| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI for proportion |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question for a proportion using normal approximation. Part (i) requires standard formula application, part (ii) is simple interpretation (checking if 0.5 is in the interval), and part (iii) involves reversing the calculation to find the critical value. All steps are routine for S2 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(p = \frac{184}{400}\) or \(0.46\) | B1 | Used |
| \(z = 1.96\) | B1 | Seen |
| \("0.46" \pm z \times \sqrt{\frac{"0.46"(1-"0.46")}{400}} = 0.411 \text{ to } 0.509\) | M1 | Using expression of correct form |
| A1 [4] | Must be an interval | |
| (ii) \(0.5\) within CI. Claim not supported or not justified | B1 [1] | Both needed. No contradictions. ft their (i) |
| (iii) \(z \times \sqrt{\frac{"0.46"(1-"0.46")}{400}} = 0.05\) | M1 | |
| \(z = 2.006\) | A1 | Allow M1 for \(z \times \sqrt{\frac{"0.46"(1-"0.46")}{400}} = 0.1\) |
| \(\Phi('2.006') = 0.9775\) | M1 | |
| \(\alpha = '0.9775' - (1 - '0.9775') = 95.5\%\) | A1 [4] | or \(1 - 2(1 - '0.9775')\) |
**(i)** $p = \frac{184}{400}$ or $0.46$ | B1 | Used
$z = 1.96$ | B1 | Seen
$"0.46" \pm z \times \sqrt{\frac{"0.46"(1-"0.46")}{400}} = 0.411 \text{ to } 0.509$ | M1 | Using expression of correct form
| A1 [4] | Must be an interval
**(ii)** $0.5$ within CI. Claim not supported or not justified | B1 [1] | Both needed. No contradictions. ft their (i)
**(iii)** $z \times \sqrt{\frac{"0.46"(1-"0.46")}{400}} = 0.05$ | M1
$z = 2.006$ | A1 | Allow M1 for $z \times \sqrt{\frac{"0.46"(1-"0.46")}{400}} = 0.1$
$\Phi('2.006') = 0.9775$ | M1
$\alpha = '0.9775' - (1 - '0.9775') = 95.5\%$ | A1 [4] | or $1 - 2(1 - '0.9775')$
---5 Mahmoud throws a coin 400 times and finds that it shows heads 184 times. The probability that the coin shows heads on any throw is denoted by $p$.\\
(i) Calculate an approximate $95 \%$ confidence interval for $p$.\\
(ii) Mahmoud claims that the coin is not fair. Use your answer to part (i) to comment on this claim.\\
(iii) Mahmoud's result of 184 heads in 400 throws gives an $\alpha \%$ confidence interval for $p$ with width 0.1 . Calculate the value of $\alpha$.
\hfill \mbox{\textit{CAIE S2 2014 Q5 [9]}}