Understand · explore · practise
Carry out a full one-sided binomial test from contextual hypotheses through a null-model tail probability to a justified conclusion.
Before you startHypotheses, binomial tails, significance and critical regions.
01 / Make the reasoning traceable
Parameter → hypotheses → null model → tail → decision → context.
State the significance level and check the model assumptions. Include the observed count in the evidence tail.
Use the probability given by H₀ for calculations. The sample proportion can describe the observation, but it must not replace the null probability. The alternative comes from the preselected contextual claim.
02 / Test a suspected increase
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Let p be the current probability an item is faulty
H₀:p=0.2; H₁:p>0.2.
Let X count faulty items in 16; under H₀, X~B(16,0.2)
The observed count is x=7. Common p and independence are assumed.
P(X≥7)=1−P(X≤6)≈0.02665733
Use the upper tail, including seven.
0.02665733≤0.05, so reject H₀
There is sufficient evidence at 5% that the process fault probability has increased.
Why does the calculation use p=0.2 rather than 7/16?
The test assesses the sample under the baseline claim.
0.2 is the null probability. 7/16 is the observed proportion, not the assumed baseline used to calculate the evidence probability.
What error results from using 1−P(X≤7)?
Which count is lost?
It computes P(X≥8), excluding the observed seven and understating the required upper-tail probability.
03 / A nearby sample can give a different decision
For the same process, observing six faulty items gives P(X≥6)≈0.08168789. This exceeds 0.05, so do not reject H₀. The observed proportion 6/16=0.375 is above 0.2, but that difference alone is not enough at 5%.
Give the full conclusion for six faults.
Avoid asserting that the old rate is correct.
There is insufficient evidence at 5% that the process fault probability has increased above 0.2, under the stated assumptions.
Would seven faults reject at 1%?
Compare its tail 0.02665733 with 0.01.
No. It is greater than 0.01, so there is insufficient evidence at 1%. The chosen level affects the decision.
04 / Test a suspected decrease
Let p be germination probability after the storage change
H₀:p=0.6; H₁:p<0.6.
Let Y count germinated seeds among 15
Under H₀, Y~B(15,0.6), with observed y=5.
P(Y≤5)≈0.03383330
Use the inclusive lower tail.
0.03383330≤0.05: reject H₀
There is sufficient evidence at 5% that germination probability has fallen below 0.6.
If six germinate, P(Y≤6)≈0.09504741. State the decision at 5%.
Compare the correct lower tail with 0.05.
Do not reject H₀. There is insufficient evidence at 5% of reduced germination probability.
Why is “p is the number of seeds that germinate” incorrect?
A count and a probability are different.
p is the probability that a seed germinates under the specified conditions. Y is the random number that germinate in the sample.
05 / A critical-region method reaches the same decision
For the fault example at 5%, P(X≥6)>0.05 while P(X≥7)≤0.05. The upper critical region is X≥7. For the germination example, P(Y≤5)≤0.05 while P(Y≤6)>0.05, so the lower region is Y≤5. Applying those regions agrees with the tail-probability method.
Explain why seven faults rejects using the critical-region method.
Use membership rather than another calculation.
The observed count 7 is in X≥7, the predefined critical region. Therefore reject H₀ and give the increase conclusion.
For the 16-item fault count, list the full rejecting range.
Respect the maximum possible count.
7≤X≤16. Writing X≥7 is sufficient when the binomial support is understood.
06 / A correct calculation needs a credible model
Checking many items from one shared faulty batch can create dependence. Mixing markedly different production lines can undermine common p. If assumptions are doubtful, explain the limitation instead of treating a precise binomial tail as automatically reliable.
All 16 items come from a single batch with a shared contamination event. What assumption is questionable?
One cause can affect several outcomes.
Independence. The binomial probability may not describe the variation between such samples.
Half the germination trials use one temperature and half another with different success chances. What is questionable?
Does one common p describe every trial?
Constant success probability. A single binomial model needs a justified common p, even if the seeds act independently.
07 / State evidence rather than proof
Why should the conclusion not say “the storage change definitely caused the reduction”?
What can this test establish about cause?
Rejection gives statistical evidence of a lower probability under the model. It does not prove causation or rule out other differences in conditions.
Seven faults reject H₀:p=0.2 against p>0.2. Does this prove p=7/16?
An alternative is a range, not the sample estimate.
No. The result supports an increase, not that the population probability equals the observed proportion.
Replace “accept H₀, so p definitely equals 0.6” with a sound statement.
Use insufficient evidence at the stated level.
Do not reject H₀; there is insufficient evidence at the stated level that the germination probability has fallen below 0.6. Equality is not proved.
Name two essential statements missing from “0.0267<0.05, reject”.
A reader needs the question and model behind the number.
For example, the defined population parameter and hypotheses; the null binomial distribution and inclusive tail; and a contextual conclusion. Any two of these identify meaningful omissions.
08 / A complete argument fits into a short chain
Define p and X in context, write hypotheses, state the null model and significance level, calculate the inclusive tail, compare, and conclude. A critical-region method is equally valid when its boundaries are justified. Mention assumptions that materially affect the result.
Section 1 of 8 · Make the reasoning traceable