| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| 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 standard Poisson hypothesis test with routine calculations: finding critical region from tables, conducting the test, and computing Type I/II error probabilities. While it requires understanding of hypothesis testing concepts, all steps follow textbook procedures with no novel problem-solving required. Slightly above average difficulty due to the multi-part nature and Type II error calculation, but well within typical S2 scope. |
| 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 |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): Pop mean (or \(\lambda\) or \(\mu\)) is 1.1; \(H_1\): Pop mean (or \(\lambda\) or \(\mu\)) is more than 1.1 | B1 | |
| \(P(X \geq 4) = 1 - e^{-1.1}\!\left(1 + 1.1 + \frac{1.1^2}{2} + \frac{1.1^3}{3!}\right)\) | M1 | Correct expression for either \(P(X \geq 4)\) or \(P(X \geq 5)\) |
| \(0.0257\) | A1 | Correct value of either \(P(X \geq 4)\) or \(P(X \geq 5)\) |
| \(P(X \geq 5) = 0.0257 - e^{-1.1} \times \frac{1.1^4}{4!} = 0.00544\) | B1 | B1 for the other value. Note use of \(P(X<4)=0.9743\) and \(P(X<5)=0.99456\) can score only if comparison with 0.99 seen |
| \(0.00544 < 0.01 < 0.0257\) | M1 | OE stated (valid comparison) |
| There is evidence mean has increased | B1 | SC \(P(X \geq 6) = 0.000968\) M1A1, Conclusion B1 |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Concluding mean has increased when it has not | B1 | In context |
| \(0.00544\) | B1FT | FT *their* \(P(X \geq 5)\), dep \(< 0.01\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-7.0}\!\left(1 + 7 + \frac{7^2}{2} + \frac{7^3}{3!} + \frac{7^4}{4!}\right)\) | M1 | Correct expression for \(P(X \leq 4 \mid \lambda = 7.0)\) |
| \(0.173\) (3 sf) | A1 | |
| 2 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Pop mean (or $\lambda$ or $\mu$) is 1.1; $H_1$: Pop mean (or $\lambda$ or $\mu$) is more than 1.1 | B1 | |
| $P(X \geq 4) = 1 - e^{-1.1}\!\left(1 + 1.1 + \frac{1.1^2}{2} + \frac{1.1^3}{3!}\right)$ | M1 | Correct expression for either $P(X \geq 4)$ or $P(X \geq 5)$ |
| $0.0257$ | A1 | Correct value of either $P(X \geq 4)$ or $P(X \geq 5)$ |
| $P(X \geq 5) = 0.0257 - e^{-1.1} \times \frac{1.1^4}{4!} = 0.00544$ | B1 | B1 for the other value. Note use of $P(X<4)=0.9743$ and $P(X<5)=0.99456$ can score only if comparison with 0.99 seen |
| $0.00544 < 0.01 < 0.0257$ | M1 | OE stated (valid comparison) |
| There is evidence mean has increased | B1 | SC $P(X \geq 6) = 0.000968$ M1A1, Conclusion B1 |
| | **6** | |
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Concluding mean has increased when it has not | B1 | In context |
| $0.00544$ | B1FT | FT *their* $P(X \geq 5)$, dep $< 0.01$ |
| | **2** | |
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-7.0}\!\left(1 + 7 + \frac{7^2}{2} + \frac{7^3}{3!} + \frac{7^4}{4!}\right)$ | M1 | Correct expression for $P(X \leq 4 \mid \lambda = 7.0)$ |
| $0.173$ (3 sf) | A1 | |
| | **2** | |6 The number of accidents per month, $X$, at a factory has a Poisson distribution. In the past the mean has been 1.1 accidents per month. Some new machinery is introduced and the management wish to test whether the mean has increased. They note the number of accidents in a randomly chosen month and carry out a hypothesis test at the 1\% significance level.\\
(i) Show that the critical region for the test is $X \geqslant 5$. Given that the number of accidents is 6 , carry out the test.\\
Later they carry out a similar test, also at the $1 \%$ significance level.\\
(ii) Explain the meaning of a Type I error in this context and state the probability of a Type I error.\\
(iii) Given that the mean is now 7.0 , find the probability of a Type II error.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S2 2019 Q6 [10]}}