CAIE S2 2015 June — Question 2 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2015
SessionJune
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 one-tailed Poisson hypothesis test requiring students to state hypotheses (H₀: λ=0.5, H₁: λ>0.5) and find P(X≥2) from tables. It's slightly above average difficulty due to the scaling (5m² sample) and one-tailed nature, but follows a standard template with no conceptual surprises—typical S2 material that's more routine than problem-solving.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.02i Poisson distribution: random events model

2 Cloth made at a certain factory has been found to have an average of 0.1 faults per square metre. Suki claims that the cloth made by her machine contains, on average, more than 0.1 faults per square metre. In a random sample of \(5 \mathrm {~m} ^ { 2 }\) of cloth from Suki's machine, it was found that there were 2 faults. Assuming that the number of faults per square metre has a Poisson distribution,
  1. state null and alternative hypotheses for a test of Suki's claim,
  2. test at the \(10 \%\) significance level whether Suki's claim is justified.

Question 2
(i)
(ii)
Hypotheses:
\(H_0: \lambda = 0.5\) B1
\(H_1: \lambda > 0.5\)
Calculation:
\(1 - e^{-0.5}(1 + 0.5)\) M1
\(= 0.0902\) (3 sf) A1
Compare to 0.1 M1
Claim justified or there is evidence to support claim A1
Guidance notes:
- Accept \(\mu\) instead of \(\lambda\)
- Accept population mean = 0.5, not just mean = 0.5
- Accept population mean (per m) = 0.1
- Allow \(1 - P(X = 0,1)\) attempted, any \(\lambda\); allow 1 end error
- Allow 0.09
- Valid comparison. NB: if \(0.9098 > 0.9\) recovers M1A1 M1
- Accept 'Reject \(H_0\)' if correctly defined or oe
- No contradictions
Total: 5
**Question 2**

**(i)**

**(ii)**

**Hypotheses:**
$H_0: \lambda = 0.5$ B1

$H_1: \lambda > 0.5$

**Calculation:**
$1 - e^{-0.5}(1 + 0.5)$ M1

$= 0.0902$ (3 sf) A1

Compare to 0.1 M1

Claim justified or there is evidence to support claim A1

**Guidance notes:**

- Accept $\mu$ instead of $\lambda$
- Accept population mean = 0.5, not just mean = 0.5
- Accept population mean (per m) = 0.1
- Allow $1 - P(X = 0,1)$ attempted, any $\lambda$; allow 1 end error
- Allow 0.09
- Valid comparison. NB: if $0.9098 > 0.9$ recovers M1A1 M1
- Accept 'Reject $H_0$' if correctly defined or oe
- No contradictions

**Total: 5**
2 Cloth made at a certain factory has been found to have an average of 0.1 faults per square metre. Suki claims that the cloth made by her machine contains, on average, more than 0.1 faults per square metre. In a random sample of $5 \mathrm {~m} ^ { 2 }$ of cloth from Suki's machine, it was found that there were 2 faults. Assuming that the number of faults per square metre has a Poisson distribution,\\
(i) state null and alternative hypotheses for a test of Suki's claim,\\
(ii) test at the $10 \%$ significance level whether Suki's claim is justified.

\hfill \mbox{\textit{CAIE S2 2015 Q2 [5]}}