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Notes/Statistics/Introduction to hypothesis testing
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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

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
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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.

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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●

5 sections · 10 key takeaways · 4 formulas · 10 mistake warnings

§ 01
§ 01

The logic of a hypothesis test#

~3 min reading time●●○StandardSPPearson 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.

§ 01

Active revision

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

Practise matching questions→

Active recall

Recall the key points — then reveal.

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

§ 02
§ 02

Hypotheses, tails and significance levels#

~2 min reading time●●○StandardSPPearson 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.

Hypotheses and tails

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​

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.

§ 02

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.

Practise matching questions→

Active recall

Recall the key points — then reveal.

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

§ 03
§ 03

Critical regions and the p-value approach#

~2 min reading time●●●AdvancedSPPearson 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.

The p-value decision rule

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 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).

§ 03

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.

Practise matching questions→

Active recall

Recall the key points — then reveal.

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

§ 04
§ 04

A test for a binomial proportion#

~3 min reading time●●●AdvancedSPPearson 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%.

Critical region for the binomial test

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

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.

§ 04

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.

Practise matching questions→

Active recall

Recall the key points — then reveal.

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

§ 05
§ 05

Type I and Type II errors#

~3 min reading time●●●AdvancedSPPearson 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.

The two error probabilities

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

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.

§ 05

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.

Practise matching questions→

Active recall

Recall the key points — then reveal.

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

Full version via the depth control — same place, same anchors

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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Introduction to hypothesis testing

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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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