S3 June 2018 Q2
2. Merchandise is sold at concerts. The manager of a concert claims that the mean value of merchandise sold to premium ticket holders is more than £6 greater than the mean value of merchandise sold to standard ticket holders.
The mean value of merchandise sold to a random sample of 60 standard ticket holders at the concert is £15 with a standard deviation of £10.
The mean value of merchandise sold to a random sample of 55 premium ticket holders at the concert is £23 with a standard deviation of £8.
| Scheme | Marks |
|---|---|
| Record / List all ticket numbers of standard and premium tickets | B1 |
| Use random numbers to select a sample of standard and a sample of premium ticket holders i.e. within strata. | B1 |
| Sample sizes in proportion to the no of standard and no of premium ticket holders at the concert. | B1 |
| (3) |
Notes
1st B1 Sampling frame in context. Accept list of all standard and premium ticket holders at the concert.
2nd B1 Use of random selection eg simple random sampling within strata
3rd B1 Accept description of \(n_s, n_p\).
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \mu_p - \mu_s = 6\) oe [ \(p\) = premium \(s\) = standard ] | B1 |
| \(\mathrm{H}_1 : \mu_p - \mu_s \gt 6\) oe | B1 |
| Standard error \(= \sqrt{\dfrac{10^2}{60} + \dfrac{8^2}{55}} = \left[\sqrt{2.83030\ldots}\right] = [1.682\ldots]\) | M1 |
| \(z = \dfrac{\pm(23 - 15 - 6)}{\text{"}\sqrt{\dfrac{10^2}{60} + \dfrac{8^2}{55}}\text{"}}\) | dM1 |
| \(= \pm 1.1888\ldots\) awrt \(\pm 1.19\) | A1 |
| cv 5% one tailed = 1.6449 | B1 |
| Not significant, insufficient evidence to reject \(\mathrm{H}_0\) | dM1 |
| Insufficient evidence to support the manager’s claim or the mean value of merchandise sold to premium ticket holders is NOT more than £6 greater than the mean value of merchandise sold to standard ticket holders. | A1cso |
| (8) |
Notes
1st & 2nd B1 for hypotheses. Accept \(\mu_1, \mu_2\) or \(\mu_A, \mu_B\) etc if it is clear which is which.
1st M1 for an attempt at se with 3 out of 4 values correct.
Condone switching 10 and 8: \(\sqrt{\dfrac{10^2 \text{ or } 8^2}{60} + \dfrac{8^2 \text{ or } 10^2}{55}}\)
2nd dM1 dependent on 1st M1 for a correct numerator (must have - 6) and ft their se.
1st A1 for awrt 1.19
3rd B1 for \(\pm\) 1.6449 seen or probability of awrt 0.117, Sign must match their test statistic.
3rd dM1 dep. on 1st M1 for a correct statement based on their normal cv and their test statistic. Ignore their hypotheses. Allow accept \(\mathrm{H}_0\) but reject \(\mathrm{H}_1\) is M0. Can be implied by correct conclusion.
2nd A1cso for correct comment in context dependent upon all other marks being awarded.
Must mention merchandise, standard and premium ticket holders and 6 or manager and belief or claim
NB Use of cv for difference in means \(D\) will have \(D = 6 + 1.6449 \times \text{s.e.} =\) awrt 8.33 and requires sight of \(d = 8\) with a comment for the 3rd M1
| Scheme | Marks |
|---|---|
| Sample size is large so Central Limit Theorem (CLT) applies so | B1 |
| do not need to assume merchandise sold has a normal distribution. | dB1 |
| (2) | |
| (13 marks) |
Notes
1st B1 for mentioning large samples and CLT
2nd dB1 dependent on 1st B1 for stating no need to assume normality. Require merchandise sold not mean merchandise sold.