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.
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
| 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 Ha:p<0.4 using 97 successes in n=210. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p<0.6 using 224 successes in n=330. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.7 using 141 successes in n=220. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p<0.7 using 274 successes in n=360. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p≠0.4 using 110 successes in n=240. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p<0.7 using 137 successes in n=180. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.4 using 115 successes in n=250. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.4 using 158 successes in n=330. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.5 using 202 successes in n=360. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p≠0.7 using 226 successes in n=290. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p≠0.5 using 66 successes in n=150. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.5 using 140 successes in n=250. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.5 using 104 successes in n=180. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.7 using 124 successes in n=200. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p<0.5 using 67 successes in n=160. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.5 using 84 successes in n=150. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p≠0.7 using 281 successes in n=370. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p≠0.7 using 186 successes in n=300. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p>0.6 using 231 successes in n=340. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha:p≠0.7 using 134 successes in n=210. Compute the one-proportion z statistic and p-value, then conclude at α=.05.
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 Ha: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 Ha: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 Ha: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 Ha: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 Ha: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 Ha: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 Ha: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 Ha: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.
