How can repeated chance trials estimate a p-value?
You will be able to: Build and interpret a simulation under a stated null proportion.
How can repeated chance trials estimate a p-value?
Suppose a fair-choice benchmark predicts success half the time. To judge an unusually large success count, repeat the same-size sample under that benchmark.
A useful starting point: What probability does a p-value measure? →
Words and symbols before equations
- Null simulation
- Repeated samples generated with the hypothesized parameter.
- Replication
- One complete sample yielding one statistic.
- Extreme count
- Replications meeting the alternative’s tail rule.
- Monte Carlo error
- Random variation from a finite number of simulated replications.
What this picture assumes
Null p₀=.50; independent Bernoulli trials in each replication. Exact binomial and finite simulated tail probabilities use inclusive count boundaries. For this symmetric null, a two-sided event uses distance from n/2.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- Simulated tail 0.02; exact binomial tail 0.028444. Equality at the observed boundary is included. The curve is only a normal visual approximation to this discrete model.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
For H₀:p=.50, generate n independent Bernoulli outcomes per replication, count successes and repeat. Do not simulate with the observed p̂, which would recenter the benchmark on the data.
For a greater alternative, count simulated successes at least as large as the observed count. For a two-sided symmetric .50 model, compare absolute distances from n/2.
Divide extreme replications by all replications to estimate the null probability. A finite estimate of zero does not mean the true tail probability is zero. This simulation assumes the Bernoulli model and a suitable design; more repetitions do not repair biased data.
A worked example, step by step
For a greater-than test, 38 of 1000 null replications have a statistic at least as large as observed. Interpret the estimate.
- Generate the replications using H₀ and the original sample size.
- Include outcomes equal to the observed statistic in the tail.
- Estimate the p-value as 38/1000=.038.
- Under the null model, the estimated chance of a statistic at least this large is 3.8%; the finite estimate fluctuates.
Repeating the simulation increases precision of a probability estimate; it does not increase the original study’s sample size.
If no simulations are extreme, is the true p-value necessarily zero?
Compare with an explanation
No. A finite simulation can miss a rare event.
Predict. Change one thing. Explain.
Change the seed and the replication count, keeping n and the observed count fixed. Compare the simulation estimate with the exact binomial tail displayed.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Simulated tail 0.02; exact binomial tail 0.028444. Equality at the observed boundary is included. The curve is only a normal visual approximation to this discrete model.
| Quantity | Value |
|---|---|
| Observed x / n | 60 / 100 |
| Extreme replications | 10 / 500 |
| Simulated p-value | 0.02 |
| Exact inclusive binomial tail | 0.028444 |
Null p₀=.50; independent Bernoulli trials in each replication. Exact binomial and finite simulated tail probabilities use inclusive count boundaries. For this symmetric null, a two-sided event uses distance from n/2.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the probability values, reference groups, graph scales or model assumptions. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA simulation uses 2000 null replications; 90 are as extreme as the observed statistic. Give the estimate and two assumptions.
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Compare with the answer and four-point rubric
- 1 point: The estimated p-value is 90/2000=.045.
- 1 point: The replications use the null parameter and original sample size.
- 1 point: The chance process and dependence structure must match the study’s justified model.
- 1 point: Finite simulation error remains; .045 is not P(H₀ is true).
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What is fixed in a null simulation?
The null parameter, sample size and generation rule.
RECALL 2What does changing the seed do?
Produces another reproducible finite run.
RECALL 3What does more replication improve?
Monte Carlo precision, not the study’s representativeness.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can repeated chance trials estimate a p-value?
- Estimated p-value=extreme null replications / all replications.
- Use p₀ to generate the null model.
- Include equally extreme outcomes.
Remember: Repeating the simulation increases precision of a probability estimate; it does not increase the original study’s sample size.
Conditions: Null p₀=.50; independent Bernoulli trials in each replication. Exact binomial and finite simulated tail probabilities use inclusive count boundaries. For this symmetric null, a two-sided event uses distance from n/2.
Refresh Kid · AP Statistics Unit 3 · Objectives 3.6.A · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 3.6, objectives 3.6.A. Framework effective Fall 2026, checked September 17, 2026. Unit 3 includes inference for one and two population proportions, errors and power, and chi-square homogeneity/independence tests; it is part of the revised five-unit course.
Examples and datasets are synthetic, independently authored teaching material. Inference requires a justified design and appropriate counts. Intervals use observed proportions; null tests use their reference-model proportions. The revised CED states chi-square expected counts should be greater than 5; this unit follows that wording even though some companion texts use at least 5. Normal and chi-square inference are approximate. The chi-square goodness-of-fit test is not included in this unit’s official scope.
The Organic Chemistry Tutor companion title and destination were located; the full video was not reviewed. Khan Academy’s destination was checked, but its lesson content was not fully readable by the research tool. OpenStax provides optional reference reading. No provider scripts, questions or graphics were copied. Refresh Kid is not affiliated with these providers.
GitHub’s 3D website collection informed optional spatial inspection. Our original categorical count grid uses self-hosted Three.js with its MIT license. Two rows and three columns organize synthetic people into categorical cells. Each stacked block represents one person. Camera rotation changes only the view; exact counts, expectations and contributions are always available in the 2D table. Inference curves and intervals remain 2D; use the exact table rather than apparent 3D size for comparisons. Complete labeled diagrams, count tables and explanations remain available without 3D.
Independent teacher review and observation of students remain pending. Technical checks do not certify statistical accuracy, accessibility or learning effectiveness. This is a review edition.
Released AP Statistics questions and scoring guides are optional. Older exams use the earlier framework, so check alignment before selecting parts. All practice on this page is original, not official AP material.
Learn → Explore → Practice → Review is informed by the IES learning guide. This implementation has not yet been evaluated with learners.
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