Find discrete critical regions and their actual significance
Choose integer boundaries that respect the level rather than forcing a continuous cutoff onto a 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
- Calculate exact count tails.
- Compare probabilities with a threshold.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which upper threshold respects level 0.05?
What you will learn
- Find a valid discrete upper threshold.
- Distinguish nominal level from actual null rejection probability.
Discrete tails jump
For X ~ B(10,0.5) and H₁: p > 0.5, P₀(X ≥ 8) = 56/1024 = 0.0546875. That is too large for a 5% test. P₀(X ≥ 9) = 11/1024 = 0.0107421875, so X ≥ 9 is the first admissible upper boundary.
The nominal level is 0.05, but the actual Type I probability for this fixed region is 0.0107421875. A non-randomised integer rule need not use its entire probability allowance.
Check the region before using it
At X = 8, a correct 5% decision is do not reject. At X = 9, reject. Rounding a calculated real-number boundary downward could create a region exceeding the level.
Retain full probability precision when choosing the boundary. The important check is the original tail inequality and its neighbouring integer, not only a rounded p-value.
Pause and explain
What is the actual Type I probability for the ≥ 9 rule?
Worked example
Find the upper 5% critical region for B(10,0.5).
Show the worked solution
- Tail at eight is 0.0546875 > 0.05.
- Tail at nine is 0.0107421875 ≤ 0.05.
- Reject for X ≥ 9; actual size is 11/1024.
Answer X ≥ 9; actual size 0.0107421875.
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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.
Find the upper 5% rejection region for B(10,0.5).
Give me a hint
Compare the tails at eight and nine before choosing.
Compare my reasoning
- P₀(X ≥ 8) = 56/1024 > 0.05.
- P₀(X ≥ 9) = 11/1024 ≤ 0.05.
- The smallest admissible threshold is nine.
X ≥ 9.
Look for these in your work
- Check both boundaries.
- Use an inclusive event.
Give the actual significance of this rule.
Give me a hint
Evaluate the rejection region under p = 0.5.
Compare my reasoning
- The region is nine or ten successes.
- Its null probability is (10 + 1)/1024.
- This equals 0.0107421875, below the nominal 0.05.
Actual Type I probability 11/1024 = 0.0107421875.
Look for these in your work
- Use the fixed region.
- Distinguish nominal and actual size.
Should observed eight be declared significant after rounding 0.0546875 to 0.05?
Give me a hint
Use the unrounded probability for the decision.
Compare my reasoning
- The exact tail exceeds 0.05.
- Rounding for display does not change the region.
- Do not reject at the original 5% level.
No; eight does not reject at 5%.
Look for these in your work
- Retain decision precision.
- Do not alter the level retrospectively.
Common mistakes
- Actual size need not equal nominal level.
- Test the neighbouring integer.
- Do not use rounded probabilities to force a boundary.
What can you explain now?
Is ≥ 8 valid at 5%?
Compare with the explanation
No.
Its actual null tail exceeds 0.05.
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Make it stick
A valid integer critical region fits the probability budget; its neighbour does not.
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.