Free Response Practice on Two-Proportion Intervals

Turn the Interval into an Exam Response

In Complete Four-Step Two-Proportion Interval Solution, you learned to connect the parameter, conditions, calculation, and conclusion. This tutorial puts those skills into exam-style practice. The goal is not just to find endpoints: a strong response shows why the interval is appropriate, uses the calculator correctly, and explains what the interval says about the two population proportions.

Keep the group order visible throughout. An interval for $p_1-p_2$ estimates the proportion in Group 1 minus the proportion in Group 2. Positive values favor a larger proportion in Group 1; negative values favor a larger proportion in Group 2. The interval’s endpoints describe plausible population differences, not the observed sample difference alone.

Formula: The two-proportion $z$-interval uses the sample difference and the unpooled standard error: $$ (\hat{p}_1-\hat{p}_2)\mathbin{\pm}z^* \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} +\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}. $$ For an interval, each group’s contribution to the standard error uses its own sample proportion.

A Quick Plan for a Free-Response Item

Before calculating, underline the confidence level, the two groups, the shared outcome, and the requested subtraction order. Then organize the response into the four parts below. This makes it easier to avoid a common exam problem: correct arithmetic attached to an unclear or unsupported conclusion.

State.
Define both population proportions using the same outcome, state the order of their difference, and identify the requested confidence level.

Plan.
Name the two-proportion $z$-interval. Explain the random process, address independence and the 10% condition when relevant, and check successes and failures separately in both groups.

Do.
Enter each group’s successes and sample size in the correct order in 2-PropZInt. Report the interval and, when useful, show the sample proportions, standard error, and margin of error.

Conclude.
Interpret the endpoints as plausible values for the population difference in context. Use the signs of the endpoints to make the requested comparison without overstating the result.

For independent random samples, say that the groups came from separate random samples and explain how each sample relates to its target population. If sampling without replacement, the 10% condition requires each sample size to be no more than 10% of its population. Separate groups must also be independent and unpaired. For Large Counts, check each group’s successes and failures, not just the combined total. These are the same conditions described in Two-Sample Independence Conditions for Proportions and Large Counts for Each Group in Two-Proportion Intervals.

Worked Example: Compost Awareness in Two Housing Communities

Exam-style item: A fictional survey uses separate random samples of 180 apartment residents and 160 detached-house residents. In the apartment sample, 126 residents know where to find community composting information. In the detached-house sample, 96 residents know where to find it. Each population has at least ten times as many residents as its sample. Construct and interpret a 95% confidence interval for the difference in awareness proportions, apartment residents minus detached-house residents. Does the interval establish which population has the higher awareness proportion?

State: Let $p_1$ be the true proportion of apartment residents in the community who know where to find community composting information, and let $p_2$ be the corresponding proportion of detached-house residents. The parameter is $p_1-p_2$, apartment residents minus detached-house residents. We will construct a 95% confidence interval.

Plan: Use a two-proportion $z$-interval. The two groups were selected using separate random samples, supporting the Random condition and inference to the two communities sampled. They are separate, unpaired groups, so observations in one sample are not paired with observations in the other. The problem states that each population has at least ten times its sample size, so the 10% condition is met for both samples. For Large Counts, the apartment sample has 126 successes and $180-126=54$ failures; the detached-house sample has 96 successes and $160-96=64$ failures. All four counts are at least 10.

Do: The sample proportions are $\hat{p}_1=126/180=0.70$ and $\hat{p}_2=96/160=0.60$, so the estimated difference is $0.70-0.60=0.10$. On a TI-84, select 2-PropZInt from the inference menu. Enter $x_1=126$, $n_1=180$, $x_2=96$, $n_2=160$, and C-Level $=0.95$, keeping the same group order as in the parameter. The calculator gives endpoints of about $-0.0012$ and $0.2012$.

The calculator result can be checked with the formula. For 95% confidence, $z^*=1.96$. The unpooled standard error is:

\begin{aligned} SE_{\hat{p}_1-\hat{p}_2} &=\sqrt{\frac{0.70(0.30)}{180}+\frac{0.60(0.40)}{160}}\\ &=\sqrt{0.0011667+0.0015000}\\ &=\sqrt{0.0026667}\approx0.051640. \end{aligned}

The margin of error is $1.96(0.051640)\approx0.101214$. Thus:

0.10\mathbin{\pm}0.101214 =(-0.001214,\ 0.201214) \approx(-0.0012,\ 0.2012).

Conclude: We are 95% confident that the true difference in compost-information awareness proportions, apartment residents minus detached-house residents, is between about $-0.0012$ and $0.2012$. In context, the apartment-resident proportion could be about 0.12 percentage points lower or as much as 20.12 percentage points higher. Because zero is in the interval, the results do not establish which population proportion is higher. This does not prove the proportions are equal; it means a zero difference is among the plausible values.

Reading the Calculator Result as a Comparison

A confidence interval calculator returns numbers, not the conclusion an exam question asks for. Once you have the endpoints, compare both with zero and then translate their signs using the stated subtraction order. An interval entirely above zero indicates plausible values in which Group 1’s proportion is higher. An interval entirely below zero indicates plausible values in which Group 2’s proportion is higher. If the interval includes zero, it does not establish a higher proportion for either group.

The endpoints are differences in proportions. For example, a difference of $0.08$ means 8 percentage points, not an 8% increase relative to the other group. State the unit clearly when converting endpoints to percentages. Keep the original order even if the result is negative.

Worked Example: Comparing Satisfaction on Two Bus Routes

Exam-style item: A fictional transit agency selects separate random samples of riders from two routes. On Route A, 142 of 200 sampled riders say they are satisfied with the service. On Route B, 108 of 180 sampled riders say they are satisfied. Each route’s rider population is at least ten times the corresponding sample size. Construct a 95% confidence interval for Route A minus Route B, and state what it suggests about the population satisfaction proportions.

State: Let $p_1$ be the true proportion of Route A riders who are satisfied with the service, and let $p_2$ be the true proportion of Route B riders who are satisfied. The parameter is $p_1-p_2$, Route A minus Route B, at 95% confidence.

Plan: Use a two-proportion $z$-interval. Separate random samples of riders support the Random condition and generalization to the respective route populations. The route samples are separate and unpaired, so the groups are independent. Each sample is at most 10% of its route’s population, as stated. For Large Counts, Route A has 142 successes and $200-142=58$ failures; Route B has 108 successes and $180-108=72$ failures. All four counts are at least 10.

Do: The sample proportions are $\hat{p}_1=142/200=0.71$ and $\hat{p}_2=108/180=0.60$. The point estimate is $0.71-0.60=0.11$. Enter $x_1=142$, $n_1=200$, $x_2=108$, $n_2=180$, and a 0.95 confidence level in 2-PropZInt. To verify the result by hand, use $z^*=1.96$:

\begin{aligned} SE_{\hat{p}_1-\hat{p}_2} &=\sqrt{\frac{0.71(0.29)}{200}+\frac{0.60(0.40)}{180}}\\ &=\sqrt{0.0010295+0.0013333}\\ &=\sqrt{0.0023628}\approx0.048609. \end{aligned}

The margin of error is $1.96(0.048609)\approx0.095274$, giving:

0.11\mathbin{\pm}0.095274 =(0.014726,\ 0.205274) \approx(0.0147,\ 0.2053).

Conclude: We are 95% confident that the true difference in rider satisfaction proportions, Route A minus Route B, is between about $0.0147$ and $0.2053$. In context, the Route A proportion is plausibly about 1.47 to 20.53 percentage points higher than the Route B proportion. Since the entire interval is above zero, it supports the conclusion that Route A’s population satisfaction proportion is higher. The samples were random, so the comparison can be generalized to the route populations described; this observational survey does not show that riding Route A causes greater satisfaction.

When the Defined Order Runs Opposite to the Result

A negative interval is not an error if the parameter was defined as Group 1 minus Group 2. It means the plausible differences are below zero in that order. A complete response preserves the order and then explains the direction in ordinary language. Do not reverse the group labels midway through the calculation just to make the point estimate positive.

Worked Example: Reusable Container Use at Two Markets

Exam-style item: In a fictional survey, separate random samples include 150 shoppers at Market North and 140 shoppers at Market South. At North, 75 shoppers bring a reusable container; at South, 91 do. Each market has at least ten times as many shoppers as its sample. Find a 90% confidence interval for the proportion at North minus the proportion at South, and make a comparison conclusion.

State: Let $p_1$ be the true proportion of Market North shoppers who bring a reusable container, and $p_2$ the corresponding proportion of Market South shoppers. The parameter is $p_1-p_2$, North minus South, at 90% confidence.

Plan: Use a two-proportion $z$-interval. Separate random samples support the Random condition and generalization to the respective market populations. The two groups are independent and unpaired. Each sample is no more than 10% of its population, because each population is at least ten times its sample size. For Large Counts, North has 75 successes and $150-75=75$ failures; South has 91 successes and $140-91=49$ failures. Each count is at least 10.

Do: The sample proportions are $\hat{p}_1=75/150=0.50$ and $\hat{p}_2=91/140=0.65$. The estimated difference is $0.50-0.65=-0.15$. For a 90% interval, use $z^*=1.645$. On the calculator, enter $x_1=75$, $n_1=150$, $x_2=91$, $n_2=140$, and C-Level $=0.90$ in 2-PropZInt.

\begin{aligned} SE_{\hat{p}_1-\hat{p}_2} &=\sqrt{\frac{0.50(0.50)}{150}+\frac{0.65(0.35)}{140}}\\ &=\sqrt{0.0016667+0.0016250}\\ &=\sqrt{0.0032917}\approx0.057373. \end{aligned}

The margin of error is $1.645(0.057373)\approx0.094379$. Therefore:

-0.15\mathbin{\pm}0.094379 =(-0.244379,\ -0.055621) \approx(-0.2444,\ -0.0556).

Conclude: We are 90% confident that the true difference in reusable-container proportions, Market North minus Market South, is between about $-0.2444$ and $-0.0556$. In context, the North proportion is plausibly about 5.56 to 24.44 percentage points lower than the South proportion. Because the interval is entirely below zero, it supports the conclusion that the South population proportion is higher.

Common Mistakes and AP Exam Tips

Swapping calculator inputs: The first successes and sample size must belong to the group named first in $p_1-p_2$. Reversing the inputs reverses the interval’s signs and changes what its endpoints mean.
Reporting calculator output without conditions: 2-PropZInt performs the arithmetic, but it cannot establish that the sampling design and Large Counts condition justify the procedure. Write the checks explicitly.
Checking only the total number sampled: List successes and failures for each group separately. Large Counts is met only when all four counts are at least 10.
Calling a difference a percent increase: A difference such as $0.10$ is 10 percentage points. Do not call it a 10% increase unless a relative percent change was specifically calculated.
Ignoring a zero endpoint: If zero is inside the interval, do not claim that one population proportion is higher or that the proportions are equal. Explain that a zero difference is plausible at that confidence level.
Making the conclusion about the sample: The interval estimates $p_1-p_2$, not merely $\hat{p}_1-\hat{p}_2$. Name the populations and outcome in the conclusion.
Overstating the study’s reach: Random sampling supports generalization to the populations sampled. It does not establish cause and effect; that depends on the study design.

AP Exam Tip: A complete comparison conclusion should name the outcome, both groups, the subtraction order, the confidence level, and the interval’s direction. If the interval is entirely on one side of zero, say which population proportion is higher; if it includes zero, say that the interval does not establish which is higher.

Key Takeaway

Treat a two-proportion interval response as one connected argument. Define the comparison first, justify the procedure with specific condition checks, enter the groups in the right order, and turn the calculator endpoints into a population-level statement in context.

Key takeaway: Use the interval’s signs to guide the comparison conclusion: entirely positive favors a higher proportion in Group 1, entirely negative favors Group 2, and an interval containing zero does not establish which population proportion is higher.

Check Your Understanding

For each item, keep the group order and population context explicit.

A random sample of 120 people in Group 1 includes 78 with the outcome, and a separate random sample of 100 people in Group 2 includes 55. What successes and failures must be checked for Large Counts?
For a 2-PropZInt estimate of $p_1-p_2$, which group’s counts should be entered as $x_1,n_1$?
An interval for $p_1-p_2$ is $(0.02,\ 0.14)$. What comparison conclusion does its position relative to zero support?
An interval for $p_1-p_2$ is $(-0.12,\ 0.03)$. What should the conclusion say about which population proportion is higher?
Why does a random sample support generalization to a population but not, by itself, a cause-and-effect conclusion?