# Question 1:

## Part 1(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Axes drawn correctly | G1 | Condone axes drawn either way; axes should show some indication of scale |
| Values of $x$ plotted | G1 | If axes scaled and only one point incorrectly plotted, allow max G2/3 |
| Values of $y$ plotted | G1 | |
| **Total: 3** | | |

## Part 1(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 60$, $\bar{y} = 4.26$ | B1 | For $\bar{x}$ and $\bar{y}$ used appropriately (SOI); allow $\bar{y} = 4.3$ |
| $b = \frac{S_{xy}}{S_{xx}} = \frac{1803 - 300 \times 21.3/5}{27000 - 300^2/5} = \frac{525}{9000} = 0.0583$ | M1 | Attempt should be correct – e.g. evidence of either of the two suggested methods |
| OR $b = \frac{1803/5 - 60 \times 4.26}{27000/5 - 60^2} = \frac{105}{1800} = 0.0583$ | A1 | Allow 0.058; condone $0.058\hat{3}$ and $\frac{7}{120}$ |
| $y - \bar{y} = b(x - \bar{x})$ $\Rightarrow y - 4.26 = 0.0583(x - 60)$ $\Rightarrow y = 0.0583x + 0.76$ | M1 | Dependent on first M1; values must be substituted; condone use of their $b$ for FT provided $b > 0$ |
| Complete equation | A1 FT | Final equation must be simplified; $b = 0.058$ leads to $y = 0.058x + 0.78$ |
| **Total: 5** | | |

## Part 1(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Regression line plotted on graph | G1 | Line must pass through their $(\bar{x}, \bar{y})$ and $y$-intercept |
| | G1 | |
| The fit is good | E1 | E0 for notably inaccurate graphs/lines |
| **Total: 3** | | |

## Part 1(iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 30 \Rightarrow$ predicted $y = 0.0583 \times 30 + 0.76 = 2.509$ | B1 | Using their equation |
| Residual $= 2.5 - 2.509 = -0.009$ | M1 | Subtraction can be 'either way' but sign of residual must be correct |
| | A1 FT | FT sensible equations only; no FT for $y = 0.071x$ leading to $+0.37$; $[c = 0.78$ leads to residual of $-0.02]$ |
| **Total: 3** | | |

## Part 1(v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| (A) For $x = 45$: $y = 0.0583 \times 45 + 0.76 = 3.4$ | M1 | For at least one prediction attempted; prediction from their equation |
| (B) For $x = 150$: $y = 0.0583 \times 150 + 0.76 = 9.5$ | A1 FT | Both answers; FT their equation provided $b > 0$ |
| **Total: 2** | | |

## Part 1(vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| This is **well** below the predicted value… | E1 | Value (8.7) is significantly below what is expected (9.5); simply saying 'below' is not sufficient |
| …suggesting that the model breaks down for larger values of $x$ | E1 | Suitable comment on extrapolation; allow other sensible comments e.g. 'data might be better modelled by a curve', 'there may be other factors affecting yield' |
| **Total: 2** | | |

---