| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Standard +0.3 This is a straightforward confidence interval calculation using coded data with standard formulas. The main steps are computing sample mean and variance from the coded summations, then applying the Central Limit Theorem (n=80 is large, so normality not required). While it requires careful arithmetic and understanding of when normality assumptions are needed, it's a routine S3 exercise with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\bar{x} = 25.0055\) | B1 | For correct sample mean, or equivalent; the 25 may be taken into account later |
| \(s^2 = \frac{1}{79}\left(0.2287 - \frac{0.44^2}{80}\right) = 0.00286\ldots\) | M1 | For correct unsimplified expression |
| A1 | For correct unbiased estimate | |
| Interval is \(25.0055 \pm 2.576\sqrt{\frac{0.00286}{80}}\) | M1 | For calculation of the form \(\bar{x} \pm z\sqrt{(s^2/n)}\) |
| B1 | For relevant use of \(z = 2.576\) | |
| Hence \(24.99(0) < \mu < 25.02(1)\) | A1 | 6 |
| (ii) The sample size of 80 is sufficiently large for the Central Limit Theorem to apply, so it is not necessary to assume a normal distribution | M1 | For mention of sample size and CLT |
| A1 | 2 | For the correct conclusion and reason |
**(i)** $\bar{x} = 25.0055$ | B1 | For correct sample mean, or equivalent; the 25 may be taken into account later
$s^2 = \frac{1}{79}\left(0.2287 - \frac{0.44^2}{80}\right) = 0.00286\ldots$ | M1 | For correct unsimplified expression
| A1 | For correct unbiased estimate
Interval is $25.0055 \pm 2.576\sqrt{\frac{0.00286}{80}}$ | M1 | For calculation of the form $\bar{x} \pm z\sqrt{(s^2/n)}$
| B1 | For relevant use of $z = 2.576$
Hence $24.99(0) < \mu < 25.02(1)$ | A1 | 6 | For correct interval, stated to an appropriate degree of accuracy
**(ii)** The sample size of 80 is sufficiently large for the Central Limit Theorem to apply, so it is not necessary to assume a normal distribution | M1 | For mention of sample size and CLT
| A1 | 2 | For the correct conclusion and reason3 A random sample of 80 precision-engineered cylindrical components is checked as part of a quality control process. The diameters of the cylinders should be 25.00 cm . Accurate measurements of the diameters, $x \mathrm {~cm}$, for the sample are summarised by
$$\Sigma ( x - 25 ) = 0.44 , \quad \Sigma ( x - 25 ) ^ { 2 } = 0.2287 .$$
(i) Calculate a $99 \%$ confidence interval for the population mean diameter of the components.\\
(ii) For the calculation in part (i) to be valid, is it necessary to assume that component diameters are normally distributed? Justify your answer.
\hfill \mbox{\textit{OCR S3 Q3 [8]}}