| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (lower tail) |
| Difficulty | Moderate -0.3 This is a straightforward hypothesis testing question with standard setup. Part (a) requires basic knowledge of sampling terminology (opportunity/convenience sampling) and understanding of bias. Part (b) is a routine one-tailed z-test with given parameters—students simply need to calculate the test statistic and compare to critical value. The question is slightly below average difficulty because it provides all necessary information explicitly, requires no problem-solving insight, and follows a standard textbook template, though it does require careful execution across multiple steps. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| 18(a)(i) | States opportunistic (sampling). | |
| Accept opportunity/convenience. | AO1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 18(a)(ii) | Explains that sample is not | |
| random. | AO3.5b | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| 18(b) | States both hypotheses correctly | |
| for one-tailed test | AO2.5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Formulates the test statistic | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| test statistic | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE | AO1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | AO2.2b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| scored). | AO3.2a | E1F |
| Total | 8 | |
| TOTAL | 100 |
Question 18:
--- 18(a)(i) ---
18(a)(i) | States opportunistic (sampling).
Accept opportunity/convenience. | AO1.2 | B1 | Opportunistic sampling
--- 18(a)(ii) ---
18(a)(ii) | Explains that sample is not
random. | AO3.5b | E1 | The sample is not random.
--- 18(b) ---
18(b) | States both hypotheses correctly
for one-tailed test | AO2.5 | B1 | H : μ = 66.5
0
H : μ < 66.5
1
65.4−66.5
𝑧𝑧 =
21.2
�
√750
= −1.42
Critical value = -1.28
z
-1.42 < -1.28
Reject H - there is sufficient
0
evidence that the advertising
campaign has reduced the
consumption of chocolate.
Formulates the test statistic | AO1.1a | M1
Obtains the correct value of the
test statistic | AO1.1b | A1
States the correct critical z-value
OE | AO1.1b | B1
Infers H rejected CSO
0 | AO2.2b | A1
Correctly concludes in context.
(FT only available if first B1 and M1
scored). | AO3.2a | E1F
Total | 8
TOTAL | 100In a region of England, the government decides to use an advertising campaign to encourage people to eat more healthily.
Before the campaign, the mean consumption of chocolate per person per week was known to be 66.5g, with a standard deviation of 21.2g
\begin{enumerate}[label=(\alph*)]
\item After the campaign, the first 750 available people from this region were surveyed to find out their average consumption of chocolate.
\begin{enumerate}[label=(\roman*)]
\item State the sampling method used to collect the survey.
[1 mark]
\item Explain why this sample should not be used to conduct a hypothesis test.
[1 mark]
\end{enumerate}
\item A second sample of 750 people revealed that the mean consumption of chocolate per person per week was 65.4g
Investigate, at the 10% level of significance, whether the advertising campaign has decreased the mean consumption of chocolate per person per week.
Assume that an appropriate sampling method was used and that the consumption of chocolate is normally distributed with an unchanged standard deviation.
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2018 Q18 [8]}}