# Question 7:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Label full time staff 1-6000, part time staff 1-4000 | M1 | For attempt at labelling full-time and part-time staff. One set of correct numbers |
| Use random numbers to select | M1 | For mentioning use of random numbers |
| Simple random sample of 120 full time staff and 80 part time staff | A1 | For s.r.s. of 120 full-time and 80 part-time |
| | **(3)** | |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Enables estimation of statistics/errors for each strata **or** "reduce variability" **or** "more representative" **or** "reflects population structure" **NOT** "more accurate" | B1 | |
| | **(1)** | |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_f = \mu_p$, $\quad H_1: \mu_f \neq \mu_p$ $\quad$ (accept $\mu_1, \mu_2$) | B1 | |
| $\text{s.e.} = \sqrt{\dfrac{21}{80} + \dfrac{19}{80}}$ | M1 | For attempt at s.e. - condone one number wrong. NB correct s.e. $= \sqrt{\frac{1}{2}}$ |
| $z = \dfrac{52-50}{\sqrt{\dfrac{21}{80}+\dfrac{19}{80}}} = (2\sqrt{2})$ | M1 | For using their s.e. in correct formula for test statistic. Must be $\dfrac{\pm(52-50)}{\sqrt{\frac{p}{q}+\frac{r}{s}}}$ |
| $= 2.828\ldots$ (awrt **2.83**) | A1 | |
| Two tailed critical value $z = 2.5758$ (or prob of awrt 0.002 (<0.005) or 0.004 (<0.01)) | B1 | |
| $[2.828 > 2.5758$ so] significant evidence to reject $H_0$ | dM1 | Dependent on 2nd M1, for correct statement based on their normal cv and their test statistic |
| There is evidence of a difference in policy awareness between full time and part time staff | A1ft | For correct comment in context. Must mention "scores" or "policy awareness" and types of "staff". Award **A0** for a one-tailed comment. Allow ft |
| | **(7)** | |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Can use mean full time and mean part time | B1 | For mention of mean(s) **or** use of $\bar{X}$, provided $\bar{X}$ clearly refers to full-time or part-time |
| $\sim$ Normal | B1 | For stating that distribution can be assumed normal. e.g. "mean score of the test is normally distributed" gets B1B1 |
| | **(2)** | |

## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| Have assumed $s^2 = \sigma^2$ or variance of sample = variance of population | B1 | |
| | **(1)** | |

## Part (f)
| Answer | Mark | Guidance |
|--------|------|----------|
| $2.53 < 2.5758$, not significant **or** do not reject $H_0$ | M1 | For correct statement (may be implied by correct contextualised comment) |
| So there is insufficient evidence of a difference in mean awareness | A1ft | For correct contextualised comment. Accept "no difference in mean scores". Allow ft |
| | **(2)** | |

## Part (g)
| Answer | Mark | Guidance |
|--------|------|----------|
| Training course has closed the gap between full time staff and part time staff's mean awareness of company policy. | B1 | For correct comment in context that implies training was effective. This must be supported by their (c) and (f). Condone one-tailed comment here. |
| | **(1)** | |
| | **Total: 17** | |