June 2022 Paper 2 Q12
12 A firm claims that no more than 2% of their packets of sugar are underweight. A market researcher believes that the actual proportion is greater than 2%. In order to test the firm’s claim, the researcher weighs a random sample of 600 packets and carries out a hypothesis test, at the 5% significance level, using the null hypothesis \(p = 0.02\).
“18 out of 600 is 3%, so there is evidence that the actual proportion of underweight bags is greater than 2%.”
Criticise this statement. [1]
| Scheme | Marks | AO |
|---|---|---|
| \(X \sim \mathrm{B}(600, 0.02)\) | M1 | 3.3 |
| Attempt \(\mathrm{P}(X \geqslant n)\) for \(17 \leqslant n \leqslant 20\) | M1 | 2.1 |
| \((\mathrm{P}(X \geqslant 18) =)\ 0.0610\) or 0.061 (2 sf) | A1 | 3.4 |
| \((\mathrm{P}(X \geqslant 19) =)\ 0.0359\) or 0.036 (2 sf) | A1 | 1.1 |
| P(concludes claim incorrect) \(= 0.0359\) (3 sf) | A1 | 2.2a |
| [5] |
Notes
M1: soi, eg \(\mathrm{H}_0: p = 0.02\) and \(\mathrm{B}(600, p)\). Allow \(n = 600\), \(p = 0.02\)
M1: May be implied by 0.0991 or 0.0202 or 0.9798 or 0.9009 or correct values
A1: or \((\mathrm{P}(X \leqslant 17) =)\ 0.939\) or 0.94 (2 sf)
A1: or \((\mathrm{P}(X \leqslant 18) =)\ 0.964\) or 0.96 (2 sf)
These two probabilities seen imply M1M1A1A1
Condone errors such as \(\mathrm{P}(X \gt 18) = 0.0610\)
A1: Ignore hypotheses and/or “Reject \(\mathrm{H}_0\)” or similar
Unsupported answers:
0.0359: M1M1A1A0A0
Critical region is \(X \geqslant 19\): M1M1A0A0A0
Alternative method (normal with no cc)
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{N}(600 \times 0.02, 600 \times 0.02 \times 0.98)\) or \(X \sim \mathrm{N}(12, 11.76)\) | M1 |
| Attempt \(\mathrm{P}(X \geqslant n)\) for \(17 \leqslant n \leqslant 20\) | M1 |
| \(\mathrm{P}(X \geqslant 17) = 0.0724\) or 0.072 (2 sf) | A1 |
| \(\mathrm{P}(X \geqslant 18) = 0.0401\) or 0.040 (2 sf) | A1 |
| P(concludes claim incorrect) \(= 0.0401\) | A0 |
M1: soi. Can be scored either for \(\mathrm{N}(12, 11.76)\) or \(\mathrm{B}(600, 0.02)\)
M1: \(\mathrm{P}(x \gt a) = 0.05 \Rightarrow a = 17.64\) only gets M1 if a probability is calculated
Alternative method (normal with cc)
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{N}(600 \times 0.02, 600 \times 0.02 \times 0.98)\) or \(X \sim \mathrm{N}(12, 11.76)\) | M1 |
| Attempt \(\mathrm{P}(X \geqslant n)\) for \(17 \leqslant n \leqslant 20\) | M1 |
| \(\mathrm{P}(X \geqslant 18) = \mathrm{P}(X \geqslant 17.5) = 0.054\) (2 sf) | A1 |
| \(\mathrm{P}(X \geqslant 19) = \mathrm{P}(X \geqslant 18.5) = 0.0290\) (2 sf) | A1 |
| P(concludes claim incorrect) \(= 0.0290\) | A0 |
M1: soi. Can be scored either for \(\mathrm{N}(12, 11.76)\) or \(\mathrm{B}(600, 0.02)\)
| Scheme | Marks | AO |
|---|---|---|
| (Incorrect because eg:) You have to consider \(\mathrm{P}(X \geqslant 18)\) or 18 is in the acceptance region (for 5% test) or critical region is \(\geqslant 19\), or CV is 19 | B1 | 2.3 |
| [1] |
Notes
B1: or 18 is under the significance level
Allow You have to do a proper hypothesis test
No other answers acceptable