| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find Type I error probability |
| Difficulty | Standard +0.3 This is a straightforward application of hypothesis testing with a Poisson distribution. Part (a) requires calculating P(X < 3) when λ = 5.7 using standard Poisson tables or formulas—pure recall and substitution. Part (b) similarly requires P(X ≥ 3) when λ = 0.9. Both parts are routine calculations with no conceptual challenges beyond knowing the definitions of Type I and Type II errors, making this slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-5.7}\!\left(1 + 5.7 + \frac{5.7^2}{2!}\right)\) or \(e^{-5.7}(1 + 5.7 + 16.245)\) or \(0.003346 + 0.01907 + 0.05436\) | M1 | Allow one end error. Must see this expression. |
| \(= 0.0768\) (3 sf) | A1 | SC B1 for unsupported answer of 0.0768. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-0.9}\!\left(1 + 0.9 + \frac{0.9^2}{2!}\right)\) | M1 | Attempted; allow one end error (must see expression). |
| \(= 1 - e^{-0.9}\!\left(1 + 0.9 + \frac{0.9^2}{2!}\right) = 1 - e^{-0.9}(1 + 0.9 + 0.405) = 1-(0.4066+3659+0.1647)\) | A1 | Correct expression \(P(X \geq 3)\) no end errors (must see expression). |
| \(= 0.0629\) (3 sf) | A1 | SC B2 for unsupported answer of 0.0629. |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-5.7}\!\left(1 + 5.7 + \frac{5.7^2}{2!}\right)$ or $e^{-5.7}(1 + 5.7 + 16.245)$ or $0.003346 + 0.01907 + 0.05436$ | M1 | Allow one end error. Must see this expression. |
| $= 0.0768$ (3 sf) | A1 | SC B1 for unsupported answer of 0.0768. |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-0.9}\!\left(1 + 0.9 + \frac{0.9^2}{2!}\right)$ | M1 | Attempted; allow one end error (must see expression). |
| $= 1 - e^{-0.9}\!\left(1 + 0.9 + \frac{0.9^2}{2!}\right) = 1 - e^{-0.9}(1 + 0.9 + 0.405) = 1-(0.4066+3659+0.1647)$ | A1 | Correct expression $P(X \geq 3)$ no end errors (must see expression). |
| $= 0.0629$ (3 sf) | A1 | SC B2 for unsupported answer of 0.0629. |
---4 The number of accidents per 3-month period on a certain road has the distribution $\operatorname { Po } ( \lambda )$. In the past the value of $\lambda$ has been 5.7. Following some changes to the road, the council carries out a hypothesis test to determine whether the value of $\lambda$ has decreased. If there are fewer than 3 accidents in a randomly chosen 3 -month period, the council will conclude that the value of $\lambda$ has decreased.
\begin{enumerate}[label=(\alph*)]
\item Find the probability of a Type I error.
\item Find the probability of a Type II error if the mean number of accidents per 3-month period is now actually 0.9 .
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q4 [5]}}