Structure of a One-Proportion z-Interval

From a Point Estimate to an Interval

In Point Estimates and Margins of Error for a Proportion, you used the sample proportion \(\hat{p}\) to estimate an unknown population proportion \(p\). A point estimate is useful, but it is only one value from one sample. A one-proportion \(z\)-interval adds a margin of error around that estimate to show a range of plausible values for \(p\), based on a stated confidence level and a suitable sampling model.

The interval has a compact structure: estimate plus or minus critical value times standard error. Each part has a distinct job. The estimate provides the center, the standard error measures estimated sampling variability, and the critical value sets how many standard errors to extend in either direction. Together, the critical value and standard error determine the margin of error.

Formula: A one-proportion \(z\)-interval for a population proportion \(p\) is

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

The lower endpoint is \(\hat{p}-z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), and the upper endpoint is \(\hat{p}+z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\).

The interval’s structure is often written as “estimate \(\pm\) margin of error.” For a one-proportion \(z\)-interval, the margin of error is the critical value multiplied by the estimated standard error. This gives two equivalent ways to display the same interval.

\text{estimate}\pm\text{margin of error} \qquad\text{where}\qquad \text{margin of error}=z^*\times\text{estimated standard error}

Labeling Each Part

Part	What it represents
\(\hat{p}\)	The sample proportion, which is the point estimate of the population proportion \(p\); it is the center of the interval.
\(z^*\)	The positive standard Normal critical value for the chosen confidence level; it determines how many estimated standard errors extend from the center.
\(\sqrt{\hat{p}(1-\hat{p})/n}\)	The estimated standard error of \(\hat{p}\), calculated using the observed sample proportion and sample size.
\(z^*\sqrt{\hat{p}(1-\hat{p})/n}\)	The margin of error, or the distance from the point estimate to either endpoint.
The two endpoints	The lower and upper limits of the interval of plausible values for \(p\), calculated by subtracting and adding the margin of error.

The value \(z^*\) depends on the confidence level, not on the particular sample proportion. Common standard Normal critical values are about 1.645 for 90% confidence, 1.96 for 95% confidence, and 2.576 for 99% confidence. A higher confidence level uses a larger critical value. If the sample and its estimated standard error stay the same, that larger multiplier makes the interval wider.

The standard error in this formula is estimated because the population proportion \(p\) is unknown. The formula uses \(\hat{p}\) in its place. The next tutorial examines the standard error of \(\hat{p}\) in more detail; here, the important point is to identify it as the factor multiplied by \(z^*\).

Conditions for Using the Interval

The formula is used when the sampling process and observed counts support a Normal-based interval. As in earlier tutorials on the 10% condition and the Large Counts condition, check the context as well as the arithmetic. For a sample taken without replacement from a finite population, compare the sample size with 10% of the population. For Large Counts, use the observed number with the characteristic and the observed number without it.

Conditions: Before using a one-proportion \(z\)-interval, check that the data come from a random sample or a suitable random process; if sampling without replacement from a finite population, check the 10% condition, \(n\leq0.10N\); and check the Large Counts condition using the sample data: at least 10 successes and at least 10 failures.

These conditions support the interval calculation; they do not guarantee that the sample is free of bias. For example, a random sample can still have nonresponse, and an unclear question can affect responses. A calculated margin of error describes sampling uncertainty under the model, not every possible source of survey error.

Worked Example: Building a 95% Interval

Worked Example: Residents Supporting a Shoreline Cleanup

A conservation group takes a random sample of 200 residents from a town of 8,000. In the sample, 118 support a proposed shoreline cleanup program. Construct and interpret an approximate 95% confidence interval for the proportion of all town residents who support the program.

State. Let \(p\) be the proportion of all town residents who support the cleanup program. The sample has \(x=118\) supporters and \(n=200\) residents, so the sample proportion is the point estimate of \(p\).

Plan and check conditions. The group took a random sample. Because it sampled without replacement, check the 10% condition:

0.10N=0.10(8{,}000)=800 \qquad\text{and}\qquad 200\leq800

The 10% condition is met. There are \(118\) sampled supporters and \(200-118=82\) sampled residents who do not support the program. Both counts are at least 10, so the Large Counts condition is met.

Do. First find the estimate, critical value, and estimated standard error. For 95% confidence, use \(z^*=1.96\):

\hat{p}=\frac{x}{n}=\frac{118}{200}=0.59

\text{estimated standard error} =\sqrt{\frac{0.59(1-0.59)}{200}} =\sqrt{\frac{0.2419}{200}} =\sqrt{0.0012095} \approx0.03478

Now multiply the critical value by the estimated standard error to find the margin of error, then subtract and add that amount to the estimate:

\text{ME}=1.96(0.03478)\approx0.0682

0.59\pm0.0682 \qquad\Longrightarrow\qquad (0.5218,\ 0.6582)

Conclude in context. We are 95% confident that the proportion of all town residents who support the shoreline cleanup program is between approximately 0.522 and 0.658, or 52.2% and 65.8%. The interval is centered at the sample estimate of 0.59, and its margin of error is approximately 0.0682, or 6.82 percentage points.

Worked Example: Comparing Confidence Levels

Worked Example: Paperless Billing

A utility company takes a random sample of 400 customers from a large service area. Of the customers sampled, 220 use paperless billing. Find approximate 90% and 99% confidence intervals for the proportion of all customers who use paperless billing. Compare their widths.

State. Let \(p\) be the proportion of all customers in the service area who use paperless billing. The sample proportion is:

\hat{p}=\frac{220}{400}=0.55

Plan and check conditions. The sample is random. Assume there are at least 4,000 customers in the service area; then 400 is no more than 10% of the population. The sample has \(220\) paperless-billing customers and \(400-220=180\) who do not use paperless billing. Both counts exceed 10, so the Large Counts condition is met.

Do. The estimate and estimated standard error are the same for both confidence levels. Calculate the estimated standard error:

\sqrt{\frac{0.55(1-0.55)}{400}} =\sqrt{\frac{0.2475}{400}} =\sqrt{0.00061875} \approx0.02487

For 90% confidence, use \(z^*=1.645\). The margin of error is \(1.645(0.02487)\approx0.0409\), so the interval is:

0.55\pm0.0409 \qquad\Longrightarrow\qquad (0.5091,\ 0.5909)

For 99% confidence, use \(z^*=2.576\). The margin of error is \(2.576(0.02487)\approx0.0641\), so the interval is:

0.55\pm0.0641 \qquad\Longrightarrow\qquad (0.4859,\ 0.6141)

Conclude in context. The approximate 90% interval is 50.91% to 59.09%, while the approximate 99% interval is 48.59% to 61.41%. The 99% interval is wider because its larger critical value multiplies the same estimated standard error. Both intervals use the same sample estimate, 55%; the confidence level changes the margin of error, not the center.

Worked Example: Reading the Interval’s Structure

Worked Example: Students Using a Study App

A school district randomly samples 160 students from a population of 3,000 students. Of those sampled, 72 report using a particular study app each week. Construct an approximate 95% confidence interval and identify the estimate, critical value, estimated standard error, margin of error, and endpoints.

State and plan. Let \(p\) be the proportion of all students in the district who use the app each week. The sample is random. Because it is taken without replacement, check \(0.10(3{,}000)=300\), and \(160\leq300\); the 10% condition is met. There are 72 app users and \(160-72=88\) nonusers in the sample, so both observed counts are at least 10.

Do. The point estimate is \(\hat{p}=72/160=0.45\). For 95% confidence, \(z^*=1.96\). The estimated standard error is:

\sqrt{\frac{0.45(1-0.45)}{160}} =\sqrt{\frac{0.2475}{160}} =\sqrt{0.001546875} \approx0.03933

Thus, the margin of error is \(1.96(0.03933)\approx0.0771\). The estimate plus or minus the margin of error gives:

0.45\pm0.0771 \qquad\Longrightarrow\qquad (0.3729,\ 0.5271)

Interpret the parts. The interval is centered at the sample estimate, 0.45. The estimated standard error is about 0.03933, and the critical value 1.96 makes the margin of error about 0.0771. The resulting interval gives plausible values from about 0.3729 to 0.5271 for the district-wide proportion. In context, we are 95% confident that between approximately 37.29% and 52.71% of district students use the study app weekly.

How to Interpret Confidence

A confidence level describes the long-run performance of the interval method. If many random samples of the same size were taken under the same conditions and a 95% interval were calculated from each, about 95% of those intervals would capture the fixed population proportion \(p\). The interval calculated from one sample either contains \(p\) or does not; after the data are collected, it is not correct to say there is a 95% probability that this particular fixed \(p\) lies in the interval.

A careful interpretation names the population and the characteristic, gives the interval in context, and uses confidence language. It does not claim that the sample proportion itself is the population proportion. It also does not claim that the interval covers 95% of individuals; the interval is about a population proportion, not a range of individual responses.

Common Mistakes and AP Exam Tips

Mixing up the estimate and the margin of error. \(\hat{p}\) is the center. The margin of error is \(z^*\) times the estimated standard error. Do not report the standard error alone as the margin of error.
Using the wrong critical value. Match \(z^*\) to the stated confidence level. A higher confidence level requires a larger critical value and, with other parts fixed, a wider interval.
Adding or subtracting the wrong quantity. Calculate the margin of error first, then subtract it from \(\hat{p}\) for the lower endpoint and add it for the upper endpoint.
Using the unknown \(p\) in the estimated standard error. For this interval, use the observed \(\hat{p}\) and the sample size \(n\).
Giving an interpretation without context. Identify the population and characteristic. Say “we are 95% confident the proportion of [population] who [characteristic] is between…” rather than describing only two decimals.
Interpreting confidence as a probability about this fixed parameter. Explain that the confidence level describes the long-run success rate of the method, not a 95% chance assigned to \(p\) after the interval is calculated.

AP Exam Tip: Make the structure visible: write \(\hat{p}\pm z^*(\text{estimated standard error})\), label the margin of error, and show both endpoints. Then state the confidence level and interpret the interval for the population proportion in context.

Key Takeaway

A one-proportion \(z\)-interval is built around the sample proportion. Its critical value times its estimated standard error gives the margin of error; subtracting and adding that margin produces the lower and upper endpoints. The confidence level, sampling conditions, and contextual interpretation all matter when explaining what the interval says.

Key takeaway: Read \(\hat{p}\pm z^*\sqrt{\hat{p}(1-\hat{p})/n}\) as estimate plus or minus critical value times estimated standard error. The center is \(\hat{p}\), the margin of error is the product \(z^*\sqrt{\hat{p}(1-\hat{p})/n}\), and the endpoints form the confidence interval for \(p\).

Check Your Understanding

Use the interval structure and show how its parts relate.

A random sample of 250 residents includes 145 who support a new bus route. Identify \(\hat{p}\) and the two observed counts used for the Large Counts condition.
For a one-proportion interval, explain what \(z^*\) represents and what determines its value.
If an interval has estimate 0.42 and margin of error 0.06, write the interval endpoints and label the center and margin of error.
Keeping the sample and estimate fixed, explain how moving from 90% confidence to 99% confidence affects \(z^*\) and the interval width.
In a sentence, explain what it means to be 95% confident in an interval for the proportion of students who ride a school bus.