Edexcel S2 2005 January — Question 6 16 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2005
SessionJanuary
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of a Poisson distribution
TypeOne-tailed test (increase or decrease)
DifficultyStandard +0.3 This is a straightforward application of Poisson hypothesis testing with standard procedures. Parts (a)-(b) are routine probability calculations, while (c)-(d) follow textbook one-tailed test methodology with clearly stated hypotheses and significance levels. The only mild challenge is recognizing the parameter scaling in part (d) for the two-year period, but this is a standard S2 technique.
Spec2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

6. Over a long period of time, accidents happened on a stretch of road at random at a rate of 3 per month. Find the probability that
  1. in a randomly chosen month, more than 4 accidents occurred,
  2. in a three-month period, more than 4 accidents occurred. At a later date, a speed restriction was introduced on this stretch of road. During a randomly chosen month only one accident occurred.
  3. Test, at the \(5 \%\) level of significance, whether or not there is evidence to support the claim that this speed restriction reduced the mean number of road accidents occurring per month. The speed restriction was kept on this road. Over a two-year period, 55 accidents occurred.
  4. Test, at the \(5 \%\) level of significance, whether or not there is now evidence that this speed restriction reduced the mean number of road accidents occurring per month.

Question 6:
AnswerMarks
Let \(X\) represent number of accidents/month, \(X \sim Po(3)\)B1
(a)
AnswerMarks Guidance
\(P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8513 = 0.1847\)M1, A1 (3 marks)
(b)
AnswerMarks Guidance
Let \(Y\) represent number of accidents in 3 months, \(Y \sim Po(9)\)B1 Can be implied
\(P(Y > 4) = 1 - 0.0550 = 0.9450\)B1 (2 marks)
(c)
AnswerMarks Guidance
\(H_0: \lambda = 3\); \(H_1: \lambda < 3\); \(\alpha = 0.05\)B1 both; [detailed, allow \(B_0B_1M_1(0.025)A_0\)]
\(P(X \leq 1 \mid \lambda = 3) = 0.1991 > 0.05\)B1, M1
\(\therefore\) Insufficient evidence to support the claim that the mean number of accidents has been reducedA1\(\checkmark\) (4 marks)
[NB: CR: \(X \leq 0\); \(X = 1\) not in CR; same conclusion \(\Rightarrow\) B1, M1, A1]
(d)
AnswerMarks Guidance
\(H_0: \lambda = 24 \times 3 = 72\); \(H_1: \lambda < 72\)B1 Can be implied
\(\alpha = 0.05 \Rightarrow\) CR: \(z < -1.6449\)B1, B1 both \(H_0\), \(H_1\); \(-1.6449\)
Using Normal approximation with \(\mu = \bar{\lambda} = 72\)B1 Can be implied
\(z = \frac{55.5 - 72}{\sqrt{72}} = -1.94454\ldots\)M1 Standardise with \(\pm 0.5\), \(\mu\), \(\sqrt{\lambda}\)
A1AWRT \(-1.945\)
Since \(-1.944\ldots\) is in the CR, \(H_0\) is rejected. There is evidence that the restriction has reduced the number of accidents.A1\(\checkmark\) Context and clear evidence; (7 marks)
Alternative (d): \(p = 0.0262 < 0.05\), AWRT 0.026B1 equate to \(-1.6449\)
## Question 6:

Let $X$ represent number of accidents/month, $X \sim Po(3)$ | B1 |

**(a)**
$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8513 = 0.1847$ | M1, A1 | (3 marks)

**(b)**
Let $Y$ represent number of accidents in 3 months, $Y \sim Po(9)$ | B1 | Can be implied
$P(Y > 4) = 1 - 0.0550 = 0.9450$ | B1 | (2 marks)

**(c)**
$H_0: \lambda = 3$; $H_1: \lambda < 3$; $\alpha = 0.05$ | B1 | both; [detailed, allow $B_0B_1M_1(0.025)A_0$]
$P(X \leq 1 \mid \lambda = 3) = 0.1991 > 0.05$ | B1, M1 |
$\therefore$ Insufficient evidence to support the claim that the mean number of accidents has been reduced | A1$\checkmark$ | (4 marks)
[NB: CR: $X \leq 0$; $X = 1$ not in CR; same conclusion $\Rightarrow$ B1, M1, A1]

**(d)**
$H_0: \lambda = 24 \times 3 = 72$; $H_1: \lambda < 72$ | B1 | Can be implied
$\alpha = 0.05 \Rightarrow$ CR: $z < -1.6449$ | B1, B1 | both $H_0$, $H_1$; $-1.6449$
Using Normal approximation with $\mu = \bar{\lambda} = 72$ | B1 | Can be implied
$z = \frac{55.5 - 72}{\sqrt{72}} = -1.94454\ldots$ | M1 | Standardise with $\pm 0.5$, $\mu$, $\sqrt{\lambda}$
| A1 | AWRT $-1.945$
Since $-1.944\ldots$ is in the CR, $H_0$ is rejected. There is evidence that the restriction has reduced the number of accidents. | A1$\checkmark$ | Context and clear evidence; (7 marks)

**Alternative (d):** $p = 0.0262 < 0.05$, AWRT 0.026 | B1 | equate to $-1.6449$

---
6. Over a long period of time, accidents happened on a stretch of road at random at a rate of 3 per month.

Find the probability that
\begin{enumerate}[label=(\alph*)]
\item in a randomly chosen month, more than 4 accidents occurred,
\item in a three-month period, more than 4 accidents occurred.

At a later date, a speed restriction was introduced on this stretch of road. During a randomly chosen month only one accident occurred.
\item Test, at the $5 \%$ level of significance, whether or not there is evidence to support the claim that this speed restriction reduced the mean number of road accidents occurring per month.

The speed restriction was kept on this road. Over a two-year period, 55 accidents occurred.
\item Test, at the $5 \%$ level of significance, whether or not there is now evidence that this speed restriction reduced the mean number of road accidents occurring per month.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2005 Q6 [16]}}