| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Standard +0.3 This is a straightforward A-level statistics question testing standard normal distribution concepts. Part (a) is trivial (probability of exact value in continuous distribution = 0). Part (b) is a routine hypothesis test with given ฯ, requiring standard z-test calculations. Part (c) asks for basic interpretation of sampling variability. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| 14(a) | Obtains 1 or 100% | 1.2 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| correctly for two-tailed test | 2.5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains 26730 or 26700 | 1.1a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| or AWRT 26 600 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| region | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| May be seen on a diagram | 3.5a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | 2.2b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| the minimum | 3.2a | R1 |
| Subtotal | 7 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 14(c) | Explains that a different sample |
| Answer | Marks | Guidance |
|---|---|---|
| or mean of the new 24 days | 3.5b | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| definite | 2.2b | E1 |
| Subtotal | 2 | |
| Question 14 Total | 10 | |
| Q | Marking instructions | AO |
Question 14:
--- 14(a) ---
14(a) | Obtains 1 or 100% | 1.2 | B1 | 1
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
States both hypotheses
correctly for two-tailed test | 2.5 | B1 | 24 500
24 500
H 0:๐๐ =
H1:๐๐ โ
X = 26 730
52002
X ~N(24500, )
24
P(X > 26 730) = 0.018
0.018 < 0.025
Reject H
0
There is sufficient evidence to
suggest that the mean daily mass
of aluminium cans recycled has
changed.
Obtains 26730 or 26700 | 1.1a | B1
States or uses correct model
PI by normal with mean 24 500
and
52002 ๏ฆ
variance or 1 126 666.6
24
5200
or standard deviation or
24
1061 or better OE
or by correct probability
AWFW [0.017, 0.02]
or test statistic
their 26 730โ24 500
(ยฑ) or
5200รท 24
or test statistic value
AWRT (ยฑ)2.1
or AWRT 22 400
or AWRT 26 600 | 1.1a | M1
Obtains AWFW [0.017, 0.02]
or the correct value of the test
statistic AWRT 2.1
or acceptance region
AWRT22 400 โคX โค AWRT26 600
or critical region โฅ AWRT 26600
ignore reference to the lower
region | 1.1b | A1
allow strict inequalities
or critical value AWRT 26 600
ignore reference to the lower
value
Correctly compares their
probability with 0.025
or correctly compares their
positive test statistic with
AWRT 1.96
or correctly compares their
negative test statistic with
AWRT โ1.96
or correctly compares 26730 or
26700 with their acceptance
region or critical region
or correctly compares 26730 or
26700 with their upper critical
value
May be seen on a diagram | 3.5a | M1
Infers H or null hypothesis
0
rejected
All figures must be correct
Ignore reference to H
1 | 2.2b | A1
Concludes correctly in context
that there is sufficient
evidence to suggest that the
mean daily mass of aluminium
cans recycled has changed.
To be awarded R1, marks
M1A1M1A1 must be scored as
the minimum | 3.2a | R1
Subtotal | 7
Q | Marking instructions | AO | Marks | Typical solution
--- 14(c) ---
14(c) | Explains that a different sample
is likely to produce a different
sample mean
OE
e.g sample mean could be the
same, sample mean could be
different, sample mean will be
different
Must refer to the sample mean
or mean of the new 24 days | 3.5b | E1 | Sample mean could be different.
The result could be different so the
claim could be wrong.
Explains that the result in part
(b) could be different so the
claim is incorrect
OE
e.g the result might be different
so the claim is invalid
the result could be the same so
the claim is invalid
Statement on the result of the
hypothesis test must not be
definite | 2.2b | E1
Subtotal | 2
Question 14 Total | 10
Q | Marking instructions | AO | Marks | Typical solutionThe mass of aluminium cans recycled each day in a city may be modelled by a normal distribution with mean 24 500 kg and standard deviation 5 200 kg.
\begin{enumerate}[label=(\alph*)]
\item State the probability that the mass of aluminium cans recycled on any given day is not equal to 24 500 kg.
[1 mark]
\item To reduce costs, the city's council decides to collect aluminium cans for recycling less frequently.
Following the decision, it was found that over a 24-day period a total mass of 641 520 kg of aluminium cans was recycled.
It can be assumed that the distribution of the mass of aluminium cans recycled is still normal with standard deviation 5 200 kg, and that the 24-day period can be regarded as a random sample.
Investigate, at the 5% level of significance, whether the mean daily mass of aluminium cans recycled has changed.
[7 marks]
\item A member of the council claims that if a different sample of 24 days had been used the hypothesis test in part (b) would have given the same result.
Comment on the validity of this claim.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2023 Q14 [10]}}