| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 This is a straightforward S2 hypothesis testing question with standard Poisson calculations. Part (a) tests definition recall, parts (b)-(c) require routine Poisson probability calculations with parameter scaling, and part (d) is a textbook one-tailed hypothesis test. All techniques are standard S2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05a 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 model5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| A range of values of a test statistic such that if a value of the test statistic obtained from a particular sample lies in the critical region, then the null hypothesis is rejected (or equivalent). | B1B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.990717599... = 0.9907\) to 4 sf | M1, A1, A1, awrt 0.991 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 4) = 0.9473\) | B1, M1A1 | (3) |
| Answer | Marks |
|---|---|
| Accept \(\mu\) & \(H_0: \lambda = \frac{1}{7}\), \(H_1: \lambda < \frac{1}{7}\) | B1B1, Implied B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 1) = 0.0916 > 0.05\), So insufficient evidence to reject null hypothesis. Number of breakdowns has not significantly decreased | M1A1, A1, A1 | (7) |
## Part (a)
A range of values of a test statistic such that if a value of the test statistic obtained from a particular sample lies in the critical region, then the null hypothesis is rejected (or equivalent). | B1B1 | (2)
## Part (b)
$P(X < 2) = P(X = 0) + P(X = 1)$
$= e^{-\frac{1}{7}} + \frac{e^{-\frac{1}{7}}}{7}$
$= 0.990717599... = 0.9907$ to 4 sf | M1, A1, A1, awrt 0.991 | (3)
## Part (c)
$X \sim P(14 \times \frac{1}{7}) = P(2)$
$P(X \leq 4) = 0.9473$ | B1, M1A1 | (3)
Correct inequality, 0.9473
## Part (d)
$H_0: \lambda = 4$, $H_1: \lambda < 4$
Accept $\mu$ & $H_0: \lambda = \frac{1}{7}$, $H_1: \lambda < \frac{1}{7}$ | B1B1, Implied B1 |
$X \sim P(4)$
$P(X \leq 1) = 0.0916 > 0.05$, So insufficient evidence to reject null hypothesis. Number of breakdowns has not significantly decreased | M1A1, A1, A1 | (7)
Inequality 0.0916\begin{enumerate}[label=(\alph*)]
\item Explain what you understand by a critical region of a test statistic. [2]
\end{enumerate}
The number of breakdowns per day in a large fleet of hire cars has a Poisson distribution with mean $\frac{1}{7}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the probability that on a particular day there are fewer than 2 breakdowns. [3]
\item Find the probability that during a 14-day period there are at most 4 breakdowns. [3]
\end{enumerate}
The cars are maintained at a garage. The garage introduced a weekly check to try to decrease the number of cars that break down. In a randomly selected 28-day period after the checks are introduced, only 1 hire car broke down.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Test, at the 5% level of significance, whether or not the mean number of breakdowns has decreased. State your hypotheses clearly. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2004 Q5 [15]}}