S4 June 2015 Q1
1. The Sales Manager of a large chain of convenience stores is studying the sale of lottery tickets in her stores. She randomly selects 8 of her stores. From these stores she collects data for the total sales of lottery tickets in the previous January and July. The data are shown below
| Store | A | B | C | D | E | F | G | H |
|---|---|---|---|---|---|---|---|---|
| January ticket sales (£) | 1080 | 1639 | 710 | 1108 | 915 | 1066 | 1322 | 819 |
| July ticket sales (£) | 1113 | 1702 | 831 | 1048 | 861 | 1090 | 1303 | 852 |
| Scheme | Marks | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B1 | ||||||||||||||||||
| \(\bar{d} = \dfrac{141}{8} = (\pm)17.625\) | M1 | ||||||||||||||||||
| \({s_d}^2 = \dfrac{8}{7}\left(\dfrac{28241}{8} - 17.625^2\right) = 3679.4\ldots\) or \({s_d}^2 = \dfrac{1}{7}\left(28241 - \dfrac{141^2}{8}\right) = 3679.4\ldots\) | M1 | ||||||||||||||||||
| To test \(\mathrm{H}_0 : \mu_d = 0\) against \(\mathrm{H}_1 : \mu_d \gt 0\) (o.e.) | B1 | ||||||||||||||||||
| Test stat \(t = \dfrac{17.625 - 0}{\sqrt{\frac{3679.4\ldots}{8}}} = 0.8218\ldots\) | M1A1cso | ||||||||||||||||||
| Critical value, \(t_7 = 1.895\) | B1 | ||||||||||||||||||
| Not in critical region therefore insufficient reason to reject \(\mathrm{H}_0\) No significant evidence that on average stores sell more lottery tickets in July than in January | A1ft | ||||||||||||||||||
| (8) |
Notes
1st B1 for differences all correct (o.e.)
1st M1 attempt to find \(\bar{d} = \dfrac{\sum \text{"their } d\text{"}}{8}\)
2nd M1 attempting \(s_d\) or \({s_d}^2\ \dfrac{1}{7}\left(\sum \text{"their } d^2\text{"} - \dfrac{\left(\sum \text{"their } d\text{"}\right)^2}{8}\right)\)
2nd B1 both correct in terms of \(\mu\) or \(\mu_d\) (allow a defined symbol) condone \(\mu_{July-Jan}\)
3rd M1 for attempting the correct test statistic \(\dfrac{\bar{d}}{s_d/\sqrt{8}}\)
1st A1cso awrt 0.822 with no errors.
3rd B1 alternate method, \(p\) value of 0.219. Allow 2.365 for 2-tail test
Final A1 need conclusion in context, need tickets July and January, ft their test stat and critical value
NB difference of 2 means test gains no marks
| Scheme | Marks |
|---|---|
| Need assumption that the underlying distribution of the difference in sales in July and in January is normally distributed. | B1 |
| (1) | |
| (9 marks) |
Notes
B1 need differences to be normally distributed, not just normal distribution