CAIE S2 2023 November — Question 5 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2023
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of a Poisson distribution
TypeOne-tailed test (increase or decrease)
DifficultyStandard +0.3 This is a straightforward hypothesis test for a Poisson mean with clear setup. Students must recognize to scale the parameter (0.31 × 5 = 1.55), set up H₀: λ = 1.55 vs H₁: λ > 1.55, and calculate P(X ≥ 5) using tables or cumulative probabilities. While it requires careful handling of the one-tailed test and comparison with 2.5%, it's a standard application of taught procedures with no conceptual surprises—slightly easier than average due to explicit guidance that Poisson is appropriate and clear statement of what changed.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

In the past the number of enquiries per minute at a customer service desk has been modelled by a random variable with distribution 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. [5]

Question 5:
AnswerMarks
5H : Population mean no. enquiries = 1.55
0
H : Population mean no. enquiries > 1.55
AnswerMarks Guidance
1B1 Or “population mean no. enquiries = 0.31 (per minute)”
oe.
Allow 'λ = 1.55’ or µ = ‘1.55’.
1.552 1.553 1.554
P(X ⩾ 5) = 1 – e-1.55(1 + 1.55 + + + )
2! 3! 4!
or 1 – e-1.55(1 + 1.55 + 1.20125 + 0.62065 + 0.24050)
AnswerMarks Guidance
or 1 – (0.21225 + 0.32898 + 0.25496 + 0.13173 + 0.05105)M1 Allow one end error, e.g. extra term: e-1.55 ×1.555 .
5!
AnswerMarks Guidance
= 0.0210 (3 sf)A1 Allow 0.021.
SC B1 no working scores B1 instead of M1A1.
AnswerMarks Guidance
0.0210 < 0.025M1 For valid comparison.
[Reject H ] There is sufficient evidence [at 2.5% level] to suggest that mean no.
0
AnswerMarks Guidance
of enquiries has increased.A1 FT In context, not definite,
e.g., not "Mean no. of enquiries has increased".
No contradictions.
AnswerMarks
51.555
Note: e-1.55× = 0.0158 < 0.025: scores max B1
5!
AnswerMarks Guidance
QuestionAnswer Marks
Question 5:
5 | H : Population mean no. enquiries = 1.55
0
H : Population mean no. enquiries > 1.55
1 | B1 | Or “population mean no. enquiries = 0.31 (per minute)”
oe.
Allow 'λ = 1.55’ or µ = ‘1.55’.
1.552 1.553 1.554
P(X ⩾ 5) = 1 – e-1.55(1 + 1.55 + + + )
2! 3! 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 ×1.555 .
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 ] There is sufficient evidence [at 2.5% level] to suggest that mean no.
0
of enquiries has increased. | A1 FT | In context, not definite,
e.g., not "Mean no. of enquiries has increased".
No contradictions.
5 | 1.555
Note: e-1.55× = 0.0158 < 0.025: scores max B1
5!
Question | Answer | Marks | Guidance
In the past the number of enquiries per minute at a customer service desk has been modelled by a random variable with distribution 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. [5]

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