| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | One-tailed test (increase or decrease) |
| Difficulty | Moderate -0.8 This is a straightforward hypothesis testing question requiring only standard recall: stating H₀ and H₁ for a Poisson test, then comparing a given p-value (0.0683) to the significance level (0.05) to reach a conclusion. No calculations are required, and the interpretation is routine—well below average difficulty for A-level. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): Pop mean \(= 3\); \(H_1\): Pop mean \(> 3\) | B1 [1] | Allow or \(\mu\) or \(\lambda\), but not just 'mean' |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.0683 > 0.05\) | M1 | For inequality stated or clearly shown on diagram |
| No evidence that pop mean increased | A1ft [2] | Allow 'No increase in mean' |
## Question 1:
**(i)**
$H_0$: Pop mean $= 3$; $H_1$: Pop mean $> 3$ | B1 [1] | Allow or $\mu$ or $\lambda$, but not just 'mean'
**(ii)**
$0.0683 > 0.05$ | M1 | For inequality stated or clearly shown on diagram
No evidence that pop mean increased | A1ft [2] | Allow 'No increase in mean'
---1 The number of new enquiries per day at an office has a Poisson distribution. In the past the mean has been 3 . Following a change of staff, the manager wishes to test, at the $5 \%$ significance level, whether the mean has increased.\\
(i) State the null and alternative hypotheses for this test.
The manager notes the number, $N$, of new enquiries during a certain 6 -day period. She finds that $N = 25$ and then, assuming that the null hypothesis is true, she calculates that $\mathrm { P } ( N \geqslant 25 ) = 0.0683$.\\
(ii) What conclusion should she draw?
\hfill \mbox{\textit{CAIE S2 2012 Q1 [3]}}