AP Statistics Topic 3.3: Constructing a Confidence Interval for a Population Proportion

01 · An estimate with sampling uncertainty

The concept in 60 seconds

A school has $10{,}000$ students. It wants to estimate the proportion who prefer an earlier lunch. A simple random sample of $500$ students without replacement contains $320$ who prefer it.

The sample proportion is $\hat{p} =\frac{320}{500}= 0.64$ . The true population proportion $p$ is unknown. Another random sample would probably give a different estimate, so we build an interval around $0.64$ rather than treating $64\%$ as an exact population answer.

Visual guide 1: from a yes/no sample to a population estimate

Observed responses $320$ yes out of $500$

Each sampled student gives $1$ yes/no response.

Point estimate $\hat{p} = 0.64$

$64\%$ is the observed sample proportion.

Interval estimate $0.5979\text{ to }0.6821$

Add a margin of error to estimate the unknown population $p$ at $95\%$ confidence.

The sample tells us $\hat{p}$ ; the interval estimates $p$ . An SRS, a sufficiently small sampling fraction and at least $10$ observed responses in each category justify this procedure.

\text{Confidence interval}=\text{point estimate}\pm\text{margin of error}

For this sample, the approximate $95\%$ interval is $(0.5979,0.6821)$ , or $59.79\%\text{ to }68.21\%$ .

We are $95\%$ confident that between about $59.79\%$ and $68.21\%$ of all students at this school prefer an earlier lunch. The interval estimates $1$ population proportion; it does not describe the percentage of individual students whose answers fall inside a range.

What does $95\%$ confidence mean? Under the method’s conditions, about $95\%$ of intervals constructed this way from repeated random samples would contain the fixed population proportion. The level describes the procedure’s long-run capture rate. The particular calculated interval either contains $p$ or it does not.

All scenarios on this page are fictional teaching examples. The charts show calculations and theoretical models, not collected school data. We develop detailed interpretation and claims from intervals in Topic 3.4.

Quick check: which number is known,

p

\hat{p}

$\hat{p} = 0.64$ is known from this sample. $p$ , the proportion in the whole school, is unknown and is what the interval estimates.

02 · Say who and what you are estimating

Choose the parameter and method

Use a one-sample $z$ -interval for a population proportion when you want to estimate the proportion in a specified category in $1$ population, using an appropriate sample with the conditions below satisfied.

The response is categorical: each sampled student either prefers an earlier lunch or does not. “Success” simply names the category being counted. It does not mean that the response is better.

Define $p$ in context: “Let $p$ be the proportion of all $10{,}000$ students at this school who prefer an earlier lunch.”

A complete definition names the proportion, the response category and the population. “ $p$ is the sample proportion” names the wrong quantity.

$1$ population, $1$ random sample, $1$ interval.
Symbol	Meaning	School example
$p$	Unknown population proportion	Unknown proportion of all school students who prefer an earlier lunch
$\hat{p}$	Observed sample proportion $\frac{x}{n}$	$\frac{320}{500}= 0.64$
$x$	Observed success count	$320$ students
$n$	Sample size	$500$ students
$N$	Population size	$10{,}000$ students
$C$	Confidence level as a decimal	$0.95$
$z^*$	Positive two-sided normal critical value	About $1.96$
$\operatorname{SE}$	Estimated sampling standard deviation	About $0.02147$
$\operatorname{MOE}$	Half-width of the interval; $z^* \times \operatorname{SE}$	About $0.04207$

Recognize when a different method is needed

Average lunch duration: this is a population mean, not a proportion.
Difference in preferences between $2$ schools: this involves $2$ population proportions, so a one-sample interval is not the right procedure.
Test whether $p$ equals a specified value: this asks for a significance test, not just an interval estimate.

In this lesson, you estimate a single unknown $p$ . You do not plug a hypothesized $p_{0}$ into the interval formula. That different input appears when you test a population proportion later.

Quick check: what is wrong with “

p

is the proportion of the

500

sampled students who said yes”?

That defines the sample statistic $\hat{p}$ . Define $p$ for all students at the school, the population you want to learn about.

03 · Verify the method before calculating

Check the conditions

A calculator can return endpoints even when a procedure is unsuitable. The justification comes from the sampling design and counts.

Visual guide 2: check the design and observed counts

Random samplingSRS is stated

A large volunteer sample would not pass simply because $n$ is large.

Sampling fraction $500 \le 1{,}000$

Without replacement, compare $n$ with $10\%$ of $N = 10{,}000$ .

Observed categories $320$ yes; $180$ no

Check that $x$ and $n – x$ are both at least $10$ .

These checks have different purposes. For this interval, the normality check uses observed sample counts, because the population proportion is unknown.

1. Random sampling

The data should come from an appropriate random sample of the intended population. The school uses an SRS, so this condition is met. A survey of the first $500$ volunteers does not become a random sample because its size is large.

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

n\le0.10N\quad\Longleftrightarrow\quad N\ge10n

School check: $500\le0.10(10{,}000)=1{,}000$ .

Sampling without replacement creates dependence. A small sampling fraction supports the usual approximately independent standard-error calculation. Meeting $10\%$ does not make observations exactly independent or establish normality.

For a genuinely independent sampling model, such as sampling with replacement, this finite-population check is unnecessary; independence still needs a reasonable justification. If the population size is not provided, explain a justified population-size assumption rather than inventing $N$ .

3. At least $10$ observed successes and $10$ observed failures

n\hat p=x\ge10\quad\text{and}\quad n(1-\hat p)=n-x\ge10

School check: $320$ successes and $500-320=180$ failures. Both are at least $10$ .

These observed counts support the normal approximation used for this interval. There is no universal “ $n \ge 30$ is enough” rule for proportions: a sample of $100$ with only $6$ successes fails this check.

The important change from Topic 3.2: A sampling model with a supplied $p$ uses expected counts $np$ and $n(1-p)$ . An interval estimates an unknown $p$ , so this procedure checks observed counts $n\hat{p}$ and $n(1-\hat p)$ .

Complete justification: “An SRS is stated. Since the sample is taken without replacement, $500 \le 1{,}000$ satisfies the $10\%$ condition. The observed counts are $320$ successes and $180$ failures, both at least $10$ . A one-sample $z$ -interval for the population proportion is appropriate.”

If a condition fails, explain the specific issue. A large sample does not repair selection bias. A substantial sampling fraction calls for a method that accounts for dependence. Very small success/failure counts require a different interval method rather than an unsupported normal interval.

Quick check: an SRS of

100

has

8

successes. Can we use this

z

-interval?

The observed success count is $8 < 10$ , so the usual normal-approximation condition fails. Do not justify it by citing the sample size alone.

04 · Keep three different quantities separate

Standard error and margin of error

In Topic 3.2, a known population proportion gave the sampling SD $\sqrt{\frac{p(1 – p)}{n}}$ . Here $p$ is unknown. We estimate that sampling SD using $\hat{p}$ .

Estimated standard error:

\operatorname{SE}(\hat p)=\sqrt{\frac{\hat p(1-\hat p)}n}

Margin of error:

\operatorname{MOE}=z^*\operatorname{SE}(\hat p)

One-sample proportion interval:

\hat p\pm z^*\sqrt{\frac{\hat p(1-\hat p)}n}

$\operatorname{SE}$ describes the estimated scale of sample-to-sample variation in proportions. It is not the size of the actual error in the observed estimate, because $p$ is unknown. $\operatorname{MOE}$ multiplies $\operatorname{SE}$ by a critical value appropriate to the chosen confidence level.

Calculate the school’s $\operatorname{SE}$ and $\operatorname{MOE}$

$\hat{p} =\frac{320}{500}= 0.64$ .
$\operatorname{SE} = \sqrt{\frac{0.64(0.36)}{500}} \approx 0.02147$ .
For $95\%$ confidence, $z^* \approx 1.96$ .
$\operatorname{MOE} \approx 1.96(0.02147) \approx 0.04207$ .

The estimated standard error is about $2.15$ percentage points. The $95\%$ margin of error is about $4.21$ percentage points. Both are on a proportion scale, not a student-count scale.

Visual guide 3: the estimate sits in the middle of the interval

This is $1$ approximate $95\%$ interval from the fictional sample. The horizontal scale is focused on $56.5\%\text{–}71.5\%$ ; it is not the full $0\%\text{–}100\%$ proportion range. Each endpoint is about $4.21$ percentage points from the estimate.

$\operatorname{MOE}=\frac{\text{width}}2$ : $\text{lower}=\hat p-\operatorname{MOE}$ ; $\text{upper}=\hat p+\operatorname{MOE}$ . The full width is $2 \times \operatorname{MOE}$ , about $0.08415$ or $8.41$ percentage points here.

Keep full calculator precision while calculating endpoints. Rounding $\operatorname{SE}$ to $0.02$ before multiplying would change the interval unnecessarily.

Quick check: if the interval is

(0.48, 0.60)

, what are its center and

\operatorname{MOE}

$\text{Center}=\frac{0.48+0.60}{2}=0.54$ . $\operatorname{MOE} =\frac{0.60 – 0.48}{2}= 0.06$ , or $6$ percentage points. The full width is $0.12$ .

05 · Put the confidence level in the middle

Choose the critical value

The positive critical value $z^*$ marks the right edge of the middle $C$ proportion of the standard normal curve. For $95\%$ confidence, $C = 0.95$ . The remaining $0.05$ is split equally, leaving $0.025$ in each tail.

Visual guide 4: a

95\%

level leaves

2.5\%

in each tail

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Constructing a Confidence Interval for a Population Proportion

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

The concept in 60 seconds

Choose the parameter and method

Recognize when a different method is needed

Check the conditions

1. Random sampling

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

3. At least $10$ observed successes and $10$ observed failures

Standard error and margin of error

Calculate the school’s $\operatorname{SE}$ and $\operatorname{MOE}$

Choose the critical value

Main Menu

Quick Links

The concept in 60 seconds

Choose the parameter and method

Recognize when a different method is needed

Check the conditions

1. Random sampling

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

3. At least 1010 observed successes and 1010 observed failures

Standard error and margin of error

Calculate the school’s SE⁡\operatorname{SE} and MOE⁡\operatorname{MOE}

Choose the critical value

Main Menu

Quick Links​

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

3. At least $10$ observed successes and $10$ observed failures

Calculate the school’s $\operatorname{SE}$ and $\operatorname{MOE}$

Quick Links