| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Find Type I error probability |
| Difficulty | Standard +0.8 This is a multi-part hypothesis testing question requiring understanding of Poisson distributions, Type I/II errors, and normal approximation. While the individual components are standard A-level Further Maths Statistics topics, the question requires careful handling of the two-year period (λ=6.6 under H₀), correct identification of critical regions, and calculation of both error types. The normal approximation in part (d) adds another layer. This is moderately challenging but well within the scope of S2, placing it somewhat above average difficulty. |
| Spec | 2.04d Normal approximation to binomial2.05a Hypothesis testing language: null, alternative, p-value, significance5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | λ = 6.6 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| or 0.001360 + 0.008978 + 0.02963 | M1 | Expression must be seen. No end errors. |
| Answer | Marks | Guidance |
|---|---|---|
| 0.105 [ > 0.05 ] | B1 | Condone unsupported 0.105. |
| P(Type I error) = 0.0400 (3 sf) | A1 | Allow 0.040 or 0.04 AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | H : λ = 6.6, H : λ < 6.6 | |
| 0 1 | B1 | May be seen in part (a) and award B1 mark here. |
| Answer | Marks | Guidance |
|---|---|---|
| [P(X < 2) = 0.0400] ` 0.04 ` < 0.05 | M1 | For comparing their P(X < 2) any λ with 0.05. |
| Answer | Marks | Guidance |
|---|---|---|
| accidents has decreased | A1 | In context, not definite. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(c) | P(X > 2) attempted, with any λ | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| or = 1 – ( 0.3012 + 0.3614 + 0.2169 ) | M1 | Expression must be seen. |
| Answer | Marks | Guidance |
|---|---|---|
| 0.121 (3 sf) or 0.120 | A1 | SC unsupported answer scores B2. |
| Answer | Marks | Guidance |
|---|---|---|
| 7(d) | N(18, 18) seen or implied | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 18 | M1 | Allow with no or incorrect continuity correction. |
| Answer | Marks | Guidance |
|---|---|---|
| P(X > ‘−1.768’) = Φ(‘1.768’) | M1 | ft their standardised value. |
| Answer | Marks |
|---|---|
| = 0.961 or 0.962 (3 sf) | A1 |
Question 7:
--- 7(a) ---
7(a) | λ = 6.6 | B1
P(X < 2) = e−6.6(1 + 6.6 + 6.62 ) [= 0.0400] [ < 0.05 ]
2
or e−6.6(1 + 6.6 + 21.78 )
or 0.001360 + 0.008978 + 0.02963 | M1 | Expression must be seen. No end errors.
Allow use of 3.3 here.
P(X < 3) = e−6.6(1 + 6.6 + 6.62 + 6.63 ) or 0.0400 + e−6.6× 6.63 =
2 3! 3!
0.105 [ > 0.05 ] | B1 | Condone unsupported 0.105.
P(Type I error) = 0.0400 (3 sf) | A1 | Allow 0.040 or 0.04 AWRT
SC unsupported ans of 0.0400 can score max B1B1B1.
4
--- 7(b) ---
7(b) | H : λ = 6.6, H : λ < 6.6
0 1 | B1 | May be seen in part (a) and award B1 mark here.
Accept µ or λ.
Accept 3.3 or 6.6.
[P(X < 2) = 0.0400] ` 0.04 ` < 0.05 | M1 | For comparing their P(X < 2) any λ with 0.05.
[Reject H ] There is evidence to suggest that mean number of
0
accidents has decreased | A1 | In context, not definite.
No contradictions.
CWO.
3
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | P(X > 2) attempted, with any λ | M1
P(X > 2) = 1 − e−1.2(1 + 1.2 + 1.22 )
2
or = 1 − e−1.2(1 + 1.2 +0.72)
or = 1 – ( 0.3012 + 0.3614 + 0.2169 ) | M1 | Expression must be seen.
Correct λ.
No end errors.
0.121 (3 sf) or 0.120 | A1 | SC unsupported answer scores B2.
3
--- 7(d) ---
7(d) | N(18, 18) seen or implied | B1
10.5−18
[= −1.768]
18 | M1 | Allow with no or incorrect continuity correction.
Their 18.
P(X > ‘−1.768’) = Φ(‘1.768’) | M1 | ft their standardised value.
Area consistent with their values.
= 0.961 or 0.962 (3 sf) | A1
4The number of accidents per year on a certain road has the distribution $\text{Po}(\lambda)$. In the past the value of $\lambda$ was $3.3$. Recently, a new speed limit was imposed and the council wishes to test whether the value of $\lambda$ has decreased. The council notes the total number, $X$, of accidents during two randomly chosen years after the speed limit was introduced and it carries out a test at the $5\%$ significance level.
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability of a Type I error. [4]
\item Given that $X = 2$, carry out the test. [3]
\item The council decides to carry out another similar test at the $5\%$ significance level using the same hypotheses and two different randomly chosen years.
Given that the true value of $\lambda$ is $0.6$, calculate the probability of a Type II error. [3]
\item Using $\lambda = 0.6$ and a suitable approximating distribution, find the probability that there will be more than $10$ accidents in $30$ years. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q7 [14]}}