A2 October 2020 Q1
1. The number of customers entering Jeff’s supermarket each morning follows a Poisson distribution.
Past information shows that customers enter at an average rate of 2 every 5 minutes.
Using this information,
A rival supermarket is opened nearby. Following its opening, the number of customers entering Jeff’s supermarket over a randomly selected 40-minute period is found to be 10
A further randomly selected 20-minute period is observed and the hypothesis test is repeated.
Given that the true rate of customers entering Jeff’s supermarket is now 1 every 5 minutes,
| Scheme | Marks | AO |
|---|---|---|
| (i) \(X \sim \mathrm{Po}(24)\) | B1 | 3.4 |
| \(\mathrm{P}(X = 26) = 0.071912\ldots\) awrt 0.0719 | B1 | 1.1b |
| (2) | ||
| (ii) \(\mathrm{P}(X \geqslant 21) = 1 - \mathrm{P}(X \leqslant 20)\ [= 1 - 0.24263\ldots]\) | M1 | 3.4 |
| \(= 0.75736\ldots\) awrt 0.757 | A1 | 1.1b |
| (2) |
Notes
(i) B1: For realising the distribution is \(\mathrm{Po}(24)\) (May be seen or implied in part (ii))
B1: awrt 0.0719
(ii) M1: Writing or using \(1 - \mathrm{P}(X \leqslant 20)\)
A1: awrt 0.757
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0: \lambda = 2 \qquad [\mu = 16]\) \(\mathrm{H}_1: \lambda \lt 2 \qquad [\mu \lt 16]\) | B1 | 2.5 |
| \(\mathrm{P}(Y \leqslant 10 \mid Y \sim \mathrm{Po}(16)) = 0.077396\ldots\) awrt 0.0774 | B1 | 1.1b |
| Not significant / Do not reject \(\mathrm{H}_0\) / 10 is not in the CR | M1 | 1.1b |
| There is not sufficient evidence to suggest a decrease/change in the rate of customers entering Jeff’s supermarket. | A1 | 2.2b |
| (4) |
Notes
B1: Both hypotheses correct (must use \(\mu\) or \(\lambda\))
B1: awrt 0.0774 Allow awrt 0.08 from a correct probability statement.
allow CR: \(X \leqslant 9\)
M1: Correct non-contextual conclusion (may be implied by correct contextual conclusion). Allow a f.t. comparison of ‘their \(p\)’ with 0.05
(Ignore any contradictory contextual comments for this mark)
A1: A fully correct solution drawing a correct inference in context with all previous marks in (b) scored.
| Scheme | Marks | AO |
|---|---|---|
| Use of \(\mathrm{Po}(8)\) to attempt critical region | M1 | 2.1 |
| Critical region is \(Y \leqslant 3\) / \(\mathrm{H}_0\) is not rejected when \(Y \geqslant 4\) | A1 | 1.1b |
| True distribution is \(W \sim \mathrm{Po}(4)\) | B1 | 2.1 |
| \(\mathrm{P}(W \geqslant 4 \mid W \sim \mathrm{Po}(4)) = 1 - \mathrm{P}(W \leqslant 3)\ [= 1 - 0.43347\ldots]\) | M1 | 1.1b |
| \(= 0.56652\ldots\) awrt 0.567 | A1 | 1.1b |
| (5) | ||
| (13 marks) |
Notes
M1: Use of \(\mathrm{Po}(8)\) to attempt critical region [\(\mathrm{P}(Y \leqslant 3) = 0.0423..\ \ \mathrm{P}(Y \leqslant 4) = 0.0996..\)]
A1: Finding critical region for the test \(Y \leqslant 3\) which must come from \(\mathrm{Po}(8)\).
B1: Identifying the need to use \(\mathrm{Po}(4)\) as the true distribution.
Allow \(\mathrm{Po}(4)\) seen or used for this mark.
M1: Writing or using \(\mathrm{P}(W \geqslant \text{‘}4\text{’})\) or \(1 - \mathrm{P}(W \leqslant \text{‘}3\text{’})\) from \(\mathrm{Po}(4)\). Allow f.t. on their identified CR but must be using \(\mathrm{Po}(4)\)
A1: awrt 0.567