Test a mean with a large sample and estimated variance
Use unbiased sample spread in an approximate mean test while keeping the null centre separate.
Section 6.5 · Hypothesis tests
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.5. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Specify population hypotheses, significance level, test statistic and tail direction before inspecting the result. Calculate tail probabilities under H₀, including the observed discrete value. A p-value is not the probability that H₀ is true. Reject H₀ when the specified tail probability is at most the stated significance threshold; otherwise do not reject. The syllabus term acceptance region means the non-rejection region, not proof of H₀. For two-sided discrete examples here, use an explicitly stated equal-tail allocation; asymmetric discrete models admit different conventions, so do not treat twice the smaller tail as a universal definition. Integer critical regions can have actual Type I probability below the nominal level. Normal approximations need a suitable model and continuity correction for counts; check exact probabilities when approximation could change a close decision. Mean tests use a normal population with known variance or an adequate large independent sample with estimated variance. Small samples with unknown variance do not automatically justify the z method. Type I means rejecting a true null; Type II means not rejecting a false null at a specified alternative. Calculate error probabilities using the fixed decision region under the relevant parameter, not by changing the test after seeing data. Power is one minus Type II probability at that alternative. Quoted normal critical values and table probabilities have rounding limits. Interpret evidence in context without claiming certainty, causation or practical importance. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Calculate unbiased variance from sample totals.
- Use a suitable CLT approximation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which divisor gives unbiased variance for n = 100?
What you will learn
- Calculate an estimated standard error from totals.
- State large-sample test limitations.
Estimate spread from the observations
For n = 100, Σx = 5000 and Σx² = 251584, observed x̄ = 50. The centred sum is 251584 − 5000²/100 = 1584, so s² = 1584/99 = 16 and estimated SE is 0.4. Test H₀: μ = 49 against H₁: μ > 49 using z = (50 − 49)/0.4 = 2.5.
With a suitable large independent sample, the upper normal tail is approximately 0.0062 using Φ(2.5) = 0.9938. Reject at 1% with evidence of increased population mean. The mean model and estimated SE make this an approximate test, not a known-variance exact normal test.
Keep the method’s conditions
For a two-sided alternative with the same result, the approximate probability is 0.0124. It exceeds 0.01, so do not reject at 1%. Direction changes the rule, not the observed variance.
A sample of four with unknown variance does not automatically justify this large-sample z method. Dependence, strong tail problems or a biased sampling frame can also undermine the calculation; sample size alone does not repair the design.
Pause and explain
What happens for the two-sided alternative at 1%?
Worked example
Carry out the 1% increase test from mean forty-nine using the stated totals.
Show the worked solution
- Calculate x̄ = 50 and s² = 16.
- Estimated SE = 0.4; z = 2.5.
- Approximate upper tail = 0.0062; reject at 1%.
Answer Approximately 0.0062; reject at 1%.
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.
Use n = 100, Σx = 5000, Σx² = 251584 to test μ = 49 against μ > 49 at 1%.
Give me a hint
Calculate the unbiased variance before the estimated SE.
Compare my reasoning
- x̄ = 50; centred sum = 1584; s² = 16.
- Estimated SE = 0.4; z = 2.5.
- Approximate upper probability = 0.0062; reject.
s² = 16; approximate p = 0.0062; reject at 1%.
Look for these in your work
- Use n − 1.
- State approximate inference.
Use the same data for H₁: μ ≠ 49 at 1%.
Give me a hint
Account for both normal tails under the symmetric approximation.
Compare my reasoning
- z remains 2.5.
- Two-sided probability ≈ 2 × 0.0062 = 0.0124.
- Do not reject at 1%.
Approximately 0.0124; do not reject.
Look for these in your work
- Use both tails.
- Keep the level at 1%.
Can a four-observation sample with unknown variance automatically use this large-sample z test?
Give me a hint
Check whether the method’s large-sample condition is supported.
Compare my reasoning
- Variance is not known.
- Four observations do not establish the required large-sample approximation.
- A different justified method and its assumptions would be needed.
No automatic justification for this z method.
Look for these in your work
- Identify the missing condition.
- Avoid an unsupported exact-normal claim.
Common mistakes
- Use n − 1 for the unbiased variance.
- Retain the null centre in z.
- State approximation and sampling conditions.
What can you explain now?
Does estimating variance make it known population variance?
Compare with the explanation
No.
The estimate remains sample-dependent.
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Make it stick
Estimate spread with n minus one, use the null centre, and label large-sample inference approximate.
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