| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 8 |
| 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 hypothesis testing question requiring calculation of critical region from Poisson tables and Type II error probability. While it involves two-part reasoning and careful probability calculations, the methodology is straightforward textbook application with no novel insight required. The Poisson distribution work is routine for Further Maths Statistics students, placing it moderately above average difficulty. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(X > 4) = 1 - P(0, 1, 2, 3, 4)\) | M1 | Adding at least 3 relevant Poisson terms |
| \(= 1 - e^{-1.8}\left(1 + 1.8 + \frac{1.8^2}{2} + \frac{1.8^3}{3!} + \frac{1.8^4}{4!}\right)\) | M1 | Poisson expression for \(P(X > 4)\) (oe implied by later working) |
| \(= 1 - 0.9635 = 0.036(4)\) | A1 | Correct prob 0.036 (or 0.96 subject to later working) |
| This is \(< 0.05\) and so \(X > 4\) is in the critical region | A1ft | Correct comparison and statement identifying CR (ft their prob \(< 0.05\)) |
| \(P(4) = e^{-1.8}\left(\frac{1.8^4}{4!}\right) = 0.0723\) | B1 [5] | Verification that \(X = 4\) is not in the cr region |
| (ii) \(P(\text{Type II error}) = P(X = 0, 1, 2, 3, 4)\) | B1 | Correct region |
| \(= e^{-2.1}\left(1 + 2.3 + \frac{2.3^2}{2} + \frac{2.3^3}{3!} + \frac{2.3^4}{4!}\right)\) | M1 | Poisson expression \(P(0, 1, 2, 3, 4)\) |
| \(= 0.916\) | A1 [3] | Correct answer |
**(i)** $P(X > 4) = 1 - P(0, 1, 2, 3, 4)$ | M1 | Adding at least 3 relevant Poisson terms
$= 1 - e^{-1.8}\left(1 + 1.8 + \frac{1.8^2}{2} + \frac{1.8^3}{3!} + \frac{1.8^4}{4!}\right)$ | M1 | Poisson expression for $P(X > 4)$ (oe implied by later working)
$= 1 - 0.9635 = 0.036(4)$ | A1 | Correct prob 0.036 (or 0.96 subject to later working)
This is $< 0.05$ and so $X > 4$ is in the critical region | A1ft | Correct comparison and statement identifying CR (ft their prob $< 0.05$)
$P(4) = e^{-1.8}\left(\frac{1.8^4}{4!}\right) = 0.0723$ | B1 [5] | Verification that $X = 4$ is not in the cr region
**(ii)** $P(\text{Type II error}) = P(X = 0, 1, 2, 3, 4)$ | B1 | Correct region
$= e^{-2.1}\left(1 + 2.3 + \frac{2.3^2}{2} + \frac{2.3^3}{3!} + \frac{2.3^4}{4!}\right)$ | M1 | Poisson expression $P(0, 1, 2, 3, 4)$
$= 0.916$ | A1 [3] | Correct answer4 The number of severe floods per year in a certain country over the last 100 years has followed a Poisson distribution with mean 1.8. Scientists suspect that global warming has now increased the mean. A hypothesis test, at the $5 \%$ significance level, is to be carried out to test this suspicion. The number of severe floods, $X$, that occur next year will be used for the test.\\
(i) Show that the rejection region for the test is $X > 4$.\\
(ii) Find the probability of making a Type II error if the mean number of severe floods is now actually 2.3.
\hfill \mbox{\textit{CAIE S2 2009 Q4 [8]}}