Try
A synthetic random sample of 140 residents contains 84 who support a proposed change. The target is the support proportion p among residents represented by the sampling frame, not the probability that this particular sample supports it. The estimate is p̂=84/140=0.60. Suppose the frame contains more than 1400 residents and responses were obtained without material nonresponse. Before calculating, distinguish “estimate p with an interval” from “test whether p exceeds 0.50.”
Write H₀:p=0.50 and Hₐ:p>0.50 before viewing the test result. Set α=0.05 as the decision threshold. A confidence level of 95% controls a different procedure: the repeated-interval coverage target. The tool returns an interval and test together for convenience; they do not use identical standard errors.
Predict
Predict whether a proportion 0.60 from 140 people provides more evidence than the same proportion from 14. Also predict which probability enters the test’s standard error: the estimate 0.60 or the null value 0.50. The test asks how unusual the data would be under the null, while the interval estimates uncertainty around the observed proportion.
Do not press Calculate until you can identify the independent unit. One resident contributes one binary response. Multiple answers by the same resident would not increase the number of independent residents.
Experiment
Choose the one-proportion method. Enter 84 successes, sample size 140, null proportion 0.50, the greater alternative and confidence 0.95. Record the estimate, interval standard error, null-based test standard error, z statistic and p-value. Numerical references below were independently calculated; small differences in final displayed digits can reflect the numerical approximation.
| Quantity | Worked calculation | Reference value |
|---|
| Estimate | 84/140 | 0.60 |
| Interval SE | √(0.60·0.40/140) | 0.04140393 |
| 95% critical z | Lower normal probability 0.975 | 1.95996398 |
| Normal interval | Estimate ± critical z·interval SE | [0.518850,0.681150] |
| Null SE | √(0.50·0.50/140) | 0.04225771 |
| Test statistic | (0.60−0.50)/null SE | 2.36643191 |
| Greater-tail p-value | P(Z≥2.36643191) | 0.00898024 |
The observed success/failure counts are 84 and 56 for the interval check. Under the null, expected counts are 70 and 70 for the test check. All exceed 10. Random sampling and a sampling fraction below 10% support the usual independence approximation, while truthful measurement remains a substantive assumption. The software sees counts; it cannot verify recruitment.
Since 0.00898 is below 0.05, reject this null in favor of majority support, with the design qualifications stated. This p-value is not the chance that p=0.50 is true. The interval concerns the population proportion, not a range containing 95% of binary responses.
Understand
Change to independent two-proportion inference. A new worked scenario has 78 successes among 120 in A and 54 among 120 in B. Define the parameter pA−pB before entering the groups; its estimate is 0.65−0.45=0.20. Assume independent random samples from large populations, one binary observation per person. All four observed success/failure counts exceed 10.
For a 95% interval, use the separate estimates in SE=√[0.65·0.35/120+0.45·0.55/120]. The interval is approximately [0.076688,0.323312]. For the equality null, pool the successes: (78+54)/(120+120)=0.55. The null standard error is √[0.55·0.45(1/120+1/120)]. The null expected counts are 66 successes and 54 failures in each group. The greater-tail statistic is approximately 3.113996 and p≈0.000922861.
The pooled test asks about equality; the unpooled interval allows the two unknown proportions to differ. Reversing the group order negates the estimate and swaps the negated interval endpoints. A two-sided p-value stays the same, but a fixed greater alternative does not. Choose the orientation and alternative for the scientific question, not after seeing which direction gives a small p-value.
These are normal-approximation methods. If counts are sparse, a computed number is not enough to justify the approximation. The tool should flag numerical count concerns, but it cannot diagnose paired data, selective recruitment or outcome-dependent exclusions. Do not clip a Wald interval at zero or one and then claim the original nominal coverage has been repaired. Appropriate exact or alternative interval methods require a separate choice.
Apply
Return to the initial one-proportion method and restore 84 successes out of 140, null proportion 0.50 and confidence 0.95. Change its alternative to two-sided. Predict the p-value change, calculate it, and explain why the interval stays fixed. Then change only confidence to 0.90 and predict which outputs should change.
Compare the workflow changes
The two-sided normal p-value is approximately 2·0.00898024=0.01796048 because it counts equally extreme deviations in both directions. Changing the alternative does not change the selected two-sided confidence interval. Reducing confidence changes the critical multiplier and narrows the interval; it does not change the data, test statistic or p-value for the stated alternative. A one-sided test at a chosen α and a displayed two-sided interval are separate outputs; do not infer exact agreement without checking the procedures and levels.
Now refuse an invalid shortcut: the same people answer before and after a workshop, but only two yes totals are available. Those columns are paired, so this independent-group method is not justified. Recover the transition counts and select an appropriate paired binary analysis. Even a correct paired analysis alone would not establish causality in an uncontrolled before/after study.
Continue
Use one-proportion intervals and two-proportion tests for the underlying reasoning. In the new practice, calculate the reference probabilities yourself and record the complete chain: parameter, hypotheses, design/count conditions, arithmetic and contextual conclusion. No calculator action automatically supplies an understanding judgment.
