AP Statistics Topic 3.10: Constructing a Confidence Interval for the Difference Between Two Population Proportions

01 · Estimate the difference, then allow for sampling variation

The concept in 60 seconds

Imagine comparing the proportion of students who prefer digital study notes at $2$ schools. Surveying every student may be impractical, so take a random sample from each school.

The sample proportions give $1$ estimate of the difference. Because another pair of samples could give another estimate, report a confidence interval for the population difference rather than treating the observed difference as exact.

$\text{Point estimate}\pm\text{margin of error}$

(\hat p_1-\hat p_2)\pm z^*\operatorname{SE}

$\hat{p}_{1}-\hat{p}_{2}$ : the observed sample-proportion difference.
$\operatorname{SE}$ : an estimate of how much sample differences vary.
$z^*$ : the multiplier for the chosen confidence level.
$z^* \times \operatorname{SE}$ : the margin of error on either side of the estimate.

The interval estimates $p_{1}-p_{2}$ . The population proportions are unknown. The sample proportions are known after collecting the data and are used to construct the interval.

Visual guide 1:

2

samples estimate

1

population difference

Group 1 · School A $\hat{p}_{1}=\frac{124}{200}=0.62$

This observed rate estimates the unknown population proportion $p_{1}$ who prefer digital notes.

Group 2 · School B $\hat{p}_{2}=\frac{69}{150}=0.46$

This observed rate estimates the unknown population proportion $p_{2}$ with the same preference.

Keep A minus BPoint estimate: $0.16$

Build an interval for $p_{1}-p_{2}$ around this sample difference, allowing for uncertainty from both groups.

A success means “prefers digital notes” in both samples. The school results are fictional. The $2$ population proportions are unknown; the sample proportions are observed estimates.

Quick check: what is the interval trying to estimate?

The difference between the $2$ population success proportions, $p_{1}-p_{2}$ , in a stated order. It is not estimating the already observed sample difference.

02 · Name the groups and the same success outcome

Define the $2$ populations

All examples on this page are fictional teaching scenarios. For the main example, $2$ independent SRSs are selected without replacement: $200$ students from School A’s $10{,}000$ students and $150$ students from School B’s $8{,}000$ students. Both samples answer the same yes/no question about preferring digital notes.

Observed results: $124$ students in the School A sample and $69$ in the School B sample prefer digital notes. The true population proportions are unknown.

Population parameters and sample statistics for the school example.
Symbol	Meaning	School example
$p_{1}$	Proportion of all School A students who prefer digital notes	Unknown population proportion
$p_{2}$	Proportion of all School B students who prefer digital notes	Unknown population proportion
$p_{1}-p_{2}$	Target population difference, A minus B	Unknown parameter to estimate
$x_{1}$ , $n_{1}$	School A success count and sample size	$124$ and $200$
$x_{2}$ , $n_{2}$	School B success count and sample size	$69$ and $150$
$\hat{p}_{1}$ , $\hat{p}_{2}$	Separate sample proportions	$0.62$ and $0.46$
$\hat{p}_{1}-\hat{p}_{2}$	Point estimate of the population difference	$0.16$ , or $16$ percentage points

\hat p_1=\frac{124}{200}=0.62

\hat p_2=\frac{69}{150}=0.46

\hat p_1-\hat p_2=0.62-0.46=0.16

The observed preference proportion is $16$ percentage points higher at School A. This sample difference estimates the unknown population difference; it does not establish that the true difference is exactly $0.16$ .

Choose the procedure before using a formula

Use a two-sample $z$ -interval for a difference between population proportions when you want to estimate the difference in success rates for $2$ appropriately independent groups, with the conditions below satisfied.

The response is categorical, coded as success or failure.
There are $2$ groups, each with a success count and sample size.
The goal is an interval estimate of $p_{1}-p_{2}$ .

If the response is a numerical measurement such as travel time, a proportion interval is not the right procedure. If the same people are measured $2$ times, the data are paired and the independent two-sample formula does not handle that dependence.

Keep one subtraction order

Let group 1 be School A and group 2 be School B throughout. A positive A-minus-B difference means A has the higher proportion; a negative difference means B has the higher proportion.

Quick check: is

p_{1}

“the proportion in the sample from School A”?

No. $p_{1}$ is the proportion among all students at School A who prefer digital notes. The sample proportion is $\hat{p}_{1}=0.62$ . Always identify the population and the success outcome when defining a parameter.

03 · Both samples contribute uncertainty

The standard error of the difference

Topic 3.9 used stated population proportions to calculate a theoretical sampling-distribution SD. Here, the population proportions are unknown, so we substitute each group’s own sample proportion to estimate that SD.

\operatorname{SE}=\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}

Visual guide 2: use

2

separate estimates inside

1

\operatorname{SE}

A contribution $0.001178$

$\frac{0.62(1 – 0.62)}{200}$ estimates the variance contribution from School A.

B contribution $0.001656$

$\frac{0.46(1 – 0.46)}{150}$ estimates the variance contribution from School B.

Add, then square root $\operatorname{SE}\approx 0.0532353$

$\sqrt{0.001178+0.001656}$ . Match each sample proportion to its own sample size.

The $2$ estimated variance terms add under the justified independent-group model. Their square root is the unpooled standard error in proportion-difference units, about $5.32$ percentage points.

School A contribution:

\frac{0.62(0.38)}{200}=0.001178

School B contribution:

\frac{0.46(0.54)}{150}=0.001656

\operatorname{SE}=\sqrt{0.001178+0.001656}=\sqrt{0.002834}\approx0.0532353

This $\operatorname{SE}$ estimates a typical sampling fluctuation of about $0.0532$ , or $5.32$ percentage points, in the sample-proportion difference around the true population difference. It is a measure of estimated spread, not the margin of error by itself.

Why do the $2$ terms add?

Variation from either sample can change the difference. For independent groups, their variance contributions add, even though the proportions subtract. Take the square root only after adding those contributions.

This interval uses an unpooled $\operatorname{SE}$ . Use $0.62$ for School A and $0.46$ for School B. Do not combine the counts into $1$ pooled proportion and substitute it into both terms.

Pooling can be relevant to a test that assumes equal population proportions. A confidence interval here estimates the difference without assuming equality. Subtracting $2$ separate one-proportion intervals is also not the construction rule for this interval.

The usual AP formula omits finite-population corrections when samples are selected without replacement. The separate $10\%$ checks justify the usual approximation.

Quick check: should you subtract the

2

standard deviations?

No. Add the $2$ estimated variance contributions inside $1$ square root. Subtracting SDs could make uncertainty cancel incorrectly.

04 · Justify the method before calculating endpoints

Check the conditions

1. Randomization and independent groups

For a sampling study, use $2$ independent random samples. The school example explicitly specifies independent SRSs. A voluntary poll does not become a random sample merely by having many responses.

For an appropriate two-group experiment, use random assignment of treatments to experimental units. Check that a suitable independent-group design is described. Observations from the same people before and after treatment are paired and need a method that accounts for that dependence.

2. Check $10\%$ separately when sampling without replacement

$n_1\le0.10N_1$ and $n_2\le0.10N_2$ .

School A: $200\le0.10(10{,}000)=1{,}000$ .

School B: $150\le0.10(8{,}000)=800$ .

Both pass. This supports treating responses within each sample as approximately independent for the usual formula. A combined sampling fraction across both populations cannot replace these $2$ checks.

The finite-population sampling check is not required merely because participants are randomly assigned to $2$ experimental groups. Random sampling and random assignment serve different purposes.

3. Check all $4$ observed counts

For this confidence interval, check observed successes and failures in each sample:

n_1\hat p_1=x_1\ge10\quad\text{and}\quad n_1(1-\hat p_1)=n_1-x_1\ge10

n_2\hat p_2=x_2\ge10\quad\text{and}\quad n_2(1-\hat p_2)=n_2-x_2\ge10

Visual guide 3: all

4

observed counts pass

Observed counts for the confidence-interval conditions.
Group	Successes $x$	Failures $n-x$	Large-count result
School A	$124$	$200-124=76$	Both $\ge10$
School B	$69$	$150-69=81$	Both $\ge10$

For an interval, use observed successes and failures. Also justify the design and separate $10\%$ checks; a count table alone cannot establish those conditions.

All $4$ counts pass. Saying “both sample sizes exceed $30$ ” is not enough: a sample of $100$ with only $3$ successes fails this count criterion.

If a condition fails

Nonrandom selection: increasing $n$ does not repair the sampling design or guarantee population generalization.
Paired groups: do not apply the independent two-sample $\operatorname{SE}$ without addressing the dependence.
A $10\%$ check fails: the usual uncorrected $\operatorname{SE}$ is not automatically justified.
A success or failure count is below $10$ : this standard $z$ -interval is not justified by the AP large-count criterion. Do not present its endpoints as a supported interval.

Quick check: do you use a hypothesized

p_{0}

for these count checks?

No. For this confidence interval, use each sample’s observed successes and failures. A hypothesized value belongs to a test setup, not these interval checks.

05 · From estimate to two endpoints

Build the confidence interval

For the main school example, construct a $95\%$ confidence interval for $p_{1}-p_{2}$ , the A-minus-B difference in population proportions preferring digital notes. The conditions have been justified above.

Step 1: calculate the point estimate

\hat p_1-\hat p_2=0.62-0.46=0.16

Step 2: calculate the unpooled $\operatorname{SE}$

\operatorname{SE}=\sqrt{\frac{0.62(0.38)}{200}+\frac{0.46(0.54)}{150}}\approx0.0532353

Step 3: choose the critical value $z^*$

For a two-sided confidence level $C$ , place $C$ in the center of a standard normal curve. The remaining area, $1-C$ , splits equally between the $2$ tails. The positive boundary is $z^*$ .

Visual guide 4: split the unused area equally

Left tail $0.025$

$\frac{1 – 0.95}{2}$ . The negative boundary is about $-1.960$ .

Center $0.95$

The chosen $95\%$ confidence level determines the central standard normal area.

Right tail $0.025$

The positive boundary is $z^*\approx 1.960$ , with cumulative area $0.975$ to its left.

These cards show how a two-sided critical value is selected from the standard normal model. The $95\%$ is not a percentage of individual students within the interval.

Common two-sided standard normal critical values.
Confidence $C$	Area in each tail	Cumulative area for $+z^*$	$z^*$ (rounded)
$90\%$	$0.050$	$0.950$	$1.645$
$95\%$	$0.025$	$0.975$	$1.960$
$99\%$	$0.005$	$0.995$	$2.576$

For $95\%$ confidence, $C=0.95$ and each tail has area $0.025$ . The cumulative area to the left of the positive boundary is $0.975$ , so $z^*\approx 1.960$ .

z^*=\Phi^{-1}\!\left(\frac{1+C}{2}\right)

$\Phi^{-1}$ gives the standard normal boundary at the stated cumulative area.

For example, use $\operatorname{invNorm}(0.975,0,1)$ if your calculator follows the cumulative-area, mean, SD input order. Using $\operatorname{invNorm}(0.95)$ would give the wrong critical value for a two-sided $95\%$ interval.

Step 4: multiply to get the margin of error

\operatorname{ME}=z^*\operatorname{SE}=z^*\sqrt{0.002834}\approx0.1043393

Using rounded factors: $\operatorname{ME}\approx1.959964(0.0532353)\approx0.1043393$ .

The margin of error is about $10.43$ percentage points. It is the distance from the estimate to either endpoint. The full interval width is $2$ times the margin of error.

Step 5: subtract and add the margin of error

Lower endpoint: $0.16-0.1043393\approx0.0557$ .

Upper endpoint: $0.16+0.1043393\approx0.2643$ .

$95\%$ confidence interval for $p_1-p_2$ : $(0.0557,0.2643)$ .

Visual guide 5: the estimate sits halfway between the endpoints

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Constructing a Confidence Interval for the Difference Between Two Population Proportions

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

The concept in 60 seconds

Define the $2$ populations