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Notes · StatisticsUK · A-Levels

Introduction to hypothesis testing

This topic sets out the logic of significance testing: the null and alternative hypotheses, one- and two-tailed tests, the significance level, the critical region and the p-value. It carries out a hypothesis test for a binomial proportion, and analyses the two kinds of error a test can make and the actual significance level of a discrete test.

5 sections·~13 min reading time·3 competencies·Level Standard 2 · Advanced 3

T·111111 / 18
Exam profile
AO1 · Set up and carry out a hypothesis test with a stated significance levelAO2 · Interpret the significance level, critical region and p-valueAO3 · Conclude a test in context and discuss the risk of error
Operators:testfinddeterminestateconcludecomment

basic level

The AS foundation expects a one-tailed binomial hypothesis test, the hypotheses, significance level and conclusion.

higher level

The full A-Level expects two-tailed tests, the p-value approach, critical regions, actual significance levels and the two error types.

Depth

Reading depth: In depth

Text

Text size: Standard

Contents · 5 sections▾
  1. Introduction to hypothesis testing
    • 01The logic of a hypothesis test◐
    • 02Hypotheses, tails and significance levels◐
    • 03Critical regions and the p-value approach●
    • 04A test for a binomial proportion●
    • 05Type I and Type II errors●
§ 01

The logic of a hypothesis test#

●●○StandardLPPearson Edexcel 9ST0, Topic 11 (Paper 2)

The hypothesis-testing procedure

Steps of a hypothesis testGraph, State H0 and H1 → Choose significance level, Choose significance level → Find critical region, Find critical region → Compare data with region, Compare data with region → Reject or do not reject H0, Reject or do not reject H0 → Conclude in contextState H0 and H1ChoosesignificancelevelFind criticalregionCompare datawith regionReject or do notreject H0Conclude incontext
Fig. 1The fixed sequence of a hypothesis test, ending always in a contextual conclusion.

Key points

A hypothesis test weighs evidence against a specific claim about a population. It begins by assuming the claim — the null hypothesis H0H_0H0​ — is true, then asks how surprising the observed data would be under that assumption. If the data would be very unlikely when H0H_0H0​ is true, that counts as evidence against H0H_0H0​ in favour of the alternative H1H_1H1​; if not, there is insufficient evidence to reject H0H_0H0​.
The reasoning is deliberately asymmetric, like a trial in which the null is presumed until the evidence is strong enough to overturn it. We never prove H0H_0H0​ true; we either reject it or fail to reject it. This is why the correct conclusion is 'there is (in)sufficient evidence to reject H0H_0H0​', never 'accept H0H_0H0​' — the test can only fail to find evidence against the null, not confirm it.
The procedure follows fixed steps, shown in the flow below: state H0H_0H0​ and H1H_1H1​; choose a significance level; find the critical region (or the test statistic and its p-value); compare the data with it; decide to reject or not; and state the conclusion in context. Following the steps in order, and never skipping the contextual conclusion, is what full-mark answers do.
Every step has a purpose. The hypotheses fix what is being tested and in which direction; the significance level fixes how much evidence is required; the critical region or p-value measures the evidence; and the contextual conclusion translates the statistical decision back into the language of the problem. Understanding the role of each step prevents the mechanical, context-free answers that lose marks.
Worked example

Framing a test

A manufacturer claims at most 10% of its items are faulty. A retailer suspects the rate is higher. Set up the hypotheses and describe the logic of the test.

  1. 01Null hypothesis

    H0H_0H0​: p=0.10p = 0.10p=0.10 (the claimed faulty rate).

  2. 02Alternative

    H1H_1H1​: p>0.10p > 0.10p>0.10 (the retailer's suspicion), a one-tailed test.

  3. 03Logic

    Assume p=0.10p = 0.10p=0.10; if the observed number of faults would be very unlikely under this assumption, reject H0H_0H0​ in favour of H1H_1H1​.

Result: H0H_0H0​: p=0.10p = 0.10p=0.10 against H1H_1H1​: p>0.10p > 0.10p>0.10; reject H0H_0H0​ only if the data are too surprising under p=0.10p = 0.10p=0.10.

Exam focus

  • State conclusions as 'reject / do not reject H0H_0H0​', never 'accept H0H_0H0​'.
  • Follow the full sequence of steps, ending with a contextual conclusion.

Typical mistakes

  • Writing 'accept H0H_0H0​' instead of 'insufficient evidence to reject H0H_0H0​'.
  • Omitting the conclusion in context, leaving only a statistical decision.

Active revision

Explain, in your own words, why a hypothesis test can never prove the null hypothesis true.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

§ 02

Hypotheses, tails and significance levels#

●●○StandardLPPearson Edexcel 9ST0, Topic 11 (Paper 2)

Key points

The null hypothesis H0H_0H0​ specifies a particular value of the parameter (for example p=0.5p = 0.5p=0.5), and the alternative H1H_1H1​ specifies the departure of interest. If H1H_1H1​ points in one direction (p>0.5p > 0.5p>0.5 or p<0.5p < 0.5p<0.5) the test is one-tailed; if it merely says the value is different (p≠0.5p \neq 0.5p=0.5) the test is two-tailed. The wording of the question — 'has increased', 'has changed', 'is biased' — fixes the tail, and choosing it correctly is essential.
The significance level α\alphaα is the probability threshold below which the data are judged too surprising to attribute to chance, commonly 5%, 1% or 10%. It is fixed BEFORE seeing the data, so the standard of evidence cannot be adjusted to suit the result. A smaller α\alphaα demands stronger evidence to reject H0H_0H0​ and so reduces the chance of a false rejection.
For a two-tailed test the significance level is split between the two tails, so a 5% two-tailed test uses 2.5%2.5\%2.5% in each tail. This matters because the same data can be significant one-tailed but not two-tailed: a two-tailed test is more conservative, reflecting that it looks for a departure in either direction and so demands more in the direction observed.
The direction of H1H_1H1​ must be justified by the context, not chosen after seeing the data. Testing 'has the proportion increased?' one-tailed because the sample happened to be high, when the original question was 'has it changed?', is a form of cheating that inflates the true error rate. Setting the hypotheses and the tail from the wording, before any calculation, is the disciplined approach the examination rewards.
one-tailed: H1:p>p0  or  p<p0;two-tailed: H1:p≠p0\text{one-tailed: } H_1: p > p_0 \;\text{or}\; p < p_0; \qquad \text{two-tailed: } H_1: p \neq p_0one-tailed: H1​:p>p0​orp<p0​;two-tailed: H1​:p=p0​

Hypotheses and tails

A two-tailed test splits α\alphaα between the tails (2.5%2.5\%2.5% each at the 5% level).

Worked example

Setting up a two-tailed test

A die is suspected of being biased towards or against sixes. It is tested at the 5% level. State the hypotheses and the tail allocation.

  1. 01Parameter

    Let p=P(six)p = P(\text{six})p=P(six); a fair die has p=16p = \tfrac{1}{6}p=61​.

  2. 02Hypotheses

    H0H_0H0​: p=16p = \tfrac{1}{6}p=61​ against H1H_1H1​: p≠16p \neq \tfrac{1}{6}p=61​ (bias in either direction), so two-tailed.

  3. 03Tail allocation

    At the 5% level, each tail carries 2.5%2.5\%2.5%.

Result: H0H_0H0​: p=16p = \tfrac{1}{6}p=61​ against H1H_1H1​: p≠16p \neq \tfrac{1}{6}p=61​, two-tailed with 2.5%2.5\%2.5% in each tail.

Exam focus

  • Choose a one- or two-tailed alternative from the wording of the question.
  • Split the significance level correctly for a two-tailed test.

Typical mistakes

  • Selecting the tail after seeing the data rather than from the question.
  • Using the full α\alphaα in one tail of a two-tailed test instead of α/2\alpha/2α/2.

Active revision

A coin is tested for bias (in either direction) at the 5% level. State H0H_0H0​, H1H_1H1​ and the probability allocated to each tail.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

§ 03

Critical regions and the p-value approach#

●●●AdvancedLPPearson Edexcel 9ST0, Topic 11 (Paper 2)

Key points

There are two equivalent ways to make the decision. The critical-region method finds, before seeing the data, the set of extreme outcomes whose total probability under H0H_0H0​ is at most α\alphaα; if the observed value falls in this region, H0H_0H0​ is rejected. The p-value method computes, after seeing the data, the probability under H0H_0H0​ of a result at least as extreme as the one observed, and rejects H0H_0H0​ if this p-value is at most α\alphaα.
The two methods always agree, because the observed value lies in the critical region exactly when its p-value is at most α\alphaα. The p-value is often preferred because it reports HOW surprising the data are, not merely whether they cross the threshold: a p-value of 0.001 is far stronger evidence against H0H_0H0​ than a p-value of 0.049, though both reject at the 5% level.
A critical p-value must be interpreted precisely. It is the probability of the observed (or more extreme) data GIVEN that H0H_0H0​ is true — it is NOT the probability that H0H_0H0​ is true. A small p-value means the data are surprising under H0H_0H0​, which is evidence against it, but it says nothing directly about the probability of the hypothesis itself.
For a discrete test the critical region is built from cumulative probabilities and rarely has probability exactly α\alphaα, a subtlety pursued in the binomial test next. In every case the decision rule is the same: reject H0H_0H0​ when the evidence (the p-value) is at most the pre-set significance level, and otherwise do not reject.
p-value=P(data at least as extreme∣H0);reject H0  ⟺  p-value≤α\text{p-value} = P(\text{data at least as extreme} \mid H_0); \quad \text{reject } H_0 \iff \text{p-value} \leq \alphap-value=P(data at least as extreme∣H0​);reject H0​⟺p-value≤α

The p-value decision rule

The p-value is a conditional probability given H0H_0H0​, not the probability that H0H_0H0​ is true.

Worked example

Deciding from a p-value

A one-tailed test at the 5% level gives a p-value of 0.032. State the decision and interpret the value.

  1. 01Compare with alpha

    0.032≤0.050.032 \leq 0.050.032≤0.05, so the data fall in the critical region.

  2. 02Decision

    Reject H0H_0H0​ at the 5% level.

  3. 03Interpret

    If H0H_0H0​ were true, data this extreme would occur with probability 0.032 — surprising enough to reject H0H_0H0​, though not overwhelming.

Result: Reject H0H_0H0​: the p-value 0.032 is below 0.05, meaning such data are unlikely (probability 0.032) under H0H_0H0​.

Exam focus

  • State and apply the p-value decision rule, and find a critical region from cumulative probabilities.
  • Interpret a p-value correctly as P(data∣H0)P(\text{data} \mid H_0)P(data∣H0​).

Typical mistakes

  • Interpreting the p-value as the probability that H0H_0H0​ is true.
  • Comparing the p-value with the wrong threshold (for example α\alphaα rather than α/2\alpha/2α/2 in a two-tailed test).

Active revision

In a one-tailed test at the 5% level the p-value is 0.032. State the decision and interpret the p-value.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

§ 04

A test for a binomial proportion#

●●●AdvancedLPPearson Edexcel 9ST0, Topic 11 (Paper 2)

Critical region on the number of heads

Critical region (number of heads)Number line, c = 15, reject H005101520reject H0c = 15
Fig. 2Critical region X >= 15 for testing H0: p = 0.5 against H1: p > 0.5 with n = 20 at the 5% level.

Key points

The archetypal test concerns a binomial proportion. To test whether a coin is biased towards heads, take H0H_0H0​: p=0.5p = 0.5p=0.5 against H1H_1H1​: p>0.5p > 0.5p>0.5, model the number of heads XXX in nnn tosses as B(n,0.5)B(n, 0.5)B(n,0.5) under H0H_0H0​, and reject H0H_0H0​ for values of XXX in the upper tail. The critical region is the smallest set of large values whose total probability under H0H_0H0​ does not exceed α\alphaα.
For n=20n = 20n=20 tosses at the 5% level, the cumulative binomial gives P(X≥15)=1−P(X≤14)=1−0.9793=0.0207P(X \geq 15) = 1 - P(X \leq 14) = 1 - 0.9793 = 0.0207P(X≥15)=1−P(X≤14)=1−0.9793=0.0207, which is below 5%, whereas P(X≥14)=0.0577P(X \geq 14) = 0.0577P(X≥14)=0.0577 exceeds 5%. The critical region is therefore X≥15X \geq 15X≥15, shown on the number line below, and the critical value is 15.
Because the binomial is discrete, the critical region's probability (0.0207) is below the nominal 5% rather than exactly equal to it; this true probability is the ACTUAL significance level of the test. The gap arises because XXX jumps in whole steps, so no region has probability exactly 0.05 — a feature of every discrete test.
The decision then compares the observed count with the region: 15 or more heads falls in the critical region and rejects H0H_0H0​, while 14 or fewer does not. The conclusion must be contextual — 'there is sufficient evidence at the 5% level that the coin is biased towards heads' — and should acknowledge that the actual significance level is 2.07%, not 5%.
Under H0,  X∼B(20,0.5);P(X≥15)=0.0207<0.05≤P(X≥14)=0.0577\text{Under } H_0,\; X \sim B(20, 0.5); \quad P(X \geq 15) = 0.0207 < 0.05 \leq P(X \geq 14) = 0.0577Under H0​,X∼B(20,0.5);P(X≥15)=0.0207<0.05≤P(X≥14)=0.0577

Critical region for the binomial test

The critical region is X≥15X \geq 15X≥15; its actual probability 0.0207 is the actual significance level.

Worked example

A one-tailed binomial test

A coin tossed 20 times shows 16 heads. Test at the 5% level whether it is biased towards heads, using P(X≤14)=0.9793P(X \leq 14) = 0.9793P(X≤14)=0.9793 under H0H_0H0​.

  1. 01Hypotheses

    H0H_0H0​: p=0.5p = 0.5p=0.5 against H1H_1H1​: p>0.5p > 0.5p>0.5; under H0H_0H0​, X∼B(20,0.5)X \sim B(20, 0.5)X∼B(20,0.5).

  2. 02Critical region

    P(X≥15)=1−0.9793=0.0207<0.05P(X \geq 15) = 1 - 0.9793 = 0.0207 < 0.05P(X≥15)=1−0.9793=0.0207<0.05, and P(X≥14)=0.0577>0.05P(X \geq 14) = 0.0577 > 0.05P(X≥14)=0.0577>0.05, so the region is X≥15X \geq 15X≥15.

  3. 03Compare

    The observed 16 lies in the region X≥15X \geq 15X≥15.

    16≥15⇒reject H016 \geq 15 \Rightarrow \text{reject } H_016≥15⇒reject H0​
  4. 04Conclude

    Reject H0H_0H0​: there is sufficient evidence at the 5% level (actual level 2.07%) that the coin is biased towards heads.

Result: Since 16 lies in the critical region X≥15X \geq 15X≥15, reject H0H_0H0​; the coin appears biased towards heads (actual significance level 2.07%).

Exam focus

  • Find a binomial critical region from cumulative probabilities and state the critical value.
  • Quote the actual significance level of the discrete test and conclude in context.

Typical mistakes

  • Using P(X≥c)≤αP(X \geq c) \leq \alphaP(X≥c)≤α but taking ccc one step too small so the region exceeds α\alphaα.
  • Reporting the nominal level (5%) as the actual significance level for a discrete test.

Active revision

Test at the 5% level whether a coin tossed 20 times, landing heads 16 times, is biased towards heads, given P(X≤14)=0.9793P(X \leq 14) = 0.9793P(X≤14)=0.9793.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

§ 05

Type I and Type II errors#

●●●AdvancedLPPearson Edexcel 9ST0, Topic 11 (Paper 2)

Key points

A test can be wrong in two distinct ways. A Type I error is rejecting H0H_0H0​ when it is actually true — a false alarm — and a Type II error is failing to reject H0H_0H0​ when it is actually false — a missed detection. The two are not symmetric, and which is worse depends on the context: a false alarm on a smoke detector is a nuisance, but a missed fire is a disaster.
The probability of a Type I error is controlled directly: for a discrete test it is the actual significance level, the probability under H0H_0H0​ of landing in the critical region. For the binomial test above, P(Type I)=P(X≥15∣p=0.5)=0.0207P(\text{Type I}) = P(X \geq 15 \mid p = 0.5) = 0.0207P(Type I)=P(X≥15∣p=0.5)=0.0207. Lowering the significance level reduces this probability but makes rejection harder, which increases the chance of a Type II error.
The probability of a Type II error, usually denoted β\betaβ, is the probability of NOT landing in the critical region when a specific alternative value of the parameter is true. It can only be computed against a stated alternative, because 'the null is false' is not specific enough. Its complement 1−β1 - \beta1−β is the power of the test, developed fully in a later topic.
There is an unavoidable trade-off: for a fixed sample size, reducing one error probability increases the other. The only way to reduce both at once is to gather more data. Recognising this trade-off, and choosing a significance level that reflects the relative costs of the two errors, is the mature judgement that hypothesis testing demands.
P(Type I)=P(reject H0∣H0 true)=αactual,P(Type II)=βP(\text{Type I}) = P(\text{reject } H_0 \mid H_0 \text{ true}) = \alpha_{\text{actual}}, \qquad P(\text{Type II}) = \betaP(Type I)=P(reject H0​∣H0​ true)=αactual​,P(Type II)=β

The two error probabilities

α\alphaα is controlled by the significance level; β\betaβ needs a specific alternative; power is 1−β1 - \beta1−β.

Worked example

Probability of a Type I error

For the test H0H_0H0​: p=0.5p = 0.5p=0.5 against H1H_1H1​: p>0.5p > 0.5p>0.5 with n=20n = 20n=20 and critical region X≥15X \geq 15X≥15, state the probability of a Type I error and describe a Type II error.

  1. 01Type I probability

    P(Type I)=P(X≥15∣p=0.5)=0.0207P(\text{Type I}) = P(X \geq 15 \mid p = 0.5) = 0.0207P(Type I)=P(X≥15∣p=0.5)=0.0207.

  2. 02Meaning

    It is the chance of wrongly declaring the coin biased when it is in fact fair.

  3. 03Type II error

    A Type II error here is failing to detect a genuinely biased coin, i.e. observing X≤14X \leq 14X≤14 when in truth p>0.5p > 0.5p>0.5.

Result: P(Type I)=0.0207P(\text{Type I}) = 0.0207P(Type I)=0.0207; a Type II error is failing to reject H0H_0H0​ (observing X≤14X \leq 14X≤14) when the coin is really biased.

Exam focus

  • Identify and distinguish Type I and Type II errors in context.
  • Compute the probability of a Type I error as the actual significance level.

Typical mistakes

  • Swapping the two error types (false alarm versus missed detection).
  • Trying to compute a Type II error probability without a specific alternative value.

Active revision

For the binomial test with critical region X≥15X \geq 15X≥15 under H0H_0H0​: p=0.5p = 0.5p=0.5 (n=20n = 20n=20), state the probability of a Type I error and describe a Type II error in context.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content (Statistics) (Department for Education / Ofqual)

Contents

Section -- / 05

    • 01The logic of a hypothesis test◐
    • 02Hypotheses, tails and significance levels◐
    • 03Critical regions and the p-value approach●
    • 04A test for a binomial proportion●
    • 05Type I and Type II errors●

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References & sources

Sources

Pearson Edexcel

  • Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification

Department for Education / Ofqual

  • GCE AS and A level subject content (Statistics)

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