| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Proportion confidence interval |
| Difficulty | Moderate -0.3 This is a straightforward application of the normal approximation for proportion confidence intervals. Part (i) requires plugging values into the standard formula with z=2.326, while part (ii) involves rearranging the width formula to solve for n. Both are routine S3 procedures with no conceptual challenges, making it slightly easier than average but still requiring correct execution of multiple steps. |
| Spec | 5.05d Confidence intervals: using normal distribution |
3 In a random sample of credit card holders, it was found that $28 \%$ of them used their card for internet purchases.\\
(i) Given that the sample size is 1200 , find a $98 \%$ confidence interval for the percentage of all credit card holders who use their card for internet purchases.\\
(ii) Estimate the smallest sample size for which a $98 \%$ confidence interval would have a width of at most $5 \%$, and state why the value found is only an estimate.
\hfill \mbox{\textit{OCR S3 2009 Q3 [8]}}