| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| 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 hypothesis testing question requiring calculation of Type I error (significance level) using binomial probability with given parameters, and finding p from Type II error probability. Both parts involve direct application of standard formulas without requiring novel insight or complex multi-step reasoning. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(1 - \frac{1}{3}\right)^{10}\) | M1 | |
| \(= 0.0173\) (3 sf) | A1 | No working scores SC B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - (1-p)^{10} = 0.8926\) | M1 | Accept \(1 - q^{10} = 0.8926\). Equation must be in \(p\) or in \(q\) but not both |
| \(1 - p = 0.1074^{0.1}\) \([= 0.800]\) | M1 | For valid attempt to solve their (binomial) equation in \(p^{10}\) or \(q^{10}\) |
| \(p = 0.200\) (3 sf) or \(0.2\) | A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(1 - \frac{1}{3}\right)^{10}$ | M1 | |
| $= 0.0173$ (3 sf) | A1 | No working scores SC B1 |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - (1-p)^{10} = 0.8926$ | M1 | Accept $1 - q^{10} = 0.8926$. Equation must be in $p$ or in $q$ but not both |
| $1 - p = 0.1074^{0.1}$ $[= 0.800]$ | M1 | For valid attempt to solve their (binomial) equation in $p^{10}$ or $q^{10}$ |
| $p = 0.200$ (3 sf) or $0.2$ | A1 | |
---6 When a child completes an online exercise called a Mathlit, they might be awarded a medal. The publishers claim that the probability that a randomly chosen child who completes a Mathlit will be awarded a medal is $\frac { 1 } { 3 }$. Asha wishes to test this claim. She decides that if she is awarded no medals while completing 10 Mathlits, she will conclude that the true probability is less than $\frac { 1 } { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Use a binomial distribution to find the probability of a Type I error.\\
The true probability of being awarded a medal is denoted by $p$.
\item Given that the probability of a Type II error is 0.8926 , find the value of $p$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q6 [5]}}