AQA S3 2009 June — Question 3 6 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2009
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentral limit theorem
TypeSample size determination
DifficultyStandard +0.8 This S3 question requires understanding of confidence intervals for proportions, knowledge of the relationship between sample size and interval width, and manipulation of the normal approximation formula. While the setup is standard, students must correctly handle the 98% confidence level (finding z = 2.326), work backwards from width to margin of error, and solve for n algebraically. The multi-step reasoning and less common confidence level make this moderately challenging for A-level.
Spec5.05d Confidence intervals: using normal distribution

3 The proportion, \(p\), of an island's population with blood type \(\mathrm { A } \mathrm { Rh } ^ { + }\)is believed to be approximately 0.35 . A medical organisation, requiring a more accurate estimate, specifies that a \(98 \%\) confidence interval for \(p\) should have a width of at most 0.1 . Calculate, to the nearest 10, an estimate of the minimum sample size necessary in order to achieve the organisation's requirement.

Question 3:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(98\%\ (0.98)\ \text{CI} \Rightarrow z = 2.32\) to \(2.33\)B1 AWFW (2.3263)
CI width is \(2 \times z \times \sqrt{\frac{p(1-p)}{n}}\)M1 Used; allow \(z \times \sqrt{\frac{p(1-p)}{n}}\)
\(p = 0.35\) or \(0.50\)B1 Or equivalent
\(2 \times 2.3263 \times \sqrt{\frac{0.35 \times 0.65}{n}} = 0.1\)A1F F on \(z\); allow no multiplier of 2 and/or \(p = 0.50\)
\(\sqrt{n} = \frac{2 \times 2.3263}{0.1} \times \sqrt{0.35 \times 0.65}\)m1 Solving for \(\sqrt{n}\) or \(n\)
\(n = 492.5\ (p=0.35)\) or \(n = 541.2\ (p=0.50)\)
Thus to nearest 10: \(n = 500\) or \(490\)A1 Either
Notes: No '\(\times 2\)' gives \(n = 123.1\); No '\(\times 2\)' and \(p = 0.50\) gives \(n = 135.3\)
Total6
# Question 3:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $98\%\ (0.98)\ \text{CI} \Rightarrow z = 2.32$ to $2.33$ | B1 | AWFW (2.3263) |
| CI width is $2 \times z \times \sqrt{\frac{p(1-p)}{n}}$ | M1 | Used; allow $z \times \sqrt{\frac{p(1-p)}{n}}$ |
| $p = 0.35$ or $0.50$ | B1 | Or equivalent |
| $2 \times 2.3263 \times \sqrt{\frac{0.35 \times 0.65}{n}} = 0.1$ | A1F | F on $z$; allow no multiplier of 2 and/or $p = 0.50$ |
| $\sqrt{n} = \frac{2 \times 2.3263}{0.1} \times \sqrt{0.35 \times 0.65}$ | m1 | Solving for $\sqrt{n}$ or $n$ |
| $n = 492.5\ (p=0.35)$ or $n = 541.2\ (p=0.50)$ | | |
| Thus to nearest 10: $n = 500$ or $490$ | A1 | Either |
| **Notes:** No '$\times 2$' gives $n = 123.1$; No '$\times 2$' and $p = 0.50$ gives $n = 135.3$ | | |
| **Total** | **6** | |
3 The proportion, $p$, of an island's population with blood type $\mathrm { A } \mathrm { Rh } ^ { + }$is believed to be approximately 0.35 .

A medical organisation, requiring a more accurate estimate, specifies that a $98 \%$ confidence interval for $p$ should have a width of at most 0.1 .

Calculate, to the nearest 10, an estimate of the minimum sample size necessary in order to achieve the organisation's requirement.

\hfill \mbox{\textit{AQA S3 2009 Q3 [6]}}