| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Standard +0.3 This is a straightforward application of standard confidence interval procedures: calculating sample statistics from summaries, constructing a confidence interval using normal distribution, and interpreting confidence level as expected frequency. All steps are routine with no conceptual challenges beyond basic understanding of confidence intervals. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(s^2 = \frac{1}{129}\left(2371.3 - \frac{555.1^2}{130}\right) = 0.00793\) | B1 | Correct mean |
| M1 | Substituting in formula from tables (or equiv) | |
| A1 [3] | Correct variance | |
| (ii) \(CI: 4.27 \pm 2.17 \times \frac{0.08905}{\sqrt{130}} = (4.25, 4.29)\) | B1 | Correct 2 used (2.169–2.171) |
| M1 | Correct form of expression \(\sqrt{130}\) in denominator | |
| A1 [3] | Correct answer (cwo) | |
| (iii) 9 | B1 [1] | c.a.o |
(i) $\bar{x} = 4.27$
$s^2 = \frac{1}{129}\left(2371.3 - \frac{555.1^2}{130}\right) = 0.00793$ | B1 | Correct mean
| M1 | Substituting in formula from tables (or equiv)
| A1 [3] | Correct variance
(ii) $CI: 4.27 \pm 2.17 \times \frac{0.08905}{\sqrt{130}} = (4.25, 4.29)$ | B1 | Correct 2 used (2.169–2.171)
| M1 | Correct form of expression $\sqrt{130}$ in denominator
| A1 [3] | Correct answer (cwo)
(iii) 9 | B1 [1] | c.a.o
---4 Diameters of golf balls are known to be normally distributed with mean $\mu \mathrm { cm }$ and standard deviation $\sigma \mathrm { cm }$. A random sample of 130 golf balls was taken and the diameters, $x \mathrm {~cm}$, were measured. The results are summarised by $\Sigma x = 555.1$ and $\Sigma x ^ { 2 } = 2371.30$.\\
(i) Calculate unbiased estimates of $\mu$ and $\sigma ^ { 2 }$.\\
(ii) Calculate a $97 \%$ confidence interval for $\mu$.\\
(iii) 300 random samples of 130 balls are taken and a $97 \%$ confidence interval is calculated for each sample. How many of these intervals would you expect not to contain $\mu$ ?
\hfill \mbox{\textit{CAIE S2 2008 Q4 [7]}}