Plan a proportion sample size for a target margin
Use anticipated or conservative proportion variance, round upward and retain the approximation assumptions.
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 the proportion margin formula.
- Take a ceiling after solving a size inequality.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the maximum of p(1 − p) for 0 ≤ p ≤ 1?
What you will learn
- Plan n for a proportion precision target.
- Explain why p = 0.5 gives a conservative variance choice.
Use a planning value, not an unknown result
For planned proportion p₀ and target margin e, the normal formula gives n ≥ z²p₀(1 − p₀)/e². If no reliable planning estimate is available, p₀ = 0.5 maximises p₀(1 − p₀) at 1/4 and gives a conservative size for this width formula.
At 95% confidence and margin 0.05, n ≥ 1.960² × 0.25/0.05² = 384.16, so use 385. At 384 the maximum formula margin slightly exceeds 0.05. A sample-size plan does not guarantee the normal approximation’s actual coverage or remove selection bias.
State the design assumptions
With a justified anticipated proportion 0.3 and target margin 0.04, the bound is 504.21, giving 505. If actual p is different, the width may differ; describe the assumed planning value.
At 99% confidence using the conservative 0.5 and margin 0.05, z = 2.576 gives n ≥ 663.5776, so use 664. Confidence, margin and sample-size tradeoffs apply to independent valid sampling with adequate outcome counts.
Pause and explain
Which integer meets n ≥ 384.16?
Worked example
Plan conservative n for a 95% proportion margin of 0.05.
Show the worked solution
- Use worst-case p₀(1 − p₀) = 1/4.
- The bound is 384.16.
- Round upward to 385.
Answer n = 385 under the planning formula.
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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.
Plan conservative n for 95% confidence and proportion margin at most 0.05.
Give me a hint
Use p₀ = 0.5 and z = 1.960.
Compare my reasoning
- n ≥ 1.960² × 0.25/0.05² = 384.16.
- Round upward to 385.
- The preceding 384 gives a maximum margin just over 0.05.
n = 385; 384 fails the conservative width target.
Look for these in your work
- Use the maximum variance.
- Check the preceding integer.
Use anticipated p₀ = 0.3, 95% confidence and margin 0.04 to plan n.
Give me a hint
Insert p₀(1 − p₀) = 0.21.
Compare my reasoning
- n ≥ 1.960² × 0.21/0.04² = 504.21.
- Take the ceiling 505.
- The result depends on the stated anticipated proportion.
n = 505 with planning p₀ = 0.3.
Look for these in your work
- State the planning value.
- Round upward.
Plan conservative n for 99% confidence and margin 0.05, using z = 2.576.
Give me a hint
Use the higher critical value but the same worst-case variance.
Compare my reasoning
- n ≥ 2.576² × 0.25/0.05² = 663.5776.
- Round upward to 664.
- Higher confidence needs more independent observations for the same margin.
n = 664 under the conservative planning formula.
Look for these in your work
- Use the 99% multiplier.
- Retain design and approximation limits.
Common mistakes
- Use a planning value with justification.
- Take a ceiling.
- A planned formula width does not guarantee model validity or actual coverage.
What can you explain now?
Why use p₀ = 0.5 without a prior estimate?
Compare with the explanation
It maximises p₀(1 − p₀).
This is conservative for the normal width formula.
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Make it stick
Plan proportion precision with a justified variance value, then round n up and keep model limits explicit.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.