| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| 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 standard procedures: calculating a mean, finding a critical region from tables, identifying Type I error probability, and making a conclusion. The calculations are routine and the concepts are core S2 material, making it slightly easier than average A-level difficulty. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x! |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | 1 in 15 months is equivalent to 4 in 60 months | B1 [1] |
| (ii) | \(P(0) = e^{-4} = 0.01831\) | M1* |
| \(P(1) = e^{-4} \times 4 = 0.07326\) | M1* | Comparing sum with 0.10 |
| \(P(2) = e^{-4} \times 4^2/2 = 0.147\) too big | M1 | Considering and rejecting P(2) |
| \(P(0) + P(1) = 0.0916\) | A1*dep [4] | Correct answer. No errors seen. |
| Rejection region at 10% level is 0 or 1. | ||
| (iii) | \(P(\text{Type I error}) = 0.0916\) | B1 [1] |
| (iv) | 1 is in rejection region | B1 |
| There is evidence that the new guitar string lasts longer | B1ft [2] | Correct conclusion ft their rejection region |
(i) | 1 in 15 months is equivalent to 4 in 60 months | B1 [1] | Or equivalent
(ii) | $P(0) = e^{-4} = 0.01831$ | M1* | Attempt to find P(0) and / or P(1)
$P(1) = e^{-4} \times 4 = 0.07326$ | M1* | Comparing sum with 0.10
$P(2) = e^{-4} \times 4^2/2 = 0.147$ too big | M1 | Considering and rejecting P(2)
$P(0) + P(1) = 0.0916$ | A1*dep [4] | Correct answer. No errors seen.
Rejection region at 10% level is 0 or 1. |
(iii) | $P(\text{Type I error}) = 0.0916$ | B1 [1] | Correct answer
(iv) | 1 is in rejection region | B1 | Identifying where 1 is
There is evidence that the new guitar string lasts longer | B1ft [2] | Correct conclusion ft their rejection region
---5 When a guitar is played regularly, a string breaks on average once every 15 months. Broken strings occur at random times and independently of each other.\\
(i) Show that the mean number of broken strings in a 5 -year period is 4 .
A guitar is fitted with a new type of string which, it is claimed, breaks less frequently. The number of broken strings of the new type was noted after a period of 5 years.\\
(ii) The mean number of broken strings of the new type in a 5 -year period is denoted by $\lambda$. Find the rejection region for a test at the $10 \%$ significance level when the null hypothesis $\lambda = 4$ is tested against the alternative hypothesis $\lambda < 4$.\\
(iii) Hence calculate the probability of making a Type I error.
The number of broken guitar strings of the new type, in a 5 -year period, was in fact 1 .\\
(iv) State, with a reason, whether there is evidence at the $10 \%$ significance level that guitar strings of the new type break less frequently.
\hfill \mbox{\textit{CAIE S2 2008 Q5 [8]}}