One-Proportion z Test: Formula, Conditions, and Example

AP Statistics Topic Guide

One-Proportion z Test: Formula, Conditions, and Example

Test a claim about one population proportion using the null value in the standard error, match the p-value tail to the alternative hypothesis, and state the conclusion about the population parameter.

StatusCurrent AP Statistics inference method

Main keywordone proportion z test

Worked analysis24 cases

Focused practice20 MCQs + 8 FRQs

Study progress0 completed

One-Proportion z Test: Formula, Conditions, and Example: direct answer

In a one proportion z test problem, begin by identifying the population target and the inferential role of the data. Test a claim about one population proportion using the null value in the standard error, match the p-value tail to the alternative hypothesis, and state the conclusion about the population parameter.

This page uses one proportion z test 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 Test: Formula, Conditions, and Example

One-Proportion z Test: Formula, Conditions, and Example quick reference
Parameter	p: one population proportion
Hypotheses	H₀:p=p₀; Hₐ according to the claim
Null standard error	√[p₀(1−p₀)/n]
Statistic	z=(p̂−p₀)/SE₀
Count check	np₀ and n(1−p₀) sufficiently large
Decision	Compare the tail p-value with α

Concept mastery: one proportion z test

State H0 and Ha for the population proportion

Define p in context, then write H0:p=p0 and the directional or two-sided alternative that matches the claim. The observed p-hat belongs in the test statistic, not in the hypotheses.

Use the null proportion in the standard error

The one-proportion z test assumes H0 while measuring how unusual the sample result would be. Therefore its standard error is sqrt[p0(1-p0)/n], not the p-hat-based standard error used for a confidence interval.

Check expected counts under H0

For the normal approximation, inspect n*p0 and n*(1-p0), because the test’s reference distribution is built under the null model. This differs from a confidence interval, which checks observed successes and failures.

Check randomization and independence

The sample or randomized process must support the inferential model. When sampling without replacement from a finite population, the usual ten-percent guideline helps justify approximate independence.

Compute z as observed minus null over null SE

The statistic z=(p-hat-p0)/sqrt[p0(1-p0)/n] tells how many null standard errors the sample proportion lies from the claimed value. Its sign records direction; its magnitude records extremeness.

Match the p-value tail to Ha

A less-than alternative uses the lower tail, a greater-than alternative uses the upper tail, and a not-equal alternative uses both tails. The tail choice is determined before looking at the sample statistic.

Use alpha only as the decision threshold

If p<=alpha, reject H0; otherwise fail to reject H0. Alpha does not measure the probability that the null is true, and the p-value is not the probability that a Type I error occurred.

Conclude about the population proportion

The conclusion should say whether the data provide convincing evidence for Ha in context. Avoid declaring that the null proportion is proven or that the sample result must exactly represent the population.

Compare with an interval carefully

A two-sided z test at alpha=.05 is closely connected to whether a 95% confidence interval includes p0, but the standard errors are constructed differently. Use each procedure’s formula rather than substituting one calculation for the other.

Distinguish significance from size of departure

A small difference between p-hat and p0 can become significant with a very large n. The z test answers an evidence question; practical importance requires attention to the magnitude and consequences of the difference.

Worked analysis for one proportion z test

z-test case 1: Regional Hospital

For patients who received follow-up within two days, test H₀:p=0.35 against Hₐ:p<0.35. A sample of n=150 gives x=40 and p̂=0.267. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0389, so z=(p̂−p₀)/SE₀=-2.14. The less alternative gives p=0.0162.

At α=0.01, the decision is to fail to reject H₀ because 0.0162 > 0.01. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 2: Community College

For students who completed the placement module, test H₀:p=0.40 against Hₐ:p>0.40. A sample of n=159 gives x=57 and p̂=0.358. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0389, so z=(p̂−p₀)/SE₀=-1.07. The greater alternative gives p=0.8573.

At α=0.05, the decision is to fail to reject H₀ because 0.8573 > 0.05. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 3: County Library

For visitors who attended a weekly program, test H₀:p=0.50 against Hₐ:p≠0.50. A sample of n=168 gives x=89 and p̂=0.530. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0386, so z=(p̂−p₀)/SE₀=0.77. The two-sided alternative gives p=0.4404.

At α=0.10, the decision is to fail to reject H₀ because 0.4404 > 0.10. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 4: Public High School

For seniors who submitted the financial-aid form, test H₀:p=0.60 against Hₐ:p<0.60. A sample of n=177 gives x=117 and p̂=0.661. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0368, so z=(p̂−p₀)/SE₀=1.66. The less alternative gives p=0.9512.

At α=0.01, the decision is to fail to reject H₀ because 0.9512 > 0.01. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 5: Municipal Water Office

For households reporting no service interruption, test H₀:p=0.65 against Hₐ:p>0.65. A sample of n=186 gives x=138 and p̂=0.742. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0350, so z=(p̂−p₀)/SE₀=2.63. The greater alternative gives p=0.0043.

At α=0.05, the decision is to reject H₀ because 0.0043 ≤ 0.05. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 6: State Park

For visitors who used the marked trail system, test H₀:p=0.35 against Hₐ:p≠0.35. A sample of n=195 gives x=53 and p̂=0.272. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0342, so z=(p̂−p₀)/SE₀=-2.29. The two-sided alternative gives p=0.0220.

At α=0.10, the decision is to reject H₀ because 0.0220 ≤ 0.10. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 7: Food Cooperative

For members who renewed before the deadline, test H₀:p=0.40 against Hₐ:p<0.40. A sample of n=204 gives x=73 and p̂=0.358. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0343, so z=(p̂−p₀)/SE₀=-1.23. The less alternative gives p=0.1095.

At α=0.01, the decision is to fail to reject H₀ because 0.1095 > 0.01. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 8: University Advising Center

For appointments starting within ten minutes, test H₀:p=0.50 against Hₐ:p>0.50. A sample of n=213 gives x=113 and p̂=0.531. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0343, so z=(p̂−p₀)/SE₀=0.89. The greater alternative gives p=0.1865.

At α=0.05, the decision is to fail to reject H₀ because 0.1865 > 0.05. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 9: Regional Manufacturer

For parts meeting the diameter specification, test H₀:p=0.60 against Hₐ:p≠0.60. A sample of n=222 gives x=147 and p̂=0.662. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0329, so z=(p̂−p₀)/SE₀=1.89. The two-sided alternative gives p=0.0587.

At α=0.10, the decision is to reject H₀ because 0.0587 ≤ 0.10. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 10: Public Health Clinic

For clients returning for the scheduled checkup, test H₀:p=0.65 against Hₐ:p<0.65. A sample of n=231 gives x=171 and p̂=0.740. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0314, so z=(p̂−p₀)/SE₀=2.88. The less alternative gives p=0.9980.

At α=0.01, the decision is to fail to reject H₀ because 0.9980 > 0.01. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 11: Urban Recreation Program

For participants completing the eight-week session, test H₀:p=0.35 against Hₐ:p>0.35. A sample of n=240 gives x=65 and p̂=0.271. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0308, so z=(p̂−p₀)/SE₀=-2.57. The greater alternative gives p=0.9949.

At α=0.05, the decision is to fail to reject H₀ because 0.9949 > 0.05. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 12: School District

For families responding to the annual survey, test H₀:p=0.40 against Hₐ:p≠0.40. A sample of n=249 gives x=90 and p̂=0.361. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0310, so z=(p̂−p₀)/SE₀=-1.24. The two-sided alternative gives p=0.2143.

At α=0.10, the decision is to fail to reject H₀ because 0.2143 > 0.10. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 13: Local Election Office

For mailed ballots returned before election day, test H₀:p=0.50 against Hₐ:p<0.50. A sample of n=258 gives x=137 and p̂=0.531. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0311, so z=(p̂−p₀)/SE₀=1.00. The less alternative gives p=0.8404.

At α=0.01, the decision is to fail to reject H₀ because 0.8404 > 0.01. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 14: Energy-Efficiency Pilot

For homes meeting the target reduction, test H₀:p=0.60 against Hₐ:p>0.60. A sample of n=267 gives x=176 and p̂=0.659. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0300, so z=(p̂−p₀)/SE₀=1.97. The greater alternative gives p=0.0242.

At α=0.05, the decision is to reject H₀ because 0.0242 ≤ 0.05. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 15: Community Broadband Project

For households achieving the advertised speed, test H₀:p=0.65 against Hₐ:p≠0.65. A sample of n=276 gives x=204 and p̂=0.739. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0287, so z=(p̂−p₀)/SE₀=3.10. The two-sided alternative gives p=0.0019.

At α=0.10, the decision is to reject H₀ because 0.0019 ≤ 0.10. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 16: Regional Bus Network

For trips arriving within the on-time window, test H₀:p=0.35 against Hₐ:p<0.35. A sample of n=285 gives x=77 and p̂=0.270. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0283, so z=(p̂−p₀)/SE₀=-2.83. The less alternative gives p=0.0024.

At α=0.01, the decision is to reject H₀ because 0.0024 ≤ 0.01. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 17: Campus Dining Service

For transactions using reusable containers, test H₀:p=0.40 against Hₐ:p>0.40. A sample of n=294 gives x=106 and p̂=0.361. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0286, so z=(p̂−p₀)/SE₀=-1.38. The greater alternative gives p=0.9164.

At α=0.05, the decision is to fail to reject H₀ because 0.9164 > 0.05. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 18: Workforce Training Program

For participants earning the credential, test H₀:p=0.50 against Hₐ:p≠0.50. A sample of n=303 gives x=161 and p̂=0.531. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0287, so z=(p̂−p₀)/SE₀=1.09. The two-sided alternative gives p=0.2750.

At α=0.10, the decision is to fail to reject H₀ because 0.2750 > 0.10. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 19: County Recycling Audit

For sampled loads meeting contamination limits, test H₀:p=0.60 against Hₐ:p<0.60. A sample of n=312 gives x=206 and p̂=0.660. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0277, so z=(p̂−p₀)/SE₀=2.17. The less alternative gives p=0.9851.

At α=0.01, the decision is to fail to reject H₀ because 0.9851 > 0.01. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 20: University Residence Halls

For rooms passing the first safety inspection, test H₀:p=0.65 against Hₐ:p>0.65. A sample of n=321 gives x=238 and p̂=0.741. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0266, so z=(p̂−p₀)/SE₀=3.43. The greater alternative gives p=0.0003.

At α=0.05, the decision is to reject H₀ because 0.0003 ≤ 0.05. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 21: Telehealth Pilot

For appointments completed without rescheduling, test H₀:p=0.35 against Hₐ:p≠0.35. A sample of n=330 gives x=89 and p̂=0.270. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0263, so z=(p̂−p₀)/SE₀=-3.06. The two-sided alternative gives p=0.0022.

At α=0.10, the decision is to reject H₀ because 0.0022 ≤ 0.10. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 22: Public Museum

For visitors using the audio guide, test H₀:p=0.40 against Hₐ:p<0.40. A sample of n=339 gives x=122 and p̂=0.360. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0266, so z=(p̂−p₀)/SE₀=-1.51. The less alternative gives p=0.0658.

At α=0.01, the decision is to fail to reject H₀ because 0.0658 > 0.01. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 23: Youth Sports League

For players completing concussion training, test H₀:p=0.50 against Hₐ:p>0.50. A sample of n=348 gives x=184 and p̂=0.529. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0268, so z=(p̂−p₀)/SE₀=1.07. The greater alternative gives p=0.1418.

At α=0.05, the decision is to fail to reject H₀ because 0.1418 > 0.05. The correct conclusion is that the data do not provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

z-test case 24: Rural Pharmacy Network

For prescriptions filled within the service target, test H₀:p=0.60 against Hₐ:p≠0.60. A sample of n=357 gives x=236 and p̂=0.661. Under H₀, SE₀=√[p₀(1−p₀)/n]=0.0259, so z=(p̂−p₀)/SE₀=2.36. The two-sided alternative gives p=0.0185.

At α=0.10, the decision is to reject H₀ because 0.0185 ≤ 0.10. The correct conclusion is that the data provide convincing evidence for Hₐ in this population. The standard error uses p₀ because the reference distribution is constructed under H₀; replacing it with p̂ would mix the confidence-interval formula into the significance test.

One Proportion Z Test multiple-choice practice

Question 1. One-Proportion z Test

A housing authority in Cedar Grove during a semester-long cohort study tests H₀:p=0.4 against H_a:p<0.4 using 97 successes in n=210. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Use p̂ in the null SE and report p=0.462.
B. Interpret the p-value as the probability that H₀ is true.
C. Always reject H₀ when the sample proportion differs numerically from p₀.
D. z=1.831, p=0.9665; fail to reject H₀.

Answer: D

p̂=97210=0.462. Under H₀, SE=√0.4(0.6)210=0.0338 and z=(0.462−0.4)0.0338=1.831. The left-tailed p-value is 0.9665. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 2. One-Proportion z Test

A recycling program in Desert County during a monthly quality review tests H₀:p=0.6 against H_a:p<0.6 using 224 successes in n=330. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Interpret the p-value as the probability that H₀ is true.
B. Use p̂ in the null SE and report p=0.679.
C. Always reject H₀ when the sample proportion differs numerically from p₀.
D. z=2.922, p=0.9983; fail to reject H₀.

Answer: D

p̂=224330=0.679. Under H₀, SE=√0.6(0.4)330=0.027 and z=(0.679−0.6)0.027=2.922. The left-tailed p-value is 0.9983. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 3. One-Proportion z Test

A regional airport authority in Riverbend during a fall 2026 audit tests H₀:p=0.7 against H_a:p>0.7 using 141 successes in n=220. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Use p̂ in the null SE and report p=0.641.
B. Always reject H₀ when the sample proportion differs numerically from p₀.
C. z=-1.913, p=0.9721; fail to reject H₀.
D. Interpret the p-value as the probability that H₀ is true.

Answer: C

p̂=141220=0.641. Under H₀, SE=√0.7(0.3)220=0.0309 and z=(0.641−0.7)0.0309=-1.913. The right-tailed p-value is 0.9721. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 4. One-Proportion z Test

A regional airport authority in Metro East during a semester-long cohort study tests H₀:p=0.7 against H_a:p<0.7 using 274 successes in n=360. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. Use p̂ in the null SE and report p=0.761.
C. z=2.53, p=0.9943; fail to reject H₀.
D. Interpret the p-value as the probability that H₀ is true.

Answer: C

p̂=274360=0.761. Under H₀, SE=√0.7(0.3)360=0.0242 and z=(0.761−0.7)0.0242=2.53. The left-tailed p-value is 0.9943. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 5. One-Proportion z Test

A city transit agency in Capital Region during a monthly quality review tests H₀:p=0.4 against H_a:p≠0.4 using 110 successes in n=240. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. z=1.845, p=0.0651; fail to reject H₀.
C. Interpret the p-value as the probability that H₀ is true.
D. Use p̂ in the null SE and report p=0.458.

Answer: B

p̂=110240=0.458. Under H₀, SE=√0.4(0.6)240=0.0316 and z=(0.458−0.4)0.0316=1.845. The two-sided p-value is 0.0651. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 6. One-Proportion z Test

A housing authority in Lakeside district during a six-week field trial tests H₀:p=0.7 against H_a:p<0.7 using 137 successes in n=180. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. z=1.789, p=0.9632; fail to reject H₀.
C. Interpret the p-value as the probability that H₀ is true.
D. Use p̂ in the null SE and report p=0.761.

Answer: B

p̂=137180=0.761. Under H₀, SE=√0.7(0.3)180=0.0342 and z=(0.761−0.7)0.0342=1.789. The left-tailed p-value is 0.9632. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 7. One-Proportion z Test

A community college in Mountain Region during a multiweek validation study tests H₀:p=0.4 against H_a:p>0.4 using 115 successes in n=250. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. Interpret the p-value as the probability that H₀ is true.
C. z=1.936, p=0.0264; reject H₀.
D. Use p̂ in the null SE and report p=0.46.

Answer: C

p̂=115250=0.46. Under H₀, SE=√0.4(0.6)250=0.031 and z=(0.46−0.4)0.031=1.936. The right-tailed p-value is 0.0264. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 8. One-Proportion z Test

A school district in New England network during a winter readiness review tests H₀:p=0.4 against H_a:p>0.4 using 158 successes in n=330. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. Use p̂ in the null SE and report p=0.479.
C. z=2.922, p=0.0017; reject H₀.
D. Interpret the p-value as the probability that H₀ is true.

Answer: C

p̂=158330=0.479. Under H₀, SE=√0.4(0.6)330=0.027 and z=(0.479−0.4)0.027=2.922. The right-tailed p-value is 0.0017. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 9. One-Proportion z Test

A solar installer in Midwest consortium during a quarterly performance study tests H₀:p=0.5 against H_a:p>0.5 using 202 successes in n=360. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Use p̂ in the null SE and report p=0.561.
B. z=2.319, p=0.0102; reject H₀.
C. Interpret the p-value as the probability that H₀ is true.
D. Always reject H₀ when the sample proportion differs numerically from p₀.

Answer: B

One-Proportion z Test: Formula, Conditions, and Example — Question 9. One-Proportion z Test: p̂=202360=0.561. Under H₀, SE=√0.5(0.5)360=0.0264 and z=(0.561−0.5)0.0264=2.319. The right-tailed p-value is 0.0102. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 10. One-Proportion z Test

A grocery cooperative in Westview during a quarterly performance study tests H₀:p=0.7 against H_a:p≠0.7 using 226 successes in n=290. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. Interpret the p-value as the probability that H₀ is true.
C. Use p̂ in the null SE and report p=0.779.
D. z=2.947, p=0.0032; reject H₀.

Answer: D

p̂=226290=0.779. Under H₀, SE=√0.7(0.3)290=0.0269 and z=(0.779−0.7)0.0269=2.947. The two-sided p-value is 0.0032. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 11. One-Proportion z Test

A city transit agency in Atlantic Corridor during a community outreach cycle tests H₀:p=0.5 against H_a:p≠0.5 using 66 successes in n=150. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. z=-1.47, p=0.1416; fail to reject H₀.
C. Use p̂ in the null SE and report p=0.44.
D. Interpret the p-value as the probability that H₀ is true.

Answer: B

p̂=66150=0.44. Under H₀, SE=√0.5(0.5)150=0.0408 and z=(0.44−0.5)0.0408=-1.47. The two-sided p-value is 0.1416. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 12. One-Proportion z Test

A city transit agency in Lakeside district during a spring 2027 pilot tests H₀:p=0.5 against H_a:p>0.5 using 140 successes in n=250. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Interpret the p-value as the probability that H₀ is true.
B. Always reject H₀ when the sample proportion differs numerically from p₀.
C. Use p̂ in the null SE and report p=0.56.
D. z=1.897, p=0.0289; reject H₀.

Answer: D

p̂=140250=0.56. Under H₀, SE=√0.5(0.5)250=0.0316 and z=(0.56−0.5)0.0316=1.897. The right-tailed p-value is 0.0289. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 13. One-Proportion z Test

A solar installer in Capital Region during a baseline measurement week tests H₀:p=0.5 against H_a:p>0.5 using 104 successes in n=180. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Interpret the p-value as the probability that H₀ is true.
B. Always reject H₀ when the sample proportion differs numerically from p₀.
C. z=2.087, p=0.0184; reject H₀.
D. Use p̂ in the null SE and report p=0.578.

Answer: C

p̂=104180=0.578. Under H₀, SE=√0.5(0.5)180=0.0373 and z=(0.578−0.5)0.0373=2.087. The right-tailed p-value is 0.0184. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 14. One-Proportion z Test

A regional airport authority in Capital Region during a two-month observation window tests H₀:p=0.7 against H_a:p>0.7 using 124 successes in n=200. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Use p̂ in the null SE and report p=0.62.
B. Interpret the p-value as the probability that H₀ is true.
C. Always reject H₀ when the sample proportion differs numerically from p₀.
D. z=-2.469, p=0.9932; fail to reject H₀.

Answer: D

One-Proportion z Test: Formula, Conditions, and Example — Question 14. One-Proportion z Test: p̂=124200=0.62. Under H₀, SE=√0.7(0.3)200=0.0324 and z=(0.62−0.7)0.0324=-2.469. The right-tailed p-value is 0.9932. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 15. One-Proportion z Test

A regional hospital in Cedar Grove during a six-week field trial tests H₀:p=0.5 against H_a:p<0.5 using 67 successes in n=160. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. z=-2.055, p=0.0199; reject H₀.
B. Interpret the p-value as the probability that H₀ is true.
C. Use p̂ in the null SE and report p=0.419.
D. Always reject H₀ when the sample proportion differs numerically from p₀.

Answer: A

p̂=67160=0.419. Under H₀, SE=√0.5(0.5)160=0.0395 and z=(0.419−0.5)0.0395=-2.055. The left-tailed p-value is 0.0199. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 16. One-Proportion z Test

A community college in Lakeside district during a summer implementation review tests H₀:p=0.5 against H_a:p>0.5 using 84 successes in n=150. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. z=1.47, p=0.0708; fail to reject H₀.
C. Use p̂ in the null SE and report p=0.56.
D. Interpret the p-value as the probability that H₀ is true.

Answer: B

p̂=84150=0.56. Under H₀, SE=√0.5(0.5)150=0.0408 and z=(0.56−0.5)0.0408=1.47. The right-tailed p-value is 0.0708. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 17. One-Proportion z Test

A community college in Cedar Grove during a multiweek validation study tests H₀:p=0.7 against H_a:p≠0.7 using 281 successes in n=370. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. z=2.496, p=0.0126; reject H₀.
C. Interpret the p-value as the probability that H₀ is true.
D. Use p̂ in the null SE and report p=0.759.

Answer: B

p̂=281370=0.759. Under H₀, SE=√0.7(0.3)370=0.0238 and z=(0.759−0.7)0.0238=2.496. The two-sided p-value is 0.0126. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 18. One-Proportion z Test

A recycling program in Desert County during a randomized pilot period tests H₀:p=0.7 against H_a:p≠0.7 using 186 successes in n=300. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. Interpret the p-value as the probability that H₀ is true.
C. z=-3.024, p=0.0025; reject H₀.
D. Use p̂ in the null SE and report p=0.62.

Answer: C

p̂=186300=0.62. Under H₀, SE=√0.7(0.3)300=0.0265 and z=(0.62−0.7)0.0265=-3.024. The two-sided p-value is 0.0025. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 19. One-Proportion z Test

A wildlife clinic in Coastal Plains during a monthly quality review tests H₀:p=0.6 against H_a:p>0.6 using 231 successes in n=340. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. z=2.989, p=0.0014; reject H₀.
C. Interpret the p-value as the probability that H₀ is true.
D. Use p̂ in the null SE and report p=0.679.

Answer: B

p̂=231340=0.679. Under H₀, SE=√0.6(0.4)340=0.0266 and z=(0.679−0.6)0.0266=2.989. The right-tailed p-value is 0.0014. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Question 20. One-Proportion z Test

A city transit agency in Prairie District during a community outreach cycle tests H₀:p=0.7 against H_a:p≠0.7 using 134 successes in n=210. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. Always reject H₀ when the sample proportion differs numerically from p₀.
B. Use p̂ in the null SE and report p=0.638.
C. Interpret the p-value as the probability that H₀ is true.
D. z=-1.958, p=0.0503; fail to reject H₀.

Answer: D

One-Proportion z Test: Formula, Conditions, and Example — Question 20. One-Proportion z Test: p̂=134210=0.638. Under H₀, SE=√0.7(0.3)210=0.0316 and z=(0.638−0.7)0.0316=-1.958. The two-sided p-value is 0.0503. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

One Proportion Z Test free-response practice

FRQ set 1: One-Proportion z Test

Scenario. A school district in Riverbend during a community outreach cycle tests H₀:p=0.4 against H_a:p<0.4 using 61 successes in n=190. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

p̂=61190=0.321. Under H₀, SE=√0.4(0.6)190=0.0355 and z=(0.321−0.4)0.0355=-2.221. The left-tailed p-value is 0.0132. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 2: One-Proportion z Test

Scenario. A food safety laboratory in Pine Ridge during a quarterly performance study tests H₀:p=0.6 against H_a:p>0.6 using 83 successes in n=160. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

p̂=83160=0.519. Under H₀, SE=√0.6(0.4)160=0.0387 and z=(0.519−0.6)0.0387=-2.098. The right-tailed p-value is 0.982. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 3: One-Proportion z Test

Scenario. A university advising center in Central County during a yearly program evaluation tests H₀:p=0.5 against H_a:p≠0.5 using 84 successes in n=190. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

p̂=84190=0.442. Under H₀, SE=√0.5(0.5)190=0.0363 and z=(0.442−0.5)0.0363=-1.596. The two-sided p-value is 0.1105. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 4: One-Proportion z Test

Scenario. A community college in Cedar Grove during a spring 2027 pilot tests H₀:p=0.6 against H_a:p>0.6 using 88 successes in n=130. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

One-Proportion z Test: Formula, Conditions, and Example — FRQ set 4: One-Proportion z Test: p̂=88130=0.677. Under H₀, SE=√0.6(0.4)130=0.043 and z=(0.677−0.6)0.043=1.79. The right-tailed p-value is 0.0367. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 5: One-Proportion z Test

Scenario. A county election office in Pine Ridge during a school-year data collection tests H₀:p=0.6 against H_a:p<0.6 using 92 successes in n=170. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

p̂=92170=0.541. Under H₀, SE=√0.6(0.4)170=0.0376 and z=(0.541−0.6)0.0376=-1.566. The left-tailed p-value is 0.0587. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 6: One-Proportion z Test

Scenario. A public high school in Midwest consortium during a yearly program evaluation tests H₀:p=0.7 against H_a:p>0.7 using 281 successes in n=370. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

p̂=281370=0.759. Under H₀, SE=√0.7(0.3)370=0.0238 and z=(0.759−0.7)0.0238=2.496. The right-tailed p-value is 0.0063. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 7: One-Proportion z Test

Scenario. A farm cooperative in Riverbend during a summer implementation review tests H₀:p=0.4 against H_a:p≠0.4 using 106 successes in n=230. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

p̂=106230=0.461. Under H₀, SE=√0.4(0.6)230=0.0323 and z=(0.461−0.4)0.0323=1.884. The two-sided p-value is 0.0595. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 8: One-Proportion z Test

Scenario. A solar installer in Prairie District during a two-month observation window tests H₀:p=0.6 against H_a:p≠0.6 using 224 successes in n=330. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

One-Proportion z Test: Formula, Conditions, and Example — FRQ set 8: One-Proportion z Test: p̂=224330=0.679. Under H₀, SE=√0.6(0.4)330=0.027 and z=(0.679−0.6)0.027=2.922. The two-sided p-value is 0.0035. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Next steps after mastering one proportion z test

Next, solve the same one-proportion problem as a confidence interval and compare the formulas line by line. The test uses p₀ in its null standard error; the interval uses p̂ in its estimated standard error. Keeping those two formulas adjacent is one of the most effective ways to prevent cross-procedure substitution.

For a tail-direction diagnostic, keep n, x, and p₀ fixed and compute lower-tail, upper-tail, and two-sided p-values from the same z statistic. Only one of them matches a given Hₐ. This exercise shows that the statistic does not determine the p-value by itself; the alternative hypothesis determines which tail area counts as evidence.