CAIE S2 2006 November — Question 3 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2006
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeCalculate CI for proportion
DifficultyModerate -0.8 Part (i) is a standard confidence interval calculation for a proportion using the normal approximation—routine application of a formula with given values. Part (ii) requires identifying basic sampling bias (e.g., non-random sample, time/location bias), which is elementary statistical reasoning. Both parts are straightforward recall and application with no problem-solving complexity.
Spec5.05d Confidence intervals: using normal distribution

3 A survey was conducted to find the proportion of people owning DVD players. It was found that 203 out of a random sample of 278 people owned a DVD player.
  1. Calculate a \(97 \%\) confidence interval for the true proportion of people who own a DVD player. A second survey to find the proportion of people owning DVD players was conducted at 10 o'clock on a Thursday morning in a shopping centre.
  2. Give one reason why this is not a satisfactory sample.

AnswerMarks Guidance
(i) \(p = \frac{203}{278} (= 0.7302 \approx 0.73)\)B1 Correct \(p\)
\(0.7302 \pm 2.17 \times \sqrt{\frac{pq}{n}} = 0.7302 \pm 2.17 \times \sqrt{\frac{0.7302 \times 0.2698}{278}}\)M1 Correct form \(p \pm z \times \sqrt{\frac{pq}{n}}\) either/both sides
Correct \(z\)B1
\(= (0.672, 0.788)\)A1 Correct answer
(ii) Mainly unemployed, retired, or mothers with children ie not representative of whole popB1 Or any sensible equivalent
**(i)** $p = \frac{203}{278} (= 0.7302 \approx 0.73)$ | B1 | Correct $p$

$0.7302 \pm 2.17 \times \sqrt{\frac{pq}{n}} = 0.7302 \pm 2.17 \times \sqrt{\frac{0.7302 \times 0.2698}{278}}$ | M1 | Correct form $p \pm z \times \sqrt{\frac{pq}{n}}$ either/both sides

Correct $z$ | B1 | 

$= (0.672, 0.788)$ | A1 | Correct answer | 4 marks total

**(ii)** Mainly unemployed, retired, or mothers with children ie not representative of whole pop | B1 | Or any sensible equivalent | 1 mark

---
3 A survey was conducted to find the proportion of people owning DVD players. It was found that 203 out of a random sample of 278 people owned a DVD player.\\
(i) Calculate a $97 \%$ confidence interval for the true proportion of people who own a DVD player.

A second survey to find the proportion of people owning DVD players was conducted at 10 o'clock on a Thursday morning in a shopping centre.\\
(ii) Give one reason why this is not a satisfactory sample.

\hfill \mbox{\textit{CAIE S2 2006 Q3 [5]}}