CAIE S2 2014 June — Question 3 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating the Poisson to the Normal distribution
TypeHypothesis test for sum of Poisson observations
DifficultyStandard +0.8 This question requires understanding that the sum of independent Poisson distributions is also Poisson, then applying a normal approximation to conduct a two-tailed hypothesis test. While the individual steps are standard (setting up hypotheses, calculating test statistic, comparing to critical value), the question demands conceptual understanding of distribution theory and careful execution of the approximation method, making it moderately harder than typical A-level questions.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion5.05c Hypothesis test: normal distribution for population mean

3 The number of calls per day to an enquiry desk has a Poisson distribution. In the past the mean has been 5 . In order to test whether the mean has changed, the number of calls on a random sample of 10 days was recorded. The total number of calls was found to be 61 . Use an approximate distribution to test at the 10\% significance level whether the mean has changed.

AnswerMarks Guidance
H0: Pop mean (or \(\mu\) or \(\lambda\)) = 50 (or 5)B1 Not just "mean"
H1: Pop mean (or \(\mu\) or \(\lambda\) ) \(\neq\) 50 (or 5)
AnswerMarks Guidance
\(\frac{60.5 - 50}{\sqrt{50}} = (\pm)1.485\) OR \(0.0687\) OR C.VM1 For standardising with \(N(50,50)\) or \(N(5,5/\sqrt{10})\). Allow M1 with wrong or no continuity correction OR no \(\sqrt{\ }\)
A1(accept c.v method M1, A1 for 61.63 or 48.868)
\(1.485 < 1.645\) or \(0.0687 > 0.05\)M1 For valid comparison (z_s or areas or cv). (S.R For cv comparison 61.63 only award final A1 if cc used)
No evidence that mean changedA1 or if H1: \(\lambda > 50\), \(1.485 < 1.96\) M1. No evid mean changed A0 (i.e. if one-tail test, max B0 M1 A1 M1 A0)
[5]
**H0:** Pop mean (or $\mu$ or $\lambda$) = 50 (or 5) | B1 | Not just "mean"

**H1:** Pop mean (or $\mu$ or $\lambda$ ) $\neq$ 50 (or 5)

$\frac{60.5 - 50}{\sqrt{50}} = (\pm)1.485$ OR $0.0687$ OR C.V | M1 | For standardising with $N(50,50)$ or $N(5,5/\sqrt{10})$. Allow M1 with wrong or no continuity correction OR no $\sqrt{\ }$

| A1 | (accept c.v method M1, A1 for 61.63 or 48.868)

$1.485 < 1.645$ or $0.0687 > 0.05$ | M1 | For valid comparison (z_s or areas or cv). (S.R For cv comparison 61.63 only award final A1 if cc used)

No evidence that mean changed | A1 | or if H1: $\lambda > 50$, $1.485 < 1.96$ M1. No evid mean changed A0 (i.e. if one-tail test, max B0 M1 A1 M1 A0)

[5]

---
3 The number of calls per day to an enquiry desk has a Poisson distribution. In the past the mean has been 5 . In order to test whether the mean has changed, the number of calls on a random sample of 10 days was recorded. The total number of calls was found to be 61 . Use an approximate distribution to test at the 10\% significance level whether the mean has changed.

\hfill \mbox{\textit{CAIE S2 2014 Q3 [5]}}