CAIE S2 2023 November — Question 5 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2023
SessionNovember
Marks5
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TopicHypothesis test of a Poisson distribution
TypeOne-tailed test (increase or decrease)
DifficultyStandard +0.3 This is a straightforward one-tailed Poisson hypothesis test requiring students to scale the parameter (0.31 × 5 = 1.55), set up H₀ and H₁, calculate P(X ≥ 5) using tables or calculator, and compare to 2.5%. The question explicitly states the test direction and significance level, making it slightly easier than average but still requiring proper hypothesis testing procedure.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05c Hypothesis test: normal distribution for population mean
5 In the past the number of enquiries per minute at a customer service desk has been modelled by a random variable with distribution \(\operatorname { Po } ( 0.31 )\). Following a change in the position of the desk, it is expected that the mean number of enquiries per minute will increase. In order to test whether this is the case, the total number of enquiries during a randomly chosen 5-minute period is noted. You should assume that a Poisson model is still appropriate. Given that the total number of enquiries is 5 , carry out the test at the \(2.5 \%\) significance level.
Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0\): Population mean no. enquiries \(= 1.55\); \(H_1\): Population mean no. enquiries \(> 1.55\)B1 Or "population mean no. enquiries \(= 0.31\) (per minute)" oe. Allow '\(\lambda = 1.55\)' or '\(\mu = 1.55\)'
\(P(X \geq 5) = 1 - e^{-1.55}(1 + 1.55 + \frac{1.55^2}{2!} + \frac{1.55^3}{3!} + \frac{1.55^4}{4!})\) or \(1 - e^{-1.55}(1 + 1.55 + 1.20125 + 0.62065 + 0.24050)\) or \(1 - (0.21225 + 0.32898 + 0.25496 + 0.13173 + 0.05105)\)M1 Allow one end error, e.g. extra term: \(e^{-1.55} \times \frac{1.55^5}{5!}\)
\(= 0.0210\) (3 sf)A1 Allow 0.021. SC B1 no working scores B1 instead of M1A1
\(0.0210 < 0.025\)M1 For valid comparison
[Reject \(H_0\)] There is sufficient evidence [at 2.5% level] to suggest that mean no. of enquiries has increasedA1 FT In context, not definite. e.g. not "Mean no. of enquiries has increased". No contradictions 
## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Population mean no. enquiries $= 1.55$; $H_1$: Population mean no. enquiries $> 1.55$ | B1 | Or "population mean no. enquiries $= 0.31$ (per minute)" oe. Allow '$\lambda = 1.55$' or '$\mu = 1.55$' |
| $P(X \geq 5) = 1 - e^{-1.55}(1 + 1.55 + \frac{1.55^2}{2!} + \frac{1.55^3}{3!} + \frac{1.55^4}{4!})$ or $1 - e^{-1.55}(1 + 1.55 + 1.20125 + 0.62065 + 0.24050)$ or $1 - (0.21225 + 0.32898 + 0.25496 + 0.13173 + 0.05105)$ | M1 | Allow one end error, e.g. extra term: $e^{-1.55} \times \frac{1.55^5}{5!}$ |
| $= 0.0210$ (3 sf) | A1 | Allow 0.021. **SC B1** no working scores B1 instead of M1A1 |
| $0.0210 < 0.025$ | M1 | For valid comparison |
| [Reject $H_0$] There is sufficient evidence [at 2.5% level] to suggest that mean no. of enquiries has increased | A1 FT | In context, not definite. e.g. not "Mean no. of enquiries has increased". No contradictions |
5 In the past the number of enquiries per minute at a customer service desk has been modelled by a random variable with distribution $\operatorname { Po } ( 0.31 )$. Following a change in the position of the desk, it is expected that the mean number of enquiries per minute will increase. In order to test whether this is the case, the total number of enquiries during a randomly chosen 5-minute period is noted. You should assume that a Poisson model is still appropriate.

Given that the total number of enquiries is 5 , carry out the test at the $2.5 \%$ significance level.\\

\hfill \mbox{\textit{CAIE S2 2023 Q5 [5]}}
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