# Question 2:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| A paired test would eliminate any differences between individual cattle | E1 | |

**[1]**

## Part (ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Must assume: Normality of population … | B1 | |
| … of differences | B1 | |

**[2]**

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_D = 10$, $H_1: \mu_D < 10$ | B1 | Both. Accept alternatives e.g. $\mu_D > -10$ for $H_1$, or $\mu_A - \mu_B$ etc provided adequately defined. Hypotheses in words only must include "population". Do NOT allow "$\bar{X} = \ldots$" or similar unless clearly stated to be a population mean |
| Where $\mu_D$ is the (population) mean increase/difference in milk yield. MUST be PAIRED COMPARISON $t$ test | B1 | For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ is used |
| Differences (increases) (after $-$ before) are: 4 9 6 13 1 8 6 7 9 12 | M1 | Allow "before $-$ after" if consistent with alternatives for hypotheses above |
| $\bar{x} = 7.5$, $s_{n-1} = 3.566(8)$ $(s_{n-1}^2 = 12.722(2))$ | A1 | Do not allow $s_n = 3.3837$ $(s_n^2 = 11.45)$ |
| Test statistic is $\dfrac{7.5 - 10}{\frac{3.5668}{\sqrt{10}}}$ | M1 | Allow candidate's $\bar{x}$ and/or $s_{n-1}$. Allow reversed numerator compared with 2.2164 |
| $= -2.2164$ | A1 | c.a.o. but ft from here in any case if wrong. Use of $10 - \bar{x}$ scores M1A0, but ft |
| Refer to $t_9$ | M1 | No ft from here if wrong |
| Single-tailed 5% point is $-1.833$ | A1 | Must be minus 1.833 unless absolute values are being compared. No ft from here if wrong. $P(t < -2.2164) = 0.0269$ |
| Significant | A1 | ft only candidate's test statistic |
| Sufficient evidence to suggest that the mean milk yield has not increased by 10 litres (per cow per week) | A1 | ft only candidate's test statistic. Conclusion in context to include "on average" o.e. Accept "Sufficient evidence to suggest that the **company's** claim is not justified." o.e. |

**[10]**

## Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| CI is given by $7.5 \pm$ | M1 | ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to $t_9$ is OK. Allow candidate's $\bar{x}$ |
| $2.262$ | B1 | 2.262 seen |
| $\times \dfrac{3.5668}{\sqrt{10}}$ | M1 | Allow candidate's $s_{n-1}$ |
| $= 7.5 \pm 2.5514 = (4.948, 10.052)$ | A1 | c.a.o. Must be expressed as an interval |

**[4]**

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