| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find actual significance level |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard Poisson distribution calculations. Part (a) is trivial recall, part (b) requires finding P(C>10) when C~Po(6) using tables, and part (c) involves reverse table lookup. All steps are routine S2 procedures with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \((H_1:)\; \lambda > 1.5\) | B1 | Must use \(\lambda\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(C \sim \text{Po}(6)\) | B1 | Writing or using Po(6); NB \(P(X\leq9)=0.9161\), \(P(X\leq11)=0.9799\) can imply B1 |
| \(P(C>10) = 1 - P(X \leq 10)\) | M1 | Writing or using \(1-P(X\leq10)\) |
| \(= 1 - 0.9574\) | ||
| \(= 0.0426\) | A1 | awrt 0.0426; do not award if response states level of significance is 5% |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(P(X \leq 10 \mid \mu = 7) = 0.9015\) | M1 | Either this or \(P(X\leq10\mid\mu=7.5)=0.8622\); award for sight of 0.9015 (or 0.0985) or 0.8622 (or 0.1378) |
| \(P(X \leq 10 \mid \mu = 7.5) = 0.8622\) | ||
| Parameter \(\mu = 7\) | A1 | NB \(\lambda=7\) scores M1A1A0 |
| \(\lambda = \frac{7}{4},\; 1.75\) | A1 | Allow awrt 1.76 from calculator to score M1A1A1 |
## Question 4:
### Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $(H_1:)\; \lambda > 1.5$ | B1 | Must use $\lambda$ |
### Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $C \sim \text{Po}(6)$ | B1 | Writing or using Po(6); NB $P(X\leq9)=0.9161$, $P(X\leq11)=0.9799$ can imply B1 |
| $P(C>10) = 1 - P(X \leq 10)$ | M1 | Writing or using $1-P(X\leq10)$ |
| $= 1 - 0.9574$ | | |
| $= 0.0426$ | A1 | awrt 0.0426; do not award if response states level of significance is 5% |
### Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $P(X \leq 10 \mid \mu = 7) = 0.9015$ | M1 | Either this or $P(X\leq10\mid\mu=7.5)=0.8622$; award for sight of 0.9015 (or 0.0985) or 0.8622 (or 0.1378) |
| $P(X \leq 10 \mid \mu = 7.5) = 0.8622$ | | |
| Parameter $\mu = 7$ | A1 | NB $\lambda=7$ scores M1A1A0 |
| $\lambda = \frac{7}{4},\; 1.75$ | A1 | Allow awrt 1.76 from calculator to score M1A1A1 |
---\begin{enumerate}
\item The number of telephone calls per hour received by a business is a random variable with distribution $\operatorname { Po } ( \lambda )$.
\end{enumerate}
Charlotte records the number of calls, $C$, received in 4 hours.
A test of the null hypothesis $\mathrm { H } _ { 0 } : \lambda = 1.5$ is carried out.\\
$\mathrm { H } _ { 0 }$ is rejected if $C > 10$\\
(a) Write down the alternative hypothesis.\\
(b) Find the significance level of the test.
Given that $\mathrm { P } ( C > 10 ) < 0.1$\\
(c) find the largest possible value of $\lambda$ that can be found by using the tables.\\
\hfill \mbox{\textit{Edexcel S2 2014 Q4 [7]}}