Approximate a binomial hypothesis tail with continuity correction
Use a suitable normal approximation under the null while preserving the observed count cell.
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 cumulative tables.
- Calculate binomial mean and variance.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the null variance for B(100,0.4)?
What you will learn
- Check expected outcome counts.
- Correct an inclusive count boundary before standardising.
Build the approximate null model
Under H₀: p = 0.4 with one count from a hundred independent trials, X ~ B(100,0.4) has mean forty and variance twenty-four. Both expected success and failure counts are substantial, so a normal approximation is suitable. Use Y ~ N(40,24).
For H₁: p > 0.4 and observed fifty successes, P₀(X ≥ 50) is approximated by P(Y > 49.5). Standardise using √24, giving z ≈ 1.93918. The approximate tail is 0.02624, or about 0.0262 using Φ(1.94) = 0.9738. Reject at 5%.
Preserve approximation limits
Observed forty-eight instead uses 47.5, z ≈ 1.53093 and approximate tail 0.06289; do not reject at 5%. The correction is tied to the inclusive count event, not selected to obtain a preferred result.
For the first observation the exact upper tail is about 0.0270992, agreeing with the 5% decision. If a decision lies close to its threshold, use exact binomial evaluation when available. A two-sided alternative also changes the tail budget and can change the decision.
Pause and explain
Which normal event approximates X ≥ 50?
Worked example
Approximate the increase-test tail for B(100,0.4), observed fifty.
Show the worked solution
- Use null mean forty and variance twenty-four.
- Correct ≥ 50 to > 49.5; z = 9.5/√24.
- Approximate tail ≈ 0.02624; reject at 5%.
Answer Approximately 0.02624; reject at 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.
Approximate the B(100,0.4) increase-test tail for fifty successes at 5%.
Give me a hint
Use the null moments and the lower cell edge 49.5.
Compare my reasoning
- Y ~ N(40,24).
- z = 9.5/√24 ≈ 1.93918.
- Upper tail ≈ 0.02624; reject at 5%.
Approximate tail 0.02624; reject.
Look for these in your work
- Apply continuity correction.
- Label approximation.
Approximate the same test for forty-eight successes.
Give me a hint
Use the cell edge 47.5 and the same null SD.
Compare my reasoning
- z = 7.5/√24 ≈ 1.53093.
- Upper normal tail ≈ 0.06289.
- Do not reject at 5%.
Approximately 0.06289; do not reject.
Look for these in your work
- Keep the null moments.
- Preserve the observed count cell.
The first observation is instead assessed for either direction using a stated symmetric normal two-tail convention. What happens at 5%?
Give me a hint
Use twice the approximate upper tail because the normal null model is symmetric.
Compare my reasoning
- Approximate one-tail probability is 0.02624.
- Two-tail probability is about 0.05248.
- This exceeds 0.05, so do not reject under that two-sided approximation.
Approximately 0.05248; do not reject for the two-sided question.
Look for these in your work
- State the changed alternative.
- Do not reuse a one-tail decision.
Common mistakes
- Use √24 as SD, not twenty-four.
- Correct before standardising.
- An approximate p-value is not exact.
What can you explain now?
Which normal boundary represents X ≥ 50?
Compare with the explanation
49.5.
It preserves the full cell centred on fifty.
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Make it stick
Check suitability, correct the integer event, then standardise under the null.
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