Test a binomial increase using an inclusive upper tail
Calculate the probability of the observed success count or a more extreme increase under the null.
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 binomial probabilities and complements.
- Check independent fixed-probability trials.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which event is the increase-test tail for observed eight?
What you will learn
- Calculate an exact upper-tail probability.
- Interpret a binomial count test.
Include the observed count
Twenty independent items each have defect probability 0.2 under H₀. For H₁: p > 0.2 and observed X = 8, the relevant probability is P₀(X ≥ 8), not P₀(X = 8) alone and not P₀(X > 8).
Evaluate Σ from k = 8 to 20 of C(20,k)0.2ᵏ0.8²⁰⁻ᵏ. The tail is approximately 0.0321426631. It is below 0.05, so reject H₀ at 5%: there is evidence of increased defect probability.
Check the model and neighbouring outcome
For observed seven defects, P₀(X ≥ 7) is approximately 0.0866925136, so do not reject at 5%. The rule X ≥ 8 is the smallest upper-tail critical threshold satisfying the level bound.
The test assumes a fixed number of independent binary observations with one common defect probability. Changing probabilities or dependent items can invalidate this binomial model.
Pause and explain
What happens with tail 0.03214 at 5%?
Worked example
Test p = 0.2 against p > 0.2 using eight defects in twenty at 5%.
Show the worked solution
- Under H₀, X ~ B(20,0.2).
- P₀(X ≥ 8) ≈ 0.0321426631.
- Reject because this is below 0.05.
Answer Reject H₀; evidence of an increase.
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 p = 0.2 against p > 0.2 with X = 8 from twenty items at 5%.
Give me a hint
Sum the inclusive binomial upper tail under p = 0.2.
Compare my reasoning
- X ~ B(20,0.2) under H₀.
- P(X ≥ 8) ≈ 0.0321426631.
- Reject H₀; evidence that the defect probability exceeds 0.2.
Tail 0.0321426631; reject at 5%.
Look for these in your work
- Include the observed count.
- Interpret increased defects.
Use the same test with X = 7.
Give me a hint
Compare the inclusive tail with 0.05.
Compare my reasoning
- Calculate P₀(X ≥ 7).
- The tail is approximately 0.0866925136.
- Do not reject H₀ at 5%.
Tail 0.0866925136; do not reject.
Look for these in your work
- Keep the null model.
- Use the same pre-specified level.
Why is testing with B(20,8/20) an incorrect null calculation for the first observation?
Give me a hint
The observed estimate and claimed parameter play different roles.
Compare my reasoning
- 8/20 estimates the observed success fraction.
- H₀ specifies probability 0.2.
- Replacing it by the estimate changes the reference model being tested.
Use B(20,0.2), not B(20,0.4), for the null tail.
Look for these in your work
- Identify the null parameter.
- Keep observation and model separate.
Common mistakes
- Include X = 8 in the tail.
- A point mass alone is not the tail.
- Use p under H₀, not 8/20.
What can you explain now?
Should 8/20 replace 0.2 in the null tail calculation?
Compare with the explanation
No.
The test compares the observation with the claimed null model.
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Make it stick
For an upper count test, sum from the observed count upward under the null parameter.
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