Compare sample sizes standard errors and dependence
Use square-root precision scaling and avoid treating duplicated data as new independent information.
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 σ/√n.
- Distinguish exact normality from a CLT approximation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What independent size increase halves SE?
What you will learn
- Compare SE at different independent sample sizes.
- Recognise the limitations of repeated or dependent data.
Precision scales with the square root
With population SD twelve, n = 9 gives SE four and n = 36 gives SE two. Four times as many independent observations halve SE; doubling n divides SE by √2, not by two.
To reduce SE from four to one under the same model needs sixteen times the original sample size: n = 144. The population mean stays unchanged and the individual SD remains twelve.
Records are not necessarily independent observations
Copying nine original values four times creates thirty-six records without adding new independent information. If each original is copied equally, the resulting average is exactly the same average of nine originals, so its SE remains four under their independent model.
A normal population gives an exact normal mean shape; an adequate large non-normal sample gives an approximate one. These shape claims and the spread formula both need a justified sampling process.
Pause and explain
What SE follows from repeating each of nine independent observations four times?
Worked example
Compare SE for n = 9 and n = 36 with σ = 12.
Show the worked solution
- SE₉ = 12/3 = 4.
- SE₃₆ = 12/6 = 2.
- Fourfold n gives half the SE.
Answer SEs 4 and 2.
Try a topic test
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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.
With σ = 12, compare SE at n = 9 and n = 36 independently.
Give me a hint
Use the square root of each sample size.
Compare my reasoning
- At nine, SE is four.
- At thirty-six, SE is two.
- The size ratio is four and the SE ratio is one half.
SEs 4 and 2.
Look for these in your work
- Scale SD through √n.
- Keep population spread unchanged.
What independent n gives SE one when σ = 12?
Give me a hint
Solve 12/√n = 1 and verify the resulting sample size.
Compare my reasoning
- √n = 12.
- Square both sides to obtain n = 144.
- Check 12/√144 = 1.
n = 144.
Look for these in your work
- Square the size relation.
- Verify the original SE target.
Each of nine original values is copied four times. What is the SE of their thirty-six-record average if the originals were independent with σ = 12?
Give me a hint
Simplify the copied-record average.
Compare my reasoning
- The total is four times the original total.
- Dividing by thirty-six gives the original nine-value average.
- Its SE remains twelve divided by three.
SE 4; the copies do not reduce it to two.
Look for these in your work
- Count independent observations.
- Use the actual average mechanism.
Common mistakes
- Sample size enters through a square root for SE.
- Duplicates do not add independent information.
- Precision does not establish absence of bias.
What can you explain now?
Does copying data four times halve uncertainty?
Compare with the explanation
No.
The average and underlying information are unchanged.
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Make it stick
More independent observations improve precision; more copies of the same observations do not.
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.