| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | One-tailed test (increase or decrease) |
| Difficulty | Standard +0.3 This is a standard S2 hypothesis testing question requiring routine application of Poisson distribution tables, normal approximation, and stating assumptions. While it has multiple parts (6 marks), each step follows textbook procedures: finding P(X≥6) for part (a), recalling standard Poisson assumptions for part (b), suggesting a sampling frame for part (c), and applying normal approximation with continuity correction for part (d). No novel insight or complex problem-solving is required, making it slightly easier than average. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (a) let X = no. absent per lesson \(\therefore X \sim \text{Po}(2.5)\); \(P(X \ge 6) = 1 - P(X \le 5) = 1 - 0.9580 = 0.0420\) | M1 A1 | |
| (b) assumes absences occur independently and at constant rate; ill students may infect others and rate may vary at different times of year but assumptions fairly reasonable | B3 | |
| (c) registers for all classes | B1 | |
| (d) let Y = no. absent per 30 lessons \(\therefore Y \sim \text{Po}(75)\); use N approx. \(A \sim N(75, 75)\); \(P(Y \ge 96) = P(A > 95.5) = P(Z > \frac{95.5-75}{\sqrt{75}}) = P(Z > 2.367) = 1 - 0.9909 = 0.0091\); less than 5% \(\therefore\) significant, there is evidence of more absent per lesson | M1 M1 M1 A1 A1 A1 | (12 marks total) |
(a) let X = no. absent per lesson $\therefore X \sim \text{Po}(2.5)$; $P(X \ge 6) = 1 - P(X \le 5) = 1 - 0.9580 = 0.0420$ | M1 A1 |
(b) assumes absences occur independently and at constant rate; ill students may infect others and rate may vary at different times of year but assumptions fairly reasonable | B3 |
(c) registers for all classes | B1 |
(d) let Y = no. absent per 30 lessons $\therefore Y \sim \text{Po}(75)$; use N approx. $A \sim N(75, 75)$; $P(Y \ge 96) = P(A > 95.5) = P(Z > \frac{95.5-75}{\sqrt{75}}) = P(Z > 2.367) = 1 - 0.9909 = 0.0091$; less than 5% $\therefore$ significant, there is evidence of more absent per lesson | M1 M1 M1 A1 A1 A1 | (12 marks total)
---6. A teacher is monitoring attendance at lessons in her department. She believes that the number of students absent from each lesson follows a Poisson distribution and wished to test the null hypothesis that the mean is 2.5 against the alternative hypothesis that it is greater than 2.5 She visits one lesson and decides on a critical region of 6 or more students absent.
\begin{enumerate}[label=(\alph*)]
\item Find the significance level of this test.
\item State any assumptions made in carrying out this test and comment on their validity.
The teacher decides to undertake a wider study by looking at a sample of all the lessons that have taken place in the department during the previous four weeks.
\item Suggest a suitable sampling frame.
She finds that there have been 96 pupils absent from the 30 lessons in her sample.
\item Using a suitable approximation, test at the $5 \%$ level of significance the null hypothesis that the mean is 2.5 students absent per lesson against the alternative hypothesis that it is greater than 2.5. You may assume that the number of absences follows a Poisson distribution.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [12]}}