| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Confidence interval for single proportion |
| Difficulty | Moderate -0.8 This is a straightforward confidence interval question requiring only standard formula manipulation: finding the midpoint (mean of endpoints), using p̂ = 87/n to solve for n, then using the interval width formula with the normal approximation to find the confidence level. All steps are routine applications of memorized formulas with no conceptual challenges or problem-solving required. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.145 = \frac{87}{n}\), \(n = 600\) | B1, M1, A1 [3] | correct mid-point; equating their mid-point with \(\frac{87}{n}\); correct answer |
| (ii) \(0.0321 = z \times \sqrt{\frac{0.145(1-0.145)}{600}}\) | B1 | 0.0321 seen or implied |
| \(z = 2.233\), \(\Phi(z) = 0.9872\) | M1 | Equating half-width with \(z \times \sqrt{\frac{pq}{n}}\) |
| width of CI is \(1 - 2 \times (1 - 0.9872)\) | M1 | Correct method to find width of CI |
| \(\alpha = 97.4\%\) | A1 [4] | Correct answer |
**(i)** $0.145 = \frac{87}{n}$, $n = 600$ | B1, M1, A1 [3] | correct mid-point; equating their mid-point with $\frac{87}{n}$; correct answer
**(ii)** $0.0321 = z \times \sqrt{\frac{0.145(1-0.145)}{600}}$ | B1 | 0.0321 seen or implied
$z = 2.233$, $\Phi(z) = 0.9872$ | M1 | Equating half-width with $z \times \sqrt{\frac{pq}{n}}$
width of CI is $1 - 2 \times (1 - 0.9872)$ | M1 | Correct method to find width of CI
$\alpha = 97.4\%$ | A1 [4] | Correct answer2 A random sample of $n$ people were questioned about their internet use. 87 of them had a high-speed internet connection. A confidence interval for the population proportion having a high-speed internet connection is $0.1129 < p < 0.1771$.\\
(i) Write down the mid-point of this confidence interval and hence find the value of $n$.\\
(ii) This interval is an $\alpha \%$ confidence interval. Find $\alpha$.
\hfill \mbox{\textit{CAIE S2 2010 Q2 [7]}}