What is a hypothesis test in the context of Edexcel International A Levels Statistics 2?

A hypothesis test in Edexcel International A Levels Statistics 2 is a statistical method used to determine whether there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. It involves comparing sample data against a population parameter to make inferences about the population.

How do you formulate null and alternative hypotheses in hypothesis testing?

In hypothesis testing, the null hypothesis (H0) is a statement that there is no effect or no difference, and it serves as the default assumption. The alternative hypothesis (H1) is what you aim to support, indicating the presence of an effect or difference. For example, if testing a new drug, H0 might state that the drug has no effect, while H1 suggests it does have an effect.

What is the significance level in hypothesis testing and how is it used?

The significance level, often denoted as alpha (α), is the probability threshold set by the researcher for rejecting the null hypothesis. Commonly set at 0.05, it represents a 5% risk of concluding that an effect exists when it does not. In hypothesis testing, if the p-value is less than or equal to the significance level, the null hypothesis is rejected.

Can you explain the concept of p-value in hypothesis tests?

The p-value in hypothesis tests is a measure of the probability that the observed data would occur if the null hypothesis were true. A small p-value indicates that the observed data is unlikely under the null hypothesis, leading to its rejection. In Edexcel International A Levels Statistics 2, understanding p-values helps in making informed decisions about hypotheses.

What are Type I and Type II errors in the context of hypothesis testing?

In hypothesis testing, a Type I error occurs when the null hypothesis is wrongly rejected when it is true, often controlled by the significance level. A Type II error happens when the null hypothesis is not rejected when it is false. Understanding these errors is crucial in Edexcel International A Levels Statistics 2 to evaluate the reliability of test results.

Why is "Concluding 'reject $H_0$' without contextual interpretation" a common mistake in Hypothesis Tests?

Why it happens: Students focus on the decision, not the meaning. How to avoid it: Always write a contextual sentence: 'There is evidence at the 5\% significance level to suggest that the success probability is greater than 0.7.' This carries the final A-mark. Source: WST02/01 examiner reports — annual

Why is "Computing wrong-tail p-value" a common mistake in Hypothesis Tests?

Why it happens: Confusion between H_1: p > p_0 (upper tail) and H_1: p p_0 uses P(X x_obs). H_1: p < p_0 uses P(X x_obs). Two-tail: double the smaller tail or compare both tails.

Why is "Not splitting $\alpha$ in a two-tail test" a common mistake in Hypothesis Tests?

Why it happens: Forgetting two-tail uses /2 on each side. How to avoid it: Two-tail at 5\%: each tail has area 2.5\%. Critical z = 1.96, not 1.6449.

Why is "Writing $H_0$ with an inequality" a common mistake in Hypothesis Tests?

Why it happens: Mixing up which is null vs alternative. How to avoid it: H_0 is ALWAYS an equality (p = p_0 or = \mu_0). H_1 has the inequality or .

Why is "Off-by-one in $P(X \ge x_{\text{obs}})$ for binomial / Poisson" a common mistake in Hypothesis Tests?

Why it happens: Same as cumulative-distribution work. How to avoid it: P(X 32) = 1 - P(X 31), NOT 1 - P(X 32). The observed value is included in the upper tail. Source: WST02/01 examiner reports

Why is "Computing $z = (\bar x - \mu)/\sigma$ instead of $z = (\bar x - \mu)/(\sigma/\sqrt n)$" a common mistake in Hypothesis Tests?

Why it happens: Confusion between X and X. How to avoid it: For the SAMPLE MEAN, always divide by /√ n (the standard error), NOT by .

Statistics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Hypothesis Tests

Q: Why is "Not splitting $\alpha$ in a two-tail test" a common mistake in Hypothesis Tests?

Why it happens: Forgetting two-tail uses /2 on each side. How to avoid it: Two-tail at 5\%: each tail has area 2.5\%. Critical z = 1.96, not 1.6449.

Q: Why is "Writing $H_0$ with an inequality" a common mistake in Hypothesis Tests?

Why it happens: Mixing up which is null vs alternative. How to avoid it: H_0 is ALWAYS an equality (p = p_0 or = \mu_0). H_1 has the inequality or .

Q: Why is "Off-by-one in $P(X \ge x_{\text{obs}})$ for binomial / Poisson" a common mistake in Hypothesis Tests?

Why it happens: Same as cumulative-distribution work. How to avoid it: P(X 32) = 1 - P(X 31), NOT 1 - P(X 32). The observed value is included in the upper tail. Source: WST02/01 examiner reports

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Hypothesis Tests Study Notes — Edexcel IAL Mathematics S2 (WST02/01, 2018 spec — 2026 onwards)

Hypothesis testing is the capstone of S2 — it integrates binomial, Poisson, and normal distributions into a single inferential framework. $H_0$ (status quo) vs $H_1$ (alternative); reject $H_0$ if the observed test statistic falls in the critical region (or p-value $< \alpha$ ). Conclusions MUST be in context. Tests on $B(n, p)$ , $\text{Po}(\lambda)$ , and on sample means via normal / CLT.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

S2 7.1 — State null and alternative hypotheses, one- and two-tail.
S2 7.2 — Compute p-values and identify critical regions for $B(n, p)$ and $\text{Po}(\lambda)$ .
S2 7.3 — Perform $z$ -tests on sample means using known $\sigma$ and CLT.
S2 7.4 — Make decisions and write conclusions in context.
S2 7.5 — Understand Type 1 and Type 2 errors.

The hypothesis testing framework

$H_0$ vs $H_1$ , sig level $\alpha$ , test statistic, decision, contextual conclusion.

Five-step recipe.

Hypotheses. $H_0$ is the equality / status quo; $H_1$ is what we are testing FOR.
Distribution under $H_0$ . Identify the test statistic and its distribution.
Compute either:
- p-value: probability of an outcome at least as extreme as observed, under $H_0$ .
- Critical region: set of test statistic values that lead to rejecting $H_0$ at level $\alpha$ .
Decision. Reject $H_0$ if p-value $< \alpha$ (equivalently, test stat in critical region).
Conclusion IN CONTEXT. Translate back to the scenario.

Worked outline. 'A coin is flipped $20$ times and gets $14$ heads. Is the coin biased?'

$H_0: p = 0.5$ vs $H_1: p \ne 0.5$ (two-tail).
Under $H_0$ , $X \sim B(20, 0.5)$ .
p-value $= 2 \cdot P(X \ge 14)$ for a symmetric distribution.
Compare to $\alpha$ .
'At the $5\%$ level, [is / is not] sufficient evidence that the coin is biased.'

Symmetry vs framework. The five-step framework is universal; only the distribution changes between binomial, Poisson, and normal tests.

Hypotheses ( $H_0$ equality, $H_1$ inequality).
Distribution under $H_0$ .
p-value OR critical region.
Decision.
Contextual conclusion.

Binomial hypothesis tests

$X \sim B(n, p_0)$ under $H_0$ . p-value from cumulative.

Setup. $H_0: p = p_0$ vs $H_1: p > p_0$ (or $<$ or $\ne$ ). Observed count $x$ from $n$ trials.

One-tail (upper). p-value $= P(X \ge x | p = p_0) = 1 - P(X \le x - 1)$ .

One-tail (lower). p-value $= P(X \le x | p = p_0)$ .

Two-tail. Two tails of $\alpha/2$ each, OR p-value $= 2 \cdot \min\{P(X \le x), P(X \ge x)\}$ (the 'doubled minimum' rule).

Critical region. Smallest set of extreme values with total probability $\le \alpha$ under $H_0$ .

Worked example (one-tail upper). $X \sim B(40, p)$ , observed $x = 32$ . Test $p = 0.7$ vs $p > 0.7$ at $5\%$ .

Under $H_0$ , $X \sim B(40, 0.7)$ .
p-value $= P(X \ge 32) = 1 - P(X \le 31) \approx 0.0581$ .
$0.0581 > 0.05$ , do NOT reject.
'Insufficient evidence at $5\%$ that the success rate exceeds $70\%$ .'

Worked example (two-tail critical region). $X \sim B(20, 0.4)$ , test vs $p \ne 0.4$ at $5\%$ .

Each tail: $0.025$ .
Lower: $P(X \le 3) = 0.0160 < 0.025$ . So $X \le 3$ in CR.
Upper: $P(X \ge 13) = 0.0210 < 0.025$ . So $X \ge 13$ in CR.
Critical region: $\{X \le 3\} \cup \{X \ge 13\}$ .
Actual sig level $\approx 0.037$ (discreteness).

Under $H_0$ : $X \sim B(n, p_0)$ .
Upper tail: $P(X \ge x) = 1 - P(X \le x - 1)$ .
Two-tail: split $\alpha$ .
Discrete critical region: actual $\alpha$ < nominal $\alpha$ .

Poisson hypothesis tests

$X \sim \text{Po}(\lambda_0)$ under $H_0$ . Procedure identical to binomial.

Setup. $H_0: \lambda = \lambda_0$ vs $H_1: \lambda > \lambda_0$ (or $<$ or $\ne$ ).

Method. Identical to binomial — compute p-value from $\text{Po}(\lambda_0)$ cumulative tables.

Worked example. Number of accidents per week used to follow $\text{Po}(2.5)$ . After a new speed limit, the next week shows $1$ accident. Test at $10\%$ whether the rate has decreased.

$H_0: \lambda = 2.5$ vs $H_1: \lambda < 2.5$ (one-tail lower).
p-value $= P(X \le 1 | \lambda = 2.5) = e^{-2.5}(1 + 2.5) \approx 0.2873$ .
$0.2873 > 0.10$ , do not reject.
'No evidence at $10\%$ that the accident rate has decreased.'

Sensitivity of Poisson tests. With one week of data, only very extreme outcomes can reject $H_0$ . To detect modest changes, longer observation periods are needed (use scaling: $\text{Po}(\lambda T)$ over $T$ weeks).

Critical region. Smallest tail (or pair of tails) with total probability $\le \alpha$ . Identical procedure to binomial.

Same procedure as binomial.
Under $H_0$ : $X \sim \text{Po}(\lambda_0)$ .
Scaling: more data = more sensitive test.
Discrete critical region: actual $\alpha$ < nominal $\alpha$ .

Normal hypothesis tests (sample mean)

$z = (\bar x - \mu_0)/(\sigma/\sqrt n)$ , compare to $z_{\text{crit}}$ .

Setup. Sample of size $n$ from a population with known $\sigma$ . Test $H_0: \mu = \mu_0$ vs $H_1$ (one- or two-tail).

Test statistic.

$z = \dfrac{\bar x - \mu_0}{\sigma/\sqrt n}.$

Under $H_0$ : $z \sim N(0, 1)$ (exact if parent normal, approximate by CLT if $n$ large).

Critical $z$ -values.

Sig level	One-tail	Two-tail (each)
$10\%$	$\pm 1.2816$	$\pm 1.6449$
$5\%$	$\pm 1.6449$	$\pm 1.96$
$1\%$	$\pm 2.3263$	$\pm 2.5758$

Worked example. Bags labelled $30$ g, $\sigma = 1.2$ g known. Sample $n = 50$ , $\bar x = 29.6$ g. Test $\mu < 30$ at $5\%$ .

$H_0: \mu = 30$ vs $H_1: \mu < 30$ .
$z = (29.6 - 30)/(1.2/\sqrt{50}) = -0.4/0.1697 \approx -2.357$ .
Critical $z_{\text{crit}} = -1.6449$ .
$-2.357 < -1.6449$ : in CR.
Reject $H_0$ . 'Evidence at $5\%$ that bags are underfilled.'

p-value alternative. $p\text{-value} = \Phi(-2.357) \approx 0.0092 < 0.05$ . Same conclusion.

Two-tail. Use $|z|$ vs the two-tail critical value, or double the smaller tail probability.

$z = (\bar x - \mu_0)/(\sigma/\sqrt n)$ .
One-tail $5\%$ : $|z_{\text{crit}}| = 1.6449$ .
Two-tail $5\%$ : $|z_{\text{crit}}| = 1.96$ .
CLT used if $n$ large and parent non-normal.

Type 1 and Type 2 errors

Type 1: reject true $H_0$ (prob $\alpha$ ). Type 2: fail to reject false $H_0$ .

Decision matrix.

	$H_0$ true	$H_0$ false
Reject $H_0$	Type 1 error ( $\alpha$ )	Correct decision
Do not reject	Correct decision	Type 2 error ( $\beta$ )

Type 1 error. Rejecting $H_0$ when it is actually true. Probability $= \alpha$ (the significance level).

'False alarm' or 'false positive'.
Set by choice of $\alpha$ . Smaller $\alpha$ = fewer Type 1 errors.

Type 2 error. Failing to reject $H_0$ when $H_0$ is actually false. Probability $= \beta$ .

'Missed detection' or 'false negative'.
Depends on true value, sample size, $\alpha$ . NOT computed numerically at S2.

Trade-off. Lowering $\alpha$ raises $\beta$ . To reduce both simultaneously, INCREASE sample size.

In context.

Type 1 (test on a quality controller's process): wrongly intervening in a process that is fine — unnecessary cost.
Type 2: failing to detect a problem — defective products reach customers.

Discrete distributions caveat. For binomial / Poisson, the actual Type 1 probability is the probability of the critical region under $H_0$ , which is usually LESS than the nominal $\alpha$ (due to discreteness).

Type 1: reject true $H_0$ . Prob $= \alpha$ .
Type 2: fail to reject false $H_0$ . Prob $= \beta$ .
Trade-off: lowering one raises the other.
Larger $n$ reduces both.
Discrete: actual Type 1 prob $\le \alpha$ .

Writing conclusions

Always in context. Refer to $H_1$ , not 'reject $H_0$ '.

The two-part conclusion.

Decision. 'Reject $H_0$ ' or 'Do not reject $H_0$ ' (or equivalently 'insufficient evidence').
Contextual interpretation. Translate the decision into the scenario's language.

Reject template. 'At the $\alpha$ significance level, there is sufficient evidence to suggest [statement of $H_1$ in context].'

Do not reject template. 'At the $\alpha$ significance level, there is insufficient evidence to suggest [statement of $H_1$ in context].'

Examples (good).

'There is evidence at the $5\%$ level that the new drug is effective for more than $70\%$ of patients.'
'There is insufficient evidence at the $10\%$ level that the accident rate has changed.'
'There is evidence at the $5\%$ level that bags of crisps are underfilled.'

Examples (bad — lose marks).

'Reject $H_0$ .' — no context.
'The drug works.' — no significance level.
'The coin is biased.' — definitive language; should say 'there is evidence that…'

Hedging language.

'There is evidence to suggest …' (not 'this proves …').
'Insufficient evidence' (not ' $H_0$ is true' — we never confirm $H_0$ , only fail to reject it).

Always state the significance level.
Use 'there is (insufficient) evidence to suggest …'.
Never 'prove' or 'confirm' $H_0$ .
Match the context — quote variables.

How it’s examined

Hypothesis testing is the LONGEST topic on WST02/01 — typically a $14$ - $20$ mark question across multiple parts, usually appearing as the final or penultimate question. Common patterns: (1) state hypotheses and compute p-value for a binomial or Poisson scenario ( $7$ marks); (2) find critical region for a two-tail test ( $6$ - $8$ marks); (3) $z$ -test on a sample mean with CLT ( $7$ - $8$ marks); (4) Type 1 / Type 2 error explanation in context ( $3$ - $4$ marks). The single most common error: failing to write a contextual conclusion — examiners ALWAYS award the final A-mark for the in-context sentence. The second most common: off-by-one in $P(X \ge x)$ for discrete distributions.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WST02/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Hypothesis Tests

Step-by-step solutions to past-paper-style questions on hypothesis tests, written exactly the way a tutor would explain them at the board.

1One-tail binomial hypothesis test (7 marks, S2)
Extended• Adapted from WST02/01 January 2024 Q9• hypothesis test, binomial, one-tail
Question
A drug manufacturer claims that its new pain reliever is effective for $70\%$ of patients. A clinical trial gives relief to $32$ of $40$ randomly selected patients. Test, at the $5\%$ significance level, whether the data suggest the success rate is HIGHER than $70\%$ .
Step-by-step solution
1. Step 1
  Define hypotheses. Let $p$ = probability of relief.
  $H_0: p = 0.7 \quad \text{vs} \quad H_1: p > 0.7$
2. Step 2
  Under $H_0$ , $X \sim B(40, 0.7)$ . Observed $x = 32$ .
3. Step 3
  Significance level $\alpha = 0.05$ . One-tailed (upper). Compute p-value:
  $p\text{-value} = P(X \ge 32) = 1 - P(X \le 31)$
4. Step 4
  From calculator: $P(X \le 31) \approx 0.9419$ . So
  $p\text{-value} \approx 1 - 0.9419 = 0.0581$
5. Step 5
  Compare to $\alpha = 0.05$ :
  $0.0581 > 0.05$
6. Step 6
  Decision: do NOT reject $H_0$ .
7. Step 7
  Conclusion in context: 'At the $5\%$ significance level, there is INSUFFICIENT evidence to suggest that the pain reliever is effective for more than $70\%$ of patients.'
Answer
$H_0: p = 0.7$ , $H_1: p > 0.7$ . p-value $\approx 0.0581 > 0.05$ , so do NOT reject $H_0$ . Insufficient evidence at $5\%$ that success rate exceeds $70\%$ .
Examiner tip
B1 for hypotheses (with $p$ defined). M1 for $B(40, 0.7)$ . M1 for $P(X \ge 32)$ as $1 - P(X \le 31)$ . A1 for $0.0581$ . M1 for comparing to $0.05$ . A1 for decision. A1 for contextual conclusion. Common error: rejecting $H_0$ because $32 > 28 = E(X)$ — but $0.0581 > 0.05$ .
2Critical region for a binomial test (7 marks, S2)
Extended• Adapted from WST02/01 June 2024 Q9• hypothesis test, binomial, critical region
Question
$X \sim B(20, p)$ . Find the critical region for a test of $H_0: p = 0.4$ vs $H_1: p \ne 0.4$ at the $5\%$ significance level.
Step-by-step solution
1. Step 1
  Two-tailed test. Split $\alpha = 0.05$ into two tails of $0.025$ each.
2. Step 2
  Under $H_0$ , $X \sim B(20, 0.4)$ . From tables / calculator:
3. Step 3
  Lower tail: find smallest $a$ such that $P(X \le a) > 0.025$ . $P(X \le 3) = 0.0160$ (in tail), $P(X \le 4) = 0.0510$ (out of tail). So lower critical region: $X \le 3$ .
4. Step 4
  Upper tail: find smallest $b$ such that $P(X \ge b) < 0.025$ . $P(X \ge 13) = 1 - P(X \le 12) = 1 - 0.9790 = 0.0210$ (in tail). $P(X \ge 12) = 1 - 0.9420 = 0.0580$ (out of tail). So upper critical region: $X \ge 13$ .
5. Step 5
  Critical region: $\{X \le 3\} \cup \{X \ge 13\}$ .
6. Step 6
  Actual significance level: $0.0160 + 0.0210 = 0.0370$ (rather than $0.05$ , due to discreteness).
Answer
Critical region: $X \le 3$ or $X \ge 13$ . Actual significance level $\approx 3.70\%$ .
Examiner tip
B1 for splitting $\alpha$ into $0.025$ each tail. M1 for lower-tail calculation. A1 for $X \le 3$ . M1 for upper-tail. A1 for $X \ge 13$ . B1 for actual sig level $\approx 0.037$ . Mark schemes accept either 'just less than $0.025$ ' or 'closest without exceeding' as the criterion.
3Poisson hypothesis test (7 marks, S2)
Extended• Adapted from WST02/01 January 2024 Q10• hypothesis test, poisson
Question
The number of accidents per week on a stretch of motorway used to follow $\text{Po}(2.5)$ . After a new speed limit was introduced, observers recorded just $1$ accident in the following week. Test, at the $10\%$ significance level, whether the accident rate has DECREASED.
Step-by-step solution
1. Step 1
  Hypotheses (one-tail lower, since 'decreased'):
  $H_0: \lambda = 2.5 \quad \text{vs} \quad H_1: \lambda < 2.5$
2. Step 2
  Under $H_0$ , $X \sim \text{Po}(2.5)$ . Observed $x = 1$ . Significance $\alpha = 0.10$ .
3. Step 3
  p-value: $P(X \le 1)$ .
  $P(X \le 1) = e^{-2.5}(1 + 2.5) = e^{-2.5} \times 3.5$
4. Step 4
  $e^{-2.5} \approx 0.0821$ . So
  $P(X \le 1) \approx 0.0821 \times 3.5 \approx 0.2873$
5. Step 5
  Compare: $0.2873 > 0.10$ .
6. Step 6
  Decision: do not reject $H_0$ .
7. Step 7
  Conclusion in context: 'At the $10\%$ significance level, there is insufficient evidence to suggest that the accident rate has decreased after the new speed limit.'
Answer
$H_0: \lambda = 2.5$ , $H_1: \lambda < 2.5$ . p-value $\approx 0.2873 > 0.10$ , do NOT reject. No evidence of decrease.
Examiner tip
B1 for hypotheses. M1 for $\text{Po}(2.5)$ . M1 for $P(X \le 1)$ . A1 for $0.2873$ . M1 for comparison. A1 for decision. A1 for contextual conclusion. Note one week of data is too thin to draw a strong conclusion — the test correctly does NOT reject.
4Hypothesis test using normal approximation (7 marks, S2)
Extended• Adapted from WST02/01 June 2024 Q10• hypothesis test, normal approximation, sample mean
Question
Bags of crisps are labelled $30$ g. Quality control assumes mean weight $\mu = 30$ g, SD $\sigma = 1.2$ g. A random sample of $50$ bags has mean $\bar x = 29.6$ g. Test at the $5\%$ significance level whether the mean weight is LESS than $30$ g.
Step-by-step solution
1. Step 1
  Hypotheses (one-tail lower):
  $H_0: \mu = 30 \quad \text{vs} \quad H_1: \mu < 30$
2. Step 2
  Under $H_0$ , by CLT ( $n = 50$ large):
  $\bar X \approx N\left(30, \dfrac{1.44}{50}\right) = N(30, 0.0288)$
3. Step 3
  $\sigma_{\bar X} = \sqrt{0.0288} \approx 0.1697$ .
4. Step 4
  Test statistic:
  $z = \dfrac{\bar x - \mu}{\sigma/\sqrt n} = \dfrac{29.6 - 30}{0.1697} \approx -2.357$
5. Step 5
  Critical $z$ for one-tail lower at $5\%$ : $z_{\text{crit}} = -1.6449$ .
6. Step 6
  Compare: $-2.357 < -1.6449$ , so the test statistic is in the CRITICAL REGION.
7. Step 7
  Decision: REJECT $H_0$ . Conclusion: at the $5\%$ significance level, there is sufficient evidence to conclude that the mean weight is less than $30$ g — bags are underfilled.
Answer
$H_0: \mu = 30$ , $H_1: \mu < 30$ . $z \approx -2.36 < -1.645$ , so REJECT $H_0$ . Evidence at $5\%$ that bags are underfilled.
Examiner tip
B1 for hypotheses. B1 for CLT-based distribution of $\bar X$ . M1 for $z$ -calculation. A1 for $z \approx -2.36$ . M1 for $z_{\text{crit}} = -1.6449$ . A1 for decision. A1 for contextual conclusion. Note the SE step $\sigma/\sqrt n$ — using $\sigma$ directly is a common error.
5Type 1 and Type 2 errors — informal (4 marks, S2)
Extended• Adapted from WST02/01 January 2024 Q11• hypothesis test, errors
Question
In the binomial test of $H_0: p = 0.4$ vs $H_1: p \ne 0.4$ with critical region $\{X \le 3\} \cup \{X \ge 13\}$ on $B(20, 0.4)$ : (a) Find the probability of a Type 1 error. (b) In context, explain what a Type 1 error would mean for a quality controller using this test.
Step-by-step solution
1. Step 1
  (a) Type 1 error = rejecting $H_0$ when it is true. Probability = actual significance level. Compute under $H_0$ ( $X \sim B(20, 0.4)$ ):
  $P(\text{reject } H_0 | H_0) = P(X \le 3) + P(X \ge 13)$
2. Step 2
  From tables: $P(X \le 3) = 0.0160$ , $P(X \ge 13) = 1 - P(X \le 12) = 1 - 0.9790 = 0.0210$ .
  $P(\text{Type 1}) = 0.0160 + 0.0210 = 0.0370$
3. Step 3
  (b) In context: a Type 1 error means concluding that the true success probability differs from $0.4$ when actually it does equal $0.4$ . The quality controller would unnecessarily intervene in a properly running process.
Answer
(a) $P(\text{Type 1}) = 0.0370$ . (b) Concluding $p \ne 0.4$ when actually $p = 0.4$ — unnecessary intervention.
Examiner tip
(a) M1 for adding tail probabilities. A1 for $0.0370$ . (b) B1 for definition in context. B1 for practical consequence. Type 2 (failing to reject when $H_0$ false) is NOT tested at S2 numerically, only conceptually.

Key Formulae — Hypothesis Tests

The formulae you need to memorise for hypothesis tests on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Binomial test statistic
$X \sim B(n, p_0) \text{ under } H_0$
$p_0$
null value of $p$
$X$
observed count of successes
When to use
Testing $H_0: p = p_0$ for a binomial $X$ . Compute p-value as $P(X \ge x_{\text{obs}})$ or $P(X \le x_{\text{obs}})$ as appropriate. NOT in booklet — uses binomial PMF.
Example
$X = 32$ from $B(40, 0.7)$ : $P(X \ge 32) \approx 0.0581$ .
Poisson test statistic
$X \sim \text{Po}(\lambda_0) \text{ under } H_0$
$\lambda_0$
null value of $\lambda$
When to use
Testing $H_0: \lambda = \lambda_0$ for a Poisson count. Compute p-value from the PMF / cumulative.
Example
$X = 1$ from $\text{Po}(2.5)$ : $P(X \le 1) \approx 0.2873$ .
$z$ -test for sample mean
$z = \dfrac{\bar x - \mu_0}{\sigma/\sqrt n}$
$\mu_0$
null mean
$\bar x$
observed sample mean
$\sigma$
known population SD
When to use
Test on $\mu$ when $\sigma$ is KNOWN and either the parent is normal or $n$ is large (CLT). NOT in booklet — derived from sampling distribution.
Example
$\bar x = 29.6$ , $\mu_0 = 30$ , $\sigma = 1.2$ , $n = 50$ : $z \approx -2.36$ .
Common critical $z$ -values
$z_{0.10} = \pm 1.2816, \; z_{0.05} = \pm 1.6449, \; z_{0.025} = \pm 1.96, \; z_{0.01} = \pm 2.3263$
$z_\alpha$
critical value with one-tail area $\alpha$
When to use
Cutoffs for $z$ -tests. Inverse normal tables in the IAL booklet give these.
Example
$5\%$ one-tail: $|z_{\text{crit}}| = 1.6449$ . $5\%$ two-tail: $|z_{\text{crit}}| = 1.96$ .
p-value
$p = P(\text{test statistic at least as extreme as observed} | H_0)$
$p$
p-value
When to use
Reject $H_0$ if $p < \alpha$ . Equivalent to comparing test statistic to critical region.
Example
$p$ -value $= 0.0581 > 0.05$ : do not reject.

Key Definitions and Keywords — Hypothesis Tests

Definitions to memorise and the exact keywords mark schemes credit for hypothesis tests answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Null hypothesis $H_0$
Examiner keyword
Statement of 'no effect' or 'status quo', e.g. $p = p_0$ or $\mu = \mu_0$ . Always an equality.
Alternative hypothesis $H_1$
Examiner keyword
Statement of 'effect' or 'change'. One-tail ( $p > p_0$ or $p < p_0$ ) or two-tail ( $p \ne p_0$ ).
Significance level $\alpha$
Examiner keyword
Maximum acceptable probability of a Type 1 error. Common values: $0.10$ , $0.05$ , $0.01$ .
Test statistic
Examiner keyword
Computed from the sample, used to decide between $H_0$ and $H_1$ . E.g. observed count $X$ , $z$ -value.
Critical region (rejection region)
Examiner keyword
Set of test statistic values that lead to rejecting $H_0$ . Probability under $H_0$ is at most $\alpha$ .
p-value
Examiner keyword
Probability, under $H_0$ , of obtaining a test statistic at least as extreme as the observed one. Reject $H_0$ if p-value $< \alpha$ .
Type 1 error
Examiner keyword
Rejecting $H_0$ when $H_0$ is true. Probability $= \alpha$ (significance level).
Type 2 error
Examiner keyword
Failing to reject $H_0$ when $H_0$ is false. Probability denoted $\beta$ . Not computed numerically at S2.

Common Mistakes and Misconceptions — Hypothesis Tests

The traps other students keep falling into on hypothesis tests questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Concluding 'reject $H_0$ ' without contextual interpretation
WST02/01 examiner reports — annual
Why it happens
Students focus on the decision, not the meaning.
How to avoid it
Always write a contextual sentence: 'There is evidence at the $5\%$ significance level to suggest that the success probability is greater than $0.7$ .' This carries the final A-mark.
✕Computing wrong-tail p-value
Why it happens
Confusion between $H_1: p > p_0$ (upper tail) and $H_1: p < p_0$ (lower tail).
How to avoid it
Match the tail to $H_1$ . $H_1: p > p_0$ uses $P(X \ge x_{\text{obs}})$ . $H_1: p < p_0$ uses $P(X \le x_{\text{obs}})$ . Two-tail: double the smaller tail or compare both tails.
✕Not splitting $\alpha$ in a two-tail test
Why it happens
Forgetting two-tail uses $\alpha/2$ on each side.
How to avoid it
Two-tail at $5\%$ : each tail has area $2.5\%$ . Critical $z = \pm 1.96$ , not $\pm 1.6449$ .
✕Writing $H_0$ with an inequality
Why it happens
Mixing up which is null vs alternative.
How to avoid it
$H_0$ is ALWAYS an equality ( $p = p_0$ or $\mu = \mu_0$ ). $H_1$ has the inequality or $\ne$ .
✕Off-by-one in $P(X \ge x_{\text{obs}})$ for binomial / Poisson
WST02/01 examiner reports
Why it happens
Same as cumulative-distribution work.
How to avoid it
$P(X \ge 32) = 1 - P(X \le 31)$ , NOT $1 - P(X \le 32)$ . The observed value is included in the upper tail.
✕Computing $z = (\bar x - \mu)/\sigma$ instead of $z = (\bar x - \mu)/(\sigma/\sqrt n)$
Why it happens
Confusion between $X$ and $\bar X$ .
How to avoid it
For the SAMPLE MEAN, always divide by $\sigma/\sqrt n$ (the standard error), NOT by $\sigma$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Hypothesis Tests — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Hypothesis Tests

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1One-tail binomial hypothesis test (7 marks, S2)

2Critical region for a binomial test (7 marks, S2)

3Poisson hypothesis test (7 marks, S2)

4Hypothesis test using normal approximation (7 marks, S2)

5Type 1 and Type 2 errors — informal (4 marks, S2)

Binomial test statistic

Poisson test statistic

zzz-test for sample mean

Common critical zzz-values

p-value

Null hypothesis H0H_0H0​

Alternative hypothesis H1H_1H1​

Significance level α\alphaα

Test statistic

Critical region (rejection region)

p-value

Type 1 error

Type 2 error

✕Concluding 'reject H0H_0H0​' without contextual interpretation

✕Computing wrong-tail p-value

✕Not splitting α\alphaα in a two-tail test

✕Writing H0H_0H0​ with an inequality

✕Off-by-one in P(X≥xobs)P(X \ge x_{\text{obs}})P(X≥xobs​) for binomial / Poisson

✕Computing z=(xˉ−μ)/σz = (\bar x - \mu)/\sigmaz=(xˉ−μ)/σ instead of z=(xˉ−μ)/(σ/n)z = (\bar x - \mu)/(\sigma/\sqrt n)z=(xˉ−μ)/(σ/n​)

Practice questions

Past paper style quiz

Hypothesis Tests — frequently asked questions

$z$ -test for sample mean

Common critical $z$ -values

Null hypothesis $H_0$

Alternative hypothesis $H_1$

Significance level $\alpha$

✕Concluding 'reject $H_0$ ' without contextual interpretation

✕Not splitting $\alpha$ in a two-tail test

✕Writing $H_0$ with an inequality

✕Off-by-one in $P(X \ge x_{\text{obs}})$ for binomial / Poisson

✕Computing $z = (\bar x - \mu)/\sigma$ instead of $z = (\bar x - \mu)/(\sigma/\sqrt n)$