01 · Null versus observation
Why is B(20,0.75) the wrong null model here?
Hint
Where does 0.75 come from?
Worked solution
15/20=0.75 is the observed proportion. H₀ specifies 0.5 for the probability used in the test.
Understand · explore · practise
Complete two-tailed hypothesis tests
Complete a two-sided binomial test with explicit equal-tail boundaries, inclusive probabilities and a careful contextual conclusion.
Before you startTwo-tailed critical regions and one-tailed tests.
01 / Either direction can challenge the claim
Choose the two-sided alternative before seeing the count.
H₀: p=p₀; H₁: p≠p₀.
Here, use an equal-tail at-most test: the lower and upper tail budgets are each α/2.
Define the event whose probability is p. A change can be either an increase or a decrease. The calculation still uses p₀, not the sample proportion.
Choose a sample, keep the rule fixedExplore
A device should select either output equally often. Test a change with 20 independent trials at nominal 5%, using equal tails at most 2.5% each.
02 / Write the full null model
An equal-output claim concerns a population probability.
A device is designed to select output A with probability 0.5. An engineer suspects the probability has changed. In 20 independent trials under comparable conditions, A appears 15 times. Test at nominal 5% using the equal-tail at-most rule.Worked example
Let p be the probability of output A
H₀: p=0.5; H₁: p≠0.5.
Let X count output A in 20 trials
Under H₀, X~B(20,0.5), assuming independent trials and a common probability.
Observed count x=15
The upper end is relevant to this observation; each end has budget 0.025.
02 · Direction
Why is p>0.5 not the given alternative?
Hint
Read the original suspicion.
Worked solution
The engineer suspected a change in either direction. Seeing a high count does not authorise rewriting that preselected two-sided alternative.
03 / Use half the level for an observed tail
Include the observed count.
Watch: both ends share the rejection rule
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Evaluate the observed upper tail.Worked example
P(X≥15) = 1 − P(X≤14) ≈ 0.02069473
This includes 15,16,…,20.
0.02069473 ≤ 0.025
The observed upper tail meets its half-level budget.
Reject H₀
There is sufficient evidence at nominal 5%, under this rule and model, that the probability of output A differs from 0.5.
03 · Off by one
Why not use 1−P(X≤15)?
Hint
Identify the first included count.
Worked solution
That gives P(X≥16) and wrongly omits the observed 15.
04 · Nearby observation
For 14 outputs A, the upper tail is 0.05765915. State the decision.
Hint
Compare with 0.025.
Worked solution
Do not reject H₀. There is insufficient evidence of a change at nominal 5% under the stated test.
04 / Both boundaries belong to one test
Construct the region independently of the observed sample.
By symmetry, P(X≤5)=P(X≥15)≈0.02069473. The next counts fail: P(X≤6)=P(X≥14)≈0.05765915. Therefore the full critical region is X≤5 or X≥15. Counts 6 through 14 do not reject.
05 · Lower observation
What conclusion follows if output A appears five times?
Hint
Check membership of the other tail.
Worked solution
Five lies in the critical region, so reject H₀. There is sufficient evidence of a change in the output probability under the stated test.
06 · Boundary check
Why is six excluded?
Hint
Use its inclusive lower tail.
Worked solution
P(X≤6)≈0.05765915 exceeds 0.025, so it does not meet the per-tail condition.
05 / Report the achieved level when asked
Discrete tails rarely use the whole budget.
The actual significance is P(X≤5)+P(X≥15)≈0.04138947, or 4.138947%. It is at most the nominal 5%. This is the probability of rejecting under H₀, not the probability that H₀ is false.
07 · Add the regions
Why add, rather than multiply, the two tails?
Hint
Can one count lie in both?
Worked solution
They are disjoint alternatives in a union. Add their probabilities to find the rejection probability.
08 · Nonrejection under the null
Find the probability of not rejecting if H₀ is true.
Hint
Complement the actual level.
Worked solution
Approximately 1−0.04138947=0.95861053. This is not the probability that H₀ is true after observing nonrejection.
06 / Specify what your two-sided calculation means
Different exact two-sided conventions need not agree for asymmetric distributions.
For this symmetric example, doubling the observed one-sided tail gives 0.04138947, which can be compared with 0.05 and gives the same decision. These lessons define the equal-tail construction explicitly. Do not assume every software package uses that same definition of an exact two-sided p-value.
09 · Wrong full-level comparison
Why not compare an observed tail with 0.05 for this equal-tail test?
Hint
There are two preselected directions.
Worked solution
The equal-tail construction gives each end 0.025. Giving each tail 0.05 can allow up to 0.10 overall.
10 · Asymmetric model
For B(20,0.2), may you reflect the lower cutoff about 10 to get the upper cutoff?
Hint
Check symmetry.
Worked solution
No. A binomial distribution is symmetric at p=0.5. Compute both tails separately for p=0.2.
07 / Make the conclusion match the evidence
Avoid proof, causal claims and retrospective direction changes.
11 · Equality after nonrejection
Rewrite “14 does not reject, therefore the device is perfectly fair”.
Hint
State what the test failed to establish.
Worked solution
There is insufficient evidence at nominal 5%, under the stated rule and assumptions, that the output probability differs from 0.5. Fairness has not been proved.
12 · Direction after rejection
Can the result with 15 outputs A prove p=0.75?
Hint
Distinguish an estimate from a hypothesis-test conclusion.
Worked solution
No. The sample proportion estimates p, but rejection supports a difference from 0.5 rather than proving a particular value.
13 · Serial dependence
The device repeats its preceding output unusually often. Which assumption needs scrutiny?
Hint
Does one result affect the next?
Worked solution
Independence. A binomial test may be inappropriate if successive trials are dependent.
14 · Full argument
List the components of a complete two-sided test.
Hint
Follow the reasoning from parameter to context.
Worked solution
Define p and X; state H₀ and H₁; justify the null binomial model; state nominal level and tail convention; calculate inclusive tails or a justified critical region; compare and conclude in context.
08 / Two possible extremes, one planned rule
Keep the model and hypotheses fixed.
Use the null probability, allocate half the level to each tail, include boundary counts and state the full rejection region. Calculate the actual total if requested. Rejection is evidence against the null within the model; nonrejection is not proof of equality.
Section 1 of 8 · Either direction can challenge the claim