| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI for proportion |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing standard procedures: (i) routine confidence interval for proportion using normal approximation with given formula, (ii) direct application of unbiased estimator formulas for mean and variance, (iii) basic description of simple random sampling. All parts are textbook exercises requiring only recall and substitution with no problem-solving or insight needed. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(P_s) = \frac{\frac{33}{150} \times \frac{150-33}{150}}{150}\) \((= 0.001144)\) | M1 | |
| \(z = 2.576\) | B1 | Accept 2.574 to 2.579 |
| \(\frac{33}{150} \pm z\sqrt{0.001144}\) | M1 | Expression of correct form. Any \(z\) |
| \(= 0.133\) to \(0.307\) (3 sf) | A1 | Must be an interval. Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{19035}{150}\) \((= 126.9 = 127\ (3\text{sf}))\) | B1 | |
| \(\frac{150}{149}\left(\frac{4054716}{150} - \left(\frac{19035}{150}\right)^2\right)\) o.e. | M1 | For use of a correct formula |
| \(= 11001.17\) or \(11000\) (3 sf) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 4-digit nos. each digit 0–9 | B1 | Some valid way of generating 4-digit random nos from valid method |
| Ignore nos \(> 9526\) | B1 | SR If zero score, full explanation of method for drawing numbers out of a hat can score B1. NB Systematic sampling follows the scheme with first B1 for some way of generating a random starting point. |
| Ignore repeats | B1 | Total: 3 |
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(P_s) = \frac{\frac{33}{150} \times \frac{150-33}{150}}{150}$ $(= 0.001144)$ | M1 | |
| $z = 2.576$ | B1 | Accept 2.574 to 2.579 |
| $\frac{33}{150} \pm z\sqrt{0.001144}$ | M1 | Expression of correct form. Any $z$ |
| $= 0.133$ to $0.307$ (3 sf) | A1 | Must be an interval. **Total: 4** |
## Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{19035}{150}$ $(= 126.9 = 127\ (3\text{sf}))$ | B1 | |
| $\frac{150}{149}\left(\frac{4054716}{150} - \left(\frac{19035}{150}\right)^2\right)$ o.e. | M1 | For use of a correct formula |
| $= 11001.17$ or $11000$ (3 sf) | A1 | **Total: 3** |
## Question 4(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| 4-digit nos. each digit 0–9 | B1 | Some valid way of generating 4-digit random nos from valid method |
| Ignore nos $> 9526$ | B1 | SR If zero score, full explanation of method for drawing numbers out of a hat can score B1. NB Systematic sampling follows the scheme with first B1 for some way of generating a random starting point. |
| Ignore repeats | B1 | **Total: 3** |
---4 In a survey a random sample of 150 households in Nantville were asked to fill in a questionnaire about household budgeting.\\
(i) The results showed that 33 households owned more than one car. Find an approximate $99 \%$ confidence interval for the proportion of all households in Nantville with more than one car. [4]\\
(ii) The results also included the weekly expenditure on food, $x$ dollars, of the households. These were summarised as follows.
$$n = 150 \quad \Sigma x = 19035 \quad \Sigma x ^ { 2 } = 4054716$$
Find unbiased estimates of the mean and variance of the weekly expenditure on food of all households in Nantville.\\
(iii) The government has a list of all the households in Nantville numbered from 1 to 9526. Describe briefly how to use random numbers to select a sample of 150 households from this list.
\hfill \mbox{\textit{CAIE S2 2014 Q4 [10]}}