Testing for a change in either direction means splitting your evidence budget in two — and remembering to check both ends before drawing a conclusion.
10 min read4 worked examples
Splitting the significance level
For $H_1: p \neq p_0$, evidence against $H_0$ could come from $X$ being unusually small or unusually large — so the significance level $\alpha$ is split equally between the two tails, giving $\tfrac{\alpha}{2}$ at each end.
The lower critical region is the largest $c_1$ with $P(X\le c_1)\le\tfrac{\alpha}{2}$; the upper critical region is the smallest $c_2$ with $P(X\ge c_2)\le\tfrac{\alpha}{2}$. The full critical region is $X\le c_1$ or $X\ge c_2$ — two separate pieces, not one continuous range.
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two-tailed-critical-region-diagram.png
The binomial distribution under $H_0$, with both tails shaded: $X\le c_1$ on the left and $X\ge c_2$ on the right, each holding probability at most $\alpha/2$.
Why halve α, rather than use the full 5% at each end
If the full $5\%$ were used at both ends of a two-tailed test, the true overall chance of wrongly rejecting a correct $H_0$ would be up to $10\%$ — double the stated significance level. Splitting $\alpha$ in half at each tail keeps the total risk of a wrongful rejection at exactly the stated $\alpha$, however the evidence points.
The same procedure, applied twice
Finding each tail of a two-tailed critical region uses exactly the one-tailed method from the previous lesson — the only new step is remembering there are two boundaries to find, each using half the significance level, and that the observed value must be checked against both.
Strategy — for a p-value, double the smaller tail
Using the p-value method for a two-tailed test: find the probability of a result at least as extreme as observed in the direction it actually occurred, then double it before comparing to the full significance level $\alpha$ (not $\alpha/2$). Doubling accounts for the fact that an equally extreme result in the other direction would also have counted as evidence.
Worked example 1
Finding both critical regions
To test $H_0: p=0.5$ against $H_1: p\neq0.5$ at the 10% level, with $n=20$, use $X\sim B(20,0.5)$. Given $P(X\le5)\approx0.0207$, $P(X\le6)\approx0.0577$, $P(X\ge14)\approx0.0577$, $P(X\ge15)\approx0.0207$, find the critical region.
MethodFind each tail separately at α/2 = 5%
1
Split significance level: $\dfrac{10\%}{2}=5\%=0.05$ for each tail.
Halve the overall significance level.
2
Lower tail: $P(X\le5)\approx0.0207\le0.05$ ✓, but $P(X\le6)\approx0.0577>0.05$ ✗. So lower critical region: $X\le5$.
Largest c with P(X≤c) ≤ 0.05.
3
Upper tail: $P(X\ge15)\approx0.0207\le0.05$ ✓, but $P(X\ge14)\approx0.0577>0.05$ ✗. So upper critical region: $X\ge15$.
Smallest c with P(X≥c) ≤ 0.05.
4
Full critical region: $X\le5$ or $X\ge15$.
Both tails combined.
▪ found
Examiner's eye
Present both tails clearly and separately, and state the final critical region as two conditions joined by "or" — writing it as a single range (like "$5\le X\le15$", which would mean the opposite of what's intended) is a common and costly notational error.
Worked example 2
Completing a two-tailed test
Using the critical region from Worked Example 1 ($X\le5$ or $X\ge15$), a sample of 20 gives $X=16$. State a conclusion.
MethodCheck the observed value against both parts of the region
1
$X=16$. Since $16\ge15$, the observed value falls in the upper critical region.
Compare against both tails.
2
There is sufficient evidence, at the 10% level, to reject $H_0$ in favour of $H_1$ — the true value of $p$ appears to differ from $0.5$.
Conclusion in context.
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The key insight
A two-tailed conclusion never states a direction — only that $p$ appears to differ from $p_0$, since evidence from either tail was treated identically when the test was designed. Concluding "$p$ is higher than $0.5$" here would overstate what the two-tailed test can actually show; a one-tailed test would be needed to make a directional claim.
Worked example 3
The p-value method, two-tailed
A die is tested for bias: $H_0: p=\tfrac16$ (probability of rolling a 6), $H_1: p\neq\tfrac16$, at the 5% level, with $n=30$ rolls. It shows a 6 exactly $10$ times. Given $P(X\ge10)\approx0.0374$ for $X\sim B(30,\tfrac16)$, use the p-value method to conclude.
MethodDouble the one-sided tail probability, then compare to α
1
The observed value ($10$) is above the expected mean ($np=30\times\tfrac16=5$), so the relevant tail is $P(X\ge10)\approx0.0374$.
Direction of the observed deviation from H₀.
2
Two-tailed p-value: $2\times0.0374=0.0748$.
Double to account for both tails.
3
Compare to $\alpha=0.05$: since $0.0748>0.05$, there is insufficient evidence to reject $H_0$.
p-value exceeds the significance level.
4
The data does not provide significant evidence, at the 5% level, that the die is biased.
Conclusion in context.
▪ concluded
Common mistake
Forgetting to double the tail probability for a two-tailed p-value test is one of the most common errors in this topic — without doubling, $0.0374<0.05$ would wrongly suggest rejecting $H_0$, reversing the correct conclusion entirely.
Worked example 4
Choosing one-tailed vs two-tailed changes the outcome · Challenge
Using the same data as Worked Example 3 ($X=10$ out of $n=30$, testing $p=\tfrac16$), suppose instead the researcher had specifically suspected the die favours 6 before collecting data, giving $H_1: p>\tfrac16$ at the 5% level. Using $P(X\ge10)\approx0.0374$, would the conclusion differ from the two-tailed test?
MethodApply the one-tailed comparison directly, without doubling
1
For a one-tailed test, the p-value is used directly, without doubling: $P(X\ge10)\approx0.0374$.
One-tailed tests don't split α across two ends.
2
Compare to $\alpha=0.05$: since $0.0374<0.05$, there is sufficient evidence to reject $H_0$.
p-value below α this time.
3
Yes, the conclusion differs — the one-tailed test finds significant evidence that the die favours 6, while the two-tailed test (Worked Example 3) did not.
Same data, opposite conclusions.
▪ compared
Look deeper — why the choice must come before the data
The exact same result ($X=10$) gives two different conclusions purely because of the test's direction, decided in advance. This is precisely why the one-tailed/two-tailed decision must be locked in before seeing the data — choosing whichever test gives the "wanted" conclusion after the fact would make the significance level meaningless.
Common mistake
Using the full significance level $\alpha$ (instead of $\alpha/2$) when finding either tail of a two-tailed critical region overstates the evidence needed and gives an incorrect region. Always halve $\alpha$ first, and remember the final critical region has two disconnected pieces, not one continuous interval.
Check yourself
To test $H_0: p=0.3$ against $H_1: p\neq0.3$ at the 10% level with $n=15$, using $X\sim B(15,0.3)$, given $P(X\le1)\approx0.0353$, $P(X\le2)\approx0.1268$, $P(X\ge8)\approx0.0500$, $P(X\ge9)\approx0.0176$: find the critical region. (Hint: use α/2 = 5% for each tail separately.)
Upper tail: $P(X\ge8)\approx0.0500\le0.05$ ✓ (equal is allowed), $P(X\ge9)$ also satisfies but 8 is smaller, so check: since P(X≥8) ≤ 0.05 holds, upper region starts at $X\ge8$.