01 · Name both objects
State the critical value and critical region separately.
Hint
One is a boundary; one is a set of counts.
Worked solution
Critical value: 7. Critical region: X≥7 within the support 0,…,12, namely 7,…,12.
Understand · explore · practise
Upper critical regions
Build an upper critical region for a binomial hypothesis test, prove its cutoff with adjacent tail probabilities and distinguish a critical value from the whole rejection region.
Before you startUpper binomial tails, significance and hypotheses.
01 / Set the rejection rule before observing the count
A critical region is a set of possible test-statistic values.
For an upper test, reject when X≥r.
Choose the smallest integer r for which the null probability P(X≥r) is at most the stated significance level α.
This gives the largest upper-tail region compatible with the at-most rule. Any smaller boundary would include too much null probability; a larger boundary would discard additional evidence unnecessarily.
Choose the rejection boundaryExplore
Under H₀, X~B(12,0.25). For H₁: p>0.25, find a 5% upper critical region.
02 / Find the first boundary that works
A working value and its failing neighbour justify the choice.
Watch: why the boundary is seven
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Under H₀, X~B(12,0.25). Construct an upper critical region at 5%.Worked example
P(X≥6)≈0.05440223>0.05
Six cannot be the rejection boundary.
P(X≥7)≈0.01425278≤0.05
Seven meets the significance requirement.
Critical region: X≥7, or {7,8,9,10,11,12}
Seven is the smallest working boundary. It is the critical value.
02 · Why not eight?
X≥8 has null probability about 0.00278151. Why is it not the usual answer to this 5% construction?
Hint
Seven already satisfies the rule.
Worked solution
It is a valid but unnecessarily smaller rejection region. The required standard construction uses the smallest working boundary, seven.
03 / Convert the upper tail correctly
The complement ends one integer before the boundary.
P(X≥r)=1−F(r−1).
F(k)=P(X≤k). For the boundary7, subtract the cumulative probability through6.
For this model F(6)≈0.98574722, so 1−F(6)≈0.01425278. F(5)≈0.94559777 gives the failing candidate P(X≥6)≈0.05440223.
03 · Off-by-one error
A learner tests r=7 by calculating 1−F(7). What event have they actually calculated?
Hint
The complement excludes counts through seven.
Worked solution
P(X≥8), equivalently P(X>7). This misses the boundary count 7.
04 · Equivalent CDF condition
Rewrite P(X≥r)≤0.05 using F.
Hint
Move the complement terms.
Worked solution
F(r−1)≥0.95. The r−1 matters.
04 / Apply the region to a new observation
Constructing the rule and applying it are separate steps.
Once the upper critical region X≥7 is fixed, an observed count of 7 leads to rejection; a count of 6 does not. The boundary itself is included. The null model and critical region do not change to fit the observed sample.
05 · On the boundary
Twelve components are checked and seven are faulty. Test H₀:p=0.25 against H₁:p>0.25 using the region above.
Hint
Check whether x belongs to the region.
Worked solution
Seven lies in the critical region. Reject H₀: there is sufficient evidence at 5% that the fault probability exceeds 0.25, assuming the binomial model is appropriate.
06 · Just outside
Repeat the conclusion when six are faulty.
Hint
Six is below the boundary.
Worked solution
Do not reject H₀. There is insufficient evidence at 5% of an increased fault probability. This does not prove p=0.25.
05 / Read the actual rejection probability
Discrete counts make the achieved level jump.
The nominal level is 5%. The actual significance of this region is P(X≥7 | p=0.25)≈0.01425278, about 1.425%. It is not automatically 5%. Including the next lower count would raise the total to about 5.440%, exceeding the target.
07 · Probability of wrong rejection
Under H₀ and its assumptions, what is the chance that this rule rejects?
Hint
Use the whole critical region.
Worked solution
About0.01425278, or 1.425%. This is conditional on the null probability and model being correct.
08 · Nonrejection probability
Under H₀, what is the probability that the count does not enter the region?
Hint
Take the complement.
Worked solution
1−0.01425278≈0.98574722. This is not the probability that H₀ is true.
06 / Show the neighbouring probability
A boundary assertion needs more than a calculator answer.
09 · Adequate evidence
Is “P(X≥9)≤0.05, so the critical value is 9” a sufficient justification?
Hint
Could a smaller boundary also work?
Worked solution
No. A working candidate alone does not establish the smallest boundary. Check the candidate one integer lower; in this example 7 already works.
10 · Rounded equality
A tail is displayed as 0.0500. Can you conclude it meets an at-most 0.05 rule?
Hint
Rounding can hide which side of the threshold it lies on.
Worked solution
Not without enough precision. For example 0.05004 rounds to 0.0500 but exceeds 0.05, while 0.04996 does not.
07 / Allow for an empty region when needed
Sometimes even the most extreme supported count is too likely.
If P(X=n)>α, there is no supported upper count with an upper tail at most α. The critical region is empty under this construction. Writing r=n+1 is a convenient representation of the impossible event X≥n+1; it is not an observable count.
11 · Empty example
For X~B(2,0.8), can a nonempty upper critical region have probability at most 0.05?
Hint
The smallest nonempty upper event is X=2.
Worked solution
No. P(X=2)=0.64>0.05. Every other nonempty upper tail is larger. The critical region is empty.
12 · Support
For a 12-trial count, what does the region X≥13 mean?
Hint
Check the possible values.
Worked solution
It is empty: no count can exceed 12.
13 · Decreasing level
For a fixed null distribution, can reducing α make the smallest upper boundary decrease?
Hint
An easier-to-enter region would have more probability.
Worked solution
No. Tightening the allowed tail probability can only keep the boundary fixed or move it upwards, possibly producing an empty region.
14 · Tail direction
Would the same upper region be appropriate for H₁:p<0.25?
Hint
Which observations support a decrease?
Worked solution
No. A decrease requires a lower-tail construction. Large success counts point in the opposite direction.
08 / Boundary, neighbour, region, conclusion
Keep the supported set visible.
Calculate using the null probability. Find the smallest upper boundary meeting α and show the adjacent lower boundary fails. State the full critical region and its actual probability. Apply the rule to the observed count and conclude in context.
Section 1 of 8 · Set the rejection rule before observing the count