| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | November |
| Marks | 10 |
| 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 straightforward application of Poisson hypothesis testing with clear structure: (i) requires standard one-tailed test mechanics with given data, (ii) tests understanding that Type I error probability equals significance level, and (iii) asks for contextual interpretation. The multi-part structure and need to work with summed Poisson distributions adds modest complexity, but all steps follow standard procedures taught in S2 with no novel problem-solving required. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): Pop mean no. defectives \(= 5.15\) | B1 | or \(`= 1.03\) (per day)' not just 'mean', but allow just '\(\lambda\)' or '\(\mu\)' |
| \(H_1\): Pop mean no. defectives \(< 5.15\) | ||
| \(P(X \leqslant 2)\) | M1 | Attempted. Any one term error/end error/incorrect \(\lambda\)/expression \(1-\ldots\) |
| \(= e^{-5.15}(1 + 5.15 + \frac{5.15^2}{2})\) | M1 | Correct expression attempted |
| \(= 0.113\) | A1 | |
| Comp with \(0.1\) | M1 | Valid comparison |
| No evidence to believe mean no. of defectives has decreased | A1 FT | Correct conclusion (FT their value). No contradictions |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| BOTH \(P(X \leqslant 1) = e^{-5.15}(1+5.15) = (0.0357)\) AND \(P(X \leqslant 2) = e^{-5.15}(1+5.15+\frac{5.15^2}{2}) = (0.113)\) | B1* | (Could be seen in (i)) |
| Comp either with \(0.1\) | DB1 | One comparison with \(0.01\) (could be seen in (i)) |
| \(P(\text{Type I error}) = 0.0357\) (3 sf) | B1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Actually mean \(= 1.03\) but conclude that mean \(< 1.03\) | B1 | Mean no. of defectives not reduced, but conclude that it is reduced. |
| 1 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Pop mean no. defectives $= 5.15$ | B1 | or $`= 1.03$ (per day)' not just 'mean', but allow just '$\lambda$' or '$\mu$' |
| $H_1$: Pop mean no. defectives $< 5.15$ | | |
| $P(X \leqslant 2)$ | M1 | Attempted. Any one term error/end error/incorrect $\lambda$/expression $1-\ldots$ |
| $= e^{-5.15}(1 + 5.15 + \frac{5.15^2}{2})$ | M1 | Correct expression attempted |
| $= 0.113$ | A1 | |
| Comp with $0.1$ | M1 | Valid comparison |
| No evidence to believe mean no. of defectives has decreased | A1 FT | Correct conclusion (FT their value). No contradictions |
| | **6** | |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| BOTH $P(X \leqslant 1) = e^{-5.15}(1+5.15) = (0.0357)$ AND $P(X \leqslant 2) = e^{-5.15}(1+5.15+\frac{5.15^2}{2}) = (0.113)$ | B1* | (Could be seen in (i)) |
| Comp either with $0.1$ | DB1 | One comparison with $0.01$ (could be seen in (i)) |
| $P(\text{Type I error}) = 0.0357$ (3 sf) | B1 | |
| | **3** | |
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Actually mean $= 1.03$ but conclude that mean $< 1.03$ | B1 | Mean no. of defectives not reduced, but conclude that it is reduced. |
| | **1** | |6 In a certain factory the number of items per day found to be defective has had the distribution $\operatorname { Po } ( 1.03 )$. After the introduction of new quality controls, the management wished to test at the $10 \%$ significance level whether the mean number of defective items had decreased. They noted the total number of defective items produced in 5 randomly chosen days. It is assumed that defective items occur randomly and that a Poisson model is still appropriate.\\
(i) Given that the total number of defective items produced during the 5 days was 2 , carry out the test.\\
(ii) Using another random sample of 5 days the same test is carried out again, with the same significance level. Find the probability of a Type I error.\\
(iii) Explain what is meant by a Type I error in this context.\\
\hfill \mbox{\textit{CAIE S2 2017 Q6 [10]}}