| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find Type II error probability |
| Difficulty | Challenging +1.2 This is a standard S2 hypothesis testing question requiring calculation of significance level (straightforward Poisson probability) and Type II error (requires understanding that Type II error = P(accept H₀|H₁ true) = P(W≥1|λ=λ₀), then solving 1-e^(-λ₀)=0.8). While it tests understanding of error types and involves solving an equation with logarithms, it follows a well-established template for this topic with no novel insight required. The multi-step nature and need to manipulate the complement probability elevates it slightly above average difficulty. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x! |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(W = 0 \mid \lambda = 3.6)\) | M1 | Use this conditional probability. *Not* just 2.5% etc |
| \(= \mathbf{0.0273}\) or \(\mathbf{2.73\%}\) | A1 | Answer a.r.t. 0.0273 or 2.73%. ISW if appropriate (e.g. "0.0273, \(\therefore 2.5\%\)") |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - e^{-\lambda_0} = 0.8\) | M1 | Use \(P(W > 0 \mid \lambda = \lambda_0)\); formula needed but allow if wrong |
| \(e^{-\lambda_0} = 0.2\) | A1 | This exact equation, or \(e^{\lambda_0} = 5\), or exact equivalent RHS |
| \(\lambda_0 = -\ln(0.2)\) | M1 | Solve using ln or otherwise [independent of first M1, e.g. \(-\ln(0.8) = 0.223\) is M1 here] |
| \(= \mathbf{1.609}\) | A1 | Final answer, exact or a.r.t. 1.61, cwo. SC: No working: 1.60 (tables etc): B0. 1.61 (T&I): SC B4. |
| [4] |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(W = 0 \mid \lambda = 3.6)$ | M1 | Use this conditional probability. *Not* just 2.5% etc |
| $= \mathbf{0.0273}$ or $\mathbf{2.73\%}$ | A1 | Answer a.r.t. 0.0273 or 2.73%. ISW if appropriate (e.g. "0.0273, $\therefore 2.5\%$") |
| | **[2]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - e^{-\lambda_0} = 0.8$ | M1 | Use $P(W > 0 \mid \lambda = \lambda_0)$; formula needed but allow if wrong |
| $e^{-\lambda_0} = 0.2$ | A1 | This exact equation, or $e^{\lambda_0} = 5$, or exact equivalent RHS |
| $\lambda_0 = -\ln(0.2)$ | M1 | Solve using ln or otherwise [independent of first M1, e.g. $-\ln(0.8) = 0.223$ is M1 here] |
| $= \mathbf{1.609}$ | A1 | Final answer, exact or a.r.t. 1.61, cwo. SC: No working: 1.60 (tables etc): B0. 1.61 (T&I): SC B4. |
| | **[4]** | |
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To extract mark scheme content, please share the **interior pages** of the document that contain the actual marking grids/tables.8 The random variable $W$ has the distribution $\operatorname { Po } ( \lambda )$. A significance test is carried out of the null hypothesis $\mathrm { H } _ { 0 } : \lambda = 3.60$, against the alternative hypothesis $\mathrm { H } _ { 1 } : \lambda < 3.60$. The test is based on a single observation of $W$. The critical region is $W = 0$.\\[0pt]
(i) Find the significance level of the test. [2]\\
(ii) It is known that, when $\boldsymbol { \lambda } = \boldsymbol { \lambda } _ { \mathbf { 0 } }$, the probability that the test results in a Type II error is $\mathbf { 0 . 8 }$. Find the value of $\lambda _ { 0 }$. [4]
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR S2 2014 Q8 [6]}}