6.5 - Hypotheses and critical regions

6.5 - Hypotheses and critical regions

An unusual result might reflect a real change, or it might be ordinary random variation. You will turn a claim about a population into hypotheses, choose a rejection rule before observing the data, and explain what the resulting decision means. The rule controls how readily chance alone can lead you to reject a true claim.

Null and alternative hypotheses

A machine has historically produced a satisfactory item with probability 0.60.6. After an adjustment, 1515 of 2020 independently chosen items are satisfactory. Has the machine improved, or is this a lucky sample? The observed proportion 15/2015/20 describes these items; the question concerns the machine's underlying probability pp.

Null and alternative hypotheses

The null hypothesis, H0H_0, specifies the population parameter used to calculate probabilities for the test. The alternative hypothesis, H1H_1, describes the departure being investigated. For an improvement here, write H0:p=0.6H_0:p=0.6 and H1:p>0.6H_1:p>0.6.

A test statistic is a quantity calculated from the observations and used to make the decision. Here, XX, the number of satisfactory items among 2020, is convenient. Under H0H_0, X∼B(20,0.6)X\sim\mathrm B(20,0.6). The statistic is random before sampling; its observed value is 1515.

The question you decide to investigate determines the tail:

Departure investigatedAlternativeResults giving evidence against H0H_0
An increasep>p0p>p_0Unusually large counts
A decreasep<p0p<p_0Unusually small counts
A change in either directionp≠p0p\ne p_0Unusually small or large counts

The first two are one-tailed tests; the third is two-tailed. Choose this direction from the investigation before looking at its result. Choosing whichever tail looks more striking after sampling changes the chance of rejection.

Worked example 1: investigating improvement

Question. An adjustment is intended to increase a machine's satisfactory-item probability from 0.60.6. State suitable hypotheses and a test statistic for 2020 independent items.

Plan. Define the population parameter, then translate “increase” into an inequality.

Let pp be the probability that an item produced after the adjustment is satisfactory. Then

H0:p=0.6,H1:p>0.6.H_0:p=0.6,\qquad H_1:p>0.6.

Use X=number satisfactory among the 20 itemsX=\text{number satisfactory among the }20\text{ items}. Under H0H_0, X∼B(20,0.6)X\sim\mathrm B(20,0.6); large values support the alternative.

Answer. The hypotheses above define an upper-tailed test based on XX.

Check. The alternative describes improved production, whereas X=15X=15 would describe only one observed sample.

Keep the population and sample size, but change the research question from improvement to any change. Now a deterioration matters too.

Worked example 2: investigating any change

Question. The machine's adjustment may change its satisfactory-item probability in either direction. State the hypotheses and identify the tail or tails.

Plan. Keep the null value but allow departures on both sides of it.

H0:p=0.6,H1:p≠0.6.H_0:p=0.6,\qquad H_1:p\ne0.6.

The same statistic XX is used, with X∼B(20,0.6)X\sim\mathrm B(20,0.6) under the null. Both unusually low and unusually high values can now give evidence of change.

Answer. A two-tailed test with the stated hypotheses.

Check. Both p=0.4p=0.4 and p=0.8p=0.8 satisfy the alternative; only the latter satisfies the previous example's alternative.

A hypothesis is about the population, so it should not say that the sample count or sample proportion equals its observed value.

Watch out

Write H0:p=0.6H_0:p=0.6, not H0:X=12H_0:X=12. The count need not equal its expected value even when the null hypothesis is true.

Translate a different investigation into a parameter statement.

Significance and rejection

A test needs a rule for deciding what counts as unusually far from the null model. Consider 88 independent tosses of a coin and let XX be the number of heads. We investigate whether the coin favours heads, so H0:p=0.5H_0:p=0.5 and H1:p>0.5H_1:p>0.5.

Under H0H_0, a high count can still occur by chance. For example,

P(X≥7)=(87)+(88)28=9256=0.03515625.\mathrm P(X\ge7)=\frac{\binom87+\binom88}{2^8}=\frac9{256}=0.03515625.

A rule that rejects for X≥7X\ge7 therefore rejects a fair coin on about 3.52%3.52\% of repeated samples of this size.

Critical region and significance level

The rejection region, or critical region, is the set of statistic values for which H0H_0 is rejected. The acceptance region is its complement: values for which H0H_0 is not rejected. A chosen significance level, such as 5%5\%, limits the probability of rejection when H0H_0 is true. For a discrete test, the actual probability may be smaller than this chosen level.

The regions concern possible results, not the unknown parameter. Here, {7,8}\{7,8\} is a rejection region and {0,1,…,6}\{0,1,\ldots,6\} is its acceptance region. “Acceptance” is a conventional region name; it does not mean the null has been proved.

[DIAGRAM: asset_name: A discrete rejection region; asset_slug: hypotheses-binomial-region; caption: Seven or eight heads form a rejection region with null probability 9 over 256; description: A bar chart of B eight comma one half shows counts zero through eight. Counts seven and eight are shaded clay and labelled rejection region. Counts zero through six are pale blue and labelled acceptance region. The total null probability of the rejection region is 0.03515625, below the chosen five percent level.]
Seven or eight heads form a rejection region with null probability 9 over 256

For a one-tailed test, choose the largest suitable tail region whose null probability does not exceed the level. A larger region would reject more often than allowed. For a two-tailed test at level α\alpha, the usual equal-tail rule allocates α/2\alpha/2 to each tail. With discrete counts the attained tails need not have exactly equal probabilities, and their total may be below α\alpha.

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