| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | November |
| 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 with standard bookwork. Part (i) requires basic understanding of sampling (common critique about telephone directories). Parts (ii) and (iii) are routine applications of the normal approximation formula for proportions with no conceptual challenges—just substituting values and rearranging. Slightly easier than average due to minimal problem-solving required. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Excludes children; Excludes people without phones; More than one person in some houses; Some ex-directory | B1 [1] | or other implying directory excludes some people |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(p) = \frac{38}{200}(1-\frac{38}{200}) \div 200 (= 0.0007695)\) | M1 | |
| \(z = 2.576\) | B1 | Seen |
| \(\frac{38}{200} \pm z\sqrt{\frac{38}{200}(1-\frac{38}{200}) \div 200}\) | M1 | For correct form of CI |
| \(0.119\) to \(0.261\) (3 sfs) | A1 [4] | Accept 0.262; Must be an interval |
| Answer | Marks | Guidance |
|---|---|---|
| \(z \times \sqrt{'\sqrt{0.0007695}' = 0.05}\); \(z = 1.802\); \(\Phi('1.802') (= 0.9642)\) | M1 A1 M1 | \(z \times\) (their sd of p) = 0.05. Allow 0.1 |
| \(('0.9642' − (1 − '0.9642')) = 0.9284)\); \(x = 93\) (2 sfs) | A1 [4] |
**(i)**
Excludes children; Excludes people without phones; More than one person in some houses; Some ex-directory | B1 [1] | or other implying directory excludes some people
**(ii)**
$\text{Var}(p) = \frac{38}{200}(1-\frac{38}{200}) \div 200 (= 0.0007695)$ | M1 |
$z = 2.576$ | B1 | Seen
$\frac{38}{200} \pm z\sqrt{\frac{38}{200}(1-\frac{38}{200}) \div 200}$ | M1 | For correct form of CI
$0.119$ to $0.261$ (3 sfs) | A1 [4] | Accept 0.262; Must be an interval
**(iii)**
$z \times \sqrt{'\sqrt{0.0007695}' = 0.05}$; $z = 1.802$; $\Phi('1.802') (= 0.9642)$ | M1 A1 M1 | $z \times$ (their sd of p) = 0.05. Allow 0.1
$('0.9642' − (1 − '0.9642')) = 0.9284)$; $x = 93$ (2 sfs) | A1 [4] |6 In order to obtain a random sample of people who live in her town, Jane chooses people at random from the telephone directory for her town.\\
(i) Give a reason why Jane's method will not give a random sample of people who live in the town.
Jane now uses a valid method to choose a random sample of 200 people from her town and finds that 38 live in apartments.\\
(ii) Calculate an approximate $99 \%$ confidence interval for the proportion of all people in Jane's town who live in apartments.\\
(iii) Jane uses the same sample to give a confidence interval of width 0.1 for this proportion. This interval is an $x \%$ confidence interval. Find the value of $x$.
\hfill \mbox{\textit{CAIE S2 2012 Q6 [9]}}