S3 June 2011 Q4
4. A shop manager wants to find out if customers spend more money when music is playing in the shop. The amount of money spent by a customer in the shop is £\(x\). A random sample of 80 customers, who were shopping without music playing, and an independent random sample of 60 customers, who were shopping with music playing, were surveyed. The results of both samples are summarised in the table below.
| \(\sum x\) | \(\sum x^2\) | Unbiased estimate of mean | Unbiased estimate of variance | |
|---|---|---|---|---|
| Customers shopping without music | 5320 | 392 000 | \(\bar{x}\) | \(s^2\) |
| Customers shopping with music | 4140 | 312 000 | 69.0 | 446.44 |
| Scheme | Marks |
|---|---|
| \(\bar{x} = \dfrac{5320}{80} = 66.5\) | M1,A1 |
| \(s^2 = \dfrac{392000 - 80 \times (66.5)^2}{79}\) | M1A1ft |
| \(= 483.797\ldots\) awrt 484 | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \mu_m = \mu_{nm}, \quad \mathrm{H}_1 : \mu_m \gt \mu_{nm}\) (accept \(\mu_1, \mu_2\) with definition) | B1B1 |
| \(z = \dfrac{69.0 - 66.5}{\sqrt{\dfrac{483.797}{80} + \dfrac{446.44}{60}}}\) | M1dM1 |
| \(= 0.6807\) awrt 0.681 | A1 |
| One tailed cv 1.6449 (Probability is awrt 0.752) | B1 |
| \(0.6807 \lt 1.6449\) (or \(0.248 \gt 0.05\)) insufficient evidence to reject \(\mathrm{H}_0\) | dM1 |
| Mean money spent is not greater with music playing. | A1ft |
| (8) | |
| (13 marks) |
Notes
No definition award B1B0.
1st M1 for attempt at s.e. - condone one number wrong or switched 60 & 80 .
2nd dM1 for using their s.e. in correct formula for test statistic.
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 “money spent” and “music playing”. Allow ft.
Critical Region for (b)
Standard error x z value for 2nd M1
Standard error x 1.6449= awrt 6.04 for 1st A1
2.5<6.04