A2 June 2019 Q4
4. Liam and Simone are studying the distribution of oak trees in some woodland. They divided the woodland into 80 equal squares and recorded the number of oak trees in each square. The results are summarised in Table 1 below.
| Number of oak trees in a square | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 or more |
|---|---|---|---|---|---|---|---|---|
| Frequency | 1 | 4 | 21 | 23 | 13 | 11 | 7 | 0 |
Table 1
Liam believes that the oak trees were deliberately planted, with 6 oak trees per square and that a constant proportion \(p\) of the oak trees survived.
Liam decides to test whether or not his model is suitable and calculates the expected frequencies given in Table 2.
| Number of oak trees in a square | 0 or 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Expected frequency | 5.53 | 14.89 | 24.26 | 22.24 | 10.87 | 2.21 |
Table 2
Simone believes that a Poisson distribution could be used to model the number of oak trees per square. She calculates the expected frequencies given in Table 3.
| Number of oak trees in a square | 0 or 1 | 2 | 3 | 4 | 5 | 6 or more |
|---|---|---|---|---|---|---|
| Expected frequency | 12.69 | 16.07 | \(s\) | 14.58 | \(t\) | 9.37 |
Table 3
The test statistic for this test is 8.749
| Scheme | Marks | AO |
|---|---|---|
| [\(T\) = no. of oak trees in a square] \(T \sim\) Binomial | M1 | 3.3 |
| \(T \sim \mathrm{B}(6, p)\) | A1 | 1.1b |
| (2) |
Notes
M1 for choosing binomial A1 for \(\mathrm{B}(6, p)\) can be in words and allow \(\mathrm{B}(6, 0.55)\)
| Scheme | Marks | AO | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Expected frequency for 6 is less than 5 so pool: new \(E_i = 13.08\) | M1 | 2.1 | ||||||||||||
| M1, A1 | 1.1b ×2 | ||||||||||||
| \(p\) needed estimating (\(\hat{p} = 0.55\)) so \(\nu = 5 - 2 = 3\); cv 7.815 | B1, B1ft | 1.1b ×2 | ||||||||||||
| Significant result, so Liam’s model is not suitable | M1, A1 | 1.1b 2.2b | ||||||||||||
| (7) |
Notes
1st M1 for pooling last 2 classes (\(E_i = 13.08\) but accept 13.1)
2nd M1 for at least 3 correct values or expressions. Either row to at least 2 sf
1st A1 for awrt 8.31 (8.31 gets 3/3) [NB no pooling gives awrt 16.8458.. and implies M0M1A0]
1st B1 for 3 degrees of freedom 2nd B1ft for critical value of 7.815 (e.g. \(\nu = 4\) use 9.488)
3rd M1 for a correct conclusion (non-contextual ignore any contradictory contextual comments for this mark) based on their cv and their test statistic
This mark can be implied by a fully correct solution ending with correct contextual conclusion
2nd A1 for correct conclusion in context with all other marks scored
| Scheme | Marks | AO |
|---|---|---|
| [\(R\) = no. of oak trees in a square for Simone’s model] \(R \sim \mathrm{Po}(3.3)\) | M1 | 3.3 |
| Correct expression for \(s\) or \(t\) using Poisson | M1 | 3.4 |
| \(s = \underline{\mathbf{17.67}}\) and \(t = \underline{\mathbf{9.62}}\) | A1, A1 | 1.1b ×2 |
| (4) |
Notes
1st M1 for selecting a correct model \(\mathrm{Po}(3.3)\) [Allow \(\mathrm{Po}(\text{awrt } 3.3)\)]
2nd M1 for use of the model with an expression or correct value for \(s\) or \(t\)
1st A1 for one correct 2nd A1 for both correct (allow awrt 2dp)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0\): Poisson is a good fit (for no. of oak trees per square) \(\mathrm{H}_1\): Poisson is not a good fit (for no. of oak trees per square) | B1 | 2.5 |
| (1) |
Notes
B1 for correct hypotheses must mention Poisson: use of \(\mathrm{Po}(3.3)\) is B0
| Scheme | Marks | AO |
|---|---|---|
| No pooling needed so degrees of freedom is \(6 - 2 = 4\) | B1 | 1.1b |
| Critical value is 9.488 (accept 9.49) | B1 | 1.1a |
| Not significant so Poisson (or Simone’s) model is suitable | B1 | 2.2b |
| (3) |
Notes
1st B1 for correct degrees of freedom \(\nu = 4\) only
2nd B1 for selecting correct critical value (9.488 only)
3rd B1 for not significant conclusion based on 8.749 vs their cv (condone use of \(\mathrm{Po}(3.3)\) here)
| Scheme | Marks | AO |
|---|---|---|
| Poisson model has better fit so suggests that oak trees occur at random Or binomial suggests deliberately planted or cultivated | B1 | 2.2b |
| Therefore the forest is likely to be wild not cultivated | B1 | 3.5a |
| (2) | ||
| (19 marks) |
Notes
1st B1 for choosing Poisson as better or stating Poisson implies wild or bino’l implies cultivated
2nd B1 (dep on rejecting bin and accepting Poisson) for clearly stating woodland is wild
If the tests give the same results then 2nd B0 automatically