2.05a, 2.05c - Hypotheses, significance and statistical decisions

2.05a, 2.05c - Hypotheses, significance and statistical decisions

How unusual must a sample be before we question a claim? Build the language of hypothesis testing through a binomial model, then distinguish a chosen significance level from the actual chance of a false rejection.

Hypotheses concern the population parameter

If only two of ten seeds germinate, is that evidence against a claimed germination probability of one half? Start with what the claim means: a probability for all seeds grown under the stated conditions, rather than the proportion in this particular tray.

You already know

For independent trials with two outcomes and a common success probability, the success count has distribution X∼B(n,p)X\sim\mathrm B(n,p). A cumulative probability includes a range of possible counts.

The sample supplies evidence; the population parameter is what we want to learn about.

Null and alternative hypotheses

The null hypothesis, H0\mathrm H_0, is the default population claim used to calculate probabilities. The alternative hypothesis, H1\mathrm H_1, states the departure being investigated. Define the parameter in context: for example, pp is the population probability that a seed germinates under the stated conditions.

An alternative p<p0p<p_0 or p>p0p>p_0 gives a one-tail test. An alternative p≠p0p\ne p_0 gives a two-tail test. Choose the direction from the question before seeing the data: a low sample result does not turn a planned test for any change into a test for a decrease.

Worked example 1: a suspected decrease

Question. A seed producer claims a germination probability of 0.50.5. A gardener suspects a decrease. State the hypotheses and the null distribution of the number germinating among ten independently selected seeds.

Plan. Define the population probability, then translate “decrease” into an inequality.

H0:p=0.5claimed probabilityH1:p<0.5suspected decreaseX∣H0∼B(10,0.5)ten independent trials\begin{aligned} \mathrm H_0&:p=0.5 &&\text{claimed probability}\\ \mathrm H_1&:p<0.5 &&\text{suspected decrease}\\ X\mid\mathrm H_0&\sim\mathrm B(10,0.5) &&\text{ten independent trials} \end{aligned}

Why this step? Probabilities of unusual samples must be calculated under the claim being tested, not with the observed sample proportion.

Answer. Here pp is the population germination probability. This is a lower-tail test; XX is the test statistic, the random success count used to assess the evidence.

Check. An observed count x=2x=2 is a value of XX; neither 22 nor 2/102/10 replaces pp in the hypotheses.

Change just the research question: the same producer now wants to investigate any change, including an improvement.

Worked example 2: a change in either direction

Question. For the same ten-seed experiment, state the hypotheses when the concern is that the probability differs from 0.50.5.

Plan. Retain the null claim and replace the directed alternative.

H0:p=0.5,H1:p≠0.5,X∣H0∼B(10,0.5).\mathrm H_0:p=0.5,\qquad \mathrm H_1:p\ne0.5,\qquad X\mid\mathrm H_0\sim\mathrm B(10,0.5).

Why this step? Either an unusually small or an unusually large count could support “differs”.

Answer. This is a two-tail test about the same population probability pp.

Check. The sampling model has not changed. Only which extremes count as evidence has changed.

A hypothesis is a statement about the population, even when the available evidence is one small sample.

Watch out

Writing H0:p^=0.5\mathrm H_0:\hat p=0.5 makes a claim about a sample proportion. Write H0:p=0.5\mathrm H_0:p=0.5 and define pp as the population probability.

Try setting up a different direction without doing the probability calculation yet.

A p-value measures extremeness under the null model

Two successes can have an individual probability below 5%5\% while the correct lower-tail probability exceeds 5%5\%. That small surprise explains why a test needs a tail, not just the probability of the exact observation.

p-value

A p-value is the probability, assuming the null hypothesis and sampling model, of a test statistic at least as extreme as the observation in the direction of the alternative.

For a lower-tail test with observed count xx, include 0,1,…,x0,1,\ldots,x. For an upper-tail test, include x,x+1,…,nx,x+1,\ldots,n. A two-tail test allows evidence at both extremes; the next lesson develops an equal-tail procedure.

Decision from a p-value

For a significance level α\alpha chosen before sampling,

p-value≤α⟹reject H0.p\text{-value}\le\alpha\quad\Longrightarrow\quad\text{reject }\mathrm H_0.

Otherwise, do not reject H0\mathrm H_0. The conclusion remains evidence about a population, not certainty.

Not in the formula sheet: learn it.

The significance level sets how demanding the decision rule is. Here is the ten-seed question answered at a familiar level.

Worked example 3: include more extreme counts

Question. In the lower-tail ten-seed test, two seeds germinate. Conduct the test at the 5%5\% level.

Plan. Calculate the inclusive lower tail under X∼B(10,0.5)X\sim\mathrm B(10,0.5).

P(X≤2)=0.0546875counts 0,1,20.0546875>0.05compare with the level\begin{aligned} \mathrm P(X\le2)&=0.0546875 &&\text{counts }0,1,2\\ 0.0546875&>0.05 &&\text{compare with the level} \end{aligned}

Why this step? Counts below two would be still stronger evidence for a decrease, so they belong in the p-value too.

Answer. Do not reject H0\mathrm H_0. There is insufficient evidence at the 5%5\% level to suggest that the population germination probability is below 0.50.5.

Check. P(X=2)=0.0439453125\mathrm P(X=2)=0.0439453125, which is smaller. Using that point probability would give the wrong decision.

Keep the sample, claim and alternative fixed. Change only the preselected significance level.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account