01 · Allocate 1%
What is the tail budget for a 1% equal-tail test?
Hint
Divide 0.01 by two.
Worked solution
0.005, or 0.5%, in each tail.
Understand · explore · practise
Two-tailed critical regions
Construct both binomial rejection tails, halve the nominal level, distinguish at-most and explicitly closest-tail instructions and calculate the actual total.
Before you startUpper and lower critical regions and actual significance.
01 / A two-sided alternative needs both ends
Use a stated construction convention.
For an equal-tail at-most test, allocate α/2 to each tail.
Find the largest lower cutoff and smallest upper cutoff whose respective null tail probabilities do not exceed α/2.
The critical region is a union: X≤a or X≥b. For the separated tails considered here, add the probabilities to obtain the actual level. Both hypotheses still concern the same population parameter.
Same target, different construction instructionExplore
Under H₀, X~B(12,0.25). Investigate a change in either direction at nominal 10%, with target 0.05 in each tail.
02 / Halve before finding the cutoffs
Do not put the whole significance level in each tail.
For H₁:p≠p₀ at nominal 10%, each tail has an at-most 0.05 budget. At nominal 5%, the budget is 0.025 at each end. The achieved probabilities need not match each other because a binomial distribution can be asymmetric.
02 · Double counting the level
A learner allocates 5% at each end of a 5% two-tailed test. What is wrong?
Hint
Add the proposed tail budgets.
Worked solution
They have allowed up to 10% overall. The equal-tail 5% construction uses 2.5% per tail.
03 / Construct each at-most tail separately
Use neighbours on the correct side.
Watch: the instruction changes one boundary
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Under H₀, X~B(12,0.25). Construct a 10% equal-tail test with each tail at most 0.05.Worked example
Lower: F(0)≈0.03167635; F(1)≈0.15838176
Zero works; one fails. The lower region is X=0.
Upper: P(X≥7)≈0.01425278; P(X≥6)≈0.05440223
Seven works; six fails the 0.05 per-tail limit.
Critical region: X=0 or X≥7
The two regions are disjoint and lie within 0,…,12.
03 · Actual total
Find the actual significance of this rule.
Hint
Add both rejecting-tail probabilities.
Worked solution
0.03167635 + 0.01425278 ≈ 0.04592913, about 4.593%. It is below the nominal 10%.
04 · Middle region
Which supported counts do not reject?
Hint
Exclude zero and counts seven upwards.
Worked solution
1,2,3,4,5,6. This nonrejection region does not prove H₀.
04 / Read an explicit closest-tail instruction literally
Closest is a different optimisation from no greater than.
If an exercise explicitly asks for each tail as close as possible to 0.05, compare absolute distances from 0.05 on both sides. The upper tail at 6 is 0.05440223, only 0.00440223 away. The upper tail at 7 is 0.01425278, about 0.03574722 away. Closest therefore chooses 6, even though its upper-tail probability exceeds 0.05.
05 · Closest lower tail
For the same distribution, compare F(0)≈0.03167635 and F(1)≈0.15838176 with target 0.05. Which is closer?
Hint
Compare absolute differences.
Worked solution
F(0): distance about 0.01832365, versus about 0.10838176 for F(1). The lower cutoff stays 0.
06 · Closest region and level
State the closest-tail region and its actual significance.
Hint
Use zero below and six above.
Worked solution
X=0 or X≥6. Actual level≈0.03167635 + 0.05440223 = 0.08607858, about 8.608%.
07 · Observation six
Would an observed 6 reject under both constructions?
Hint
Check the two upper boundaries.
Worked solution
No. Six rejects under this explicitly closest-tail rule, but not under the at-most 0.05-per-tail rule. The construction instruction matters.
05 / Closest does not guarantee an at-most nominal level
Always report the achieved probability.
Under H₀, Y~B(10,0.5). Choose each tail closest to 0.05.Worked example
P(Y≤1) = 11/1024 ≈ 0.01074219
Its distance from 0.05 is about 0.03925781.
P(Y≤2) = 56/1024 = 0.0546875
This is closer, so choose the lower cutoff 2 and, by symmetry, upper cutoff 8.
Total = 112/1024 = 0.109375
Actual level is 10.9375%, which exceeds nominal 10%. This closest-tail construction is not an at-most 10% test.
08 · At-most alternative
What cutoffs would meet an at-most 0.05 requirement in each tail for that Y?
Hint
Reject the candidate with probability 0.0546875.
Worked solution
Lower≤1 and upper≥9. Actual total 22/1024 = 0.021484375, about 2.148%.
09 · Ambiguous wording
If no closest-tail exception is stated in these lessons, which convention should you use?
Hint
Use the stated default.
Worked solution
The equal-tail at-most construction. If a specific question prescribes closest or a strict comparison, follow and clearly identify that instruction.
06 / One side may have no rejecting count
Asymmetry can make the smallest lower event too likely.
For X~B(10,0.1), P(X=0) = 0.9¹⁰ ≈ 0.34868. A 5% equal-tail at-most test cannot include any lower count because even zero exceeds 0.025. Its lower rejection region is empty; the upper tail must still meet its own 0.025 budget.
10 · Do not transfer a budget silently
May the unused lower-tail allowance automatically be added to the upper tail?
Hint
That changes the specified equal-tail rule.
Worked solution
No. Under the stated equal-tail method, retain 0.025 at the upper end. A differently allocated test would need to be explicitly defined.
11 · Empty means what?
Does an empty lower tail make the alternative one-sided?
Hint
The chosen hypothesis and the attainable discrete region are different.
Worked solution
No. H₁ remains two-sided. This sample size and null distribution simply provide no sufficiently rare lower outcome under the chosen rule.
07 / Check support, union and context
Make the construction reproducible.
12 · Union notation
Should the region be written X≤a AND X≥b when a<b?
Hint
Could a count satisfy both?
Worked solution
No. Write OR: either extreme can lead to rejection. AND would describe an empty intersection.
13 · Symmetry assumption
Can you always reflect a lower cutoff in n/2 to obtain the upper cutoff?
Hint
A binomial model is symmetric only when p=0.5.
Worked solution
No. For p≠0.5, compute both tails separately. Equal probability budgets do not imply symmetric integer cutoffs.
14 · Evidence statement
What conclusion accompanies rejection for H₁:p≠0.25?
Hint
Use the two-sided claim and stated level.
Worked solution
There is sufficient evidence, under the specified test rule and model, that the population probability differs from 0.25. State the actual level if requested.
08 / Two boundaries and one actual total
The wording determines the rule.
State the null distribution and per-tail target. Verify adjacent candidates for both cutoffs. Distinguish at-most from explicitly closest instructions. Write the union of supported rejecting counts, calculate the actual level and apply the rule without changing it after observing the data.
Section 1 of 8 · A two-sided alternative needs both ends