| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Explain Type I or II error |
| Difficulty | Moderate -0.3 This is a straightforward application of Type I and II error definitions to a binomial hypothesis test. Part (a) requires recalling the definition, while parts (b) and (c) involve standard binomial probability calculations with clearly specified parameters. The question is slightly easier than average because it's a direct textbook-style exercise with no problem-solving insight required, though it does test understanding of hypothesis testing concepts. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05a Hypothesis testing language: null, alternative, p-value, significance |
| Answer | Marks | Guidance |
|---|---|---|
| Conclude more than 10% of the students are left handed when this is not true | B1 | OE. Must be in context (accept use of \(p\)). Need the context of one tail test. |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - (0.9^{20} + 20 \times 0.9^{19} \times 0.1 + {}^{20}C_2 \times 0.9^{18} \times 0.1^2 + {}^{20}C_3 \times 0.9^{17} \times 0.1^3 + {}^{20}C_4 \times 0.9^{16} \times 0.1^4)\) | M2 | M2: fully correct. M1: attempt \(1 - P(X = 0, 1, 2, 3, 4)\); allow \(1 - P(X = 0,1,2,3,4,5)\) or \(1 - P(X = 0,1,2,3)\); need \(1 - \ldots\); the method mark cannot be implied. |
| \(0.0432\) (3 s.f.) | A1 | If M0 awarded allow SC B2 for \(0.0432\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.7^{20} + 20 \times 0.7^{19} \times 0.3 + {}^{20}C_2 \times 0.7^{18} \times 0.3^2 + {}^{20}C_3 \times 0.7^{17} \times 0.3^3 + {}^{20}C_4 \times 0.7^{16} \times 0.3^4\) | M1 | Attempt to find \(P(\leq 4)\) using B(20,0.3); allow one end error; method mark cannot be implied |
| 0.238 or 0.237 (3 s.f.) | A1 | If M0 awarded allow SC B1 for 0.238 or 0.237 |
**Question 2(a):**
Conclude more than 10% of the students are left handed when this is not true | **B1** | OE. Must be in context (accept use of $p$). Need the context of one tail test.
---
**Question 2(b):**
$1 - (0.9^{20} + 20 \times 0.9^{19} \times 0.1 + {}^{20}C_2 \times 0.9^{18} \times 0.1^2 + {}^{20}C_3 \times 0.9^{17} \times 0.1^3 + {}^{20}C_4 \times 0.9^{16} \times 0.1^4)$ | **M2** | M2: fully correct. M1: attempt $1 - P(X = 0, 1, 2, 3, 4)$; allow $1 - P(X = 0,1,2,3,4,5)$ or $1 - P(X = 0,1,2,3)$; need $1 - \ldots$; the method mark cannot be implied.
$0.0432$ (3 s.f.) | **A1** | If M0 awarded allow SC B2 for $0.0432$
## Question 2(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.7^{20} + 20 \times 0.7^{19} \times 0.3 + {}^{20}C_2 \times 0.7^{18} \times 0.3^2 + {}^{20}C_3 \times 0.7^{17} \times 0.3^3 + {}^{20}C_4 \times 0.7^{16} \times 0.3^4$ | M1 | Attempt to find $P(\leq 4)$ using B(20,0.3); allow one end error; method mark cannot be implied |
| 0.238 or 0.237 (3 s.f.) | A1 | If M0 awarded allow SC B1 for 0.238 or 0.237 |
---2 Anton believes that $10 \%$ of students at his college are left-handed. Aliya believes that this is an underestimate. She plans to carry out a hypothesis test of the null hypothesis $p = 0.1$ against the alternative hypothesis $p > 0.1$, where $p$ is the actual proportion of students at the college that are left-handed. She chooses a random sample of 20 students from the college. She will reject the null hypothesis if at least 5 of these students are left-handed.
\begin{enumerate}[label=(\alph*)]
\item Explain what is meant by a Type I error in this context.
\item Find the probability of a Type I error in the test.
\item Given that the true value of $p$ is 0.3 , find the probability of a Type II error in the test.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q2 [6]}}