June 2023 Paper 3 Q14
14 The 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.
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]
Comment on the validity of this claim. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains 1 or 100% | B1 | 1.2 |
| (1) |
Typical solution
1
| Scheme | Marks | AO |
|---|---|---|
| States both hypotheses correctly for two-tailed test | B1 | 2.5 |
| Obtains 26 730 or 26 700 | B1 | 1.1a |
| States or uses correct model PI by normal with mean 24 500 and variance \(\dfrac{5200^2}{24}\) or \(1\,126\,666.\dot{6}\) or standard deviation \(\dfrac{5200}{\sqrt{24}}\) or 1061 or better OE or by correct probability AWFW [0.017, 0.02] or test statistic \((\pm)\dfrac{\text{their } 26\,730 - 24\,500}{5200 \div \sqrt{24}}\) or or test statistic value AWRT \((\pm)\)2.1 or AWRT 22 400 or AWRT 26 600 | M1 | 1.1a |
| Obtains AWFW [0.017, 0.02] or the correct value of the test statistic AWRT 2.1 or acceptance region AWRT \(22\,400 \leqslant \bar{X} \leqslant\) AWRT \(26\,600\) or critical region \(\geqslant\) AWRT 26 600 ignore reference to the lower region allow strict inequalities or critical value AWRT 26 600 ignore reference to the lower value | A1 | 1.1b |
| 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 \(-1.96\) or correctly compares 26 730 or 26 700 with their acceptance region or critical region or correctly compares 26 730 or 26 700 with their upper critical value May be seen on a diagram | M1 | 3.5a |
| Infers \(\mathrm{H}_0\) or null hypothesis rejected All figures must be correct Ignore reference to \(\mathrm{H}_1\) | A1 | 2.2b |
| 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 | R1 | 3.2a |
| (7) |
Typical solution
\[\mathrm{H}_0 : \mu = 24\,500\]\[\mathrm{H}_1 : \mu \neq 24\,500\]\[\bar{X} = 26\,730\]\[\bar{X} \sim \mathrm{N}\left(24500, \frac{5200^2}{24}\right)\]\[\mathrm{P}(\bar{X} \gt 26\,730) = 0.018\]\[0.018 \lt 0.025\]Reject \(\mathrm{H}_0\)
There is sufficient evidence to suggest that the mean daily mass of aluminium cans recycled has changed.
| Scheme | Marks | AO |
|---|---|---|
| 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 | E1 | 3.5b |
| 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 | E1 | 2.2b |
| (2) | ||
| (10 marks) |
Typical solution
Sample mean could be different.
The result could be different so the claim could be wrong.