Test a normal population mean in a specified direction
Compare a sample mean with its null standard error when population variance is known.
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
- Use normal sample-mean moments.
- Choose one-tailed alternatives.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the SE for known σ = 12 and n = 36?
What you will learn
- Construct a one-tailed normal mean test.
- Use the sampling distribution rather than individual spread.
The statistic is the sample mean
Independent observations come from a normal population with known SD twelve. For n = 36 and H₀: μ = 100, X̄ ~ N(100,4), so SE is two. An increase test observing x̄ = 104 uses z = (104 − 100)/2 = 2, upper tail about 0.0228 from Φ(2) = 0.9772.
Reject at 5% in favour of an increased population mean. Using individual SD twelve would test the wrong spread. No continuity correction is needed for a continuous normal sample mean.
Direction belongs to the question
A pre-specified decrease test observing x̄ = 96 gives z = −2 and lower tail about 0.0228, so reject for a decrease. An increase test observing ninety-six would have a large upper tail and would not reject in favour of an increase.
The exact normal mean shape depends on the stated normal population and known variance. Estimated spread or non-normal data require the separate large-sample method and its conditions.
Pause and explain
Does mean ninety-six support the pre-specified increase alternative?
Worked example
Test μ = 100 against μ > 100 for x̄ = 104, n = 36, known σ = 12.
Show the worked solution
- SE = 12/√36 = 2.
- z = 2, giving upper tail ≈ 0.0228.
- Reject at 5%; evidence that the population mean exceeds 100.
Answer Reject H₀ at 5%.
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.
Test μ = 100 against μ > 100 with x̄ = 104, n = 36, known σ = 12 at 5%.
Give me a hint
Standardise the sample mean using SE two.
Compare my reasoning
- SE = 2.
- z = 2 and upper tail ≈ 0.0228.
- Reject; evidence of a population mean increase.
0.0228; reject for an increase.
Look for these in your work
- Use SE.
- Interpret population μ.
Test μ = 100 against μ < 100 with x̄ = 96 and the same n and σ.
Give me a hint
Choose the lower tail for the pre-specified decrease question.
Compare my reasoning
- SE remains two.
- z = −2 gives lower tail ≈ 0.0228.
- Reject at 5% in favour of a decrease.
0.0228; reject for a decrease.
Look for these in your work
- Retain the direction.
- Use continuous normal probability.
Would x̄ = 96 reject the pre-specified increase test?
Give me a hint
Use the upper tail even though the observation is below the centre.
Compare my reasoning
- z = −2.
- The upper tail is Φ(2) ≈ 0.9772.
- Do not reject in favour of an increase.
No; upper tail approximately 0.9772.
Look for these in your work
- Keep the chosen alternative.
- Do not swap tails afterwards.
Common mistakes
- Standardise by SE.
- Do not use count continuity correction.
- Keep the selected tail direction.
What can you explain now?
Does the sample-mean test use σ or σ/√n?
Compare with the explanation
σ/√n.
The statistic is an average of n independent observations.
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Make it stick
A normal mean test standardises with the null mean and the statistic’s SE.
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