June 2023 Paper 2 Q10
10 The mass, in kilograms, of a species of fish in the UK has population mean 4.2 and standard deviation 0.25.
An environmentalist believes that the fish in a particular river are smaller, on average, than those in other rivers in the UK.
A random sample of 100 fish of this species, taken from the river, has sample mean 4.16 kg.
Stating a necessary assumption, test at the 5% significance level whether the environmentalist is correct. [8]
| Scheme | Marks | AO |
|---|---|---|
| Assumption: sd for this river is 0.25 | B1 | 3.3 |
| Allow 2 sf throughout \(\mathrm{H}_0: \mu = 4.2;\ \mathrm{H}_1: \mu \lt 4.2\) where \(\mu\) = (population) mean (mass) (of this river) | B1B1 | 1.1 2.5 |
| \(\mathrm{N}\left(4.2, \dfrac{0.25^2}{100}\right)\) & \(\bar{X} \lt 4.16\) | M1* | 3.3 |
| \(\mathrm{P}(\bar{X} \lt 4.16) = 0.0548\) | A1 | 3.4 |
| \(0.0548 \gt 0.05\) | A1FT | 1.1 |
| Do not reject \(\mathrm{H}_0\) Allow Accept \(\mathrm{H}_0\) | M1dep* | 1.1 |
| Insufficient evidence (at 5%) that (mean) mass is less than in UK | A1 | 2.2b |
| [8] |
Notes
B1: For a correct assumption, in context, that is necessary and specific to ‘the population’ or ‘the river’.
Acceptable answers include:
- sd for this river is the same as for the UK
- assume the population of this river is normally distributed
- mass of fish in this river is normally distributed
Do not accept answers that are:
- referencing the sample (e.g. bias) – question states random
- not about this river e.g. ‘in the UK’
- generic e.g. ‘masses of fish are normally distributed’
Candidates must show recognition that the assumption is needed about the (sub-)population of this river. The mean and sd for the whole of the UK are given in the question.
Condone the sample mean of fish from this river \(\bar{X} \sim N\) but references to \(X\) or \(\bar{X}\) alone are not enough unless defined in context.
Ignore all else (e.g. ignore any statements about the mean)
B1B1: Subtract B1 for each error eg:
- 2-tail: B1B0
- undefined \(\mu\): B1B0
- not in terms of parameter: B1B0
- \(\mu\) = sample mean implied: B1B0
- Not include value 4.2: B0B0
- eg \(\mathrm{H}_0 = 4.2\) etc: B0B0
Allow any letter for \(\mu\) (except \(X, \bar{X}\): B0B0)
Condone ‘average’ for ‘mean’ but do not accept definitions for \(\mu\) that are clearly not about this river e.g. ‘the UK’ B1B0
M1*: This mark may be implied by the correct value of 0.0548 or 0.945 (correct to 2sf) (even if within incorrect statement eg \(\mathrm{P}(X = 4.16) = 0.0548\)) Condone \(\gt, =, \geqslant, \leqslant\)
A1: BC (awrt 0.055 2sf)
A1FT: FT correct comparison for their value as long as consistent with their test (e.g. 2-tail 0.025 or 0.975) Must be seen, allow on diag
A0 if the comparison is not for their value (e.g. if miscopied)
Alternative (critical value)
| Scheme | Marks |
|---|---|
| \(\dfrac{a - 4.2}{0.25/10} = 1.645\) | M1* |
| \(a = 4.159\) or CV is 4.159 | A1* |
| \(4.16 \gt 4.159\) | A1dep* |
A1*: cao
A1dep*: dep A1* Must be seen
Alternative (test statistic)
| Scheme | Marks |
|---|---|
| \(\dfrac{4.16 - 4.2}{0.25/10}\) | M1* |
| \(= -1.6\) | A1* |
| \(-1.6 \gt -1.645\) or \(1.6 \lt 1.645\) | A1dep* |
A1*: cao
A1dep*: dep A1* Must be seen
M1dep*: dep M1* Correct conclusion about \(\mathrm{H}_0\) for their comparison, provided they have compared with an appropriate value.
Accept Reject \(\mathrm{H}_1\) or Insufficient (or No) evidence to reject \(\mathrm{H}_0\)
A1: www (all preceding calculations must be correct, i.e. dependent on all previous M and A marks)
Must be in context and not definite.
Acceptable answers include:
- ‘no evidence that (mean) mass is less’
- ‘insufficient evidence to say that the environmentalist is correct’
- ‘insufficient evidence that the fish in this river are smaller’
Do not accept:
- ‘therefore the mean is 4.2’ (definite)
- ‘the mean has decreased’ (incorrect reference to change over time)
- ‘the mean is not less’ or ‘the mean is the same’ (insufficient evidence does not mean that the statement for \(\mathrm{H}_0\) is true)