Construct an approximate population proportion interval
Use successes divided by sample size and the estimated binomial standard error under suitable large-sample conditions.
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.4. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Distinguish population parameters, sample statistics and repeated-sample random variables. Random selection needs a relevant frame and does not automatically remove undercoverage or non-response. Particular named sampling methods are not required. Sample-mean formulae here assume independent observations from one population with finite variance: E(X̄) = μ, Var(X̄) = σ²/n and SE = σ/√n. A mean of independent normal observations is normal; the CLT gives an approximate normal mean for an adequate large independent sample, not normal individual data. No universal sample-size cutoff guarantees that approximation. Unbiased sample variance uses n − 1, with n > 1. Mean intervals use the normal model with known variance or a large-sample approximation; a small sample with unknown variance does not justify the z method taught here. Proportion intervals use the large-sample normal approximation with adequate success and failure counts; extreme or sparse counts may invalidate it. Quoted two-sided critical values are 1.645 for 90%, 1.960 for 95% and 2.576 for 99%; retain other working unrounded. Confidence describes repeated-sampling coverage, not posterior probability for a fixed realised interval, individual observations or a guarantee. A narrower interval does not repair a biased sample. Hypothesis testing belongs to section 6.5. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Use a binomial success model.
- Calculate an estimated standard error.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is p̂ for 72 successes among 200 observations?
What you will learn
- Estimate a population proportion.
- Calculate a large-sample approximate confidence interval.
Estimate p from binary outcomes
With 72 successes in 200 independent observations, p̂ = 72/200 = 0.36. Estimated SE is √(0.36 × 0.64/200) ≈ 0.0339411. The approximate 95% interval p̂ ± 1.960SE is [0.2935,0.4265], retaining unrounded working until the endpoints.
The target p is the population probability of the consistently defined success event. The observed p̂ is a sample statistic. Do not confuse the proportion with the success count or use a mean-measurement SD formula.
Use the large-sample approximation honestly
For 120 successes in 400 observations, p̂ = 0.3 and the approximate 95% interval is [0.2551,0.3449]. This normal interval assumes the sampling model and enough expected success and failure outcomes; observed counts provide a practical adequacy check, not a universal accuracy guarantee.
Express endpoints as proportions or percentages consistently. The first interval corresponds to about 29.35% through 42.65%, not 0.2935% through 0.4265%. A narrow computed interval does not repair voluntary-response bias.
Pause and explain
Which estimated SE belongs to the large-sample proportion interval?
Worked example
Find a 95% interval for 72 successes from 200 observations.
Show the worked solution
- p̂ = 0.36.
- Estimated SE = √(0.36 × 0.64/200).
- Margin ≈ 0.0665246; add and subtract.
Answer Approximately [0.2935,0.4265].
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find an approximate 95% p interval for 72 successes in 200 suitable independent observations.
Give me a hint
Estimate p and its SE, then use z = 1.960.
Compare my reasoning
- p̂ = 0.36.
- SE ≈ 0.0339411 and margin ≈ 0.0665246.
- Endpoints ≈ 0.2935 and 0.4265.
Approximately [0.2935,0.4265].
Look for these in your work
- Use both success and failure proportions.
- State approximation.
Find the corresponding 95% interval for 120 successes in 400 observations.
Give me a hint
Apply the same formula to the new proportion and n.
Compare my reasoning
- p̂ = 0.3.
- SE = √(0.21/400); margin ≈ 0.0449092.
- Endpoints ≈ 0.2551 and 0.3449.
Approximately [0.2551,0.3449].
Look for these in your work
- Recalculate both components.
- Keep unrounded working.
Express the first interval as percentages and describe its target.
Give me a hint
Multiply proportion endpoints by one hundred.
Compare my reasoning
- Lower endpoint is about 29.35%.
- Upper endpoint is about 42.65%.
- The interval estimates the population success proportion, not an individual binary result.
About 29.35%–42.65% for the population success proportion.
Look for these in your work
- Use percentage units correctly.
- State the population parameter.
Common mistakes
- Divide success count by n.
- Use p̂(1 − p̂) inside SE.
- Keep proportion and percentage units consistent.
What can you explain now?
Is p̂ the fixed population parameter?
Compare with the explanation
No; it varies with the sample.
It estimates the unknown p.
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Make it stick
A proportion interval uses p̂ ± z√(p̂(1 − p̂)/n), with a justified large-sample binary model.
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