S5 Statistical hypothesis testing — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in S5, read straight through: 3 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

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.

How this checking works

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.

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.

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.

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.

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.

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.