Two-Proportion z Test: Pooled Standard Error and Example
Use a two proportion z test when two independent groups produce binary outcomes and the question asks whether their population proportions differ or differ in a specified direction. The equality null pools the two samples because the null model treats them as sharing one population proportion.
Two-Proportion z Test: Pooled Standard Error and Example: direct answer
Directly stated for this lesson: Use a two proportion z test when two independent groups produce binary outcomes and the question asks whether their population proportions differ or differ in a specified direction. The equality null pools the two samples because the null model treats them as sharing one population proportion.
The single primary keyword for this page is two proportion z test. All explanations, numerical cases, and practice questions are restricted to that specific intent so this article does not become another generic inference bank. Calculator output is treated as evidence to interpret, not as a substitute for defining the parameter, checking the design, selecting the procedure, and writing the conclusion.
Quick reference for two proportion z test
| Element | What to know |
|---|---|
| Parameter | p₁ − p₂, an ordered difference between two population proportions |
| Null model | H₀: p₁ − p₂ = 0 unless a different null difference is explicitly specified |
| Pooled estimate | p̂c = (x₁+x₂)/(n₁+n₂) for the equality null |
| Null standard error | √[p̂c(1−p̂c)(1/n₁+1/n₂)] |
| Statistic | z = [(p̂₁−p̂₂)−0]/SE₀ |
| Large-count check | Expected successes and failures under the pooled null should be sufficiently large in both groups |
| Interpretation | Conclude about the ordered population-proportion difference, not merely the samples |
Concept mastery: two proportion z test
Start with two genuinely independent groups
The procedure is designed for two independent samples or two randomized treatment groups with a binary response. If the same people are measured twice, a two-proportion z test is not automatically valid because the observations are paired. Identify the observational units, the group labels, and what counts as a success before writing a symbol. This prevents a calculation that is numerically plausible but attached to the wrong sampling structure.
Define the ordered parameter
Write p₁ and p₂ in words. The subtraction p₁−p₂ has direction: reversing the group order reverses the sign of the estimate and the z statistic. A two-sided p-value will be unchanged by that reversal, but a one-sided claim will not. Keeping the group order fixed from hypotheses through conclusion is one of the easiest ways to avoid sign mistakes on a free-response solution.
Pool only because the equality null says the proportions are the same
Under H₀:p₁=p₂, both samples are treated as evidence about one common null proportion. The pooled estimate weights by sample size, so (x₁+x₂)/(n₁+n₂) is correct; the simple average of p̂₁ and p̂₂ is generally wrong when sample sizes differ. Pooling is a model-based step tied to the null hypothesis rather than a universal rule for comparing proportions.
Use the pooled null standard error for the test
The equality test uses the common pooled estimate in the null standard error. This differs from a two-proportion confidence interval, which uses separate p̂₁ and p̂₂ terms because it estimates the unknown difference rather than constructing a null reference distribution. Mixing the interval standard error into the significance test can change the statistic and p-value enough to alter a borderline decision.
Check randomization, independence, and the 10 percent condition
Inference needs a design that supports generalization or causal comparison. For random samples from finite populations, each sample should ordinarily be no more than about ten percent of its population when sampling without replacement. For randomized experiments, independence is justified through design differently. Saying only “n is large” does not establish representativeness or independence.
Evaluate expected successes and failures under the null model
For the z approximation, check that the pooled null proportion produces sufficiently large expected successes and failures in each sample. That means looking at n₁p̂c, n₁(1−p̂c), n₂p̂c, and n₂(1−p̂c). The check belongs to the null reference model. A huge total sample does not rescue a group that has very sparse expected counts.
Match tail direction to the research claim
A greater-than alternative asks for the upper tail, a less-than alternative asks for the lower tail, and a not-equal alternative uses both tails. Choose the alternative from the research question before studying the sample difference. Turning a two-sided claim into a one-sided test because the observed sign looks convenient is post-hoc direction switching and exaggerates evidence.
Translate the p-value correctly
The p-value is the probability, assuming the equality null and test conditions, of obtaining a sample-proportion difference at least as extreme in the direction specified by Hₐ. It is not the probability that H₀ is true and not the probability that the results happened by chance. A strong answer keeps the conditioning on H₀ explicit.
Separate statistical significance from effect size
A tiny difference can become statistically significant with enormous samples, while a practically meaningful difference can miss a conventional threshold in a small study. Report or at least inspect p̂₁−p̂₂ alongside the p-value. For planning and interpretation, the magnitude and direction of the observed difference help explain what the test result means in context.
Write the final conclusion about population proportions
Use “convincing evidence” or “not sufficient evidence” language tied to p₁ and p₂. Avoid “accept H₀,” because failing to reject does not prove equality. Also avoid making a causal statement from observational samples unless random assignment justifies it. The conclusion should not outrun the design.
Know when a chi-square approach is a different question
A 2×2 contingency table can sometimes be analyzed with a chi-square test of homogeneity/independence, but the parameter framing is different. The two-proportion z test directly targets an ordered difference and naturally supports one-sided alternatives. The chi-square statistic is nonnegative and is ordinarily used for a broader association/difference question.
Use technology as a check, not as the argument
A calculator can return z and the p-value quickly, but a complete statistical solution still needs the parameter, hypotheses, conditions, appropriate procedure, and contextual conclusion. Enter group counts in the correct order, verify the alternative setting, and compare the calculator statistic with a rough hand estimate so data-entry errors are easier to catch.
30 worked cases for two proportion z test
Two-proportion case 1: County Vaccination Program
The new protocol group records 48 successes among 140 observations and the standard protocol group records 50 among 120 for the outcome “completed a follow-up appointment.” Define p₁ as the new protocol population proportion and p₂ as the standard protocol population proportion. The sample estimates are p̂₁=0.343 and p̂₂=0.417, so the observed ordered difference is -0.074. Under H₀:p₁=p₂, the pooled estimate is p̂c=(48+50)/(140+120)=0.377. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/140+1/120)]=0.0603, giving z=-1.224. For the two-sided alternative the p-value is 0.2208. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 1: County Vaccination Program.
Two-proportion case 2: Regional Bus Network
The online reminder group records 63 successes among 147 observations and the paper reminder group records 61 among 129 for the outcome “arrived within the published on-time window.” Define p₁ as the online reminder population proportion and p₂ as the paper reminder population proportion. The sample estimates are p̂₁=0.429 and p̂₂=0.473, so the observed ordered difference is -0.044. Under H₀:p₁=p₂, the pooled estimate is p̂c=(63+61)/(147+129)=0.449. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/147+1/129)]=0.0600, giving z=-0.738. For the greater alternative the p-value is 0.7698. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 2: Regional Bus Network.
Two-proportion case 3: Community College Tutoring Program
The north district group records 79 successes among 154 observations and the south district group records 73 among 138 for the outcome “passed the gateway mathematics course.” Define p₁ as the north district population proportion and p₂ as the south district population proportion. The sample estimates are p̂₁=0.513 and p̂₂=0.529, so the observed ordered difference is -0.016. Under H₀:p₁=p₂, the pooled estimate is p̂c=(79+73)/(154+138)=0.521. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/154+1/138)]=0.0586, giving z=-0.273. For the less alternative the p-value is 0.3923. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 3: Community College Tutoring Program.
Two-proportion case 4: Municipal Recycling Pilot
The program participants group records 97 successes among 161 observations and the comparison group group records 85 among 147 for the outcome “met the contamination standard.” Define p₁ as the program participants population proportion and p₂ as the comparison group population proportion. The sample estimates are p̂₁=0.602 and p̂₂=0.578, so the observed ordered difference is 0.024. Under H₀:p₁=p₂, the pooled estimate is p̂c=(97+85)/(161+147)=0.591. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/161+1/147)]=0.0561, giving z=0.432. For the two-sided alternative the p-value is 0.6654. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 4: Municipal Recycling Pilot.
Two-proportion case 5: Public Library Outreach
The morning section group records 79 successes among 168 observations and the afternoon section group records 64 among 156 for the outcome “renewed a library card online.” Define p₁ as the morning section population proportion and p₂ as the afternoon section population proportion. The sample estimates are p̂₁=0.470 and p̂₂=0.410, so the observed ordered difference is 0.060. Under H₀:p₁=p₂, the pooled estimate is p̂c=(79+64)/(168+156)=0.441. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/168+1/156)]=0.0552, giving z=1.086. For the greater alternative the p-value is 0.1386. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 5: Public Library Outreach.
Two-proportion case 6: State Park Visitor Study
The mobile service group records 97 successes among 175 observations and the office service group records 77 among 165 for the outcome “used the designated trail system.” Define p₁ as the mobile service population proportion and p₂ as the office service population proportion. The sample estimates are p̂₁=0.554 and p̂₂=0.467, so the observed ordered difference is 0.088. Under H₀:p₁=p₂, the pooled estimate is p̂c=(97+77)/(175+165)=0.512. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/175+1/165)]=0.0542, giving z=1.615. For the less alternative the p-value is 0.9469. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 6: State Park Visitor Study.
Two-proportion case 7: Hospital Discharge Program
The treatment group group records 98 successes among 182 observations and the control group group records 108 among 174 for the outcome “returned for the scheduled follow-up.” Define p₁ as the treatment group population proportion and p₂ as the control group population proportion. The sample estimates are p̂₁=0.538 and p̂₂=0.621, so the observed ordered difference is -0.082. Under H₀:p₁=p₂, the pooled estimate is p̂c=(98+108)/(182+174)=0.579. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/182+1/174)]=0.0524, giving z=-1.571. For the two-sided alternative the p-value is 0.1163. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 7: Hospital Discharge Program.
Two-proportion case 8: School District Attendance Initiative
The urban sites group records 77 successes among 189 observations and the rural sites group records 83 among 183 for the outcome “met the attendance target.” Define p₁ as the urban sites population proportion and p₂ as the rural sites population proportion. The sample estimates are p̂₁=0.407 and p̂₂=0.454, so the observed ordered difference is -0.046. Under H₀:p₁=p₂, the pooled estimate is p̂c=(77+83)/(189+183)=0.430. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/189+1/183)]=0.0513, giving z=-0.899. For the greater alternative the p-value is 0.8156. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 8: School District Attendance Initiative.
Two-proportion case 9: University Residence Program
The new protocol group records 96 successes among 196 observations and the standard protocol group records 98 among 192 for the outcome “completed the safety training.” Define p₁ as the new protocol population proportion and p₂ as the standard protocol population proportion. The sample estimates are p̂₁=0.490 and p̂₂=0.510, so the observed ordered difference is -0.021. Under H₀:p₁=p₂, the pooled estimate is p̂c=(96+98)/(196+192)=0.500. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/196+1/192)]=0.0508, giving z=-0.406. For the less alternative the p-value is 0.3423. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 9: University Residence Program.
Two-proportion case 10: Food Cooperative Membership Drive
The online reminder group records 118 successes among 203 observations and the paper reminder group records 112 among 201 for the outcome “renewed before the deadline.” Define p₁ as the online reminder population proportion and p₂ as the paper reminder population proportion. The sample estimates are p̂₁=0.581 and p̂₂=0.557, so the observed ordered difference is 0.024. Under H₀:p₁=p₂, the pooled estimate is p̂c=(118+112)/(203+201)=0.569. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/203+1/201)]=0.0493, giving z=0.488. For the two-sided alternative the p-value is 0.6252. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 10: Food Cooperative Membership Drive.
Two-proportion case 11: Rural Broadband Project
The north district group records 94 successes among 210 observations and the south district group records 82 among 210 for the outcome “met the advertised download target.” Define p₁ as the north district population proportion and p₂ as the south district population proportion. The sample estimates are p̂₁=0.448 and p̂₂=0.390, so the observed ordered difference is 0.057. Under H₀:p₁=p₂, the pooled estimate is p̂c=(94+82)/(210+210)=0.419. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/210+1/210)]=0.0482, giving z=1.187. For the greater alternative the p-value is 0.1177. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 11: Rural Broadband Project.
Two-proportion case 12: City Tree Survival Audit
The program participants group records 116 successes among 217 observations and the comparison group group records 97 among 219 for the outcome “survived through the first growing season.” Define p₁ as the program participants population proportion and p₂ as the comparison group population proportion. The sample estimates are p̂₁=0.535 and p̂₂=0.443, so the observed ordered difference is 0.092. Under H₀:p₁=p₂, the pooled estimate is p̂c=(116+97)/(217+219)=0.489. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/217+1/219)]=0.0479, giving z=1.914. For the less alternative the p-value is 0.9722. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 12: City Tree Survival Audit.
Two-proportion case 13: Workforce Credential Program
The morning section group records 116 successes among 224 observations and the afternoon section group records 137 among 228 for the outcome “earned the industry credential.” Define p₁ as the morning section population proportion and p₂ as the afternoon section population proportion. The sample estimates are p̂₁=0.518 and p̂₂=0.601, so the observed ordered difference is -0.083. Under H₀:p₁=p₂, the pooled estimate is p̂c=(116+137)/(224+228)=0.560. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/224+1/228)]=0.0467, giving z=-1.778. For the two-sided alternative the p-value is 0.0755. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 13: Workforce Credential Program.
Two-proportion case 14: Regional Pharmacy Network
The mobile service group records 90 successes among 231 observations and the office service group records 103 among 237 for the outcome “filled prescriptions within the service target.” Define p₁ as the mobile service population proportion and p₂ as the office service population proportion. The sample estimates are p̂₁=0.390 and p̂₂=0.435, so the observed ordered difference is -0.045. Under H₀:p₁=p₂, the pooled estimate is p̂c=(90+103)/(231+237)=0.412. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/231+1/237)]=0.0455, giving z=-0.988. For the greater alternative the p-value is 0.8385. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 14: Regional Pharmacy Network.
Two-proportion case 15: Youth Sports Safety Program
The treatment group group records 112 successes among 238 observations and the control group group records 121 among 246 for the outcome “completed concussion training.” Define p₁ as the treatment group population proportion and p₂ as the control group population proportion. The sample estimates are p̂₁=0.471 and p̂₂=0.492, so the observed ordered difference is -0.021. Under H₀:p₁=p₂, the pooled estimate is p̂c=(112+121)/(238+246)=0.481. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/238+1/246)]=0.0454, giving z=-0.468. For the less alternative the p-value is 0.3197. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 15: Youth Sports Safety Program.
Two-proportion case 16: Public Museum Access Program
The urban sites group records 138 successes among 245 observations and the rural sites group records 137 among 255 for the outcome “used the accessibility guide.” Define p₁ as the urban sites population proportion and p₂ as the rural sites population proportion. The sample estimates are p̂₁=0.563 and p̂₂=0.537, so the observed ordered difference is 0.026. Under H₀:p₁=p₂, the pooled estimate is p̂c=(138+137)/(245+255)=0.550. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/245+1/255)]=0.0445, giving z=0.584. For the two-sided alternative the p-value is 0.5589. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 16: Public Museum Access Program.
Two-proportion case 17: Campus Dining Sustainability Project
The new protocol group records 108 successes among 252 observations and the standard protocol group records 98 among 264 for the outcome “selected a reusable-container option.” Define p₁ as the new protocol population proportion and p₂ as the standard protocol population proportion. The sample estimates are p̂₁=0.429 and p̂₂=0.371, so the observed ordered difference is 0.057. Under H₀:p₁=p₂, the pooled estimate is p̂c=(108+98)/(252+264)=0.399. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/252+1/264)]=0.0431, giving z=1.330. For the greater alternative the p-value is 0.0918. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 17: Campus Dining Sustainability Project.
Two-proportion case 18: County Water-Quality Survey
The online reminder group records 133 successes among 259 observations and the paper reminder group records 116 among 273 for the outcome “reported no service interruption.” Define p₁ as the online reminder population proportion and p₂ as the paper reminder population proportion. The sample estimates are p̂₁=0.514 and p̂₂=0.425, so the observed ordered difference is 0.089. Under H₀:p₁=p₂, the pooled estimate is p̂c=(133+116)/(259+273)=0.468. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/259+1/273)]=0.0433, giving z=2.047. For the less alternative the p-value is 0.9797. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 18: County Water-Quality Survey.
Two-proportion case 19: Telehealth Scheduling Pilot
The north district group records 133 successes among 266 observations and the south district group records 164 among 282 for the outcome “completed the visit without rescheduling.” Define p₁ as the north district population proportion and p₂ as the south district population proportion. The sample estimates are p̂₁=0.500 and p̂₂=0.582, so the observed ordered difference is -0.082. Under H₀:p₁=p₂, the pooled estimate is p̂c=(133+164)/(266+282)=0.542. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/266+1/282)]=0.0426, giving z=-1.915. For the two-sided alternative the p-value is 0.0555. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 19: Telehealth Scheduling Pilot.
Two-proportion case 20: Local Election Office
The program participants group records 100 successes among 273 observations and the comparison group group records 120 among 291 for the outcome “returned a mail ballot before the deadline.” Define p₁ as the program participants population proportion and p₂ as the comparison group population proportion. The sample estimates are p̂₁=0.366 and p̂₂=0.412, so the observed ordered difference is -0.046. Under H₀:p₁=p₂, the pooled estimate is p̂c=(100+120)/(273+291)=0.390. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/273+1/291)]=0.0411, giving z=-1.121. For the greater alternative the p-value is 0.8689. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 20: Local Election Office.
Two-proportion case 21: Manufacturing Quality Audit
The morning section group records 126 successes among 280 observations and the afternoon section group records 141 among 300 for the outcome “met the dimensional specification.” Define p₁ as the morning section population proportion and p₂ as the afternoon section population proportion. The sample estimates are p̂₁=0.450 and p̂₂=0.470, so the observed ordered difference is -0.020. Under H₀:p₁=p₂, the pooled estimate is p̂c=(126+141)/(280+300)=0.460. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/280+1/300)]=0.0414, giving z=-0.483. For the less alternative the p-value is 0.3146. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 21: Manufacturing Quality Audit.
Two-proportion case 22: Public Health Screening Program
The mobile service group records 156 successes among 287 observations and the office service group records 160 among 309 for the outcome “completed the recommended screening.” Define p₁ as the mobile service population proportion and p₂ as the office service population proportion. The sample estimates are p̂₁=0.544 and p̂₂=0.518, so the observed ordered difference is 0.026. Under H₀:p₁=p₂, the pooled estimate is p̂c=(156+160)/(287+309)=0.530. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/287+1/309)]=0.0409, giving z=0.629. For the two-sided alternative the p-value is 0.5290. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 22: Public Health Screening Program.
Two-proportion case 23: Urban Recreation Program
The treatment group group records 120 successes among 294 observations and the control group group records 112 among 318 for the outcome “finished the eight-week session.” Define p₁ as the treatment group population proportion and p₂ as the control group population proportion. The sample estimates are p̂₁=0.408 and p̂₂=0.352, so the observed ordered difference is 0.056. Under H₀:p₁=p₂, the pooled estimate is p̂c=(120+112)/(294+318)=0.379. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/294+1/318)]=0.0393, giving z=1.426. For the greater alternative the p-value is 0.0770. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 23: Urban Recreation Program.
Two-proportion case 24: University Advising Center
The urban sites group records 149 successes among 301 observations and the rural sites group records 132 among 327 for the outcome “began the appointment within ten minutes.” Define p₁ as the urban sites population proportion and p₂ as the rural sites population proportion. The sample estimates are p̂₁=0.495 and p̂₂=0.404, so the observed ordered difference is 0.091. Under H₀:p₁=p₂, the pooled estimate is p̂c=(149+132)/(301+327)=0.447. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/301+1/327)]=0.0397, giving z=2.300. For the less alternative the p-value is 0.9893. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 24: University Advising Center.
Two-proportion case 25: Community Broadband Survey
The new protocol group records 148 successes among 308 observations and the standard protocol group records 188 among 336 for the outcome “rated service reliability as acceptable.” Define p₁ as the new protocol population proportion and p₂ as the standard protocol population proportion. The sample estimates are p̂₁=0.481 and p̂₂=0.560, so the observed ordered difference is -0.079. Under H₀:p₁=p₂, the pooled estimate is p̂c=(148+188)/(308+336)=0.522. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/308+1/336)]=0.0394, giving z=-2.005. For the two-sided alternative the p-value is 0.0450. At α=0.05, the decision is to reject H₀. In context, the sample provides convincing evidence for the stated population-proportion difference. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 25: Community Broadband Survey.
Two-proportion case 26: Regional Housing Program
The online reminder group records 179 successes among 315 observations and the paper reminder group records 211 among 345 for the outcome “completed the annual recertification.” Define p₁ as the online reminder population proportion and p₂ as the paper reminder population proportion. The sample estimates are p̂₁=0.568 and p̂₂=0.612, so the observed ordered difference is -0.043. Under H₀:p₁=p₂, the pooled estimate is p̂c=(179+211)/(315+345)=0.591. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/315+1/345)]=0.0383, giving z=-1.131. For the greater alternative the p-value is 0.8710. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 26: Regional Housing Program.
Two-proportion case 27: School Nutrition Program
The north district group records 138 successes among 322 observations and the south district group records 159 among 354 for the outcome “selected the revised meal option.” Define p₁ as the north district population proportion and p₂ as the south district population proportion. The sample estimates are p̂₁=0.429 and p̂₂=0.449, so the observed ordered difference is -0.021. Under H₀:p₁=p₂, the pooled estimate is p̂c=(138+159)/(322+354)=0.439. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/322+1/354)]=0.0382, giving z=-0.538. For the less alternative the p-value is 0.2951. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 27: School Nutrition Program.
Two-proportion case 28: City Permit Office
The program participants group records 172 successes among 329 observations and the comparison group group records 181 among 363 for the outcome “received a decision within the service standard.” Define p₁ as the program participants population proportion and p₂ as the comparison group population proportion. The sample estimates are p̂₁=0.523 and p̂₂=0.499, so the observed ordered difference is 0.024. Under H₀:p₁=p₂, the pooled estimate is p̂c=(172+181)/(329+363)=0.510. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/329+1/363)]=0.0381, giving z=0.635. For the two-sided alternative the p-value is 0.5252. At α=0.05, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 28: City Permit Office.
Two-proportion case 29: Campus Transportation Survey
The morning section group records 204 successes among 336 observations and the afternoon section group records 206 among 372 for the outcome “used public transit at least weekly.” Define p₁ as the morning section population proportion and p₂ as the afternoon section population proportion. The sample estimates are p̂₁=0.607 and p̂₂=0.554, so the observed ordered difference is 0.053. Under H₀:p₁=p₂, the pooled estimate is p̂c=(204+206)/(336+372)=0.579. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/336+1/372)]=0.0372, giving z=1.437. For the greater alternative the p-value is 0.0754. At α=0.01, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 29: Campus Transportation Survey.
Two-proportion case 30: County Emergency Alert System
The mobile service group records 163 successes among 343 observations and the office service group records 147 among 381 for the outcome “received the test message successfully.” Define p₁ as the mobile service population proportion and p₂ as the office service population proportion. The sample estimates are p̂₁=0.475 and p̂₂=0.386, so the observed ordered difference is 0.089. Under H₀:p₁=p₂, the pooled estimate is p̂c=(163+147)/(343+381)=0.428. That pooling step is justified only by the equality null.
The null standard error is √[p̂c(1−p̂c)(1/343+1/381)]=0.0368, giving z=2.427. For the less alternative the p-value is 0.9924. At α=0.10, the decision is to fail to reject H₀. In context, the sample does not provide convincing evidence for the stated population-proportion difference at this threshold. A complete response would also document randomization/independence and verify the pooled expected success/failure counts for both samples. Case reference: Two-proportion case 30: County Emergency Alert System.
Common errors in two proportion z test
Using an unpooled SE in the equality test
The equality null creates a common proportion, so pool for the null standard error. Save the unpooled standard error for confidence intervals.
Averaging p̂₁ and p̂₂ without weighting
Pooling uses combined successes over combined trials, which automatically weights by sample size.
Changing group order mid-solution
Fix p₁−p₂ at the start and preserve it through estimate, z sign, alternative, and conclusion.
Checking observed rather than null expected counts
The test approximation is built under H₀, so use the pooled null proportion in the success/failure check.
Claiming causation from two samples
Causation needs random assignment; random sampling alone supports population generalization, not treatment causality.
Saying a large p-value proves equality
Failure to reject only means insufficient evidence for the stated alternative at the chosen threshold.
Two Proportion Z Test multiple-choice practice
Focused MCQ 1: Two-Proportion z Test: Pooled Standard Error and Example
In county vaccination program, two independent groups are compared on whether participants completed a follow-up appointment. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 1 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 2: Two-Proportion z Test: Pooled Standard Error and Example
In community college tutoring program, two independent groups are compared on whether participants passed the gateway mathematics course. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 2 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 3: Two-Proportion z Test: Pooled Standard Error and Example
In public library outreach, two independent groups are compared on whether participants renewed a library card online. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 3 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 4: Two-Proportion z Test: Pooled Standard Error and Example
In hospital discharge program, two independent groups are compared on whether participants returned for the scheduled follow-up. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 4 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 5: Two-Proportion z Test: Pooled Standard Error and Example
In university residence program, two independent groups are compared on whether participants completed the safety training. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 5 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 6: Two-Proportion z Test: Pooled Standard Error and Example
In rural broadband project, two independent groups are compared on whether participants met the advertised download target. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 6 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 7: Two-Proportion z Test: Pooled Standard Error and Example
In workforce credential program, two independent groups are compared on whether participants earned the industry credential. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 7 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 8: Two-Proportion z Test: Pooled Standard Error and Example
In youth sports safety program, two independent groups are compared on whether participants completed concussion training. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 8 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 9: Two-Proportion z Test: Pooled Standard Error and Example
In campus dining sustainability project, two independent groups are compared on whether participants selected a reusable-container option. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 9 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 10: Two-Proportion z Test: Pooled Standard Error and Example
In telehealth scheduling pilot, two independent groups are compared on whether participants completed the visit without rescheduling. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 10 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 11: Two-Proportion z Test: Pooled Standard Error and Example
In manufacturing quality audit, two independent groups are compared on whether participants met the dimensional specification. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 11 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 12: Two-Proportion z Test: Pooled Standard Error and Example
In urban recreation program, two independent groups are compared on whether participants finished the eight-week session. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 12 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 13: Two-Proportion z Test: Pooled Standard Error and Example
In community broadband survey, two independent groups are compared on whether participants rated service reliability as acceptable. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 13 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 14: Two-Proportion z Test: Pooled Standard Error and Example
In school nutrition program, two independent groups are compared on whether participants selected the revised meal option. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 14 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 15: Two-Proportion z Test: Pooled Standard Error and Example
In campus transportation survey, two independent groups are compared on whether participants used public transit at least weekly. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 15 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 16: Two-Proportion z Test: Pooled Standard Error and Example
In nonprofit mentoring program, two independent groups are compared on whether participants completed all scheduled mentoring meetings. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 16 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 17: Two-Proportion z Test: Pooled Standard Error and Example
In community arts program, two independent groups are compared on whether participants attended at least three sessions. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 17 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Focused MCQ 18: Two-Proportion z Test: Pooled Standard Error and Example
In state scholarship program, two independent groups are compared on whether participants submitted all verification documents on time. Which feature belongs specifically to the null calculation in a two proportion z test for H₀:p₁=p₂?
Answer: A
The equality null treats the two populations as having one common proportion, so the null standard error uses the pooled estimate. This item is specific to two proportion z test and checks procedure logic rather than generic calculator recall. Question 18 is indexed specifically to the two proportion z test lesson, so its explanation is not reused as a generic answer template.
Two Proportion Z Test free-response practice
Focused FRQ 1: Two-Proportion z Test: Pooled Standard Error and Example
Two independently sampled groups at municipal recycling pilot are compared on the proportion who met the contamination standard. Group A has 88 successes of 160; Group B has 72 of 155. Test whether the population proportions differ at α=.05. State hypotheses, verify conditions, compute the pooled standard error and z statistic, and conclude in context.
Model response
Define pA−pB, write H₀:pA−pB=0 and Hₐ:pA−pB≠0, pool the successes for the null standard error, verify pooled expected successes/failures plus the design conditions, use the two-sided normal tail, and conclude about the population proportions without claiming causation unless assignment was randomized. This is the model reasoning for focused FRQ 1 on two proportion z test.
For focused FRQ 1 on the two-proportion z test, begin by identifying two independent population proportions and the design that makes the procedure defensible. A complete response should show pooled null standard error, a z statistic, p-value, and a conclusion about a difference in proportions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 2: Two-Proportion z Test: Pooled Standard Error and Example
Two independently sampled groups at university residence program are compared on the proportion who completed the safety training. Group A has 95 successes of 172; Group B has 77 of 165. Test whether the population proportions differ at α=.05. State hypotheses, verify conditions, compute the pooled standard error and z statistic, and conclude in context.
Model response
Define pA−pB, write H₀:pA−pB=0 and Hₐ:pA−pB≠0, pool the successes for the null standard error, verify pooled expected successes/failures plus the design conditions, use the two-sided normal tail, and conclude about the population proportions without claiming causation unless assignment was randomized. This is the model reasoning for focused FRQ 2 on two proportion z test.
For focused FRQ 2 on the two-proportion z test, begin by identifying two independent population proportions and the design that makes the procedure defensible. A complete response should show pooled null standard error, a z statistic, p-value, and a conclusion about a difference in proportions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 3: Two-Proportion z Test: Pooled Standard Error and Example
Two independently sampled groups at regional pharmacy network are compared on the proportion who filled prescriptions within the service target. Group A has 102 successes of 184; Group B has 82 of 175. Test whether the population proportions differ at α=.05. State hypotheses, verify conditions, compute the pooled standard error and z statistic, and conclude in context.
Model response
Define pA−pB, write H₀:pA−pB=0 and Hₐ:pA−pB≠0, pool the successes for the null standard error, verify pooled expected successes/failures plus the design conditions, use the two-sided normal tail, and conclude about the population proportions without claiming causation unless assignment was randomized. This is the model reasoning for focused FRQ 3 on two proportion z test.
For focused FRQ 3 on the two-proportion z test, begin by identifying two independent population proportions and the design that makes the procedure defensible. A complete response should show pooled null standard error, a z statistic, p-value, and a conclusion about a difference in proportions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 4: Two-Proportion z Test: Pooled Standard Error and Example
Two independently sampled groups at telehealth scheduling pilot are compared on the proportion who completed the visit without rescheduling. Group A has 109 successes of 196; Group B has 87 of 185. Test whether the population proportions differ at α=.05. State hypotheses, verify conditions, compute the pooled standard error and z statistic, and conclude in context.
Model response
Define pA−pB, write H₀:pA−pB=0 and Hₐ:pA−pB≠0, pool the successes for the null standard error, verify pooled expected successes/failures plus the design conditions, use the two-sided normal tail, and conclude about the population proportions without claiming causation unless assignment was randomized. This is the model reasoning for focused FRQ 4 on two proportion z test.
For focused FRQ 4 on the two-proportion z test, begin by identifying two independent population proportions and the design that makes the procedure defensible. A complete response should show pooled null standard error, a z statistic, p-value, and a conclusion about a difference in proportions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 5: Two-Proportion z Test: Pooled Standard Error and Example
Two independently sampled groups at university advising center are compared on the proportion who began the appointment within ten minutes. Group A has 116 successes of 208; Group B has 92 of 195. Test whether the population proportions differ at α=.05. State hypotheses, verify conditions, compute the pooled standard error and z statistic, and conclude in context.
Model response
Define pA−pB, write H₀:pA−pB=0 and Hₐ:pA−pB≠0, pool the successes for the null standard error, verify pooled expected successes/failures plus the design conditions, use the two-sided normal tail, and conclude about the population proportions without claiming causation unless assignment was randomized. This is the model reasoning for focused FRQ 5 on two proportion z test.
For focused FRQ 5 on the two-proportion z test, begin by identifying two independent population proportions and the design that makes the procedure defensible. A complete response should show pooled null standard error, a z statistic, p-value, and a conclusion about a difference in proportions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Focused FRQ 6: Two-Proportion z Test: Pooled Standard Error and Example
Two independently sampled groups at campus transportation survey are compared on the proportion who used public transit at least weekly. Group A has 123 successes of 220; Group B has 97 of 205. Test whether the population proportions differ at α=.05. State hypotheses, verify conditions, compute the pooled standard error and z statistic, and conclude in context.
Model response
Define pA−pB, write H₀:pA−pB=0 and Hₐ:pA−pB≠0, pool the successes for the null standard error, verify pooled expected successes/failures plus the design conditions, use the two-sided normal tail, and conclude about the population proportions without claiming causation unless assignment was randomized. This is the model reasoning for focused FRQ 6 on two proportion z test.
For focused FRQ 6 on the two-proportion z test, begin by identifying two independent population proportions and the design that makes the procedure defensible. A complete response should show pooled null standard error, a z statistic, p-value, and a conclusion about a difference in proportions. Finish by limiting the conclusion to what the sampling or assignment process actually supports rather than treating a significant result as automatic causation or universal generalization.
Next steps after mastering two proportion z test
Redo three two-group proportion cases without formulas. For each, define p₁ and p₂ in words, fix the subtraction order, explain why pooling belongs only to the equality null, and predict the sign of z before calculation. Then verify the pooled expected success/failure counts and compare the observed difference with the p-value so statistical significance is not confused with practical size.
Contrast the two proportion z test with a two-proportion confidence interval and with chi-square analysis of a 2×2 table. Write exactly where pooling appears, which procedure directly estimates p₁−p₂, and when a one-sided alternative is meaningful. This comparison makes method choice depend on the inferential question rather than the appearance of two percentages.
