| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find Type I error probability |
| Difficulty | Standard +0.8 This question requires understanding of hypothesis testing with Poisson distributions, including calculating Type I and Type II error probabilities. Students must recognize that a 5-day period means λ = 5×0.9 = 4.5 for the null hypothesis, find P(X < 3) under H₀, then repeat for H₁ with λ = 5×0.2 = 1.0. While the calculations are straightforward once set up, the conceptual understanding of error types and the multi-day period scaling makes this moderately challenging for A-level. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\lambda =) 4.5\) | B1 | |
| \(e^{-4.5}(1 + 4.5 + \frac{4.5^2}{2!})\) | M1 | Allow any \(\lambda\). Allow one end error |
| \(= 0.174\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Accept reduction in mean no. of missed appts although untrue | B1 | or Mean is \(0.9\) (or \(4.5\)) but \(< 3\) missed appts. In context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X \geqslant 3)\) | M1 | Attempted |
| \(= 1 - e^{-1}(1 + 1 + \frac{1^2}{2!})\) | M1 | Allow any \(\lambda\) except \(4.5\) or \(0.9\). Allow one end error |
| \(= 0.0803\) (3 sfs) | A1 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\lambda =) 4.5$ | B1 | |
| $e^{-4.5}(1 + 4.5 + \frac{4.5^2}{2!})$ | M1 | Allow any $\lambda$. Allow one end error |
| $= 0.174$ | A1 | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Accept reduction in mean no. of missed appts although untrue | B1 | or Mean is $0.9$ (or $4.5$) but $< 3$ missed appts. In context |
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X \geqslant 3)$ | M1 | Attempted |
| $= 1 - e^{-1}(1 + 1 + \frac{1^2}{2!})$ | M1 | Allow any $\lambda$ except $4.5$ or $0.9$. Allow one end error |
| $= 0.0803$ (3 sfs) | A1 | |
---4 At a doctors' surgery, the number of missed appointments per day has a Poisson distribution. In the past the mean number of missed appointments per day has been 0.9 . Following some publicity, the manager carries out a hypothesis test to determine whether this mean has decreased. If there are fewer than 3 missed appointments in a randomly chosen 5-day period, she will conclude that the mean has decreased.\\
(i) Find the probability of a Type I error.\\
(ii) State what is meant by a Type I error in this context.\\
(iii) Find the probability of a Type II error if the mean number of missed appointments per day is 0.2 .\\
\hfill \mbox{\textit{CAIE S2 2017 Q4 [7]}}