Mathematics 9709 · Probability & Statistics 2

One/two-tailed & the normal test

One-tailed vs two-tailed is just which way you suspect the claim is wrong. Suspect a direction (more / less) → one tail. Suspect only “different / changed” → the surprise can come from either end, so split your 5% into $2.5\%+2.5\%$ .

Two tails, and the large-sample normal engine

The wording chooses the tail. “Increase / higher / better / biased towards” fixes a direction → one tail. “Different / changed / not working / incorrect” admits both → two tails, with $H_1{:}\ p\ne p_0$ , and you compare each tail with $\alpha/2$ , not $\alpha$ .

Two-tailed splits the level: each tail gets α/2. Comparing a tail with the full α is the classic two-tailed slip.

Worked example

A two-tailed test with both critical values

On average 35% of rocks contain fossils. Jamila breaks 12 rocks and finds fossils in 2. Test at 10% whether 35% is wrong.

Let $X=$ rocks with fossils, $X\sim B(12,0.35)$ .
$H_0{:}\ p=0.35$ , $H_1{:}\ p\ne0.35$ (two-tailed).
Level 10% → each tail 5%.
Low result, so lower tail: $P(X\le2)=15.1\%$ .
$15.1\%>5\%$ .
Accept $H_0$ : insufficient evidence 35% is wrong.

Answer

\text{Critical values: reject if } X\le1 \text{ or } X\ge8.

Lower: P(X≤1) = 4.24% < 5%, P(X≤2) = 15.1% > 5% → cut-off 1. Upper: P(X≥8) = 2.55% < 5%, P(X≥7) = 8.46% > 5% → cut-off 8. A two-tailed test has TWO critical values.

Two-tailed B(12, 0.35) at the 10% level: 5% is shaded in each tail. The rejection region is X ≤ 1 or X ≥ 8; everything between is the acceptance region.

The normal-approximation engine (large n)

When $n$ is large the binomial sum is brutal, so approximate $B(n,p)\approx N(np,\ npq)$ and standardise — with a continuity correction, because you are wrapping a continuous curve around discrete bars.

Worked example

Normal-approximation test (a decrease)

Arra had 52% support; one year on, 91 of 200 members back him. Test at 5% for a fall.

$X\sim B(200,0.52)$ ; $\mu=104,\ \sigma^2=npq=49.92$ , so $X\approx N(104,49.92)$ .
$H_0{:}\ p=0.52$ , $H_1{:}\ p<0.52$ .
Lower tail, so correct $P(X\le91)\to P(X<91.5)$ : $z=\tfrac{91.5-104}{\sqrt{49.92}}=-1.769$ .
$\Phi(-1.769)=0.0384=3.84\%<5\%$ .
Reject $H_0$ : some evidence support has fallen.

Answer

3.84\%<5\%\Rightarrow\text{reject }H_0.

Worked example

Normal-approximation test + critical value (an increase)

60% react to a cream; a new cream gives 79 of 120 reacting. Evidence of an increase at 5%? Then find the critical value.

$X\sim B(120,0.6)$ : $\mu=72,\ \sigma^2=28.8$ , $H_1{:}\ p>0.6$ . Upper tail, correct $P(X\ge79)\to P(X>78.5)$ : $z=\tfrac{78.5-72}{\sqrt{28.8}}=1.211$ , $1-\Phi(1.211)=0.113$ .

Answer

11.3\%>5\%\Rightarrow\text{accept }H_0.

Critical value backwards: reject when $x>72+1.645\sqrt{28.8}=80.8$ , i.e. from $81$ upward.

The 5% rejection region is x > 80.8, i.e. 81 or more. The observed 79 is high but lands inside the acceptance region — not surprising enough.

Drag $z$ below to see how a critical $z$ carves off its tail area — that is exactly what “compare with the level” does.

z = 1.00·Φ(z) = P(Z < z) = 0.8413·right tail 0.1587

Continuity-correct the count first (≥ x → x−½, ≤ x → x+½), then standardise with √(npq). Keep variance in N(·,·); divide by its root.

Level	one-tailed z	two-tailed z
10%	1.282	1.645
5%	1.645	1.960
2%	2.054	2.326
1%	2.326	2.576

Examiner note

A sample-mean z-test uses

\bar X\sim N(\mu,\sigma^2/n)

(estimate

\sigma^2

with the unbiased

\tfrac{n}{n-1}

form if it isn't given) and no continuity correction; the CLT is only needed if the population is not already Normal. (9709/62/M/J/23 Q3) (9709/62/O/N/23 Q4)

Where the critical z come from

Each is the $z$ whose tail area equals the level: $\Phi(1.645)=0.95$ , $\Phi(1.960)=0.975$ , $\Phi(2.326)=0.99$ . Two-tailed 5% splits into 2.5% per tail, so it uses the $0.975$ value — 1.960, the same number a one-tailed 2.5% test would use.

Common mistake

Forgetting the continuity correction (

P(X\ge79)\to78.5

); for two tails, comparing with the full

\alpha

instead of

\alpha/2

; the wrong critical

z

(1.645 one-tailed 5%, 1.960 two-tailed 5%, 2.326 one-tailed 1%); and putting SD instead of variance in

N(\cdot,\cdot)

Your turn— tap to reveal the worked answer (one-tailed normal-approximation test)

A coin is flipped 180 times to test bias towards tails; 102 tails appear. Test at 5%. $X\sim B(180,0.5)$ , $\mu=90,\ \sigma^2=45$ , $H_1{:}\ p>0.5$ .

$P(X\ge102)\to P(X>101.5)$ : $z=\tfrac{101.5-90}{\sqrt{45}}=1.714$ , $1-\Phi(1.714)=0.0433$ .

Answer

4.33\%<5\%\ \Rightarrow\ \text{reject }H_0

There is some evidence of bias towards tails.

PreviousThe hypothesis-test machine NextType I and Type II errors

One/two-tailed & the normal test

Two tails, and the large-sample normal engine

Two-tailed splits the level: each tail gets α/2. Comparing a tail with the full α is the classic two-tailed slip.

Worked example

A two-tailed test with both critical values

On average 35% of rocks contain fossils. Jamila breaks 12 rocks and finds fossils in 2. Test at 10% whether 35% is wrong.

Let $X=$ rocks with fossils, $X\sim B(12,0.35)$ .
$H_0{:}\ p=0.35$ , $H_1{:}\ p\ne0.35$ (two-tailed).
Level 10% → each tail 5%.
Low result, so lower tail: $P(X\le2)=15.1\%$ .
$15.1\%>5\%$ .
Accept $H_0$ : insufficient evidence 35% is wrong.

Answer

\text{Critical values: reject if } X\le1 \text{ or } X\ge8.

Lower: P(X≤1) = 4.24% < 5%, P(X≤2) = 15.1% > 5% → cut-off 1. Upper: P(X≥8) = 2.55% < 5%, P(X≥7) = 8.46% > 5% → cut-off 8. A two-tailed test has TWO critical values.

Two-tailed B(12, 0.35) at the 10% level: 5% is shaded in each tail. The rejection region is X ≤ 1 or X ≥ 8; everything between is the acceptance region.

The normal-approximation engine (large n)

Worked example

Normal-approximation test (a decrease)

Arra had 52% support; one year on, 91 of 200 members back him. Test at 5% for a fall.

$X\sim B(200,0.52)$ ; $\mu=104,\ \sigma^2=npq=49.92$ , so $X\approx N(104,49.92)$ .
$H_0{:}\ p=0.52$ , $H_1{:}\ p<0.52$ .
Lower tail, so correct $P(X\le91)\to P(X<91.5)$ : $z=\tfrac{91.5-104}{\sqrt{49.92}}=-1.769$ .
$\Phi(-1.769)=0.0384=3.84\%<5\%$ .
Reject $H_0$ : some evidence support has fallen.

Answer

3.84\%<5\%\Rightarrow\text{reject }H_0.

Worked example

Normal-approximation test + critical value (an increase)

60% react to a cream; a new cream gives 79 of 120 reacting. Evidence of an increase at 5%? Then find the critical value.

$X\sim B(120,0.6)$ : $\mu=72,\ \sigma^2=28.8$ , $H_1{:}\ p>0.6$ . Upper tail, correct $P(X\ge79)\to P(X>78.5)$ : $z=\tfrac{78.5-72}{\sqrt{28.8}}=1.211$ , $1-\Phi(1.211)=0.113$ .

Answer

11.3\%>5\%\Rightarrow\text{accept }H_0.

Critical value backwards: reject when $x>72+1.645\sqrt{28.8}=80.8$ , i.e. from $81$ upward.

The 5% rejection region is x > 80.8, i.e. 81 or more. The observed 79 is high but lands inside the acceptance region — not surprising enough.

Drag $z$ below to see how a critical $z$ carves off its tail area — that is exactly what “compare with the level” does.

z = 1.00·Φ(z) = P(Z < z) = 0.8413·right tail 0.1587

Continuity-correct the count first (≥ x → x−½, ≤ x → x+½), then standardise with √(npq). Keep variance in N(·,·); divide by its root.

Level	one-tailed z	two-tailed z
10%	1.282	1.645
5%	1.645	1.960
2%	2.054	2.326
1%	2.326	2.576

Examiner note

A sample-mean z-test uses

\bar X\sim N(\mu,\sigma^2/n)

(estimate

\sigma^2

with the unbiased

\tfrac{n}{n-1}

form if it isn't given) and no continuity correction; the CLT is only needed if the population is not already Normal. (9709/62/M/J/23 Q3) (9709/62/O/N/23 Q4)

Where the critical z come from

Common mistake

Forgetting the continuity correction (

P(X\ge79)\to78.5

); for two tails, comparing with the full

\alpha

instead of

\alpha/2

; the wrong critical

z

(1.645 one-tailed 5%, 1.960 two-tailed 5%, 2.326 one-tailed 1%); and putting SD instead of variance in

N(\cdot,\cdot)

Your turn— tap to reveal the worked answer (one-tailed normal-approximation test)

A coin is flipped 180 times to test bias towards tails; 102 tails appear. Test at 5%. $X\sim B(180,0.5)$ , $\mu=90,\ \sigma^2=45$ , $H_1{:}\ p>0.5$ .

$P(X\ge102)\to P(X>101.5)$ : $z=\tfrac{101.5-90}{\sqrt{45}}=1.714$ , $1-\Phi(1.714)=0.0433$ .

Answer

4.33\%<5\%\ \Rightarrow\ \text{reject }H_0

There is some evidence of bias towards tails.