Calculate normal test errors and compare sample-size power
Translate the fixed mean decision region under a specified alternative and compare its sensitivity.
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 the normal distribution of a sample mean.
- Standardise using the relevant true mean.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What mean is used to standardise β at true μ = 104?
What you will learn
- Calculate normal Type I and Type II areas.
- Compare power while holding level and effect fixed.
Fix the mean boundary under the null
For a normal population with known σ = 12, n = 36 gives SE two. The upper 5% test of μ = 100 uses quoted z = 1.645 and rejects when X̄ > 103.29. Under μ = 100 its actual continuous-tail probability is approximately 0.0499849, reflecting critical-value rounding.
At true μ = 104, not rejecting means X̄ ≤ 103.29. Standardise under that true mean: z = (103.29 − 104)/2 = −0.355. Thus β ≈ 0.3612948 and power ≈ 0.6387052. Keep the boundary 103.29 rather than moving it to 107.29.
More independent data can improve this test
With n = 144, SE is one. Keeping the same one-sided 5% level and alternative μ = 104, the new designed null boundary is 101.645. At the alternative, β = Φ(−2.355) ≈ 0.00926135 and power ≈ 0.99073865. This comparison assumes the same normal model and independent observations.
For the n = 36 two-sided 5% test, the non-rejection interval is [96.08,103.92]. At μ = 104, β = Φ(−0.04) − Φ(−3.96) ≈ 0.4840091. Both limits must be used. These calculated powers are specific to the stated models and alternatives, not general guarantees of learner or experimental success.
Pause and explain
What is the two-sided non-rejection event for the n = 36 rule?
Worked example
Find β at μ = 104 for the fixed upper boundary 103.29 and SE two.
Show the worked solution
- The non-rejection event is X̄ ≤ 103.29.
- Under μ = 104, z = −0.355.
- β ≈ 0.3612948; power ≈ 0.6387052.
Answer β ≈ 0.3612948.
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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.
For rejection X̄ > 103.29, SE two and true μ = 104, find β and power.
Give me a hint
Standardise the fixed non-rejection boundary under mean 104.
Compare my reasoning
- Non-rejection is X̄ ≤ 103.29.
- z = −0.355; β ≈ 0.3612948.
- Power ≈ 0.6387052.
β 0.3612948; power 0.6387052.
Look for these in your work
- Keep the boundary fixed.
- Use the alternative mean.
With n = 144, σ = 12 and the same upper 5% test, calculate β at μ = 104.
Give me a hint
First design the null boundary with SE one, then use the alternative model.
Compare my reasoning
- SE = 1; boundary = 100 + 1.645 = 101.645.
- At μ = 104, z = −2.355.
- β ≈ 0.00926135; power ≈ 0.99073865.
β 0.00926135; power 0.99073865.
Look for these in your work
- Hold the level fixed.
- Distinguish redesign from evaluating a fixed rule.
For n = 36 and two-sided rejection outside [96.08,103.92], calculate β at μ = 104.
Give me a hint
Use the central non-rejection interval under N(104,4).
Compare my reasoning
- Lower z = −3.96; upper z = −0.04.
- β = Φ(−0.04) − Φ(−3.96) ≈ 0.4840091.
- Power ≈ 0.5159909.
β 0.4840091; power 0.5159909.
Look for these in your work
- Use both limits.
- Retain the original null-centred region.
Common mistakes
- The original boundary stays fixed for β.
- Use the alternative mean when standardising β.
- Two-sided β uses both non-rejection limits.
What can you explain now?
For β, should the upper boundary move with the alternative mean?
Compare with the explanation
No.
That would change the tested procedure instead of evaluating its sensitivity.
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Make it stick
Keep a test’s region fixed, change to the specified true distribution, and integrate the appropriate error event.
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