MH4–MH6 - Testing a binomial proportion

MH4–MH6 - Testing a binomial proportion

Use exact binomial probabilities to investigate a population probability. Set up the hypotheses, find critical and acceptance regions, and express what the evidence suggests without treating a statistical decision as certainty.

Use the null probability for the observed tail

If only one of twenty-five cuttings roots, is a claimed rooting probability of 0.20.2 still plausible? The sample is not the population: test its extremeness using the probability claimed for all cuttings under the stated conditions.

You already know

A binomial count needs a fixed number of independent trials with two outcomes and a common success probability. Choose the alternative and significance level before seeing the sample; a p-value includes outcomes more extreme than the observation.

Define pp as the population rooting probability. If trays share a disease or receive different treatments, independence or a common probability may fail, undermining the test even if the calculation is correct.

Binomial probability model

For X∼B(n,p)X\sim\mathrm B(n,p) and integers r=0,…,nr=0,\ldots,n,

P(X=r)=(nr)pr(1−p)n−r.\mathrm P(X=r)=\binom nr p^r(1-p)^{n-r}.

Use a cumulative probability, or add these probabilities, for the required tail.

In the formula sheet.

A calculator supplies a probability. The written model and event explain which claim that probability tests.

Worked example 1: a low success count

Question. A nursery claims a population rooting probability of 0.20.2. One cutting roots among twenty-five independent trials. Test for a decrease at 5%5\%.

Plan. Define the population probability and calculate the inclusive lower tail under the null.

H0:p=0.2,H1:p<0.2X∣H0∼B(25,0.2)P(X≤1)=0.0273897256…<0.05\begin{aligned} \mathrm H_0&:p=0.2,\qquad \mathrm H_1:p<0.2\\ X\mid\mathrm H_0&\sim\mathrm B(25,0.2)\\ \mathrm P(X\le1)&=0.0273897256\ldots<0.05 \end{aligned}

Why this step? Zero rooted cuttings would be stronger evidence for a decrease, so it belongs with the observed one in the p-value.

Answer. Reject the null. There is sufficient evidence at 5%5\% to suggest that the population rooting probability is below 0.20.2.

Check. The expected null count is 25(0.2)=525(0.2)=5, so one is in the lower direction being tested.

Change only the observed count from one to two, leaving the planned test fixed.

Worked example 2: a less extreme observation

Question. What conclusion would two rooted cuttings give in the same investigation?

Plan. Include counts zero, one and two in the lower tail.

P(X≤2)=0.0982252228…>0.05.\mathrm P(X\le2)=0.0982252228\ldots>0.05.

Why this step? Replacing the null probability with 2/252/25 would fit the sample instead of testing the nursery's claim.

Answer. Do not reject the null. There is insufficient evidence at 5%5\% to suggest that the population rooting probability is below 0.20.2.

Check. The cumulative probability increased when the observed count increased, so evidence for a decrease weakened.

Use the exact binomial distribution for these tests; a Normal approximation does not replace it here.

Watch out

A small value of P(X=x)\mathrm P(X=x) alone is not the relevant evidence. Use P(X≤x)\mathrm P(X\le x) for a decrease, or P(X≥x)\mathrm P(X\ge x) for an increase, always including the observation.

Try a lower-tail calculation for a new setting.

Verify the critical boundary and its neighbour

For an increase, identify the smallest success count that would lead to rejection. Every larger count then belongs to the critical region too.

Critical and acceptance regions

The critical region is the set of counts causing rejection. The acceptance region contains all other possible counts. Its name describes a decision rule, not proof that the null claim is true.

An upper tail beginning at kk is P(X≥k)=1−P(X≤k−1)\mathrm P(X\ge k)=1-\mathrm P(X\le k-1). This inclusive-tail rule is not printed as a separate formula in the sheet: learn it. The k−1k-1 in the complement keeps the boundary count kk in the rejection tail.

Worked example 3: identifying both regions

Question. A process has claimed population fault probability 0.40.4. For fourteen independent items, determine the critical and acceptance regions for an increase at 5%5\%, then conclude for nine faulty items.

Plan. Use H0:p=0.4\mathrm H_0:p=0.4, H1:p>0.4\mathrm H_1:p>0.4, and X∼B(14,0.4)X\sim\mathrm B(14,0.4) under the null. Check adjacent upper tails.

P(X≥10)=0.0175095415≤0.05P(X≥9)=0.0583189450>0.05\begin{aligned} \mathrm P(X\ge10)&=0.0175095415\le0.05\\ \mathrm P(X\ge9)&=0.0583189450>0.05 \end{aligned}

Why this step? Ten is admissible, but nine would make the region's null probability exceed the intended level.

Answer. The critical region is X≥10X\ge10; the acceptance region is 0≤X≤90\le X\le9. Nine faulty items are outside the critical region: there is insufficient evidence at 5%5\% to suggest that the population fault probability exceeds 0.40.4.

Check. Both regions together contain every possible count from zero to fourteen, without overlap.

Change only the intended level from 5%5\% to 10%10\%; the larger level allows a less extreme boundary.

Worked example 4: applying a changed rule

Question. Determine the regions at 10%10\% for the same process, then conclude for nine faulty items.

Plan. Check whether nine is now admissible and whether eight could be included.

P(X≥9)=0.0583189450≤0.10P(X≥8)=0.1501401029>0.10\begin{aligned} \mathrm P(X\ge9)&=0.0583189450\le0.10\\ \mathrm P(X\ge8)&=0.1501401029>0.10 \end{aligned}

Why this step? A critical-value calculation needs both the admissible tail and evidence that extending it would break the level.

Answer. Reject for X≥9X\ge9, with acceptance region 0≤X≤80\le X\le8. Nine faults give sufficient evidence at 10%10\% to suggest that the population fault probability exceeds 0.40.4.

Check. Nine was not significant at 5%5\%, but is significant under the less demanding, preselected 10%10\% rule.

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