S3 June 2010 Q7
7. A large company surveyed its staff to investigate the awareness of company policy. The company employs 6000 full time staff and 4000 part time staff.
A random sample of 80 full time staff and an independent random sample of 80 part time staff were given a test of policy awareness. The results are summarised in the table below.
| Mean score (\(\bar{x}\)) | Variance of scores (\(s^2\)) | |
|---|---|---|
| Full time staff | 52 | 21 |
| Part time staff | 50 | 19 |
After all the staff had completed a training course the 80 full time staff and the 80 part time staff were given another test of policy awareness. The value of the test statistic \(z\) was 2.53
| Scheme | Marks |
|---|---|
| Label full time staff 1-6000, part time staff 1-4000 | M1 |
| Use random numbers to select | M1 |
| Simple random sample of 120 full time staff and 80 part time staff | A1 |
| (3) |
Notes
1st M1 for attempt at labelling full-time and part-time staff. One set of correct numbers.
2nd M1 for mentioning use of random numbers
1st A1 for s.r.s. of 120 full-time and 80 part-time
| Scheme | Marks |
|---|---|
| Enables estimation of statistics / errors for each strata or “reduce variability” or “more representative” or “reflects population structure” NOT “more accurate” | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \mu_f = \mu_p, \quad \mathrm{H}_1 : \mu_f \ne \mu_p\) (accept \(\mu_1, \mu_2\)) | B1 |
| s.e. \(= \sqrt{\dfrac{21}{80} + \dfrac{19}{80}}\), \(z = \dfrac{52 - 50}{\sqrt{\frac{21}{80} + \frac{19}{80}}} = \left(2\sqrt{2}\right)\) | M1,M1 |
| \(= 2.828\ldots\) (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 |
| [\(2.828 \gt 2.5758\) so] significant evidence to reject \(\mathrm{H}_0\) | dM1 |
| There is evidence of a difference in policy awareness between full time and part time staff | A1ft |
| (7) |
Notes
1st M1 for attempt at s.e. - condone one number wrong . NB correct s.e. \(= \sqrt{\tfrac{1}{2}}\)
2nd M1 for using their s.e. in correct formula for test statistic. Must be \(\dfrac{\pm(52 - 50)}{\sqrt{\frac{p}{q} + \frac{r}{s}}}\)
3rd dM1 dep. on 2nd M1 for a correct statement based on their normal cv and their test statistic
2nd A1 for correct comment in context. Must mention “scores” or “ policy awareness” and types of “staff”. Award A0 for a one-tailed comment. Allow ft
| Scheme | Marks |
|---|---|
| Can use mean full time and mean part time | B1 |
| ~ Normal | B1 |
| (2) |
Notes
1st B1 for mention of mean(s) or use of \(\overline{X}\), provided \(\overline{X}\) clearly refers to full-time or part-time
2nd B1 for stating that distribution can be assumed normal
e.g. “mean score of the test is normally distributed” gets B1B1
| Scheme | Marks |
|---|---|
| Have assumed \(s^2 = \sigma^2\) or variance of sample = variance of population | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(2.53 \lt 2.5758\), not significant or do not reject \(\mathrm{H}_0\) | M1 |
| So there is insufficient evidence of a difference in mean awareness | A1ft |
| (2) |
Notes
M1 for correct statement (may be implied by correct contextualised comment)
A1 for correct contextualised comment. Accept “no difference in mean scores”. Allow ft
| Scheme | Marks |
|---|---|
| Training course has closed the gap between full time staff and part time staff’s mean awareness of company policy. | B1 |
| (1) | |
| (17 marks) |
Notes
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.