Estimate population moments from summarised sample totals
Use n, the sum and sum of squares, and reject summaries that imply negative squared spread.
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 sample mean.
- Expand the sum of squared deviations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which expression is the centred squared sum?
What you will learn
- Calculate unbiased estimates from summary statistics.
- Check consistency of a supplied summary.
Recover the centred sum
The centred squared sum is Q = Σx² − (Σx)²/n. With n = 5, Σx = 20 and Σx² = 90, the mean estimate is four and Q = 90 − 80 = 10. Unbiased variance is Q/(n − 1) = 2.5.
Square the complete sum before dividing by n. Σx² is the sum of individually squared values; it is not (Σx)². Those two summaries serve different roles in the formula.
Reject inconsistent totals
With n = 8, Σx = 48 and Σx² = 316, mean is six and Q = 316 − 288 = 28, giving variance estimate four. If the supplied sum of squares were 280 instead, Q would be negative, which no real sample can have.
This check identifies inconsistency; it does not tell you which reported total is wrong. Do not set negative variance to zero to make the calculation appear valid. Verify the data or correct the summary with evidence.
Pause and explain
What should you do if the summaries imply negative variance?
Worked example
Use n = 5, Σx = 20, Σx² = 90 to estimate mean and variance.
Show the worked solution
- Mean = 20/5 = 4.
- Q = 90 − 20²/5 = 10.
- s² = 10/4 = 2.5.
Answer Mean 4; variance 2.5.
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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.
Estimate mean and variance for n = 5, Σx = 20, Σx² = 90.
Give me a hint
Find the centred squared sum before applying n − 1.
Compare my reasoning
- Mean = 4.
- Q = 90 − 400/5 = 10.
- s² = 10/4 = 2.5.
Mean 4; variance 2.5.
Look for these in your work
- Keep the sums distinct.
- Use the unbiased divisor.
Estimate mean and variance for n = 8, Σx = 48, Σx² = 316.
Give me a hint
Square forty-eight before dividing by eight.
Compare my reasoning
- Mean = 6.
- Q = 316 − 2304/8 = 28.
- s² = 28/7 = 4.
Mean 6; variance 4.
Look for these in your work
- Calculate the centred total.
- Use exact working.
Check n = 8, Σx = 48, Σx² = 280 for consistency.
Give me a hint
A squared-deviation sum must be non-negative.
Compare my reasoning
- The centre contribution is 2304/8 = 288.
- Q = 280 − 288 = −8.
- These totals cannot describe a real sample.
Inconsistent summary; Q = −8 is impossible.
Look for these in your work
- Flag the source inconsistency.
- Do not silently alter variance.
Common mistakes
- Distinguish sum of squares from square of sum.
- Divide centred squared sum by n − 1.
- Reject negative centred spread.
What can you explain now?
Can a real sample have negative Q?
Compare with the explanation
No.
Q is a sum of squared deviations.
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Make it stick
Use the full sample totals to reconstruct centred spread, then divide by n − 1.
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.