One-Proportion z Interval: Formula, Conditions, and Example
Estimate one population proportion from a random sample, check the conditions that justify a normal interval, calculate the margin of error, and interpret the confidence statement in context.
One-Proportion z Interval: Formula, Conditions, and Example: direct answer
In a one proportion z interval problem, begin by identifying the population target and the inferential role of the data. Estimate one population proportion from a random sample, check the conditions that justify a normal interval, calculate the margin of error, and interpret the confidence statement in context.
This page uses one proportion z interval as its single primary search focus. The lesson, numerical cases, and retained questions are restricted to that intent so the page does not function as a generic inference question bank.
Quick reference for One-Proportion z Interval: Formula, Conditions, and Example
| Parameter | p: one population proportion |
|---|---|
| Point estimate | p̂=x/n |
| Standard error | √[p̂(1−p̂)/n] |
| Interval | p̂ ± z* × SE |
| Core count check | Observed successes and failures are each at least 10 |
| Interpretation target | The population proportion p |
Concept mastery: one proportion z interval
Identify the parameter before calculating
The target is a single population proportion p. The sample proportion p-hat is the point estimate, not the parameter itself. State the population and the success condition in words before opening a calculator; this prevents a confidence interval from becoming an interval for the sample rather than an estimate of the population proportion.
Use the observed sample proportion in the interval standard error
For a confidence interval, the standard error is based on p-hat because the unknown population proportion is being estimated. The expression is SE = sqrt[p-hat(1-p-hat)/n]. This is different from a one-proportion significance test, where the null value p0 belongs in the null standard error.
Check randomization and independence
The data should come from a random sample, randomized process, or another design that justifies treating observations as representative and approximately independent. When sampling without replacement from a finite population, the sample should ordinarily be no more than about ten percent of that population.
Check observed successes and failures
A normal-based interval requires enough observed successes and observed failures. In the common AP Statistics condition check, both n*p-hat and n*(1-p-hat) should be at least 10. Write the actual counts rather than only saying that the condition is met.
Choose the critical value from the confidence level
The confidence level determines z-star. Common values are about 1.645 for 90%, 1.960 for 95%, and 2.576 for 99%. A higher confidence level uses a larger critical value and therefore produces a wider interval when the sample is unchanged.
Interpret confidence as a procedure statement
A 95% confidence interval does not mean there is a 95% probability that a fixed parameter lies inside the already-computed interval. The defensible wording is that we are 95% confident that the true population proportion is between the endpoints, with the population and response identified in context.
Read margin of error as precision
Margin of error is z-star times the standard error. It describes the half-width of the interval. Larger samples shrink the standard error; greater confidence enlarges z-star. These effects explain interval width without needing to recompute every example.
Do not use p0 unless a test supplies it
A confidence interval has no null value unless the problem separately asks you to compare the interval with a benchmark. Putting a claimed p0 into the interval standard error silently changes the method into test-style reasoning and gives the wrong uncertainty estimate.
Use the interval to evaluate plausible parameter values
Values inside the interval are reasonably compatible with the data under the confidence procedure; values outside are not supported at the corresponding two-sided level. This connection can help compare an interval with a two-sided significance test without claiming the two tasks are identical.
Keep scope tied to the sampling design
A numerical interval cannot repair a biased sample. If the sample is voluntary, undercovers important groups, or uses a response mechanism that creates bias, a narrow interval can still estimate the wrong target. Statistical precision and study validity must be considered separately.
Worked analysis for one proportion z interval
Interval case 1: City Transit Survey
A random sample of 120 observations found 41 riders who used the mobile ticket option. The sample proportion is p̂=0.342. For a 90% one proportion z interval, the observed success and failure counts are 41 and 79, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0433; with z*=1.645, the margin of error is 0.0712. The resulting interval is (0.270, 0.413).
The contextual conclusion is that we are 90% confident the population proportion of riders who used the mobile ticket option is between 0.270 and 0.413, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.071 quantifies precision rather than the spread of individual binary outcomes.
Interval case 2: Regional Hospital
A random sample of 131 observations found 54 patients who received follow-up within two days. The sample proportion is p̂=0.412. For a 95% one proportion z interval, the observed success and failure counts are 54 and 77, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0430; with z*=1.960, the margin of error is 0.0843. The resulting interval is (0.328, 0.497).
The contextual conclusion is that we are 95% confident the population proportion of patients who received follow-up within two days is between 0.328 and 0.497, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.084 quantifies precision rather than the spread of individual binary outcomes.
Interval case 3: Community College
A random sample of 142 observations found 68 students who completed the placement module. The sample proportion is p̂=0.479. For a 99% one proportion z interval, the observed success and failure counts are 68 and 74, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0419; with z*=2.576, the margin of error is 0.1080. The resulting interval is (0.371, 0.587).
The contextual conclusion is that we are 99% confident the population proportion of students who completed the placement module is between 0.371 and 0.587, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.108 quantifies precision rather than the spread of individual binary outcomes.
Interval case 4: County Library
A random sample of 153 observations found 84 visitors who attended a weekly program. The sample proportion is p̂=0.549. For a 90% one proportion z interval, the observed success and failure counts are 84 and 69, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0402; with z*=1.645, the margin of error is 0.0662. The resulting interval is (0.483, 0.615).
The contextual conclusion is that we are 90% confident the population proportion of visitors who attended a weekly program is between 0.483 and 0.615, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.066 quantifies precision rather than the spread of individual binary outcomes.
Interval case 5: Public High School
A random sample of 164 observations found 102 seniors who submitted the financial-aid form. The sample proportion is p̂=0.622. For a 95% one proportion z interval, the observed success and failure counts are 102 and 62, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0379; with z*=1.960, the margin of error is 0.0742. The resulting interval is (0.548, 0.696).
The contextual conclusion is that we are 95% confident the population proportion of seniors who submitted the financial-aid form is between 0.548 and 0.696, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.074 quantifies precision rather than the spread of individual binary outcomes.
Interval case 6: Municipal Water Office
A random sample of 175 observations found 121 households reporting no service interruption. The sample proportion is p̂=0.691. For a 99% one proportion z interval, the observed success and failure counts are 121 and 54, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0349; with z*=2.576, the margin of error is 0.0899. The resulting interval is (0.601, 0.781).
The contextual conclusion is that we are 99% confident the population proportion of households reporting no service interruption is between 0.601 and 0.781, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.090 quantifies precision rather than the spread of individual binary outcomes.
Interval case 7: State Park
A random sample of 186 observations found 141 visitors who used the marked trail system. The sample proportion is p̂=0.758. For a 90% one proportion z interval, the observed success and failure counts are 141 and 45, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0314; with z*=1.645, the margin of error is 0.0517. The resulting interval is (0.706, 0.810).
The contextual conclusion is that we are 90% confident the population proportion of visitors who used the marked trail system is between 0.706 and 0.810, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.052 quantifies precision rather than the spread of individual binary outcomes.
Interval case 8: Food Cooperative
A random sample of 197 observations found 79 members who renewed before the deadline. The sample proportion is p̂=0.401. For a 95% one proportion z interval, the observed success and failure counts are 79 and 118, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0349; with z*=1.960, the margin of error is 0.0684. The resulting interval is (0.333, 0.469).
The contextual conclusion is that we are 95% confident the population proportion of members who renewed before the deadline is between 0.333 and 0.469, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.068 quantifies precision rather than the spread of individual binary outcomes.
Interval case 9: University Advising Center
A random sample of 208 observations found 98 appointments starting within ten minutes. The sample proportion is p̂=0.471. For a 99% one proportion z interval, the observed success and failure counts are 98 and 110, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0346; with z*=2.576, the margin of error is 0.0892. The resulting interval is (0.382, 0.560).
The contextual conclusion is that we are 99% confident the population proportion of appointments starting within ten minutes is between 0.382 and 0.560, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.089 quantifies precision rather than the spread of individual binary outcomes.
Interval case 10: Regional Manufacturer
A random sample of 219 observations found 118 parts meeting the diameter specification. The sample proportion is p̂=0.539. For a 90% one proportion z interval, the observed success and failure counts are 118 and 101, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0337; with z*=1.645, the margin of error is 0.0554. The resulting interval is (0.483, 0.594).
The contextual conclusion is that we are 90% confident the population proportion of parts meeting the diameter specification is between 0.483 and 0.594, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.055 quantifies precision rather than the spread of individual binary outcomes.
Interval case 11: Public Health Clinic
A random sample of 230 observations found 140 clients returning for the scheduled checkup. The sample proportion is p̂=0.609. For a 95% one proportion z interval, the observed success and failure counts are 140 and 90, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0322; with z*=1.960, the margin of error is 0.0631. The resulting interval is (0.546, 0.672).
The contextual conclusion is that we are 95% confident the population proportion of clients returning for the scheduled checkup is between 0.546 and 0.672, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.063 quantifies precision rather than the spread of individual binary outcomes.
Interval case 12: Urban Recreation Program
A random sample of 241 observations found 164 participants completing the eight-week session. The sample proportion is p̂=0.680. For a 99% one proportion z interval, the observed success and failure counts are 164 and 77, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0300; with z*=2.576, the margin of error is 0.0774. The resulting interval is (0.603, 0.758).
The contextual conclusion is that we are 99% confident the population proportion of participants completing the eight-week session is between 0.603 and 0.758, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.077 quantifies precision rather than the spread of individual binary outcomes.
Interval case 13: School District
A random sample of 252 observations found 189 families responding to the annual survey. The sample proportion is p̂=0.750. For a 90% one proportion z interval, the observed success and failure counts are 189 and 63, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0273; with z*=1.645, the margin of error is 0.0449. The resulting interval is (0.705, 0.795).
The contextual conclusion is that we are 90% confident the population proportion of families responding to the annual survey is between 0.705 and 0.795, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.045 quantifies precision rather than the spread of individual binary outcomes.
Interval case 14: Local Election Office
A random sample of 263 observations found 103 mailed ballots returned before election day. The sample proportion is p̂=0.392. For a 95% one proportion z interval, the observed success and failure counts are 103 and 160, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0301; with z*=1.960, the margin of error is 0.0590. The resulting interval is (0.333, 0.451).
The contextual conclusion is that we are 95% confident the population proportion of mailed ballots returned before election day is between 0.333 and 0.451, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.059 quantifies precision rather than the spread of individual binary outcomes.
Interval case 15: Energy-Efficiency Pilot
A random sample of 274 observations found 126 homes meeting the target reduction. The sample proportion is p̂=0.460. For a 99% one proportion z interval, the observed success and failure counts are 126 and 148, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0301; with z*=2.576, the margin of error is 0.0776. The resulting interval is (0.382, 0.537).
The contextual conclusion is that we are 99% confident the population proportion of homes meeting the target reduction is between 0.382 and 0.537, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.078 quantifies precision rather than the spread of individual binary outcomes.
Interval case 16: Community Broadband Project
A random sample of 285 observations found 151 households achieving the advertised speed. The sample proportion is p̂=0.530. For a 90% one proportion z interval, the observed success and failure counts are 151 and 134, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0296; with z*=1.645, the margin of error is 0.0486. The resulting interval is (0.481, 0.578).
The contextual conclusion is that we are 90% confident the population proportion of households achieving the advertised speed is between 0.481 and 0.578, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.049 quantifies precision rather than the spread of individual binary outcomes.
Interval case 17: Regional Bus Network
A random sample of 296 observations found 178 trips arriving within the on-time window. The sample proportion is p̂=0.601. For a 95% one proportion z interval, the observed success and failure counts are 178 and 118, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0285; with z*=1.960, the margin of error is 0.0558. The resulting interval is (0.546, 0.657).
The contextual conclusion is that we are 95% confident the population proportion of trips arriving within the on-time window is between 0.546 and 0.657, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.056 quantifies precision rather than the spread of individual binary outcomes.
Interval case 18: Campus Dining Service
A random sample of 307 observations found 206 transactions using reusable containers. The sample proportion is p̂=0.671. For a 99% one proportion z interval, the observed success and failure counts are 206 and 101, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0268; with z*=2.576, the margin of error is 0.0691. The resulting interval is (0.602, 0.740).
The contextual conclusion is that we are 99% confident the population proportion of transactions using reusable containers is between 0.602 and 0.740, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.069 quantifies precision rather than the spread of individual binary outcomes.
Interval case 19: Workforce Training Program
A random sample of 318 observations found 235 participants earning the credential. The sample proportion is p̂=0.739. For a 90% one proportion z interval, the observed success and failure counts are 235 and 83, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0246; with z*=1.645, the margin of error is 0.0405. The resulting interval is (0.698, 0.780).
The contextual conclusion is that we are 90% confident the population proportion of participants earning the credential is between 0.698 and 0.780, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.041 quantifies precision rather than the spread of individual binary outcomes.
Interval case 20: County Recycling Audit
A random sample of 329 observations found 125 sampled loads meeting contamination limits. The sample proportion is p̂=0.380. For a 95% one proportion z interval, the observed success and failure counts are 125 and 204, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0268; with z*=1.960, the margin of error is 0.0524. The resulting interval is (0.327, 0.432).
The contextual conclusion is that we are 95% confident the population proportion of sampled loads meeting contamination limits is between 0.327 and 0.432, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.052 quantifies precision rather than the spread of individual binary outcomes.
Interval case 21: University Residence Halls
A random sample of 340 observations found 153 rooms passing the first safety inspection. The sample proportion is p̂=0.450. For a 99% one proportion z interval, the observed success and failure counts are 153 and 187, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0270; with z*=2.576, the margin of error is 0.0695. The resulting interval is (0.381, 0.519).
The contextual conclusion is that we are 99% confident the population proportion of rooms passing the first safety inspection is between 0.381 and 0.519, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.069 quantifies precision rather than the spread of individual binary outcomes.
Interval case 22: Telehealth Pilot
A random sample of 351 observations found 183 appointments completed without rescheduling. The sample proportion is p̂=0.521. For a 90% one proportion z interval, the observed success and failure counts are 183 and 168, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0267; with z*=1.645, the margin of error is 0.0439. The resulting interval is (0.478, 0.565).
The contextual conclusion is that we are 90% confident the population proportion of appointments completed without rescheduling is between 0.478 and 0.565, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.044 quantifies precision rather than the spread of individual binary outcomes.
Interval case 23: Public Museum
A random sample of 362 observations found 214 visitors using the audio guide. The sample proportion is p̂=0.591. For a 95% one proportion z interval, the observed success and failure counts are 214 and 148, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0258; with z*=1.960, the margin of error is 0.0506. The resulting interval is (0.541, 0.642).
The contextual conclusion is that we are 95% confident the population proportion of visitors using the audio guide is between 0.541 and 0.642, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.051 quantifies precision rather than the spread of individual binary outcomes.
Interval case 24: Youth Sports League
A random sample of 373 observations found 246 players completing concussion training. The sample proportion is p̂=0.660. For a 99% one proportion z interval, the observed success and failure counts are 246 and 127, so the large-count check is satisfied. The estimated standard error is √[p̂(1−p̂)/n]=0.0245; with z*=2.576, the margin of error is 0.0632. The resulting interval is (0.596, 0.723).
The contextual conclusion is that we are 99% confident the population proportion of players completing concussion training is between 0.596 and 0.723, assuming the sampling design supports inference. Notice that the interval uses p̂ in the standard error; substituting a benchmark p₀ would be test logic, not interval logic. Here the half-width of 0.063 quantifies precision rather than the spread of individual binary outcomes.
One Proportion Z Interval multiple-choice practice
Question 1. One-Proportion z Interval
In an SRS of 120 appointments from a university advising center in Midwest consortium during a yearly program evaluation, 57 meet the appointment wait time criterion. Construct a 95% one-proportion z interval and interpret it.
Answer: C
p̂=57120=0.475. Conditions include randomization/independence and 57 successes and 63 failures, both at least 10. SE=√p̂(1−p̂)n=0.0456; z*=1.96; ME=0.089. The interval is (0.386, 0.564). We are 95% confident that the true population proportion lies in this interval.
Question 2. One-Proportion z Interval
In an SRS of 150 parts from a regional manufacturer in Atlantic Corridor during a winter readiness review, 90 meet the part diameter criterion. Construct a 95% one-proportion z interval and interpret it.
Answer: A
p̂=90150=0.6. Conditions include randomization/independence and 90 successes and 60 failures, both at least 10. SE=√p̂(1−p̂)n=0.04; z*=1.96; ME=0.078. The interval is (0.522, 0.678). We are 95% confident that the true population proportion lies in this interval.
Question 3. One-Proportion z Interval
In an SRS of 200 students from a public high school in Metro East during a community outreach cycle, 136 meet the algebra benchmark completion criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: D
p̂=136200=0.68. Conditions include randomization/independence and 136 successes and 64 failures, both at least 10. SE=√p̂(1−p̂)n=0.033; z*=1.645; ME=0.054. The interval is (0.626, 0.734). We are 90% confident that the true population proportion lies in this interval.
Question 4. One-Proportion z Interval
In an SRS of 370 enrolled learners from a community college in Midwest consortium during a spring 2027 pilot, 169 meet the course completion criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: B
p̂=169370=0.457. Conditions include randomization/independence and 169 successes and 201 failures, both at least 10. SE=√p̂(1−p̂)n=0.0259; z*=1.645; ME=0.043. The interval is (0.414, 0.499). We are 90% confident that the true population proportion lies in this interval.
Question 5. One-Proportion z Interval
In an SRS of 200 visitors from a county library in Mountain Region during a monthly quality review, 98 meet the weekly program attendance criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: C
p̂=98200=0.49. Conditions include randomization/independence and 98 successes and 102 failures, both at least 10. SE=√p̂(1−p̂)n=0.0353; z*=1.645; ME=0.058. The interval is (0.432, 0.548). We are 90% confident that the true population proportion lies in this interval.
Question 6. One-Proportion z Interval
In an SRS of 290 samples from a food safety laboratory in Mountain Region during a two-month observation window, 156 meet the sample concentration criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: C
p̂=156290=0.538. Conditions include randomization/independence and 156 successes and 134 failures, both at least 10. SE=√p̂(1−p̂)n=0.0293; z*=1.645; ME=0.048. The interval is (0.49, 0.586). We are 90% confident that the true population proportion lies in this interval.
Question 7. One-Proportion z Interval
In an SRS of 260 visitors from a state park in Riverbend during a service-improvement study, 150 meet the trail-use duration criterion. Construct a 99% one-proportion z interval and interpret it.
Answer: D
p̂=150260=0.577. Conditions include randomization/independence and 150 successes and 110 failures, both at least 10. SE=√p̂(1−p̂)n=0.0306; z*=2.576; ME=0.079. The interval is (0.498, 0.656). We are 99% confident that the true population proportion lies in this interval.
Question 8. One-Proportion z Interval
In an SRS of 220 students from a school district in Atlantic Corridor during a semester-long cohort study, 123 meet the lunch-program participation criterion. Construct a 95% one-proportion z interval and interpret it.
Answer: A
p̂=123220=0.559. Conditions include randomization/independence and 123 successes and 97 failures, both at least 10. SE=√p̂(1−p̂)n=0.0335; z*=1.96; ME=0.066. The interval is (0.493, 0.625). We are 95% confident that the true population proportion lies in this interval.
Question 9. One-Proportion z Interval
In an SRS of 380 ballots from a county election office in Sunbelt district during a two-month observation window, 123 meet the ballot-processing time criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: A
p̂=123380=0.324. Conditions include randomization/independence and 123 successes and 257 failures, both at least 10. SE=√p̂(1−p̂)n=0.024; z*=1.645; ME=0.039. The interval is (0.284, 0.363). We are 90% confident that the true population proportion lies in this interval.
Question 10. One-Proportion z Interval
In an SRS of 190 accounts from a municipal water office in Midwest consortium during a multiweek validation study, 133 meet the monthly household use criterion. Construct a 95% one-proportion z interval and interpret it.
Answer: B
p̂=133190=0.7. Conditions include randomization/independence and 133 successes and 57 failures, both at least 10. SE=√p̂(1−p̂)n=0.0332; z*=1.96; ME=0.065. The interval is (0.635, 0.765). We are 95% confident that the true population proportion lies in this interval.
Question 11. One-Proportion z Interval
In an SRS of 250 customers from a grocery cooperative in Pine Ridge during a fall 2026 audit, 134 meet the checkout time criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: D
One-Proportion z Interval: Formula, Conditions, and Example — Question 11. One-Proportion z Interval: p̂=134250=0.536. Conditions include randomization/independence and 134 successes and 116 failures, both at least 10. SE=√p̂(1−p̂)n=0.0315; z*=1.645; ME=0.052. The interval is (0.484, 0.588). We are 90% confident that the true population proportion lies in this interval.
Question 12. One-Proportion z Interval
In an SRS of 350 enrolled learners from a community college in Mountain Region during a fall 2026 audit, 105 meet the course completion criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: C
p̂=105350=0.3. Conditions include randomization/independence and 105 successes and 245 failures, both at least 10. SE=√p̂(1−p̂)n=0.0245; z*=1.645; ME=0.04. The interval is (0.26, 0.34). We are 90% confident that the true population proportion lies in this interval.
Question 13. One-Proportion z Interval
In an SRS of 300 customers from a community bank in Atlantic Corridor during a two-month observation window, 139 meet the mobile-deposit adoption criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: D
p̂=139300=0.463. Conditions include randomization/independence and 139 successes and 161 failures, both at least 10. SE=√p̂(1−p̂)n=0.0288; z*=1.645; ME=0.047. The interval is (0.416, 0.511). We are 90% confident that the true population proportion lies in this interval.
Question 14. One-Proportion z Interval
In an SRS of 350 plots from a farm cooperative in Capital Region during a semester-long cohort study, 193 meet the crop yield criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: D
p̂=193350=0.551. Conditions include randomization/independence and 193 successes and 157 failures, both at least 10. SE=√p̂(1−p̂)n=0.0266; z*=1.645; ME=0.044. The interval is (0.508, 0.595). We are 90% confident that the true population proportion lies in this interval.
Question 15. One-Proportion z Interval
In an SRS of 230 installations from a solar installer in Lakeside district during a six-week field trial, 111 meet the daily energy output criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: B
p̂=111230=0.483. Conditions include randomization/independence and 111 successes and 119 failures, both at least 10. SE=√p̂(1−p̂)n=0.0329; z*=1.645; ME=0.054. The interval is (0.428, 0.537). We are 90% confident that the true population proportion lies in this interval.
Question 16. One-Proportion z Interval
In an SRS of 300 enrolled learners from a community college in Desert County during a summer implementation review, 120 meet the course completion criterion. Construct a 99% one-proportion z interval and interpret it.
Answer: A
p̂=120300=0.4. Conditions include randomization/independence and 120 successes and 180 failures, both at least 10. SE=√p̂(1−p̂)n=0.0283; z*=2.576; ME=0.073. The interval is (0.327, 0.473). We are 99% confident that the true population proportion lies in this interval.
Question 17. One-Proportion z Interval
In an SRS of 370 visitors from a county library in Sunbelt district during a summer implementation review, 135 meet the weekly program attendance criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: A
p̂=135370=0.365. Conditions include randomization/independence and 135 successes and 235 failures, both at least 10. SE=√p̂(1−p̂)n=0.025; z*=1.645; ME=0.041. The interval is (0.324, 0.406). We are 90% confident that the true population proportion lies in this interval.
Question 18. One-Proportion z Interval
In an SRS of 250 installations from a solar installer in Capital Region during a six-week field trial, 151 meet the daily energy output criterion. Construct a 90% one-proportion z interval and interpret it.
Answer: B
p̂=151250=0.604. Conditions include randomization/independence and 151 successes and 99 failures, both at least 10. SE=√p̂(1−p̂)n=0.0309; z*=1.645; ME=0.051. The interval is (0.553, 0.655). We are 90% confident that the true population proportion lies in this interval.
Question 19. One-Proportion z Interval
In an SRS of 150 appointments from a university advising center in Great Lakes during a pre-exam training cycle, 63 meet the appointment wait time criterion. Construct a 95% one-proportion z interval and interpret it.
Answer: D
p̂=63150=0.42. Conditions include randomization/independence and 63 successes and 87 failures, both at least 10. SE=√p̂(1−p̂)n=0.0403; z*=1.96; ME=0.079. The interval is (0.341, 0.499). We are 95% confident that the true population proportion lies in this interval.
Question 20. One-Proportion z Interval
In an SRS of 340 appointments from a university advising center in Westview during a baseline measurement week, 124 meet the appointment wait time criterion. Construct a 95% one-proportion z interval and interpret it.
Answer: A
p̂=124340=0.365. Conditions include randomization/independence and 124 successes and 216 failures, both at least 10. SE=√p̂(1−p̂)n=0.0261; z*=1.96; ME=0.051. The interval is (0.314, 0.416). We are 95% confident that the true population proportion lies in this interval.
One Proportion Z Interval free-response practice
FRQ set 1: One-Proportion z Interval
Scenario. In an SRS of 270 ballots from a county election office in Pacific Northwest during a service-improvement study, 156 meet the ballot-processing time criterion. Construct a 90% one-proportion z interval and interpret it.
- Define the population parameter and preserve the stated group order.
- Verify randomization, independence, and procedure-specific success/failure or shape conditions.
- Construct the interval from the statistic, critical value, and standard error.
- Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.
Model response
p̂=156270=0.578. Conditions include randomization/independence and 156 successes and 114 failures, both at least 10. SE=√p̂(1−p̂)n=0.0301; z*=1.645; ME=0.049. The interval is (0.528, 0.627). We are 90% confident that the true population proportion lies in this interval.
FRQ set 2: One-Proportion z Interval
Scenario. In an SRS of 150 visitors from a county library in Desert County during a six-week field trial, 53 meet the weekly program attendance criterion. Construct a 90% one-proportion z interval and interpret it.
- Define the population parameter and preserve the stated group order.
- Verify randomization, independence, and procedure-specific success/failure or shape conditions.
- Construct the interval from the statistic, critical value, and standard error.
- Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.
Model response
p̂=53150=0.353. Conditions include randomization/independence and 53 successes and 97 failures, both at least 10. SE=√p̂(1−p̂)n=0.039; z*=1.645; ME=0.064. The interval is (0.289, 0.418). We are 90% confident that the true population proportion lies in this interval.
FRQ set 3: One-Proportion z Interval
Scenario. In an SRS of 310 patients from a regional hospital in Westview during a yearly program evaluation, 191 meet the appointment completion criterion. Construct a 95% one-proportion z interval and interpret it.
- Define the population parameter and preserve the stated group order.
- Verify randomization, independence, and procedure-specific success/failure or shape conditions.
- Construct the interval from the statistic, critical value, and standard error.
- Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.
Model response
p̂=191310=0.616. Conditions include randomization/independence and 191 successes and 119 failures, both at least 10. SE=√p̂(1−p̂)n=0.0276; z*=1.96; ME=0.054. The interval is (0.562, 0.67). We are 95% confident that the true population proportion lies in this interval.
FRQ set 4: One-Proportion z Interval
Scenario. In an SRS of 240 parts from a regional manufacturer in North Valley during a monthly quality review, 85 meet the part diameter criterion. Construct a 99% one-proportion z interval and interpret it.
- Define the population parameter and preserve the stated group order.
- Verify randomization, independence, and procedure-specific success/failure or shape conditions.
- Construct the interval from the statistic, critical value, and standard error.
- Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.
Model response
p̂=85240=0.354. Conditions include randomization/independence and 85 successes and 155 failures, both at least 10. SE=√p̂(1−p̂)n=0.0309; z*=2.576; ME=0.08. The interval is (0.275, 0.434). We are 99% confident that the true population proportion lies in this interval.
FRQ set 5: One-Proportion z Interval
Scenario. In an SRS of 260 parts from a regional manufacturer in Midwest consortium during a weekday operations study, 101 meet the part diameter criterion. Construct a 95% one-proportion z interval and interpret it.
- Define the population parameter and preserve the stated group order.
- Verify randomization, independence, and procedure-specific success/failure or shape conditions.
- Construct the interval from the statistic, critical value, and standard error.
- Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.
Model response
p̂=101260=0.388. Conditions include randomization/independence and 101 successes and 159 failures, both at least 10. SE=√p̂(1−p̂)n=0.0302; z*=1.96; ME=0.059. The interval is (0.329, 0.448). We are 95% confident that the true population proportion lies in this interval.
FRQ set 6: One-Proportion z Interval
Scenario. In an SRS of 180 visitors from a county library in North Valley during a pre-exam training cycle, 113 meet the weekly program attendance criterion. Construct a 95% one-proportion z interval and interpret it.
- Define the population parameter and preserve the stated group order.
- Verify randomization, independence, and procedure-specific success/failure or shape conditions.
- Construct the interval from the statistic, critical value, and standard error.
- Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.
Model response
p̂=113180=0.628. Conditions include randomization/independence and 113 successes and 67 failures, both at least 10. SE=√p̂(1−p̂)n=0.036; z*=1.96; ME=0.071. The interval is (0.557, 0.698). We are 95% confident that the true population proportion lies in this interval.
FRQ set 7: One-Proportion z Interval
Scenario. In an SRS of 180 calls from a municipal emergency dispatch center in Capital Region during a school-year data collection, 113 meet the response time criterion. Construct a 99% one-proportion z interval and interpret it.
- Define the population parameter and preserve the stated group order.
- Verify randomization, independence, and procedure-specific success/failure or shape conditions.
- Construct the interval from the statistic, critical value, and standard error.
- Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.
Model response
p̂=113180=0.628. Conditions include randomization/independence and 113 successes and 67 failures, both at least 10. SE=√p̂(1−p̂)n=0.036; z*=2.576; ME=0.093. The interval is (0.535, 0.721). We are 99% confident that the true population proportion lies in this interval.
FRQ set 8: One-Proportion z Interval
Scenario. In an SRS of 210 parts from a regional manufacturer in Coastal Plains during a multiweek validation study, 120 meet the part diameter criterion. Construct a 99% one-proportion z interval and interpret it.
- Define the population parameter and preserve the stated group order.
- Verify randomization, independence, and procedure-specific success/failure or shape conditions.
- Construct the interval from the statistic, critical value, and standard error.
- Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.
Model response
p̂=120210=0.571. Conditions include randomization/independence and 120 successes and 90 failures, both at least 10. SE=√p̂(1−p̂)n=0.0341; z*=2.576; ME=0.088. The interval is (0.483, 0.659). We are 99% confident that the true population proportion lies in this interval.
Next steps after mastering one proportion z interval
After this interval page, compare the same sample with a one-proportion z test. Write down exactly which quantity changes in the standard error: the interval estimates uncertainty with p̂, whereas the null test constructs its reference model with p₀. Then practice translating a desired margin of error into a required sample size so interval construction and planning are not confused.
For error review, separate four failure types: using a benchmark proportion in the interval SE, skipping observed success/failure counts, treating confidence as probability for a fixed parameter, and generalizing beyond the sampling design. Rework one interval at 90%, 95%, and 99% confidence to see how z* changes width while p̂ stays fixed.
