**Part (a)(i):**
$H_0: \mu_D = 0$ | B1 | Both. Accept alternatives e.g. $\mu_D < 0$ for $H_1$, or $\mu_A - \mu_B$ etc provided adequately defined.

$H_1: \mu_D > 0$ | B1 | Allow absence of "population" if correct notation $\mu$ is used, but do NOT allow "$\bar{x} = ...$" or similar unless $\bar{x}$ is clearly and explicitly stated to be a population mean. Hypotheses in words only must include "population".

Where $\mu_D$ is the (population) mean reduction in absenteeism. | B1 | 

Must assume Normality ... ... of differences. | B1 | | 4 marks

**Part (a)(ii):**
Differences (reductions) (before – after)
1.7, 0.7, 0.6, –1.3, 0.1, –0.9, 0.6, –0.7, 0.4, 2.7, 0.9

$\bar{x} = 0.4364$, $s_{n1} = 1.1518$ ($s_n^2 = 1.3265$) | B1, M1 | Do not allow $s_n = 1.098$ ($s_n^2 = 1.205$). Allow c's $\bar{x}$ and/or $s_{n1}$. Allow alternative: $0 \pm$ (c's $1.812) \times \frac{1.1518}{\sqrt{11}} (= -0.6293, 0.6293)$ for subsequent comparison with $\bar{x}$. (Or $\bar{x} \pm$ (c's $1.812) \times \frac{1.1518}{\sqrt{11}} (= -0.1929, 1.0657)$ for comparison with 0.)

Test statistic is $\frac{0.4364 - 0}{1.1518 / \sqrt{11}} = 1.256(56...)$ | A1 | c.a.o. but ft from here in any case if wrong. Use of $0 - \bar{x}$ scores M1A0, but ft.

Refer to $t_{10}$. Upper 5% point is 1.812. | M1, A1 | No ft from here if wrong. No ft from here if wrong. For alternative $H_1$ expect $-1.812$ unless it is clear that absolute values are being used.

$1.256 < 1.812$, $\therefore$ Result is not significant. Seems there has been no reduction in mean absenteeism. | E1, E1 | ft only c's test statistic. ft only c's test statistic. Special case: ($t_{11}$ and 1.796) can score 1 of these last 2 marks if either form of conclusion is given. | 7 marks

**Part (b):**
For "days lost after"
$\bar{x} = 4.6182$, $s_{n1} = 1.4851$ ($s_n^2 = 2.2056$) | B1, M1, B1, M1 | Do not allow $s_n = 1.4160$ ($s_n^2 = 2.0051$). ft c's $\bar{x} \pm$. ft c's $s_{n1}$.

CI is given by $4.6182 \pm \frac{2.228}{} \times \frac{1.4851}{\sqrt{11}} = 4.6182 \pm 0.9976 = (3.620(6), 5.615(8))$ | A1 | c.a.o. Must be expressed as an interval. ZERO if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to $t_{10}$ is OK.

Assume Normality of population of "days lost after". Since 3.5 lies outside the interval it seems that the target has not been achieved. | E1, E1 | | 7 marks