| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | One-tailed test (increase or decrease) |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution properties and hypothesis testing. Part (i) uses the sum of independent Poisson variables, parts (ii)-(iii) involve standard Poisson probability calculations and a routine one-tailed hypothesis test. All techniques are standard S2 content with no novel insight required, making it slightly easier than average A-level difficulty. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(Po(1.0)\) | B1 | Seen or implied |
| \(e^{-1}\left(1 + 1 + \frac{1^2}{2}\right)\) | M1 | Allow any \(\lambda\). Allow one end error. |
| \(= 0.920\) (3 sfs) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 3) = 1 - e^{-1.5}\left(1 + 1.5 + \frac{1.5^2}{2} + \frac{1.5^3}{3!}\right)\) | M1 | Allow any \(\lambda\). Allow one end error |
| \(= 0.0656\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Incorrectly concluding that more absences than usual when there are not oe | B1 | In context |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \lambda = 1.5\) (or 0.3), \(H_1: \lambda > 1.5\) (or 0.3) | B1 | Or \(\mu\); Both |
| \(P(X > 4) = \text{"0.0656"} - e^{-1.5} \times \frac{1.5^4}{4!} = 0.0186\) (3 sf) | M1 | or \(1 - e^{-1.5}\left(1+1.5+\frac{1.5^2}{2}+\frac{1.5^3}{3!}+\frac{1.5^4}{4!}\right)\) |
| \(P(\text{Type I}) = 0.0186\) or \(0.0185\) | A1ft | Ft their \(P(X > 4)\) if less than 0.05 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 3) = \text{"0.0656"}\) | B1ft | Ft their (ii) |
| \(0.0656 > 0.05\) | M1 | |
| No evidence of more than usual male absences | A1ft | Ft their \(P(X>3)\). Correct conclusion. No contradictions. |
## Question 7(i):
$Po(1.0)$ | B1 | Seen or implied
$e^{-1}\left(1 + 1 + \frac{1^2}{2}\right)$ | M1 | Allow any $\lambda$. Allow one end error.
$= 0.920$ (3 sfs) | A1 |
**Total: 3 marks**
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## Question 7(ii):
$P(X > 3) = 1 - e^{-1.5}\left(1 + 1.5 + \frac{1.5^2}{2} + \frac{1.5^3}{3!}\right)$ | M1 | Allow any $\lambda$. Allow one end error
$= 0.0656$ | A1 |
**Total: 2 marks**
---
## Question 7(iii)(a):
Incorrectly concluding that more absences than usual when there are not oe | B1 | In context
**Total: 1 mark**
---
## Question 7(iii)(b):
$H_0: \lambda = 1.5$ (or 0.3), $H_1: \lambda > 1.5$ (or 0.3) | B1 | Or $\mu$; Both
$P(X > 4) = \text{"0.0656"} - e^{-1.5} \times \frac{1.5^4}{4!} = 0.0186$ (3 sf) | M1 | or $1 - e^{-1.5}\left(1+1.5+\frac{1.5^2}{2}+\frac{1.5^3}{3!}+\frac{1.5^4}{4!}\right)$
$P(\text{Type I}) = 0.0186$ or $0.0185$ | A1ft | Ft their $P(X > 4)$ if less than 0.05
**Total: 3 marks**
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## Question 7(iii)(c):
$P(X > 3) = \text{"0.0656"}$ | B1ft | Ft their **(ii)**
$0.0656 > 0.05$ | M1 |
No evidence of more than usual male absences | A1ft | Ft their $P(X>3)$. Correct conclusion. No contradictions.
**Total: 3 marks**7 The number of absences by girls from a certain class on any day is modelled by a random variable with distribution $\operatorname { Po } ( 0.2 )$. The number of absences by boys from the same class on any day is modelled by an independent random variable with distribution $\operatorname { Po } ( 0.3 )$.\\
(i) Find the probability that, during a randomly chosen 2-day period, the total number of absences is less than 3 .\\
(ii) Find the probability that, during a randomly chosen 5-day period, the number of absences by boys is more than 3.\\
(iii) The teacher claims that, during the football season, there are more absences by boys than usual. In order to test this claim at the 5\% significance level, he notes the number of absences by boys during a randomly chosen 5-day period during the football season.
\begin{enumerate}[label=(\alph*)]
\item State what is meant by a Type I error in this context.
\item State appropriate null and alternative hypotheses and find the probability of a Type I error.
\item In fact there were 4 absences by boys during this period. Test the teacher's claim at the 5\% significance level.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2018 Q7 [12]}}