Statistical hypothesis testing

Section S5

Both years

3 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9MA0 section S5

Checked against Edexcel 9MA0 section S5. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

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S5.1

Apply the language of hypothesis testing via a binomial model: null/alternative hypothesis, significance level, test statistic, 1- and 2-tail tests, critical value/region, acceptance region, p-value; extend to correlation coefficients.

Notes

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Explanation

The null hypothesis $H_0$ gives the reference parameter value; the alternative $H_1$ states the direction or difference supported by the claim being tested. Use a one-tailed test for a pre-specified directional alternative and a two-tailed test for any change; choose this before observing the data.
The critical region contains outcomes sufficiently unlikely under $H_0$ ; the acceptance region is its complement, and the p-value is the probability under $H_0$ of an outcome at least as extreme as observed.
For a correlation test, use $H_0:\rho=0$ and compare the sample product-moment correlation coefficient with the supplied critical value; significance does not establish causation.
The correlation coefficient satisfies $|r|\leq1$ : $r=1$ or $-1$ means perfect positive or negative linear correlation.
For a supplied calculator test, compare the p-value with the significance level, or $|r|$ with the critical value, in the stated tail.

Worked example

Under $H_0$ , $X\sim\operatorname{B}(20,0.2)$ . For an upper-tailed test at the $5\%$ level, $P(X\geq7)=0.0867$ and $P(X\geq8)=0.0321$ . State the critical region, the critical value and the acceptance region.

1.Choose the smallest upper-tail boundary whose probability under $H_0$ does not exceed $0.05$ .
2.The boundary $7$ is too liberal because $0.0867>0.05$ , while $P(X\geq8)=0.0321<0.05$ .
3.Hence the critical region starts at $8$ and its complement is $X\leq7$ .

Answer: Critical region: $X\geq8$ .; Critical value: $8$ .; Acceptance region: $X\leq7$ .

Common mistakes

Don't place non-extreme outcomes in the critical region while excluding more extreme outcomes from the same tail.
Don't choose a boundary whose tail probability exceeds the significance level, so the critical region is too large.

Exam tip

For a critical region, select the most extreme outcomes whose total null probability does not exceed the stated level.

Tier 1 · Easy

ORIGINAL

A company claims that the probability

p

of a customer choosing its premium plan has increased from

0.40

. State suitable hypotheses and identify the number of tails.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

A hypothesis test produces a p-value of

0.037

. State the conclusion at the

5\%

significance level and at the

1\%

significance level. Explain what the significance level represents.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

For a sample of

18

paired observations, a two-tailed

5\%

correlation test has critical values

-0.468

and

0.468

. The sample product-moment correlation coefficient is

r=-0.520

, with p-value

0.026

. State the hypotheses, carry out the test and interpret the p-value without claiming causation.

(6)

(Total for Question 1 is 6 marks)

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Answer conventions

Follow the wording on the question and its mark scheme. awrt means an appropriately rounded value is accepted; an exact answer must stay as a fraction, surd, logarithm or multiple of π when required, and a rounded decimal may be disallowed. Include requested units and forms. A cso tag protects that accuracy mark, while earlier method marks follow the question-specific dependencies.

S5.2

Conduct a hypothesis test for the proportion in the binomial distribution and interpret results in context; a sample makes an inference about the population; the significance level is the probability of incorrectly rejecting H0.

Notes

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Your confidence:

Explanation

Model the number of sample successes by $X\sim\operatorname{B}(n,p_0)$ under $H_0:p=p_0$ , provided the binomial assumptions are defensible.
Calculate the probability, under $H_0$ , of the observed result or one more extreme in the direction specified by $H_1$ ; double an appropriate tail for a symmetric two-tailed binomial test.
Reject $H_0$ when the p-value is at most the significance level; otherwise say there is insufficient evidence to reject $H_0$ , not that $H_0$ has been proved.
The significance level is the probability of incorrectly rejecting $H_0$ when it is true; discreteness often makes the actual probability of the critical region smaller than the nominal level.

Worked example

A process is tested using $H_0:p=0.5$ against $H_1:p\neq0.5$ . In $20$ independent trials there are $5$ successes. Given $P(X\leq5)=0.02069$ under $X\sim\operatorname{B}(20,0.5)$ , conduct a two-tailed test at the $5\%$ level.

1.The null distribution is symmetric because $p_0=0.5$ .
2.Outcomes at least as extreme as $5$ lie in the two tails, so the p-value is $2P(X\leq5)=2(0.02069)=0.04138$ .
3.This is less than $0.05$ , so reject $H_0$ and infer a difference in the population proportion.

Answer: Two-tailed p-value $=2(0.02069)=0.04138$ .; Reject $H_0$ .; There is sufficient evidence that the population success probability differs from $0.5$ .

Common mistakes

Don't conclude that $H_0$ is true when the result is not significant rather than stating insufficient evidence against it.
Don't double a one-tail probability without checking that the alternative is two-tailed and the opposite tail is symmetric.

Exam tip

State both hypotheses, compare the full p-value with the significance level and phrase the conclusion about evidence, not proof.

Tier 1 · Easy

ORIGINAL

A coin is tested with

H_0:p=0.5

against

H_1:p>0.5

. It lands heads

10

times in

12

tosses. Given

P(X\geq10)=0.0193

for

X\sim\operatorname{B}(12,0.5)

, conduct the test at the

5\%

level.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

A seed supplier claims that the germination probability is

0.8

. In an independent sample of

30

seeds,

19

germinate. Test

H_0:p=0.8

against

H_1:p<0.8

at the

5\%

level, given that

P(X\leq19)=0.0256

for

X\sim\operatorname{B}(30,0.8)

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

A one-tailed test uses

H_0:p=0.2

against

H_1:p>0.2

with a sample of

15

. Under

H_0

P(X\geq6)=0.0611

and

P(X\geq7)=0.0181

. Find the

5\%

critical region. If

6

successes are observed, state the conclusion and explain the actual probability of a Type I error.

(6)

(Total for Question 1 is 6 marks)

S5.3

Conduct a statistical hypothesis test for the mean of a Normal distribution with known, given or assumed variance and interpret the results in context.

Notes

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Your confidence:

Explanation

For a Normal population with known or assumed standard deviation $\sigma$ , the sample mean satisfies $\overline{X}\sim\operatorname{N}(\mu,\sigma^2/n)$ .
Under $H_0:\mu=\mu_0$ , standardise the observed mean using $Z=\frac{\overline{x}-\mu_0}{\sigma/\sqrt n}$ .
Use the tail or tails specified by $H_1$ , compare the p-value with the significance level, and give the conclusion in the language of the population mean.
The method relies on a random, independent sample from a Normal population and uses a known, given or assumed variance rather than estimating it within this specified test.
The standard error is $5/\sqrt{25}=1$ .

Worked example

A Normal population has known standard deviation $5$ . For a sample of $25$ , test $H_0:\mu=50$ against $H_1:\mu\neq50$ at the $1\%$ level. The critical standard Normal values are $\pm2.576$ . Find the critical values of the sample mean and decide what to conclude if $\overline{x}=47.3$ .

1.The standard error is $5/\sqrt{25}=1$ .
2.The acceptance interval is $50\pm2.576(1)$ , namely $47.424\leq\overline{X}\leq52.576$ .
3.Since $47.3$ lies below the lower boundary, it is in the critical region, so reject $H_0$ and conclude that the mean differs from $50$ .

Answer: Critical sample-mean values are approximately $47.424$ and $52.576$ .; Reject $H_0$ because $47.3<47.424$ .; There is sufficient evidence at the $1\%$ level that the population mean differs from $50$ .

Common mistakes

Don't use a two-tailed critical value for a directional alternative hypothesis.
Don't use the population standard deviation directly as the spread of the sample mean instead of dividing by the square root of sample size.

Exam tip

Standardise the sample mean with its standard error, then compare with the correct one- or two-tailed critical values.

Tier 1 · Easy

ORIGINAL

A Normal population has known standard deviation

12

. Test

H_0:\mu=100

against

H_1:\mu>100

using a random sample of

36

with mean

104

, at the

5\%

level.

(5)

(Total for Question 1 is 5 marks)

Tier 2 · Standard

ORIGINAL

A Normal population has known standard deviation

10

. A random sample of

25

has mean

54.4

. Test

H_0:\mu=50

against

H_1:\mu\neq50

at the

2\%

significance level. Use the critical values

z=\pm2.326

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

A Normal population has known standard deviation

6

. To test

H_0:\mu=80

against

H_1:\mu>80

at the

5\%

level, find the smallest sample size

n

for which an observed mean of

82

would lead to rejection. Use the critical value

1.645

, then find the p-value for this minimum

n

(7)

(Total for Question 1 is 7 marks)

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