June 2019 Paper 3 Q3
3. Barbara is investigating the relationship between average income (GDP per capita), \(x\) US dollars, and average annual carbon dioxide (CO2) emissions, \(y\) tonnes, for different countries.
She takes a random sample of 24 countries and finds the product moment correlation coefficient between average annual CO2 emissions and average income to be 0.446
Barbara believes that a non-linear model would be a better fit to the data.
She codes the data using the coding \(m = \log_{10} x\) and \(c = \log_{10} y\) and obtains the model \(c = -1.82 + 0.89m\)
The product moment correlation coefficient between \(c\) and \(m\) is found to be 0.882
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0 : \rho = 0 \qquad \mathrm{H}_1 : \rho \gt 0\) | B1 | 2.5 |
| Critical value 0.3438 | M1 | 1.1a |
| \((0.446 \gt 0.3438)\) so there is evidence that the product moment correlation coefficient (pmcc) is greater than 0/there is positive correlation | A1 | 2.2b |
| (3) |
Notes
B1: for both hypotheses correct in terms of \(\rho\)
M1: for the critical value: sight of 0.3438 or any cv such that \(0.25 \lt |\text{cv}| \lt 0.45\)
A1: a comment suggesting a significant result/\(\mathrm{H}_0\) is rejected on the basis of seeing +0.3438 and which mentions “pmcc/correlation/relationship” and “greater than 0/positive” (not just \(\rho \gt 0\))
or an answer in context e.g. ‘as “income”(o.e.) increases, “CO2/emissions”(o.e.) increases’
A contradictory statement scores A0 e.g. ‘Accept H0, therefore positive correlation’
| Scheme | Marks | AO |
|---|---|---|
| The value is close(r) to 1 or there is strong(er) (positive) correlation | B1 | 2.4 |
| (1) |
Notes
B1: for suitable reason e.g. \(r\) is close(r) to 1 or “strong(er)”/“near perfect” “correlation”
Do not allow ‘association’
| Scheme | Marks | AO |
|---|---|---|
| Method 1 \(\log_{10} y = -1.82 + 0.89(\log_{10} x)\) | M1 | 1.1b |
| \(y = 10^{-1.82 + 0.89(\log_{10} x)}\) | M1 | 2.1 |
| \(y = 10^{-1.82} \times 10^{0.89(\log_{10} x)}\) \(\left[= 10^{-1.82} \times 10^{(\log_{10} x)^{0.89}}\right]\) | M1 | 1.1b |
| \(y = 0.015x^{0.89}\) | A1A1 | 1.1b 1.1b |
| (5) | ||
| (9 marks) |
Notes
Method 2
| Scheme | Marks | AO |
|---|---|---|
| \(y = ax^n \to\) \(\log_{10} y = \log_{10}(ax^n)\) | M1 | 1.1b |
| \(\log_{10} y = \log_{10} a + \log_{10} x^n\) | M1 | 2.1 |
| \(\log_{10} y = \log_{10} a + n\log_{10} x\) \([\log_{10} a = -1.82,\ n = 0.89]\) | M1 | 1.1b |
| \(y = 0.015x^{0.89}\) | A1A1 | 1.1b 1.1b |
| (5) |
For both methods, once an M0 is scored, no further marks can be awarded and condone missing base 10 throughout
Method 1: (working to the model)
M1: Correct substitution for both \(c\) and \(m\) (may be implied by 2nd M1 mark)
M1: Making \(y\) the subject to give an equation in the form \(y = 10^{a + b(\log_{10} x)}\) (may be implied by 3rd M1 mark)
M1: Correct multiplication to give an equation in the form \(y = 10^a \times 10^{b(\log_{10} x)}\) (this line implies M1M1M1 provided no previous incorrect working seen)
Method 2: (working from the model)
M1: Taking the log of both sides (may be implied by 2nd M1 mark)
M1: Correct use of addition rule (may be implied by 3rd M1 mark)
M1: Correct multiplication of power (this line implies M1M1M1 provided no previous incorrect working seen)
Both methods
A1: \(n = 0.89\) or \(a = \text{awrt } 0.015\) or \(y = ax^{0.89}\) or \(y = \text{awrt } 0.015x^n\) (dep on M3)
A1: \(n = 0.89\) and \(a = \text{awrt } 0.015\) / \(y = \text{awrt } 0.015x^{0.89}\) (dep on M3)
do not award the final A1 if answer is given in an incorrect form e.g. \(y = 0.015 + x^{0.89}\)