| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Describe or suggest sampling method |
| Difficulty | Moderate -0.8 This question tests basic sampling concepts (bias, random sampling methods) and routine confidence interval calculations using standard formulas. Parts (i)-(ii) require simple recall of sampling theory, while parts (iii)-(iv) involve straightforward application of the CI formula with no novel problem-solving—all standard S2 material with minimal computational or conceptual challenge. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.01d Select/critique sampling: in context5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Biased towards people who like tennis; Excludes people who don't like tennis | B1 | or other sensible |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain a list of all people in the town | B1 | |
| Use random numbers | B1 | or, e.g. pick numbers from a hat or other sensible |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(p) = \dfrac{\frac{47}{350}\left(1-\frac{47}{350}\right)}{350}\) \((= 0.000332152)\) | M1 | |
| \(z = 1.645\) | B1 | |
| \(\dfrac{47}{350} \pm z\sqrt{\dfrac{\frac{47}{350}\left(1-\frac{47}{350}\right)}{350}}\) | M1 | Must be a \(z\) value |
| \(0.104\) to \(0.164\) (3 sf) | A1 | Must be an interval |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1.25 \times 1.645\) \((= 2.056)\) | M1 | or \(1.25 \times\) their width \(\div 2 \div\) their \(\sqrt{\dfrac{\frac{47}{350}\left(1-\frac{47}{350}\right)}{350}}\) (Complete method) |
| \(\Phi('2.056')\) \((= 0.980)\) | M1 | Attempt \(\Phi(\text{their } z)\) |
| \(x = 96\) (2 sf) | A1 | Allow \(0.96\) (2 sf) CWO |
| Total: 3 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Biased towards people who like tennis; Excludes people who don't like tennis | B1 | or other sensible |
| **Total: 1** | | |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain a list of all people in the town | B1 | |
| Use random numbers | B1 | or, e.g. pick numbers from a hat or other sensible |
| **Total: 2** | | |
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(p) = \dfrac{\frac{47}{350}\left(1-\frac{47}{350}\right)}{350}$ $(= 0.000332152)$ | M1 | |
| $z = 1.645$ | B1 | |
| $\dfrac{47}{350} \pm z\sqrt{\dfrac{\frac{47}{350}\left(1-\frac{47}{350}\right)}{350}}$ | M1 | Must be a $z$ value |
| $0.104$ to $0.164$ (3 sf) | A1 | Must be an interval |
| **Total: 4** | | |
---
## Question 6(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1.25 \times 1.645$ $(= 2.056)$ | M1 | or $1.25 \times$ their width $\div 2 \div$ their $\sqrt{\dfrac{\frac{47}{350}\left(1-\frac{47}{350}\right)}{350}}$ (Complete method) |
| $\Phi('2.056')$ $(= 0.980)$ | M1 | Attempt $\Phi(\text{their } z)$ |
| $x = 96$ (2 sf) | A1 | Allow $0.96$ (2 sf) CWO |
| **Total: 3** | | |6 Ramesh plans to carry out a survey in order to find out what adults in his town think about local sports facilities. He chooses a random sample from the adult members of a tennis club and gives each of them a questionnaire.\\
(i) Give a reason why this will not result in Ramesh having a random sample of adults who live in the town.\\
(ii) Describe briefly a valid method that Ramesh could use to choose a random sample of adults in the town.\\
Ramesh now uses a valid method to choose a random sample of 350 adults from the town. He finds that 47 adults think that the local sports facilities are good.\\
(iii) Calculate an approximate $90 \%$ confidence interval for the proportion of all adults in the town who think that the local sports facilities are good.\\
(iv) Ramesh calculates a confidence interval whose width is 1.25 times the width of this $90 \%$ confidence interval. Ramesh's new interval is an $x \%$ confidence interval. Find the value of $x$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S2 2019 Q6 [10]}}