# Question 1:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Suitably labelled scatter diagram with 11 points plotted | G1 | For suitably labelled axes. Condone absence of scale here. |
| 11 points correctly plotted relative to a suitable linear scale | G2,1,0 | G1 if 9 or 10 correctly plotted. G0 if 3 or more incorrectly plotted/omitted. Allow axes interchanged. |
| **[3]** | | |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Points do not appear to be roughly elliptical | E1 | For "not elliptical" |
| The population may not have a bivariate Normal distribution | E1 | For not **underlying** bivariate Normal. Do not allow "the data" in place of population/underlying. Allow "data is not from a bivariate Normal distribution". Do not allow "Normal bivariate…" |
| **[2]** | | |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Rankings calculated correctly | M1 | For ranking (allow ranks reversed). **NB No ranking scores 0/5** |
| $d^2$ values: 16, 25, 9, 4, 100, 9, 9, 81, 64, 0, 49 | M1 | For $d^2$ |
| $\Sigma d^2 = 366$ | A1 | For $\Sigma d^2$ (may be embedded in calculation) |
| $r_s = 1 - \dfrac{6\Sigma d^2}{n(n^2-1)} = 1 - \dfrac{6 \times 366}{11 \times 120} = 1 - \dfrac{2196}{1320} = 1 - 1.6636$ | M1 | For method for $r_s$ |
| $= -0.664$ (to 3 s.f.) [allow $-0.66$ to 2 s.f. or $-73/110$] | A1 | FT their $\Sigma d^2$ provided $-1 < r_s < 0$, and ranking used. **NB No ranking scores 0/5** |
| **[5]** | | |

## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: no association between percentage of population living in rural areas and fertiliser use (in the population of countries) | B1 | For null hypothesis in context. **NB** $H_0$, $H_1$ not in terms of $\rho$ |
| $H_1$: **negative** association between percentage of population living in rural areas and fertiliser use (in the population of countries) | B1 | For alternative hypothesis in context. Context needed in at least one hypothesis. |
| | B1 | For **population of countries** or **underlying population** |
| One tail test critical value at 1% level is $-0.7091$ | B1 | For $\pm 0.7091$. **No further marks from here if incorrect** |
| Since $-0.664 > -0.7091$ [or $0.664 < 0.7091$] there is… | M1 | For sensible comparison of "$-0.664$" with $\pm 0.7091$ seen, leading to conclusion, only if $-1 <$ their $r_s < 0$ |
| …insufficient evidence to reject $H_0$. There is insufficient evidence to suggest that there is **negative** association between percentage of population living in rural areas and fertiliser use (in the population of countries) | A1 | For not significant, oe, and correct conclusion in context. FT their $r_s$ with correct cv. |
| **[6]** | | |

## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| It means that the probability of rejecting $H_0$ given that it is correct is 1% o.e. | E1 | Allow "the probability of a false positive is 1%", "the probability of a Type I Error is 1%". Do not allow "It means that the probability rejecting $H_0$ when it should have been accepted is 1%" |
| **[1]** | | |

## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| None | E1 | |
| **[1]** | | |

---