June 2022 Paper 2 Q12
12 A retailer sells bags of flour which are advertised as containing 1.5 kg of flour. A trading standards officer is investigating whether there is enough flour in each bag. He collects a random sample and uses software to carry out a hypothesis test at the 5% level. The analysis is shown in the software printout below.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H_0}: \mu = 1.5\) \(\mathrm{H_1}: \mu \lt 1.5\) | B1 | 1.1 |
| \(\mu\) is the population mean weight of flour in a bag | B1 | 2.5 |
| [2] |
Notes
B1: both hypotheses in terms of \(\mu\)
B1: allow \(\mu\) is the population mean weight of a bag of flour
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{N}(a, b)\) | M1 | 3.3 |
| \(a = 1.5\) or \(b = \frac{0.24^2}{32}\) or 0.0018 | A1 | 2.2a |
| \(\mathrm{N}\left(1.5, \frac{0.24^2}{32}\right)\) isw or \(\mathrm{N}(1.5, 0.0018)\) isw | A1 | 3.1a |
| [3] |
Notes
M1: \(a\) and \(b\) are numerical values
A1: allow \(0.0424^2\) for variance
A0 if answer spoiled by wrong variable quoted eg \(\mu \sim \mathrm{N}\left(1.5, \frac{0.24^2}{32}\right)\) or \(X \sim \mathrm{N}\left(1.5, \frac{0.24^2}{32}\right)\);
allow only \(\bar{X}\) oe if variable included
| Scheme | Marks | AO |
|---|---|---|
| \(0.0786 \gt 0.05\) or \(-1.4142 \gt -1.645\) | M1 | 3.4 |
| do not reject \(\mathrm{H_0}\) | A1 | 1.1 |
| there is insufficient evidence at the 5% level to suggest that the mean weight of the flour in the bags is less than 1.5 kg | A1 | 2.2b |
| [3] |
Notes
M1: or \(1.44 \gt 1.43(02586\ldots)\)
NB 1.43(02586…) is from InvNorm(0.05, 1.5, 0.0424)
A1: allow accept \(\mathrm{H_0}\) or not significant or reject \(\mathrm{H_1}\)
A1: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest
if calculated values are used full marks may be awarded for awrt 0.07865 or 0.0786, or \(-1.415 \leqslant z \leqslant -1.414\);
otherwise award a maximum of M1A1 for 0.07…or −1.4…
other calculated values score M0