# Question 2:

## Part 2(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $z = 1.96$ | B1 | Understanding sample mean modelled by Normal with $z = 1.96$; allow 1.959... or better |
| $57.2 \pm 1.96 \times \frac{1.2}{\sqrt{120}}$ | M1 | Setting up confidence interval |
| $(56.985..., \ 57.414...)$ awrt $(57.0, \ 57.4)$ | A1 | Must come from correct $z$ value $= 1.96$ or better |

## Part 2(b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\bar{Y} - \bar{W} \sim N\ldots$ | M1 | Translating context into Normal distribution model |
| Mean $= 0$, Variance $= \frac{1.2^2}{120} + \frac{0.9^2}{140}$ | A1A1 | A1 correct mean; A1 correct variance (allow awrt 0.0178 or exact fraction $\frac{249}{14000}$) |

## Part 2(b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Central limit theorem applies so we do not need to know the distributions of $Y$ and $W$. Allows us to assume that sample means ($\bar{Y}$ and $\bar{W}$) are normally distributed. | B1 | Correct explanation about distributions of $Y$ and $W$ or $\bar{Y}$ and $\bar{W}$ |

## Part 2(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $H_0: \mu_{y(ellow)} = \mu_{w(hite)}$, $\quad H_1: \mu_{y(ellow)} > \mu_{w(hite)}$ | B1 | Both hypotheses correct with correct notation; if using $\mu_x$ and $\mu_y$ these must be defined |
| $z = \frac{57.2 - 56.9}{\sqrt{\frac{1.2^2}{120} + \frac{0.9^2}{140}}} = 2.24949...$ | M1, A1 | M1: Standardising using normal distribution test statistic for difference of two means with known variance; A1: awrt 2.25 |
| $CV = 1.6449$ or $p\text{-value} = 0.01224...$ awrt $0.01$ | B1 | Correct critical value 1.6449 or better; or $p = 0.01$ or better from correct working |
| Reject $H_0$. Significant evidence to support Jamie's claim/mean weight of the population of yellow tennis balls is greater than mean weight of the population of white tennis balls. | A1 | Correct inference in context; do not allow contradictory statements e.g. 'Do not reject $H_0$, so Jamie's claim is supported' |

---