O1, O2 - Binomial hypothesis tests | AQA A-Level Mathematics

Use a sample to investigate a claim about a population proportion. O1 supplies the language of hypotheses, tails and significance; O2 connects exact binomial probabilities to a cautious conclusion about the population.

Population proportions and hypotheses

A manufacturer claims that only $10\%$ of its sensors fail a test. You randomly select $20$ sensors and $5$ fail. Is that enough evidence to challenge the claim, or could it be ordinary sampling variation?

The sample failure proportion is $5/20=0.25$ . That describes the sample exactly. It does not establish that exactly a quarter of all the manufacturer's sensors fail. A hypothesis test weighs the sample evidence against a population claim.

You already know

A random sample can support an inference about its population. For $n$ independent trials with two outcomes and a constant success probability $p$ , the number of successes has distribution $\mathrm B(n,p)$ . “Success” means the event counted, even when that event is a failure.

Let $p$ be the probability that a randomly chosen sensor from the current production fails. This is the population proportion we want to investigate. Specify the question about $p$ before looking at the result.

Null and alternative hypotheses

The null hypothesis, $H_0$ , specifies the population value used to calculate reference probabilities. The alternative hypothesis, $H_1$ , describes the departure being investigated. For the sensor claim, $H_0:p=0.10$ . An increase uses $H_1:p>0.10$ , a decrease uses $H_1:p<0.10$ , and a change in either direction uses $H_1:p\ne0.10$ .

An increase or decrease calls for a one-tailed test. An unspecified change calls for a two-tailed test. The equality belongs in $H_0$ . Neither hypothesis is a statement about the already known sample proportion. These hypothesis conventions are not in the formula booklet: learn them.

Worked example 1: an intended improvement

Question. A service previously delivered $80\%$ of orders on time. A new system is intended to improve this proportion. Define the parameter and state suitable hypotheses.

Plan. Define a population proportion, then translate “improve” into a direction.

\begin{aligned} p&=\text{population proportion delivered on time under the new system} \\ H_0&:p=0.80 &&\text{reference claim} \\ H_1&:p>0.80 &&\text{improvement means more on time} \end{aligned}

Why this direction? On-time delivery is the event counted, so improvement makes its probability larger.

Answer. Use $H_0:p=0.80$ , $H_1:p>0.80$ , a one-tailed upper-tail test.

Check. A proportion below $0.80$ would contradict improvement, so it should not be the rejection direction.

Change only the investigation in example 2: the same new system might improve or worsen performance. The alternative must allow both possibilities.

Worked example 2: a change without a direction

Question. The service now investigates whether its new system has changed the on-time proportion from $0.80$ . Define the parameter and state suitable hypotheses.

Plan. Keep the population parameter and reference value, but include departures on either side.

\begin{aligned} p&=\text{population proportion delivered on time under the new system} \\ H_0&:p=0.80 &&\text{same reference claim} \\ H_1&:p\ne0.80 &&\text{either direction is relevant} \end{aligned}

Why two tails? Both unusually few and unusually many on-time deliveries could provide evidence of change.

Answer. Use $H_0:p=0.80$ , $H_1:p\ne0.80$ , a two-tailed test.

Check. Both $p=0.70$ and $p=0.90$ satisfy this alternative; $p=0.80$ does not.

Choose the direction from the purpose of the investigation. Selecting a favourable tail after seeing the sample changes the test being carried out.

Watch out

Wrong: $H_0:p=5/20$ . This replaces the population claim with the sample evidence. Right: test $H_0:p=0.10$ , and use the observed count $5$ to judge that claim.

The next question changes the event from timely deliveries to complaints. Define the event clearly before writing an inequality.

The null model and the relevant tail

To judge the sensor claim, temporarily assume it is true. Under $H_0$ , the number $X$ of failures in the sample has distribution $\mathrm B(20,0.10)$ . This tells us which counts would be unusual if the population failure proportion really were $0.10$ .

A binomial model needs independent trials with a common probability. Sampling without replacement from a much larger population can make independence a reasonable approximation. Twenty related sensors from one faulty batch need not represent twenty independent chances from all production.

Test statistic and one-tailed p-value

A test statistic is a quantity calculated from the sample; here its observed value is the number of failures. Under the null, the random variable $X$ describes its possible values. A one-tailed p-value is the probability, assuming $H_0$ , of the observed count or a count further in the direction of $H_1$ .

For an increase, include the observed count and every larger count. For a decrease, include it and every smaller count. A calculator's cumulative binomial function usually returns a lower inclusive probability, so the discrete boundary matters.

Inclusive binomial tails

For an observed count $x$ ,

\begin{aligned} H_1:p>p_0&:\quad \mathrm P(X\ge x)=1-\mathrm P(X\le x-1) \\ H_1:p<p_0&:\quad \mathrm P(X\le x) \end{aligned}

Use $X\sim\mathrm B(n,p_0)$ under $H_0$ . The binomial probability formula is supplied separately; this box gives the tail choices and complement boundaries. Not in the formula booklet: learn it.

A significance level provides the threshold for deciding that evidence is sufficiently unusual. At a one-tailed $5\%$ level, reject $H_0$ if the relevant p-value is at most $0.05$ . Keep unrounded probabilities until the comparison.

Worked example 3: five failures

Question. For $20$ independent sensors, $5$ fail. Test $H_0:p=0.10$ against $H_1:p>0.10$ at $5\%$ .

Plan. Calculate the inclusive upper tail under the claimed failure proportion.

\begin{aligned} X&\sim\mathrm B(20,0.10) &&\text{under }H_0 \\ \mathrm P(X\ge5)&=1-\mathrm P(X\le4) &&\text{include the observed five} \\ &=0.0431745\ldots<0.05 &&\text{calculator and comparison} \end{aligned}

Why the upper tail? More failures support the proposed increase in the population failure proportion.

Answer. Reject $H_0$ . There is sufficient evidence at $5\%$ to suggest that the population failure proportion exceeds $0.10$ .

Check. The complement removes counts $0$ to $4$ , leaving exactly $5$ to $20$ .

Example 4 changes only the observed count. A sample proportion can exceed the claim without giving sufficiently strong evidence against it.

Worked example 4: four failures

Question. Repeat the same $5\%$ upper-tail test when $4$ of the $20$ sensors fail.

Plan. Keep the null model, but move the inclusive boundary to four.

\begin{aligned} X&\sim\mathrm B(20,0.10) &&\text{same null model} \\ \mathrm P(X\ge4)&=1-\mathrm P(X\le3) &&\text{include four this time} \\ &=0.132953\ldots>0.05 &&\text{not sufficiently unusual} \end{aligned}

Why does the probability increase? The tail now includes the additional possible count $4$ .

Answer. Do not reject $H_0$ . There is insufficient evidence at $5\%$ to suggest that the population failure proportion exceeds $0.10$ .

Check. Although $4/20=0.20>0.10$ , the relevant null probability is about $13.3\%$ , above the chosen threshold.

The p-value concerns a set of possible outcomes, rather than just the exact count observed.

Watch out

Wrong: $\mathrm P(X\ge5)=1-\mathrm P(X\le5)$ . This omits the observed count itself. Right: subtract $\mathrm P(X\le4)$ . Using only $\mathrm P(X=5)$ also omits more extreme counts.

Use your calculator for exact binomial tails. A Normal approximation is unnecessary here, and the current Mathematics formula booklet contains no statistical tables.

Critical regions for one and two tails

Instead of waiting for a sample and then calculating its p-value, decide in advance which counts would lead to rejection. This makes the decision rule visible.

Critical and acceptance regions

The critical region is the set of possible test-statistic values leading to rejection of $H_0$ . Its boundary count or counts are called critical values. The remaining possible counts form the acceptance region, where the test does not reject $H_0$ . The name does not mean that the null has been proved.

For a lower-tail test with nominal level $\alpha$ , choose the largest integer $c$ satisfying $\mathrm P(X\le c)\le\alpha$ . Check that $c+1$ would exceed the limit. For an upper tail, choose the smallest $c$ satisfying $\mathrm P(X\ge c)\le\alpha$ , and check $c-1$ . These rules are not in the formula booklet: learn them.

Worked example 5: a lower critical region

Question. Under $H_0$ , $X\sim\mathrm B(25,0.20)$ . Find the lower-tail $5\%$ critical region and its actual rejection probability.

Plan. Find consecutive counts whose cumulative probabilities fall either side of $0.05$ .

\begin{aligned} \mathrm P(X\le1)&=0.0273897\ldots\le0.05 &&\text{allowed} \\ \mathrm P(X\le2)&=0.0982252\ldots>0.05 &&\text{next count is too much} \end{aligned}

Why check the next count? It proves that the chosen lower region is as large as the significance limit permits.

Answer. The critical region is $X\le1$ , with critical value $1$ . Its actual rejection probability under $H_0$ is $0.0273897\ldots$ , about $2.74\%$ . The acceptance region is $2\le X\le25$ .

Check. Adding count $2$ would increase the rejection probability beyond $5\%$ .

Example 6 changes only the alternative's direction. The same distribution now needs an upper tail; reflecting the old boundary would be wrong because this binomial distribution is not symmetric.

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Pure Mathematics

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Statistics

O1, O2 - Binomial hypothesis tests

Population proportions and hypotheses

The null model and the relevant tail

Critical regions for one and two tails

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