Complete Four-Step Confidence Interval Procedure for a Proportion

From a Formula to a Complete Inference

In What Margin of Error Does Not Account For, you considered what a margin of error can and cannot tell you. Now the focus is how to present a complete confidence interval procedure. A strong response does more than report two endpoints: it identifies the population proportion of interest, explains why the method is appropriate, shows the calculation, and interprets the result in context.

For a population proportion \(p\), the one-proportion \(z\)-interval uses the sample proportion \(\hat p=x/n\) as its center. Its margin of error is a critical value multiplied by the estimated standard error. You have already constructed intervals in earlier tutorials; here, the new task is to organize the full reasoning using the four steps State, Plan, Do, Conclude.

Definition: A complete four-step confidence interval response identifies the population parameter and confidence level, names the interval procedure and justifies its conditions, carries out the calculation, and interprets the resulting interval in context.

The Four Steps

State.
Name the population proportion \(p\) you want to estimate, including the population and characteristic. State the requested confidence level.

Plan.
Name the one-proportion \(z\)-interval. Check that the data come from a random sample or suitable random process, check the 10% condition when sampling without replacement, and check the Large Counts condition using the observed success and failure counts.

Do.
Calculate \(\hat p\), the estimated standard error, and the interval endpoints. Show the formula, substitutions, and appropriately rounded results.

Conclude.
Interpret the interval as a range of plausible values for the population proportion, in context. Include the confidence level and do not treat the interval as a range for individual responses.

The condition checks in Plan should be arguments, not unexplained labels. As in Writing a Justification for Model Appropriateness, connect each condition to evidence from the situation and explain what that evidence allows you to do. If a condition is not met, do not simply proceed as if it were.

Conditions: For a one-proportion \(z\)-interval, the data should come from a random sample or suitable random process. If sampling without replacement from a finite population, check the 10% condition, \(n\leq0.10N\). Check the Large Counts condition using the observed counts: \(x\geq10\) and \(n-x\geq10\). When these conditions support the method, use \(\hat p\pm z^*\sqrt{\hat p(1-\hat p)/n}\).

Worked Examples

Worked Example: Estimating Module Completion in a School District

A fictional school district has 4,200 students. A simple random sample of 240 students is selected from the district roster. Of those sampled, 156 have completed a new online safety module. Construct and interpret a 95% confidence interval for the proportion of all students in the district who have completed the module.

State. Let \(p\) be the true proportion of all 4,200 students in this district who have completed the online safety module. We want a 95% confidence interval for \(p\).

Plan. Use a one-proportion \(z\)-interval. The district selected a simple random sample, so the random condition is met. Because sampling was without replacement, check the 10% condition: \(0.10(4{,}200)=420\), and \(240\leq420\), so the sample is no more than 10% of the student population. There are \(156\) observed successes and \(240-156=84\) observed failures; both counts are at least 10, so the Large Counts condition is met. The interval procedure is appropriate.

Do. The sample proportion is \(\hat p=156/240=0.65\). For a 95% confidence level, \(z^*=1.96\). The estimated standard error is \(\sqrt{0.65(1-0.65)/240}=\sqrt{0.2275/240}=\sqrt{0.00094792}\approx0.03079\). The margin of error is \(1.96(0.03079)\approx0.06034\). Therefore,

\hat p\pm z^*\sqrt{\frac{\hat p(1-\hat p)}{n}} =0.65\pm1.96\sqrt{\frac{0.65(0.35)}{240}} =0.65\pm0.06034 \approx(0.5897,\ 0.7103)

Conclude. We are 95% confident that between about 58.97% and 71.03% of all students in this district have completed the online safety module. The interval estimates the district’s population proportion, not the percentage of students in any particular future sample.

Notice that the conclusion does not say there is a 95% probability that \(p\) falls in this particular interval. As explained in Interpreting a 95% Confidence Level Correctly, the confidence level describes the long-run success rate of the method when it is used repeatedly under appropriate conditions. The endpoints from this one sample either contain the fixed population proportion or they do not.

Worked Example: A 90% Interval for Community Garden Participation

A fictional city has 2,000 households. A random sample of 150 households is asked whether anyone in the household participated in a community garden during the past year. Eighty-four households answer yes. Construct and interpret a 90% confidence interval for the proportion of city households with participation.

State. Let \(p\) be the true proportion of the city’s 2,000 households in which someone participated in a community garden during the past year. We want a 90% confidence interval for \(p\).

Plan. Use a one-proportion \(z\)-interval. The households were randomly sampled. The 10% check is \(0.10(2{,}000)=200\), and \(150\leq200\). There are \(84\) yes responses and \(150-84=66\) no responses, both at least 10, so the Large Counts condition is met.

Do. Here, \(\hat p=84/150=0.56\). For 90% confidence, \(z^*\approx1.645\). The estimated standard error is \(\sqrt{0.56(0.44)/150}=\sqrt{0.2464/150}=\sqrt{0.00164267}\approx0.04053\). The margin of error is \(1.645(0.04053)\approx0.06667\). The interval is

0.56\pm1.645\sqrt{\frac{0.56(0.44)}{150}} =0.56\pm0.06667 \approx(0.4933,\ 0.6267)

A calculator using 1-PropZInt with \(x=84\), \(n=150\), and confidence level \(0.90\) gives the same interval to rounding.

Conclude. We are 90% confident that between about 49.33% and 62.67% of the city’s households had someone participate in a community garden during the past year. This interval estimates the proportion of households with the stated characteristic, not the proportion of individual residents who gardened.

The same four steps work at different confidence levels. A lower confidence level uses a smaller critical value and, for the same sample, produces a narrower interval. The choice of confidence level does not change the population parameter being estimated; it changes the interval method’s long-run capture rate and the width of the resulting interval.

Worked Example: Estimating Use of a Neighborhood Water Station

A fictional neighborhood has 1,200 households. A random sample of 80 households is asked whether anyone used a new public water-refill station during its first month. Thirty-two households say yes. Find and interpret a 95% confidence interval for the proportion of neighborhood households that used the station.

State. Let \(p\) be the true proportion of the neighborhood’s 1,200 households in which someone used the water-refill station during its first month. We want a 95% confidence interval for \(p\).

Plan. Use a one-proportion \(z\)-interval. The sample is random. For sampling without replacement, \(0.10(1{,}200)=120\), and \(80\leq120\), so the 10% condition is met. The sample includes \(32\) households with use and \(80-32=48\) without use; both counts are at least 10. The Large Counts condition is met.

Do. The sample proportion is \(\hat p=32/80=0.40\). For 95% confidence, use \(z^*=1.96\). The estimated standard error is \(\sqrt{0.40(0.60)/80}=\sqrt{0.24/80}=\sqrt{0.003}\approx0.05477\). The margin of error is \(1.96(0.05477)\approx0.10735\). Thus,

0.40\pm1.96\sqrt{\frac{0.40(0.60)}{80}} =0.40\pm0.10735 \approx(0.2926,\ 0.5074)

Conclude. We are 95% confident that between about 29.26% and 50.74% of neighborhood households had someone use the water-refill station during its first month.

When a Condition Is Not Met

A complete response also makes clear when the usual interval procedure is not justified. Suppose a random sample of 32 households from a fictional population of 900 is asked about home composting, and 3 report having a compost bin. The sample proportion is \(3/32\approx0.0938\), and the 10% condition is met because \(32\leq0.10(900)=90\). However, the observed success count is only 3, which is less than 10. The Large Counts condition is not met.

In that situation, do not present the usual one-proportion \(z\)-interval as though its conditions were satisfied. State which condition fails and explain that the Normal-based interval is not supported by these data. A good response does not hide a failed condition just to produce endpoints.

Common Mistakes and AP Exam Tips

Leaving the parameter vague. Writing only “estimate the proportion” does not identify whose proportion or what characteristic is being measured. State the population and the yes/no characteristic in context.
Listing conditions without evidence. “Random and independent” is not a complete justification. Name how the sample was selected and, for sampling without replacement, show the 10% comparison. The reasoning should connect evidence to the condition, as emphasized in Writing a Justification for Model Appropriateness.
Checking counts with the wrong values. For a one-proportion interval, use the observed counts \(x\) and \(n-x\) to check Large Counts. Do not substitute a hypothesized population proportion from a test for the sample’s observed counts.
Mixing up standard deviation and standard error. The interval’s estimated standard error uses \(\hat p\), not the unknown \(p\): \(SE_{\hat p}=\sqrt{\hat p(1-\hat p)/n}\).
Reporting endpoints without units or context. If the result is written as percentages, describe percentages of the named population. If it is written as proportions, make clear what population characteristic those proportions represent.
Giving an incorrect confidence-level interpretation. Do not say that there is a 95% probability that the fixed \(p\) lies in the interval. State that you are 95% confident in the interval, and understand that the confidence level describes the method’s long-run capture rate.
Proceeding after a failed condition. If the Large Counts condition fails, identify the issue and do not claim that the usual \(z\)-interval is justified.

AP Exam Tip: Make each step easy to find. A full-credit response names \(p\) in context, identifies the one-proportion \(z\)-interval, gives evidence for the random, 10%, and Large Counts conditions as applicable, shows the calculation, and concludes with a confidence statement about the population proportion.

Key Takeaway

The four-step structure turns an interval calculation into a complete inference. State the population proportion and confidence level, Plan with the correct procedure and condition checks, Do the calculation, and Conclude in context. If the conditions support the method, the interval gives plausible values for the population proportion; it does not guarantee accuracy or describe individual outcomes.

Key takeaway: For a one-proportion confidence interval, connect the parameter, conditions, calculation, and contextual interpretation. The endpoints are meaningful as an estimate only when the interval procedure is justified.

Check Your Understanding

For each situation, identify the relevant four-step reasoning or explain what is missing.

A random sample of 180 students from a school of 1,500 includes 108 who ride the bus. Identify \(p\), check the 10% condition, and check the observed success and failure counts.
In a sample of 100 households, 92 report using a recycling service. What are the success and failure counts, and does the Large Counts condition hold?
A 95% confidence interval for the proportion of residents supporting a park renovation is \((0.41,0.57)\). Write an appropriate contextual conclusion using the stated population and characteristic.
Explain why “There is a 95% probability that \(p\) is in this interval” is not the standard interpretation of a 95% confidence interval.
A random sample of 40 people includes only 2 who have a particular characteristic. Which condition should be examined before using a one-proportion \(z\)-interval, and what should you conclude if it fails?