Statistics · Hypothesis Testing · Year 1

Critical Values

A hypothesis test needs a hard boundary — a line drawn through the distribution that separates "unremarkable" from "so unlikely it counts as evidence". That line is the critical value.

10 min read 3 worked examples

Critical region and acceptance region

Under $H_0$, the test statistic $X$ follows a known distribution — here, always binomial. The critical region is the set of values of $X$ so extreme that, if observed, they would cause $H_0$ to be rejected. Everything outside it is the acceptance region: values consistent enough with $H_0$ that there's no reason to doubt it.

A bar chart of the probability distribution of X~B(10,0.4). Bars for x=0 to 7 are shaded teal and labelled the acceptance region; bars for x=8, 9 and 10 are shaded orange and labelled the critical region.
For $X\sim B(10,0.4)$ tested at the 5% level against $H_1:p>0.4$: $P(X\ge8)=0.0123$ is under 5%, so $x=8,9,10$ form the critical region and $x=0$ to $7$ form the acceptance region.

Critical value and actual significance level

The critical value is the first value of $X$ that falls inside the critical region — the boundary itself. In the diagram above, the critical value is $8$.

Because $X$ is discrete, the critical region's total probability almost never lands exactly on the stated significance level $\alpha$ — it's built to be as close as possible to $\alpha$ without exceeding it. The actual probability of the test statistic falling in the critical region, assuming $H_0$ is true, is called the actual significance level of the test, and it's usually slightly below the stated level.

Why this matters

The stated significance level (5%, say) is a target, not a guarantee — quoting the actual significance level shows you understand that a discrete test can only approximate it, and lets you judge exactly how cautious (or how permissive) a particular test really is.

Finding a critical region

To find the critical region for $H_1:p>p_0$ at significance level $\alpha$: find the smallest value $c$ such that $P(X\ge c)\le\alpha$. Everything from $c$ upward is the critical region. For $H_1:p<p_0$, the logic mirrors this from the other end: find the largest value $c$ such that $P(X\le c)\le\alpha$.

Worked example 1

Critical region for an upper-tail test
A manufacturer claims that 30% of a certain component are defective. To test $H_0:p=0.3$ against $H_1:p>0.3$ at the 5% significance level, a sample of $n=8$ components is checked. Find the critical region, and state the actual significance level of the test.
MethodTest successive upper-tail probabilities against 5%
1
Under $H_0$: $X\sim B(8, 0.3)$.
The claimed value of p, under the null hypothesis.
2
$P(X\ge5) = 0.0580$, $P(X\ge6) = 0.0113$.
From a calculator's binomial cumulative distribution function.
3
$P(X\ge5)=0.0580 > 0.05$, but $P(X\ge6)=0.0113 \le 0.05$.
Compare each against the stated 5% significance level.
4
The critical region is $X\ge6$; the critical value is $6$.
The smallest c with P(X≥c) ≤ 0.05.
5
The actual significance level is $P(X\ge6) = 0.0113 = 1.13\%$.
The true probability of the critical region under H₀.
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Examiner's eye

Always show the probability just outside the critical region too (here, $P(X\ge5)=0.0580$) — this proves the region you've chosen is the largest possible one still under 5%, which is exactly what a mark scheme checks for.

Worked example 2

Critical region for a lower-tail test
A charity's regular donors have historically given at a rate of 60%. To test whether this rate has fallen, $H_0:p=0.6$ is tested against $H_1:p<0.6$ at the 5% level, using a sample of $n=15$ donors. Find the critical region and the actual significance level.
MethodTest successive lower-tail probabilities against 5%
1
Under $H_0$: $X\sim B(15, 0.6)$.
The claimed donation rate under H₀.
2
$P(X\le5) = 0.0338$, $P(X\le6) = 0.0950$.
Cumulative probabilities from the low end.
3
$P(X\le5)=0.0338 \le 0.05$, but $P(X\le6)=0.0950 > 0.05$.
Test against 5% starting near the mean and working outward.
4
The critical region is $X\le5$; the critical value is $5$.
The largest c with P(X≤c) ≤ 0.05.
5
The actual significance level is $P(X\le5) = 0.0338 = 3.38\%$.
Somewhat below the stated 5%, as expected for a discrete test.
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The key insight

The method is a mirror image of the upper-tail case: instead of building the region up from the largest values of $X$, it's built up from the smallest. Deciding which end to search from comes entirely from the direction stated in $H_1$.

Common mistake

Do not stop at the first value where the cumulative probability looks "close to" $\alpha$ — always check the boundary value and its neighbour to confirm the region is as large as possible without exceeding $\alpha$. A critical region built from a value with $P>\alpha$ is simply wrong, however close it looks.

Check yourself

$X\sim B(12, 0.25)$ is used to test $H_0:p=0.25$ against $H_1:p<0.25$ at the 5% significance level. Given $P(X\le0)=0.0317$ and $P(X\le1)=0.1584$, find the critical region and the actual significance level.
(Hint: find the largest c with P(X≤c) ≤ 0.05.)

Method
1
$P(X\le0)=0.0317\le0.05$, but $P(X\le1)=0.1584>0.05$.
2
So the critical region is $X\le0$ (i.e. $X=0$ only); the critical value is $0$.
3
The actual significance level is $P(X\le0)=0.0317=3.17\%$.
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