| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Standard +0.3 This is a straightforward application of standard hypothesis testing procedures for binomial distributions. Part (a) requires a routine one-tailed test with given values, part (b) tests understanding of Type I error definition (which equals the significance level), and part (c) involves a standard Poisson approximation to binomial—all textbook procedures with no novel problem-solving required. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 2.04c Calculate binomial probabilities5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution5.02n Sum of Poisson variables: is Poisson5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: p = 0.3\), \(H_1: p < 0.3\) | B1 | |
| \(B(31, 0.3),\ P(X \leqslant 4) = 0.7^{31} + 31\times0.7^{30}\times0.3 + {}^{31}C_2\times0.7^{29}\times0.3^2 + {}^{31}C_3\times0.7^{28}\times0.3^3 + {}^{31}C_4\times0.7^{27}\times0.3^4\) \(= 0.00001577 + 0.0002096 + 0.0013475 + 0.0055826 + 0.016748\) | M1 | No end errors. |
| \(= 0.0239\) (3sf) | A1 | SC \(0.0239\) with no working scores B1. |
| \(\text{'0.0239'} < 0.05\) | M1 | Valid comparison. |
| [Reject \(H_0\)] There is sufficient evidence (at 5% level) to support Rita's suspicion, or 'There is sufficient evidence to suggest the probability of seeing this type of bird has decreased' | A1FT | In context. Not definite. No contradictions. FT *their* \(0.0239\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X \leq 5) = [\text{'0.0239'} + {}^{31}C_5 \times 0.7^{26} \times 0.3^5] = 0.0627\ [\text{which is} > 0.05]\) | B1FT | Attempt \(P(X \leqslant 5)\). Only FT if \(> 0.05\). Only FT *their* \(0.0239\) if \(P(X \leqslant 4)\) attempted in (a); arithmetic error only. |
| \(P(\text{Type I error}) = \text{'0.0239'}\) | B1FT | Only FT *their* \(0.0239\) if \(P(X \leqslant 4)\) attempted in (a); arithmetic error only and *their* \(0.0239 < 0.05\). |
| Answer | Marks | Guidance |
|---|---|---|
| \([\lambda =]\ 3.65\) | B1 | Stated or implied. |
| \(e^{-3.65} \times \frac{3.65^4}{4!}\) | M1 | Must see expression. Any \(\lambda\). |
| \(= 0.192\) (3sf) | A1 | SC: Use of Binomial. \(0.193\) scores B1. SC: \(0.192\) with no working scores B1 B1. |
| Answer | Marks | Guidance |
|---|---|---|
| \(np = 3.65 < 5\) or \(p = 0.01 < 0.1\) | B1 | Explicit. Both needed. Note: and '\(n\) large, \(p\) small' is insufficient. |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = 0.3$, $H_1: p < 0.3$ | B1 | |
| $B(31, 0.3),\ P(X \leqslant 4) = 0.7^{31} + 31\times0.7^{30}\times0.3 + {}^{31}C_2\times0.7^{29}\times0.3^2 + {}^{31}C_3\times0.7^{28}\times0.3^3 + {}^{31}C_4\times0.7^{27}\times0.3^4$ $= 0.00001577 + 0.0002096 + 0.0013475 + 0.0055826 + 0.016748$ | M1 | No end errors. |
| $= 0.0239$ (3sf) | A1 | SC $0.0239$ with no working scores **B1**. |
| $\text{'0.0239'} < 0.05$ | M1 | Valid comparison. |
| [Reject $H_0$] There is sufficient evidence (at 5% level) to support Rita's suspicion, or 'There is sufficient evidence to suggest the probability of seeing this type of bird has decreased' | A1FT | In context. Not definite. No contradictions. FT *their* $0.0239$. |
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X \leq 5) = [\text{'0.0239'} + {}^{31}C_5 \times 0.7^{26} \times 0.3^5] = 0.0627\ [\text{which is} > 0.05]$ | B1FT | Attempt $P(X \leqslant 5)$. Only FT if $> 0.05$. Only FT *their* $0.0239$ if $P(X \leqslant 4)$ attempted in (a); arithmetic error only. |
| $P(\text{Type I error}) = \text{'0.0239'}$ | B1FT | Only FT *their* $0.0239$ if $P(X \leqslant 4)$ attempted in (a); arithmetic error only and *their* $0.0239 < 0.05$. |
## Question 7(c)(i):
$[\lambda =]\ 3.65$ | B1 | Stated or implied.
$e^{-3.65} \times \frac{3.65^4}{4!}$ | M1 | Must see expression. Any $\lambda$.
$= 0.192$ (3sf) | A1 | SC: Use of Binomial. $0.193$ scores **B1**. SC: $0.192$ with no working scores **B1 B1**.
**Total: 3 marks**
---
## Question 7(c)(ii):
$n = 365 > 50$
$np = 3.65 < 5$ or $p = 0.01 < 0.1$ | B1 | Explicit. Both needed. Note: and '$n$ large, $p$ small' is insufficient.
**Total: 1 mark**7 Every July, as part of a research project, Rita collects data about sightings of a particular kind of bird. Each day in July she notes whether she sees this kind of bird or not, and she records the number $X$ of days on which she sees it. She models the distribution of $X$ by $\mathrm { B } ( 31 , p )$, where $p$ is the probability of seeing this kind of bird on a randomly chosen day in July.
Data from previous years suggests that $p = 0.3$, but in 2022 Rita suspected that the value of $p$ had been reduced. She decided to carry out a hypothesis test.
In July 2022, she saw this kind of bird on 4 days.
\begin{enumerate}[label=(\alph*)]
\item Use the binomial distribution to test at the $5 \%$ significance level whether Rita's suspicion is justified.\\
In July 2023, she noted the value of $X$ and carried out another test at the $5 \%$ significance level using the same hypotheses.
\item Calculate the probability of a Type I error.\\
Rita models the number of sightings, $Y$, per year of a different, very rare, kind of bird by the distribution $B ( 365,0.01 )$.
\item \begin{enumerate}[label=(\roman*)]
\item Use a suitable approximating distribution to find $\mathrm { P } ( Y = 4 )$.
\item Justify your approximating distribution in this context.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q7 [11]}}