Test a binomial decrease using an inclusive lower tail
Use cumulative binomial probability for an unexpectedly small success count.
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
- Sum binomial masses.
- Recognise the decrease direction.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What event corresponds to observed five in a decrease test?
What you will learn
- Calculate an exact lower tail.
- Check an integer critical boundary.
A lower alternative changes the event
A process claims success probability 0.5. With n = 20 and H₁: p < 0.5, observed X = 5 uses P₀(X ≤ 5) = 21700/1048576 ≈ 0.0206947327. Reject at 5% and conclude there is evidence of a decreased success probability.
The lower tail includes zero through five. Using the upper tail answers a different directional question. A small point mass is insufficient unless its whole relevant tail also meets the criterion.
Find the largest acceptable lower boundary
P₀(X ≤ 6) ≈ 0.0576591492 exceeds 0.05. Therefore X ≤ 5 is the largest lower-tail region fitting the 5% bound; observed six does not reject.
A non-rejection at six does not prove p = 0.5. The result could reflect sampling variation or a test with limited sensitivity to a particular alternative.
Pause and explain
Which is the 5% lower critical region here?
Worked example
Test a decrease from p = 0.5 with five successes among twenty.
Show the worked solution
- Use X ~ B(20,0.5) under H₀.
- P₀(X ≤ 5) ≈ 0.0206947327.
- Reject at 5% in favour of a decrease.
Answer Reject H₀ at 5%.
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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.
Test p = 0.5 against p < 0.5 with five successes among twenty at 5%.
Give me a hint
Sum all masses from zero through the observed count.
Compare my reasoning
- Use the null B(20,0.5).
- The inclusive lower tail is 0.0206947327.
- Reject; evidence that success probability is below 0.5.
0.0206947327; reject.
Look for these in your work
- Use the lower tail.
- State the contextual direction.
Would six successes reject under the same rule?
Give me a hint
Check the next lower-tail cumulative value.
Compare my reasoning
- P₀(X ≤ 6) ≈ 0.0576591492.
- This is greater than 0.05.
- Do not reject H₀.
No; tail 0.0576591492 exceeds 0.05.
Look for these in your work
- Include six.
- Retain the level.
Explain why do not reject at six does not prove the process success probability is 0.5.
Give me a hint
Non-rejection and proof of equality are different statements.
Compare my reasoning
- The test measures evidence using a specific rule.
- A false null can still produce a non-rejection outcome.
- The result supplies insufficient evidence for the stated decrease, not certainty.
Insufficient evidence of a decrease; no proof of equality.
Look for these in your work
- State the limited conclusion.
- Recognise possible Type II error.
Common mistakes
- Use ≤ for the inclusive lower tail.
- Do not reverse the chosen alternative.
- Check the neighbouring integer boundary.
What can you explain now?
Does observed six reject using the stated 5% lower-tail rule?
Compare with the explanation
No.
P₀(X ≤ 6) exceeds the level.
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Make it stick
A lower discrete test includes the observed count and every smaller count.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.