June 2022 Paper 3 Q6
6. Anna is investigating the relationship between exercise and resting heart rate.
She takes a random sample of 19 people in her year at school and records for each person
- their resting heart rate, \(h\) beats per minute
- the number of minutes, \(m\), spent exercising each week
Her results are shown on the scatter diagram.
Anna codes the data using the formulae
\[\begin{gathered}x = \log_{10} m\\ y = \log_{10} h\end{gathered}\]The product moment correlation coefficient between \(x\) and \(y\) is \(-0.897\)
You should
- state your hypotheses clearly
- use a 5% level of significance
- state the critical value used
The equation of the line of best fit of \(y\) on \(x\) is
\[y = -0.05x + 1.92\]| Scheme | Marks | AO |
|---|---|---|
| eg As the number of minutes exercise (\(m\)) increases the resting heart rate (\(h\)) decreases or the gradient of the curve is becoming flatter with increasing \(m\): diminishing effect of each additional minute of exercise | B1 | 2.4 |
| (1) |
Notes
B1: eg Idea as one increases the other decreases (in context). Allow use of \(m\) and \(h\) eg As \(m\) increases \(h\) decreases. Do not allow negative correlation with no context or \(\rho \lt 0\)
Allow there is a negative correlation/association/relationship/exponential between minutes exercise(\(m\)) and resting heart rate (\(h\)) oe
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0: \rho = 0 \quad \mathrm{H}_1: \rho \lt 0\) | B1 | 2.5 |
| Critical value \(-0.3887\) (Allow \(\pm\)) | M1 | 1.1b |
| There is evidence that the product moment correlation is less than 0/ there is a negative correlation | A1 | 2.2b |
| (3) |
Notes
B1: Both hypotheses correct in terms of \(\rho\) (allow p)
M1: For the cv of \(-0.3887\) or any cv such that \(0.3 \lt \lvert\text{cv}\rvert \lt 0.5\)
A1: Independent of hypotheses. Correct conclusion that implies reject \(\mathrm{H}_0\) on basis of seeing \(-0.3887\) or if they give 0.3887 we must see the comparison \(0.3887 \lt 0.897\) and which mentions “pmcc/correlation/relationship” and less than 0/ negative or \(\boldsymbol{\rho \lt 0}\)
A contradictory statement scores A0 eg Accept \(\mathrm{H}_0\) therefore negative correlation
| Scheme | Marks | AO |
|---|---|---|
| \(\log_{10} h = -0.05\,\log_{10} m + 1.92\) or \(h = am^k \to \log_{10} h = \log_{10} am^k\) | M1 | 1.1b |
| \(\log_{10} h = -\log_{10} m^{0.05} + 1.92\) or \(\log_{10} h = \log_{10} m^{-0.05} + 1.92\) or \(h = 10^{1.92 - 0.05\log_{10} m}\) oe or \(\log_{10} h = \log_{10} a + \log_{10} m^k\) or \(\log_{10} a = 1.92\) | M1 | 2.1 |
| \(\log_{10} hm^{0.05} = 1.92\) or \(\log_{10}\left(\dfrac{h}{m^{-0.05}}\right) = 1.92\) or \(h = 10^{1.92} \times 10^{-0.05\log_{10} m}\) oe or \(\log_{10} h = \log_{10} a + k\,\log_{10} m\) | M1 | 1.1b |
| \(hm^{0.05} = 10^{1.92}\) or \(\dfrac{h}{m^{-0.05}} = 10^{1.92}\) or \(h = 10^{1.92} \times 10^{\log_{10} m^{-0.05}}\) or \(\log_{10} a = 1.92\) and \(k = -0.05\) | M1 | 1.1b |
| \(h = 10^{1.92}m^{-0.05}\) or \(h = 83.17\ldots m^{-0.05}\) or \(a =\) awrt 83.17 and \(k = -0.05\) | A1 | 1.1b |
| (5) | ||
| (9 marks) |
Notes
In this part once M0 is scored no more marks can be scored. Condone no base
M1: May be implied by 2nd M1 mark
Method 1: Correct substitution for both \(x\) and \(y\) Method 2: Taking the log of both sides
M1: May be implied by 3rd M1 mark
Method 1: Correct use of the power log rule or making \(h\) the subject
Method 2: Correct use of the addition/subtraction log rule
M1: This line implies M1M1M1
Method 1: Correct use of the addition/subtraction log rule or eqn in the form \(h = 10^{1.92} \times 10^{-0.05\log m}\)
Method 2: A second correct step for correct use of the power log rule
M1: This line implies M1M1M1M1
Method 1: Correct removal of logs or \(h = 10^{1.92} \times 10^{\log m^{-0.05}}\) Method 2: Log \(a\) (or \(a\)) and \(k\) correct
A1: Allow \(h =\) awrt \(83.2m^{-0.05}\)
NB award 5/5 for \(a =\) awrt 83.2 and \(k = -0.05\) or \(h =\) awrt \(83.2\ldots m^{-0.05}\) or \(h = 10^{1.92}m^{-0.05}\)