# Question 6:

## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| 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 | 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 $\rho<0$. Allow there is a negative correlation/association/relationship/exponential between minutes exercise($m$) and resting heart rate ($h$) oe |

## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \rho=0 \quad H_1: \rho<0$ | B1 | Both hypotheses correct in terms of $\rho$ (allow p) |
| Critical value $-0.3887$ (Allow $\pm$) | M1 | For the cv of $-0.3887$ or any cv such that $0.3<|cv|<0.5$ |
| There is evidence that the product moment **correlation** is **less than 0** / **there is a negative correlation** | A1 | Independent of hypotheses. Correct conclusion that implies reject $H_0$ on basis of seeing $-0.3887$ or if they give 0.3887 we must see comparison $0.3887<0.897$ **and** which mentions "pmcc/correlation/relationship" and less than 0/negative or $\rho<0$. A contradictory statement scores A0 eg Accept $H_0$ therefore negative correlation |

## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_{10}h = -0.05\log_{10}m + 1.92$ | M1 | $h=am^k \rightarrow \log_{10}h = \log_{10}am^k$. Correct substitution for both $x$ and $y$. Method 2: Taking the log of both sides |
| $\log_{10}h = -\log_{10}m^{0.05}+1.92$ or $\log_{10}h = \log_{10}m^{-0.05}+1.92$ or $h=10^{1.92-0.05\log_{10}m}$ oe | M1 | $\log_{10}h = \log_{10}a + \log_{10}m^k$ or $\log_{10}a = 1.92$. Correct use of the power log rule **or** making $h$ the subject. Method 2: Correct use of the addition/subtraction log rule |
| $\log_{10}hm^{0.05}=1.92$ or $\log_{10}\left(\frac{h}{m^{-0.05}}\right)=1.92$ or $h=10^{1.92}\times10^{-0.05\log_{10}m}$ oe | M1 | $\log_{10}h = \log_{10}a + k\log_{10}m$. Method 1: Correct use of addition/subtraction log rule or equation in form $h=10^{1.92}\times10^{-0.05\log m}$. Method 2: A second correct step for correct use of power log rule |
| $hm^{0.05}=10^{1.92}$ or $\frac{h}{m^{-0.05}}=10^{1.92}$ or $h=10^{1.92}\times10^{\log_{10}m^{-0.05}}$ | M1 | $\log_{10}a=1.92$ and $k=-0.05$. Method 1: Correct removal of logs or $h=10^{1.92}\times10^{\log m^{-0.05}}$. Method 2: $\log a$ (or $a$) and $k$ correct |
| $h=10^{1.92}m^{-0.05}$ or $h=83.17...m^{-0.05}$ or $a=$ awrt 83.17 **and** $k=-0.05$ | A1 | Allow $h=$ awrt $83.2m^{-0.05}$. NB award 5/5 for $a=$ awrt 83.2 and $k=-0.05$ or $h=$ awrt $83.2...m^{-0.05}$ or $h=10^{1.92}m^{-0.05}$ |

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Could you please share the correct pages that contain the actual mark scheme questions and answers? I'd be happy to format them as requested once I can see the relevant content.