| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI for proportion |
| Difficulty | Moderate -0.3 This is a straightforward application of confidence interval formulas for proportions with standard normal approximation. Part (a) requires plugging values into the formula with z=2.576, part (b) is simple interpretation (checking if 1/6 is in the interval), and part (c) involves algebraic manipulation of the width formula. All steps are routine for S2 level with no novel problem-solving required. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(p = \frac{70}{500}\) or \(0.14\) | B1 | |
| \(z = 2.576\) | B1 | |
| \(\text{"0.14"} \pm z \times \sqrt{\frac{\text{"0.14"}(1 - \text{"0.14"})}{500}}\) | M1 | |
| \(0.100\) to \(0.180\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.1666...\) is within confidence interval; belief supported or justified | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z \times \sqrt{\frac{\text{"0.14"}(1 - \text{"0.14"})}{500}} = 0.02\) | M1 | |
| \(z = 1.289\) | A1 | |
| \(\Phi(\text{'1.289'}) = 0.9013\) | M1 | |
| \(\alpha = \text{'0.9013'} - (1 - \text{'0.9013'})\) | M1 | |
| \(80.3\%\) (3 sf) | A1 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $p = \frac{70}{500}$ or $0.14$ | B1 | |
| $z = 2.576$ | B1 | |
| $\text{"0.14"} \pm z \times \sqrt{\frac{\text{"0.14"}(1 - \text{"0.14"})}{500}}$ | M1 | |
| $0.100$ to $0.180$ | A1 | |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.1666...$ is within confidence interval; belief supported or justified | B1 | |
---
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z \times \sqrt{\frac{\text{"0.14"}(1 - \text{"0.14"})}{500}} = 0.02$ | M1 | |
| $z = 1.289$ | A1 | |
| $\Phi(\text{'1.289'}) = 0.9013$ | M1 | |
| $\alpha = \text{'0.9013'} - (1 - \text{'0.9013'})$ | M1 | |
| $80.3\%$ (3 sf) | A1 | |
---5 Sunita has a six-sided die with faces marked $1,2,3,4,5,6$. The probability that the die shows a six on any throw is $p$. Sunita throws the die 500 times and finds that it shows a six 70 times.
\begin{enumerate}[label=(\alph*)]
\item Calculate an approximate $99 \%$ confidence interval for $p$.
\item Sunita believes that the die is fair. Use your answer to part (a) to comment on her belief.
\item Sunita uses the result of her 500 throws to calculate an $\alpha \%$ confidence interval for $p$. This interval has width 0.04 .
Find the value of $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q5 [10]}}