10$ .

The counts are based on $p_{0}$ , because the test model assumes $H_{0}$ is true. They need not equal the observed counts $x$ and $n – x$ . Conditions for a confidence interval use $\hat{p}$ in the success–failure check; the hypothesis test uses $p_{0}$ .

If a condition fails: do not present the normal $z$ -test as justified. State the problem. Depending on the situation, a different method, a larger planned sample or a better sampling design may be needed. A calculator output cannot repair the conditions.

Quick check: a sample has

9

successes, but

np_{0} = 16

. Does

9

automatically invalidate the

z

-test?

No. The test’s expected-success condition uses $np_{0} = 16$ , not the observed $9$ . Check the expected failures too, as well as the sampling conditions. The quality-control example later has $9$ observed defects and satisfies both expected-count checks.

04 · Measure the departure in null standard deviations

Calculate the $z$ -statistic

The test statistic tells us how far the sample proportion is from the null benchmark, measured in standard deviations under $H_{0}$ .

\hat p=\frac xn

\operatorname{SD}_0=\sqrt{\frac{p_0(1-p_0)}n}

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

Visual guide 3: see what each part of

z

contributes

Numerator $0.64 – 0.60 = 0.04$

The sample is $4$ percentage points above the null benchmark.

Denominator $\operatorname{SD}_0 \approx 0.0219089$

Expected sample-to-sample variation under the $60\%$ null model.

Ratio $z \approx +1.8257$

The sample sits about $1.83$ null standard deviations above the benchmark.

Use the unrounded denominator and $z$ for the tail probability. The statistic is unitless; it is not the $p$ -value.

The school calculation, one line at a time

Observed proportion: $\hat{p} =\frac{320}{500}= 0.64$ .
Departure from the benchmark: $\hat{p} – p_{0} = 0.64 – 0.60 = 0.04$ . This is a difference of $4$ percentage points.
Null standard deviation: $\operatorname{SD}_0 = \sqrt{\frac{0.60(0.40)}{500}} \approx 0.0219089$ .
Standardize: $z=\frac{0.64-0.60}{\sqrt{\frac{0.60(0.40)}{500}}}\approx1.8257$ .

The sample proportion is approximately $1.83$ null standard deviations above $0.60$ . A negative $z$ would place it below the null benchmark. The $z$ -statistic is unitless; it is not a percentage or a probability.

Why use $p_{0}$ in the denominator?

A test asks how unusual the data would be if the null population proportion were $p_{0}$ . That is why both factors in the standard deviation use $p_{0}$ .

For a confidence interval, the estimated standard error is $\sqrt{\frac{\hat p(1 – \hat p)}{n}}$ . That expression estimates variability from the sample. It serves a different purpose. Do not copy the interval formula into the hypothesis test.

Keep precision: use unrounded $\hat{p}$ , the unrounded denominator and the unrounded $z$ for the tail calculation. Round the values you report at the end. Early rounding matters most when the $p$ -value is close to $\alpha$ .

Quick check: if

\hat{p} < p_{0}

, what sign should

z

have?

Negative. The numerator is negative and the square-root denominator is positive. The $p$ -value still lies between $0$ and $1$ ; its tail is determined by $H_{a}$ .

05 · Let the alternative choose the area

Find the $p$ -value from the standard normal model

When the conditions are justified, use $Z\mathrel{\overset{\text{approx.}}{\sim}}\mathcal{N}(0,1)$ as the null reference for the standardized statistic. The $p$ -value measures results at least as extreme as the observed $z$ , in the direction specified by $H_{a}$ .

Choose a normal tail from the alternative.
Alternative	$p$ -value region	Using left-area function $\Phi$
$H_{a}: p > p_{0}$	$P(Z \ge z_{\mathrm{obs}})$	$1 – \Phi (z_{\mathrm{obs}})$
$H_{a}: p < p_{0}$	$P(Z \le z_{\mathrm{obs}})$	$\Phi (z_{\mathrm{obs}})$
$H_{a}: p \ne p_{0}$	$P(Z \le -\lvert z_{\mathrm{obs}}\rvert ) + P(Z \ge \lvert z_{\mathrm{obs}}\rvert )$	$2[1 – \Phi (\lvert z_{\mathrm{obs}}\rvert )]$

Here $z_{\mathrm{obs}}$ denotes the unrounded observed test statistic. The function $\Phi (z)$ means the standard normal area to the left of $z$ . For the school’s $H_{a}: p > 0.60$ , use the right area:

\text{p-value}=P(Z\ge1.825741\ldots)\approx0.0339446\approx0.0339

Interpret it conditionally: if the school’s true preference proportion is $0.60$ , the approximate probability of a random sample of $500$ producing a proportion of $0.64$ or higher is $0.0339$ .

Using a standard normal table

A common $z$ -table reports left-tail areas. If you round the school statistic to $z = 1.83$ , that table gives an area near $0.9664$ . Subtract from $1$ to get a right-tail estimate near $0.0336$ . The small difference from $0.0339$ is due to rounding $z$ for the table. For this example, both lead to the same decision at $\alpha = 0.05$ .

For a two-sided normal $z$ -test, add both equally distant outer tails. A convenient equivalent is $2[1 – \Phi (\lvert z\rvert )]$ . Do not multiply a large left-tail area by $2$ . The rule here applies to this symmetric normal test.

Using technology

A built-in one-proportion $z$ -test typically needs $4$ entries: $p_{0}$ , the success count $x$ , the sample size $n$ , and the alternative $>$ , $<$ or $\ne$ . For the school, enter $0.60, 320, 500$ , $>$ . The test output should show $z \approx 1.8257$ and $\text{p-value} \approx 0.0339$ .

Alternatively, calculate $z$ first and use a standard normal cumulative-probability tool with mean $0$ and standard deviation $1$ . Specify the required left tail, right tail or both tails. Menu names differ across calculators, so check what area your tool returns.

Technology reports the calculation; you supply the reasoning. Write the hypotheses, justify the conditions, compare with $\alpha$ and conclude in context. Do not enter $\hat{p}$ in the $p_{0}$ field or enter the number of failures as $x$ unless failures are your defined success outcome.

Quick check:

z = -1.8

, but

H_{a}: p > p_{0}

. Which area is the

p

-value?

The right area, $P(Z \ge -1.8) \approx 0.9641$ . The result points opposite to the greater-than claim. The negative sign does not change the alternative to a left-tailed test.

06 · Compare evidence with the chosen threshold

Use $\alpha$ to make the formal decision

The significance level, written $\alpha$ and read “alpha,” sets the rejection threshold before the data are examined. Common choices include $0.01, 0.05,\text{ and }0.10$ . Use the value specified for the study or question; $0.05$ is not an automatic choice for every test.

Under the null model, $\alpha$ is the predetermined probability of rejecting $H_{0}$ when $H_{0}$ is true. For a proportion $z$ -test this calibration uses an approximate continuous normal model, so the actual probability for discrete sample counts need not equal $\alpha$ exactly. We examine the consequences of test errors in Topic 3.8.

If $\text{p-value}\le\alpha$ : reject $H_0$ . The result is statistically significant at that level.

If $\text{p-value}>\alpha$ : fail to reject $H_0$ . The result is not statistically significant at that level.

$\alpha$ and the $p$ -value have different jobs

$\alpha$ is selected in advance. It states the decision rule. The $p$ -value is calculated from the observed data using the specified null model and alternative.

Visual guide 4: observed

p

-value versus predetermined

\alpha

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Carrying Out a Test for a Population Proportion

By the end of this lesson, you should be able to:

The concept in 60 seconds

The complete test workflow