Construct a large-sample mean interval with estimated variance
Use an unbiased variance estimate in an approximate large-sample interval and flag small unknown-variance cases.
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
- Estimate s² with n − 1.
- Use the CLT and normal critical values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What estimated SE follows from s² = 100 and n = 100?
What you will learn
- Form a large-sample approximate mean interval.
- Identify when the taught z approximation is unsupported.
Estimated variance adds an approximation
For n = 100, x̄ = 80 and unbiased variance estimate s² = 100, estimated SE is √100/√100 = 1. The approximate 95% mean interval is 80 ± 1.960, or [78.04,81.96], assuming an adequate independent large-sample setting.
If σ² is unknown, substituting s² does not make a known-variance exact formula. Describe the normal approximation and its conditions. The population may remain non-normal while an adequate large-sample mean is approximately normal.
Use summaries without losing the unbiased divisor
For n = 64, Σx = 1280 and Σx² = 26167, x̄ = 20 and s² = (26167 − 1280²/64)/63 = 9. Estimated SE is 3/8, giving an approximate 95% interval [19.265,20.735].
A sample of four from a population with unknown variance does not justify this large-sample z method automatically, even if normality is asserted. Other inference methods may be needed but are not introduced as required section 6.4 content here.
Pause and explain
Does n = 4 with unknown variance justify the large-sample z interval automatically?
Worked example
Find an approximate 95% interval for n = 100, x̄ = 80, s² = 100.
Show the worked solution
- Estimated SD = 10.
- Estimated SE = 10/√100 = 1.
- Add and subtract 1.960.
Answer Approximately [78.04,81.96].
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Test this topic →From guided practice to a new situation
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% mean interval for n = 100, x̄ = 80, s² = 100.
Give me a hint
Use estimated SE and z = 1.960 under suitable large-sample assumptions.
Compare my reasoning
- Estimated SE = 1.
- Margin = 1.960.
- Endpoints are 78.04 and 81.96.
Approximately [78.04,81.96].
Look for these in your work
- Use s, not s², for SE.
- State assumptions and approximation.
Use n = 64, Σx = 1280, Σx² = 26167 to form a suitable large-sample 95% mean interval.
Give me a hint
Calculate the unbiased variance from the summary first.
Compare my reasoning
- x̄ = 20 and s² = 567/63 = 9.
- Estimated SE = 3/8 = 0.375; margin = 0.735.
- Endpoints are 19.265 and 20.735.
Approximately [19.265,20.735].
Look for these in your work
- Retain n − 1.
- Use estimated SE.
Can the same large-sample z procedure be justified from four observations with unknown population variance?
Give me a hint
Check known variance or adequate large-sample justification.
Compare my reasoning
- The population variance is unknown.
- Four is not a large-sample setting.
- The taught z procedure is not justified automatically; an appropriate other method would need its own assumptions.
No automatic justification for the large-sample z interval.
Look for these in your work
- Do not turn an estimate into known variance.
- Flag the method limitation.
Common mistakes
- Square-root s² before forming SE.
- Keep the approximation label.
- Small unknown-variance samples do not automatically justify z intervals.
What can you explain now?
Is substituting s² the same as knowing σ²?
Compare with the explanation
No.
The estimate varies and the large-sample justification remains necessary.
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Make it stick
Estimated variance gives an approximate large-sample mean interval, with its assumptions stated.
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