AQA S3 2016 June — Question 3 7 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2016
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of a Poisson distribution
TypeCompare two Poisson means
DifficultyStandard +0.3 This is a standard two-sample Poisson hypothesis test requiring students to set up hypotheses, calculate sample means (315/30 and 747/60), use the normal approximation for the difference of Poisson means, compute a test statistic, and compare to critical value. While it involves multiple steps and careful interpretation of context, it follows a routine S3 procedure with no novel insights required, making it slightly easier than average.
Spec5.05b Unbiased estimates: of population mean and variance
Car parking in a market town's high street was, until 31 May 2014, limited to one hour free of charge between 8 am and 6 pm. Records show that, during a period of 30 days prior to this date, a total of 315 penalty tickets were issued. Car parking in the high street later became limited to thirty minutes free of charge between 8 am and 6 pm. A subsequent investigation revealed that, during a period of 60 days from 1 October 2014, a total of 747 penalty tickets were issued. The daily numbers of penalty tickets issued may be modelled by independent Poisson distributions with means \(\lambda_A\) until 31 May 2014 and \(\lambda_B\) from 1 October 2014. Investigate, at the 1\% level of significance, a claim by traders on the high street that \(\lambda_B > \lambda_A\). [7 marks]
AnswerMarks Guidance
\(H_0: \lambda_B = \lambda_A\)B1 Both
\(H_1: \lambda_B > \lambda_A\)
\(CV(1\%) \Rightarrow z = \mathbf{2.32 \text{ to } 2.33}\)B1 AWFW (2.3263)
\(\hat{\lambda}_A = \frac{315}{30} = \mathbf{10.5}\) and \(\hat{\lambda}_B = \frac{747}{60} = \mathbf{12.45}\)B1 Both CAO; \(\lambda = \frac{1062}{90} = 11.8\)
\(z = \frac{12.45 - 10.5}{\sqrt{\frac{12.45}{60} + \frac{10.5}{30}}} = \frac{2.61}{\sqrt{\frac{12.45}{60} + \frac{10.5}{30}}}\)M1 Correct numerator
M1Correct denominator
Adep1AWRT; dep on M1 M1 (2.61163)
or
AnswerMarks Guidance
\(z = \frac{12.45 - 10.5}{\sqrt{11.8(\frac{1}{60} + \frac{1}{30})}} = \frac{2.54}{\sqrt{11.8(\frac{1}{60} + \frac{1}{30})}}\)(M1) Correct numerator
(M1)Correct denominator
(A1)AWRT; dep on M1 M1 (2.53868)
Thus evidence, at 1% level, to support the claim that \(\lambda_B > \lambda_A\)Adep1 Dep on z-value and CV
7
Total 7 
$H_0: \lambda_B = \lambda_A$ | B1 | Both
$H_1: \lambda_B > \lambda_A$ | | 

$CV(1\%) \Rightarrow z = \mathbf{2.32 \text{ to } 2.33}$ | B1 | AWFW (2.3263)

$\hat{\lambda}_A = \frac{315}{30} = \mathbf{10.5}$ and $\hat{\lambda}_B = \frac{747}{60} = \mathbf{12.45}$ | B1 | Both CAO; $\lambda = \frac{1062}{90} = 11.8$

$z = \frac{12.45 - 10.5}{\sqrt{\frac{12.45}{60} + \frac{10.5}{30}}} = \frac{2.61}{\sqrt{\frac{12.45}{60} + \frac{10.5}{30}}}$ | M1 | Correct numerator
| M1 | Correct denominator
| Adep1 | AWRT; dep on M1 M1 (2.61163)

or

$z = \frac{12.45 - 10.5}{\sqrt{11.8(\frac{1}{60} + \frac{1}{30})}} = \frac{2.54}{\sqrt{11.8(\frac{1}{60} + \frac{1}{30})}}$ | (M1) | Correct numerator
| (M1) | Correct denominator
| (A1) | AWRT; dep on M1 M1 (2.53868)

Thus evidence, at 1% level, to **support the claim** that $\lambda_B > \lambda_A$ | Adep1 | Dep on z-value and CV

| | **7** |

**Total** | | **7** |

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Car parking in a market town's high street was, until 31 May 2014, limited to one hour free of charge between 8 am and 6 pm. Records show that, during a period of 30 days prior to this date, a total of 315 penalty tickets were issued.

Car parking in the high street later became limited to thirty minutes free of charge between 8 am and 6 pm. A subsequent investigation revealed that, during a period of 60 days from 1 October 2014, a total of 747 penalty tickets were issued.

The daily numbers of penalty tickets issued may be modelled by independent Poisson distributions with means $\lambda_A$ until 31 May 2014 and $\lambda_B$ from 1 October 2014.

Investigate, at the 1\% level of significance, a claim by traders on the high street that $\lambda_B > \lambda_A$. [7 marks]

\hfill \mbox{\textit{AQA S3 2016 Q3 [7]}}
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Q1 8 Q2 15 Q3 7 Q4 13 Q5 10 Q6 22