| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Critical region determination |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring a z-test with known population standard deviation, calculation of critical values, and interpretation of test results. The calculations are routine (z = (127-123)/(70/√12144) ≈ 6.3), and the interpretation questions in part (b) test basic understanding of what hypothesis tests can and cannot conclude. While multi-part, each component is standard A-level statistics with no novel problem-solving required. |
| Spec | 2.05e Hypothesis test for normal mean: known variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| 14(a)(i) | States both hypotheses using | |
| correct language | AO2.5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Finds test statistic | AO1.1a | M1 |
| Obtains correct test statistic | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| of ts with cv | AO2.2b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| including ‘some evidence’ | AO3.2a | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| ALT | States both hypotheses using | |
| correct language | AO2.5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Attempts to find p value for | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3.045×10−10 < 0.025 | From calculator, | |
| z-test | P(X < 127 )=3.045×10−10 | |
| Finds correct p value | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | AO2.2b | AO2.2b |
| Answer | Marks | Guidance |
|---|---|---|
| including ‘some evidence’ | Concludes correctly in context, | AO3.2a |
| Answer | Marks |
|---|---|
| including ‘some evidence’ | 5% level) to suggest the mean |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(ii) | Uses Normal model to find | |
| critical values PI | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Disallow integer answers | AO1.1b | A1 |
| (b)(i) | States valid reason for | |
| statement 1 ‘not supported’ | AO2.4 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| would not imply the statement | AO2.2b | R1 |
| (b)(ii) | States valid reason for | |
| statement 2 ‘not supported’ | AO2.4 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| would not imply the statement | AO2.2b | R1 |
| Total | 11 |
| Answer | Marks | Guidance |
|---|---|---|
| Q | Marking Instructions | AO |
Question 14:
--- 14(a)(i) ---
14(a)(i) | States both hypotheses using
correct language | AO2.5 | B1 | H :µ=123
0
H :µ≠123
1
127−123
Test statistic =
70
12144
= 6.30
Critical z values ±1.96
6.30 > 1.96
Finds test statistic | AO1.1a | M1
Obtains correct test statistic | AO1.1b | A1
Infers H rejected by comparison
0
of ts with cv | AO2.2b | A1
Concludes correctly in context,
including ‘some evidence’ | AO3.2a | E1 | Reject H there is evidence (at the
0
5% level) to suggest the mean
expenditure on bread had changed
from 2012 to 2013
ALT | States both hypotheses using
correct language | AO2.5 | B1 | H :µ=123
0
H :µ≠123
0
Attempts to find p value for | AO1.1a | M1 | From calculator,
P(X < 127 )=3.045×10−10
3.045×10−10 < 0.025 | From calculator,
z-test | P(X < 127 )=3.045×10−10
Finds correct p value | AO1.1b | A1 | 3.045×10−10 < 0.025
Infers H rejected by comparison
0 | AO2.2b | AO2.2b | A1 | A1
of p with 0.025
Concludes correctly in context,
including ‘some evidence’ | Concludes correctly in context, | AO3.2a | E1 | Reject H - there is evidence (at the
0
including ‘some evidence’ | 5% level) to suggest the mean
expenditure on bread had changed
from 2012 to 2013
(a)(ii) | Uses Normal model to find
critical values PI | AO3.4 | M1 | 70
123±1.96×
12144
min=121.75 and max=124.25
Obtains correct critical values
Correct accuracy required for
this mark
Disallow integer answers | AO1.1b | A1
(b)(i) | States valid reason for
statement 1 ‘not supported’ | AO2.4 | R1 | The conclusion implies that the
mean changed, not that it
increased by a specific amount, so
the statement is not supported
Infers that model/test used
would not imply the statement | AO2.2b | R1
(b)(ii) | States valid reason for
statement 2 ‘not supported’ | AO2.4 | R1 | The conclusion implies that there is
evidence that the mean has
changed, but expenditure increase
may be due to price changes, so
statement is not supported
Infers that model/test used
would not imply the statement | AO2.2b | R1
Total | 11
States both hypotheses using
correct language
Q | Marking Instructions | AO | Marks | Typical SolutionA survey during 2013 investigated mean expenditure on bread and on alcohol.
The 2013 survey obtained information from 12 144 adults.
The survey revealed that the mean expenditure per adult per week on bread was 127p.
\begin{enumerate}[label=(\alph*)]
\item For 2012, it is known that the expenditure per adult per week on bread had mean 123p, and a standard deviation of 70p.
\begin{enumerate}[label=(\roman*)]
\item Carry out a hypothesis test, at the 5% significance level, to investigate whether the mean expenditure per adult per week on bread changed from 2012 to 2013.
Assume that the survey data is a random sample taken from a normal distribution. [5 marks]
\item Calculate the greatest and least values for the sample mean expenditure on bread per adult per week for 2013 that would have resulted in acceptance of the null hypothesis for the test you carried out in part (a)(i).
Give your answers to two decimal places. [2 marks]
\end{enumerate}
\item The 2013 survey revealed that the mean expenditure per adult, per week on alcohol was 324p.
The mean expenditure per adult per week on alcohol for 2009 was 307p.
A test was carried out on the following hypotheses relating to mean expenditure per adult per week on alcohol in 2013.
$H_0 : \mu = 307$
$H_1 : \mu \neq 307$
This test resulted in the null hypothesis, $H_0$, being rejected.
State, with a reason, whether the test result supports the following statements:
\begin{enumerate}[label=(\roman*)]
\item the mean UK expenditure on alcohol per adult per week increased by 17p from 2009 to 2013; [2 marks]
\item the mean UK consumption of alcohol per adult per week changed from 2009 to 2013. [2 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 Q14 [11]}}