S5 Statistical hypothesis testing — revision question pack

3 specification points · notes, questions, answers and worked methods

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.

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

1.
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)
2.
For a test based on a binomial count $X$ , where $X$ is the number of successes, the critical region is $X\leq2$ or $X\geq14$ . State whether the test is one-tailed or two-tailed, and state the acceptance region.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
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)
2.
Under $H_0$ , $X\sim\operatorname{B}(18,0.45)$ . For a lower-tailed test at the $5\%$ level, $P(X\leq4)=0.0411$ and $P(X\leq5)=0.1077$ . State the critical region, critical value, acceptance region and actual significance level.
(4)
(Total for Question 2 is 4 marks)
3.
Under $H_0$ , the number $X$ of successes in a sample of $20$ has distribution $\operatorname{B}(20,0.5)$ . For a two-tailed test at the $5\%$ level, $P(X\leq5)=0.0207$ and $P(X\leq6)=0.0577$ . State the critical region, acceptance region and actual significance level, using equal tail allocations.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
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)
2.
For a particular sample size, the $5\%$ critical value for a one-tailed positive-correlation test is $0.497$ , while the two-tailed $5\%$ critical values are $\pm0.576$ . A pre-registered test of $H_0:\rho=0$ against $H_1:\rho>0$ gives $r=0.532$ . Carry out this test. A second researcher chose a positive alternative only after seeing the scatter diagram; explain how this changes the appropriate conclusion.
(5)
(Total for Question 2 is 5 marks)
3.
Under $H_0:p=0.30$ , the number $X$ of successes in a sample of $12$ has distribution $\operatorname{B}(12,0.30)$ . A two-tailed test of $H_0:p=0.30$ against $H_1:p\neq0.30$ is to have no more than $2.5\%$ in either tail. The following probabilities are rounded to $4$ decimal places: $P(X\leq0)=0.0138$ , $P(X\leq1)=0.0850$ , $P(X\geq7)=0.0386$ and $P(X\geq8)=0.0095$ . Find the critical region and actual significance level. Conduct the test when $X=8$ .
(6)
(Total for Question 3 is 6 marks)
4.
Two independent studies both obtain a sample product-moment correlation coefficient $r=0.460$ . Study A has sample size $18$ and two-tailed $5\%$ critical values $\pm0.468$ . Study B has sample size $40$ and two-tailed $5\%$ critical values $\pm0.312$ . For each study, test $H_0:\rho=0$ against $H_1:\rho\neq0$ . Explain why the decisions differ and state what neither result establishes.
(5)
(Total for Question 4 is 5 marks)
5.
Under $H_0:p=0.25$ , the number $X$ of successes in a sample of $16$ has distribution $\operatorname{B}(16,0.25)$ . For an upper-tailed test, $P(X\geq7)=0.07956$ , $P(X\geq8)=0.02713$ and $P(X\geq9)=0.00747$ . Find the critical region and actual significance level for a nominal $5\%$ test. Conduct the test when $X=7$ . State how the critical region and conclusion change at the $10\%$ level.
(6)
(Total for Question 5 is 6 marks)

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.

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

1.
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)
2.
A call centre claims that the proportion $p$ of callers who abandon a queue has fallen from $0.12$ . State suitable null and alternative hypotheses and identify the relevant tail.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
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)
2.
A service has historically completed a request successfully with probability $0.25$ . After a change, only $1$ of $18$ independent requests succeeds. Test $H_0:p=0.25$ against $H_1:p<0.25$ at the $5\%$ level. Give the p-value to $4$ decimal places.
(5)
(Total for Question 2 is 5 marks)
3.
A delivery service claims that the proportion $p$ of orders arriving on time is $0.40$ . In a random sample of $40$ orders, let $X$ be the number arriving on time. A lower-tailed $5\%$ test uses the critical region $X\leq10$ , whose probability under $H_0$ is $0.0352$ . State the hypotheses and conduct the test when $10$ orders arrive on time. State the actual significance level.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
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)
2.
A voice assistant is claimed to recognise a spoken command with probability $0.65$ . A researcher tests $H_0:p=0.65$ against $H_1:p<0.65$ using $25$ attempts and observes $11$ recognitions. Calculate the p-value and conduct the test at the $5\%$ level. The attempts consisted of five commands from each of five speakers; explain why this sampling detail may weaken the conclusion. Give the p-value to $4$ decimal places.
(6)
(Total for Question 2 is 6 marks)
3.
A provider claims that the population proportion $p$ of successful installations is $0.60$ . One random sample contains $8$ successes from $20$ installations and a second independent random sample contains $15$ successes from $30$ installations. Assuming both samples concern the same population probability, test $H_0:p=0.60$ against $H_1:p<0.60$ at the $5\%$ level. You may use $P(X\leq23)=0.03141$ for $X\sim\operatorname{B}(50,0.60)$ . Explain why pooling the samples would be invalid if they came from customer groups with different success probabilities.
(6)
(Total for Question 3 is 6 marks)
4.
A platform claims that the population probability $p$ of a user completing a task is $0.40$ . In a random sample of $25$ independent users, $16$ complete the task. Test $H_0:p=0.40$ against $H_1:p>0.40$ at the $5\%$ level. You may use $P(X=16)=0.008843$ and $P(X\geq16)=0.013169$ for $X\sim\operatorname{B}(25,0.40)$ . State which probability is the p-value and explain why.
(5)
(Total for Question 4 is 5 marks)
5.
A provider claims that the population success probability is $0.75$ . A fixed random sample of $60$ independent cases is used to test $H_0:p=0.75$ against $H_1:p<0.75$ at the $5\%$ level. The preliminary record shows $39$ successes, but a source check confirms that one of these cases was misclassified and the verified number of successes is $38$ . Calculate the exact lower-tail p-value using each count, giving both values to $4$ decimal places. Conduct the test using the verified data and explain how the confirmed correction changes the conclusion. The sample size remains $60$ throughout.
(7)
(Total for Question 5 is 7 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.

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

1.
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)
2.
A Normal population has known standard deviation $4$ . A random sample of $25$ has mean $51.2$ . Find the test statistic for testing $H_0:\mu=50$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
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)
2.
A Normal population has known standard deviation $2.4$ . A random sample of $16$ has mean $73.6$ . Test $H_0:\mu=75$ against $H_1:\mu\neq75$ at the $5\%$ level. Give the test statistic to $3$ significant figures and the p-value to $4$ decimal places.
(5)
(Total for Question 2 is 5 marks)
3.
A Normal population has known standard deviation $8$ . A random sample of $64$ is used to test $H_0:\mu=100$ against $H_1:\mu>100$ at the $1\%$ level, using critical value $z=2.326$ . The recorded sample mean is $102.4$ , correct to the nearest $0.1$ . Find the critical value of the sample mean and determine whether the test conclusion is unaffected by the rounding.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
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)
2.
Brightness readings from a display are modelled as independent Normal variables with known standard deviation $4$ units. The display is tested using $H_0:\mu=50$ against $H_1:\mu<50$ at the $1\%$ level with a sample of $36$ readings. Using the critical value $-2.326$ , find the critical value of the sample mean and the corresponding critical value of the total of the readings, giving each to $3$ decimal places. The observed total is $1740$ units. Conduct the test and find the p-value to $4$ decimal places.
(6)
(Total for Question 2 is 6 marks)
3.
A random sample of $25$ from a Normal population has mean $21.8$ . For the test $H_0:\mu=20$ against $H_1:\mu>20$ at the $5\%$ level, the population standard deviation $\sigma$ is assumed known and the critical value is $1.645$ . Find the greatest value of $\sigma$ for which the result is significant, giving your answer to $3$ significant figures. Hence state the conclusion when $\sigma=5.0$ and when $\sigma=5.5$ . Use unrounded values in your working.
(6)
(Total for Question 3 is 6 marks)
4.
A random sample of $49$ observations is taken from a Normal population with known standard deviation $14$ . For each observation define $y=(x-100)/2$ , and the sample satisfies $\sum y=112$ . Test $H_0:\mu=100$ against $H_1:\mu>100$ . Calculate the sample mean, test statistic and p-value. State the conclusions at the $5\%$ and $1\%$ significance levels. Give the p-value to $4$ decimal places.
(6)
(Total for Question 4 is 6 marks)
5.
A random sample of $36$ observations from a Normal population with known standard deviation $6$ is used to test $H_0:\mu=48$ against $H_1:\mu>48$ at the $1\%$ level. The recorded total is $1836$ , but one recorded value $74$ is confirmed to be an error for $47$ . Calculate the test statistic before and after correction. Conduct the test using the corrected data and find its p-value to $4$ decimal places. Explain how the error would have changed the conclusion.
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

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.

Tier 1 · Easy

Mark scheme for S5.1 Tier 1 · Easy
Question	Scheme	Marks
1	$H_0:p=0.40$ . $H_1:p>0.40$ . This is a one-tailed test.	2
(2 marks)		2
Notes
The null uses the established value $0.40$ . The word 'increased' gives the directional alternative $p>0.40$ , so only the upper tail is relevant.
2	Two-tailed Acceptance region: $3\leq X\leq13$	2
(2 marks)		2
Notes
There is a critical region in each tail, so the test is two-tailed. The integer values not in either critical region run from $3$ to $13$ inclusive.

Tier 2 · Standard

Mark scheme for S5.1 Tier 2 · Standard
Question	Scheme	Marks
1	Reject $H_0$ at the $5\%$ level. Do not reject $H_0$ at the $1\%$ level. The significance level is the probability, under $H_0$ , of rejecting $H_0$ when it is true.	4
(4 marks)		4
Notes
Compare the p-value with each significance level. Since $0.037<0.05$ , reject $H_0$ at $5\%$ . Since $0.037>0.01$ , the evidence is not sufficient to reject $H_0$ at $1\%$ . The significance level controls the probability of a Type I error: rejecting a true null hypothesis.
2	Critical region: $X\leq4$ Critical value: $4$ Acceptance region: $X\geq5$ Actual significance level: $0.0411$ (or $4.11\%$ )	4
(4 marks)		4
Notes
The boundary $5$ would make the lower-tail probability exceed $0.05$ , while the probability up to $4$ is below $0.05$ . Hence the critical region is $X\leq4$ , its largest value is the critical value, and the complement is $X\geq5$ . The null probability of the chosen region is $0.0411$ .
3	Critical region: $X\leq5$ or $X\geq15$ Acceptance region: $6\leq X\leq14$ Actual significance level $=2(0.0207)=0.0414$ , or $4.14\%$	4
(4 marks)		4
Notes
Each tail may contain at most $0.025$ . The boundary $6$ is too large because its lower-tail probability is $0.0577$ , while $X\leq5$ is admissible. Symmetry about $10$ makes the corresponding upper region $X\geq15$ . The complement is $6\leq X\leq14$ , and the total null probability of the two critical regions is $2(0.0207)=0.0414$ .

Tier 3 · Hard

Mark scheme for S5.1 Tier 3 · Hard
Question	Scheme	Marks
1	$H_0:\rho=0$ and $H_1:\rho\neq0$ . Reject $H_0$ because $-0.520<-0.468$ (equivalently $0.026<0.05$ ). There is sufficient evidence of negative correlation in the population. If $\rho=0$ , the probability of a sample correlation at least this extreme in either direction is $0.026$ ; this does not prove causation.	6
(6 marks)		6
Notes
A two-tailed association test uses $H_0:\rho=0$ against $H_1:\rho\neq0$ . The observed coefficient lies in the lower critical region, and its p-value is below $0.05$ , so reject $H_0$ . State the conclusion as evidence of population correlation, not as proof that either variable causes the other.
2	Reject $H_0$ for the pre-registered one-tailed test because $0.532>0.497$ ; there is sufficient evidence of positive population correlation. Choosing the direction after seeing the data is not a valid pre-specified one-tailed test. The second researcher should use the two-tailed comparison, for which $\|0.532\|<0.576$ , so there is insufficient evidence of population correlation at the $5\%$ level. Neither test establishes causation.	5
(5 marks)		5
Notes
The planned directional alternative places the rejection region only in the positive tail, and $r$ exceeds its supplied critical value. A direction selected after inspection gives an unfair second opportunity to choose the favourable tail, so the non-directional two-tailed threshold is appropriate. The observed magnitude does not reach that threshold.
3	Critical region: $X=0$ or $X\geq8$ Actual significance level $=0.0138+0.0095=0.0233$ , or approximately $2.33\%$ . Since $8$ is in the critical region, reject $H_0$ . There is sufficient evidence that the population success probability differs from $0.30$ .	6
(6 marks)		6
Notes
In the lower tail, adding $X=1$ would raise the probability above $0.025$ , so only $X=0$ is included. In the upper tail, $X\geq7$ is too large but $X\geq8$ is admissible. Using the supplied probabilities, the regions have total null probability $0.0138+0.0095=0.0233$ . The observation $8$ lies on the upper critical boundary, so reject the null and state the conclusion about the population parameter $p$ .
4	Study A: do not reject $H_0$ because $\|0.460\|<0.468$ ; there is insufficient evidence of population correlation. Study B: reject $H_0$ because $\|0.460\|>0.312$ ; there is sufficient evidence of a correlation. The larger sample has a smaller critical magnitude, so the same sample correlation supplies stronger evidence against $H_0$ . Neither result establishes that one variable causes the other.	5
(5 marks)		5
Notes
Use the population parameter $\rho$ in the hypotheses and compare $\|r\|$ with the supplied two-tailed critical magnitude. The coefficient misses A's boundary by $0.008$ but exceeds B's by $0.148$ . Increased sample size reduces the magnitude needed for significance, while a correlation test still addresses association rather than causation.
5	At the $5\%$ level, the critical region is $X\geq8$ and the actual significance level is $0.02713$ . With $X=7$ , do not reject $H_0$ ; there is insufficient evidence that the population success probability exceeds $0.25$ . At the $10\%$ level, the critical region is $X\geq7$ and its actual significance level is $0.07956$ . The observation $X=7$ is then in the critical region, so reject $H_0$ and conclude that there is sufficient evidence that $p>0.25$ .	6
(6 marks)		6
Notes
For $5\%$ , including $7$ would make the null tail probability exceed $0.05$ , while $X\geq8$ is admissible. At $10\%$ , $X\geq7$ is admissible and is larger than the region starting at $8$ , so it is used. The observed boundary value is excluded from the first critical region but included in the second, producing different evidence statements at the two pre-specified levels.

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.

Tier 1 · Easy

Mark scheme for S5.2 Tier 1 · Easy
Question	Scheme	Marks
1	Reject $H_0$ because $0.0193<0.05$ . There is sufficient evidence that the probability of heads is greater than $0.5$ .	4
(4 marks)		4
Notes
Under $H_0$ , the number of heads is $X\sim\operatorname{B}(12,0.5)$ . The upper-tail p-value for $10$ observed heads is $0.0193$ . Since this is below $0.05$ , reject $H_0$ and give the directional conclusion in context.
2	$H_0:p=0.12$ $H_1:p<0.12$ Lower-tailed test	3
(3 marks)		3
Notes
The null hypothesis uses the established proportion. The word 'fallen' specifies values below $0.12$ , so the alternative is directional and the lower tail is used.

Tier 2 · Standard

Mark scheme for S5.2 Tier 2 · Standard
Question	Scheme	Marks
1	p-value $=0.0256$ Reject $H_0$ . There is sufficient evidence that the germination probability is less than $0.8$ .	5
(5 marks)		5
Notes
Under $H_0$ , $X\sim\operatorname{B}(30,0.8)$ . The alternative is lower-tailed, so the p-value for observing $19$ germinations is $P(X\leq19)=0.0256$ . Since $0.0256<0.05$ , reject $H_0$ . The sample provides sufficient evidence, at the $5\%$ level, that the supplier's germination probability is below $0.8$ .
2	Reject $H_0$ because the p-value is $P(X\leq1)=0.0395<0.05$ . There is sufficient evidence at the $5\%$ level that the success probability has decreased from $0.25$ .	5
(5 marks)		5
Notes
Under $H_0$ , $X\sim\operatorname{B}(18,0.25)$ . The lower-tailed p-value is $P(X\leq1)=0.75^{18}+18(0.25)(0.75)^{17}=0.0394639\ldots$ . Since this is less than $0.05$ , reject $H_0$ and make the population-proportion inference in context.
3	$H_0:p=0.40$ and $H_1:p<0.40$ Reject $H_0$ because $10$ lies in the critical region $X\leq10$ . There is sufficient evidence at the $5\%$ level that the population proportion arriving on time is below $0.40$ . Actual significance level $=0.0352$ , or $3.52\%$ .	5
(5 marks)		5
Notes
The claim is tested against a decrease, so the alternative is lower-tailed. Under $H_0$ , $X\sim\operatorname{B}(40,0.40)$ . The observed count equals the critical value and is therefore inside the rejection region. Reject $H_0$ and make the inference about the population proportion. The null probability of that region is the actual significance level $0.0352$ .

Tier 3 · Hard

Mark scheme for S5.2 Tier 3 · Hard
Question	Scheme	Marks
1	Critical region: $X\geq7$ . With $X=6$ , do not reject $H_0$ ; there is insufficient evidence that $p>0.2$ . The actual probability of rejecting a true $H_0$ is $P(X\geq7)=0.0181$ , which is below the nominal $5\%$ level.	6
(6 marks)		6
Notes
A critical region must have probability at most $0.05$ under $H_0$ . Since $P(X\geq6)=0.0611$ is too large but $P(X\geq7)=0.0181$ is acceptable, use $X\geq7$ . The observed value $6$ is outside this region, so there is insufficient evidence to reject $H_0$ . Because the binomial distribution is discrete, the attainable Type I error probability is $0.0181$ , not exactly $0.05$ .
2	Reject $H_0$ because the p-value is $P(X\leq11)=0.0255<0.05$ ; the binomial calculation gives sufficient evidence that the recognition probability is below $0.65$ . Attempts by the same speaker may share voice characteristics or recording conditions, so their outcomes may not be independent. The effective amount of independent information may be less than $25$ attempts, so the binomial p-value and conclusion may be unreliable.	6
(6 marks)		6
Notes
Assuming the binomial model, $X\sim\operatorname{B}(25,0.65)$ under $H_0$ . The lower-tail probability is $P(X\leq11)=0.0254606\ldots$ , so it is below $0.05$ and the model-based decision is to reject. However, repeated attempts from one speaker can be correlated, violating the independence assumption used to calculate that p-value.
3	The pooled count is $23$ successes from $50$ installations, so the p-value is $0.03141$ . Reject $H_0$ because $0.03141<0.05$ . There is sufficient evidence that the common population success probability is below $0.60$ . Pooling requires one constant success probability across all $50$ independent trials. If the customer groups have different probabilities, the pooled count is not binomial with one parameter, and the p-value and conclusion need not be valid; analyse the groups separately.	6
(6 marks)		6
Notes
Under the stated common-probability assumption, independent binomial counts combine: the number of successes is $X\sim\operatorname{B}(20+30,0.60)$ and the observed total is $8+15=23$ . The lower-tail p-value $0.03141$ is below $0.05$ , so reject. If group membership changes $p$ , the constant-probability condition fails even though each result is still success or failure, so a single pooled binomial test conceals the group structure.
4	The p-value is $P(X\geq16)=0.0132$ to $4$ decimal places, not $P(X=16)$ . The upper-tailed p-value includes the observed result and every result more extreme in the direction of $H_1$ . Reject $H_0$ because $0.013169<0.05$ . There is sufficient evidence at the $5\%$ level that the population completion probability exceeds $0.40$ .	5
(5 marks)		5
Notes
Under $H_0$ , $X\sim\operatorname{B}(25,0.40)$ . The directional alternative makes large counts evidence against the null, so the tail from $16$ through $25$ is required. Its probability is $0.013169$ , whereas the single-point probability omits more extreme outcomes. Comparison with $0.05$ gives rejection and the conclusion is stated about the population parameter.
5	For the preliminary count, $P(X\leq39)=0.0541$ to $4$ decimal places, so it would not lead to rejection at the $5\%$ level. For the verified count, $P(X\leq38)=0.0298$ to $4$ decimal places. Reject $H_0$ because $0.0298<0.05$ ; there is sufficient evidence that the population success probability is below $0.75$ . The confirmed correction changes $X$ from $39$ to $38$ without changing $n=60$ , moving the exact lower-tail p-value from above to below the significance level and reversing the decision.	7
(7 marks)		7
Notes
Under $H_0$ , the fixed sample count is $X\sim\operatorname{B}(60,0.75)$ . Since the alternative is lower-tailed, include the observed count and every smaller count. Thus $P(X\leq39)=\sum_{r=0}^{39}\binom{60}{r}(0.75)^r(0.25)^{60-r}=0.0541439\ldots$ , whereas $P(X\leq38)=0.0298009\ldots$ . Only the verified count gives a p-value below $0.05$ , so the evidence statement is about the population parameter $p$ using the corrected data.

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.

Tier 1 · Easy

Mark scheme for S5.3 Tier 1 · Easy
Question	Scheme	Marks
1	Test statistic $z=2.00$ and p-value $0.0228$ . Reject $H_0$ ; there is sufficient evidence that the population mean exceeds $100$ .	5
(5 marks)		5
Notes
Under $H_0$ , $\overline{X}\sim\operatorname{N}(100,12^2/36)$ , so the standard error is $12/6=2$ . Thus $z=(104-100)/2=2.00$ . The upper-tail p-value is $P(Z\geq2)=0.0228<0.05$ , so reject $H_0$ and state the conclusion about the population mean.
2	$z=1.50$	2
(2 marks)		2
Notes
The standard error is $4/\sqrt{25}=0.8$ . Hence $z=(51.2-50)/0.8=1.50$ .

Tier 2 · Standard

Mark scheme for S5.3 Tier 2 · Standard
Question	Scheme	Marks
1	$z=2.20$ . Do not reject $H_0$ . There is insufficient evidence at the $2\%$ level that the population mean differs from $50$ .	5
(5 marks)		5
Notes
The standard error is $10/\sqrt{25}=2$ . Hence $z=(54.4-50)/2=2.20$ . This lies between the two critical values $-2.326$ and $2.326$ , so it is not in the critical region. Do not reject $H_0$ ; there is insufficient evidence at the $2\%$ level that the population mean differs from $50$ .
2	Reject $H_0$ because the two-tailed p-value is $0.0196<0.05$ . $z=-2.33$ . There is sufficient evidence at the $5\%$ level that the population mean differs from $75$ .	5
(5 marks)		5
Notes
The standard error is $2.4/\sqrt{16}=0.6$ , so $z=(73.6-75)/0.6=-2.333\ldots$ . The two-tailed p-value is $2P(Z\leq-2.333\ldots)=0.019630\ldots$ . Since this is below $0.05$ , reject $H_0$ and state the conclusion about the population mean.
3	Critical sample mean $=100+2.326(8/\sqrt{64})=102.326$ . The unrounded sample mean lies in $102.35\leq\bar{x}<102.45$ . Every value in this interval exceeds $102.326$ , so reject $H_0$ regardless of the rounding. There is sufficient evidence at the $1\%$ level that the population mean exceeds $100$ .	5
(5 marks)		5
Notes
The standard error is $8/\sqrt{64}=1$ , so the upper critical boundary is $100+2.326=102.326$ . A value recorded as $102.4$ to the nearest tenth represents the half-open interval from $102.35$ to $102.45$ . Its lower endpoint already lies above the critical boundary, making the rejection decision robust to every possible unrounded value.

Tier 3 · Hard

Mark scheme for S5.3 Tier 3 · Hard
Question	Scheme	Marks
1	Smallest sample size $n=25$ . For $n=25$ , p-value $=0.0478$ .	7
(7 marks)		7
Notes
Rejection requires $\frac{82-80}{6/\sqrt n}\geq1.645$ . Hence $\sqrt n\geq\frac{1.645(6)}{2}=4.935$ , so $n\geq24.354\ldots$ and the smallest integer is $25$ . For $n=25$ , $z=2/(6/5)=1.6667$ , giving the upper-tail p-value $1-\Phi(1.6667)=0.0478$ .
2	Reject $H_0$ because $1740<1744.176$ (equivalently $\bar{x}=48.333\ldots<48.449\ldots$ ). Critical sample mean $=48.449$ units and critical total $=1744.176$ units. p-value $=0.0062$ . There is sufficient evidence at the $1\%$ level that the population mean brightness reading is below $50$ units.	6
(6 marks)		6
Notes
The standard error is $4/\sqrt{36}=2/3$ units. The lower critical mean is $50-2.326(2/3)=48.4493\ldots$ units, corresponding to total $36(48.4493\ldots)=1744.176$ units. The observed mean is $1740/36=48.3333\ldots$ units, which is in the critical region. Its test statistic is $(48.3333\ldots-50)/(2/3)=-2.5$ , so the lower-tail p-value is $P(Z\leq-2.5)=0.006209\ldots=0.0062$ .
3	Significance requires $\sigma\leq5.47112\ldots$ , so the greatest value is $5.47$ to $3$ significant figures. For $\sigma=5.0$ , $z=1.80>1.645$ : reject $H_0$ ; there is sufficient evidence that $\mu>20$ . For $\sigma=5.5$ , $z=1.636\ldots<1.645$ : do not reject $H_0$ ; there is insufficient evidence that $\mu>20$ .	6
(6 marks)		6
Notes
The test statistic is $(21.8-20)/(\sigma/\sqrt{25})=9/\sigma$ . Rejection requires $9/\sigma\geq1.645$ , so $\sigma\leq9/1.645=5.47112\ldots$ . At $\sigma=5.0$ , the statistic is $1.8$ and lies in the critical region. At $\sigma=5.5$ , it is $1.63636\ldots$ and lies outside, so the correct non-significant conclusion uses insufficient-evidence language.
4	$\bar{x}=104.571\ldots$ and $z=2.2857\ldots$ Upper-tail p-value $=0.0111$ . At the $5\%$ level, reject $H_0$ ; there is sufficient evidence that the population mean exceeds $100$ . At the $1\%$ level, do not reject $H_0$ ; there is insufficient evidence that the population mean exceeds $100$ .	6
(6 marks)		6
Notes
Since $x=2y+100$ , $\sum x=2(112)+49(100)=5124$ and $\bar{x}=5124/49=104.5714\ldots$ . The standard error is $14/\sqrt{49}=2$ , so $z=(104.5714\ldots-100)/2=2.285714\ldots$ . The upper-tail probability is $0.011135\ldots=0.0111$ , which lies below $0.05$ but above $0.01$ .
5	Before correction, $\bar{x}=51$ and $z=3.00$ . The corrected total is $1809$ , so $\bar{x}=50.25$ and $z=2.25$ . The corrected upper-tail p-value is $0.0122$ . Do not reject $H_0$ at the $1\%$ level; there is insufficient evidence that the population mean exceeds $48$ . The erroneous value would have produced rejection because $z=3.00$ has p-value below $0.01$ ; the data error therefore reverses the decision.	7
(7 marks)		7
Notes
The standard error is $6/\sqrt{36}=1$ . Before correction the mean is $1836/36=51$ , giving $z=(51-48)/1=3$ . Correct the total by $1836-74+47=1809$ , so the mean is $50.25$ and $z=2.25$ . The upper-tail probability is $1-\Phi(2.25)=0.012224\ldots=0.0122$ , above $0.01$ , so the corrected result is not significant at the stated level.