| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Sample size determination |
| Difficulty | Standard +0.3 This is a straightforward application of CLT-based confidence intervals and sample size formula. Part (i) is routine calculation, part (ii) uses the standard formula n = (z*σ/E)², and part (iii) tests understanding that CLT justifies normality for large samples. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(2.18 \pm z(1.58/\sqrt{500})\) | M1 | allow \(t\) |
| \(z = 2.326\) | B1 | |
| \((2.02,\ 2.34)\) seconds | A1 | Interval seen somewhere, allow reversed. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Requires \(P( | \bar{x} - \mu | < 0.05) = 0.95\) |
| \(P(z < 0.05/(1.58/\sqrt{n})) = 0.95\) | M1 | Standardising. Expression not involving 2.18. Can have \(\times 2\). Allow use of same incorrect sd as part (i) for 2nd M1 |
| \(0.05\sqrt{n}/1.58 > 1.96\) or \(= 1.96\) | A1 | All but \(n\) numeric |
| \(n = [3836,\ 3840]\) | A1 | NOT \(<\) or \(>\). Integer required. |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(n\) large enough for sample mean to approximate to normal | B1 | Must be mean. Allow \(n\) large enough for CLT to apply. NOT just '\(n\) large'. |
| [1] |
# Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $2.18 \pm z(1.58/\sqrt{500})$ | M1 | allow $t$ |
| $z = 2.326$ | B1 | |
| $(2.02,\ 2.34)$ seconds | A1 | Interval seen somewhere, allow reversed. |
| | **[3]** | |
---
# Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Requires $P(|\bar{x} - \mu| < 0.05) = 0.95$ | M1 | Setting up an equality or equation, even if not $z$ |
| $P(z < 0.05/(1.58/\sqrt{n})) = 0.95$ | M1 | Standardising. Expression not involving 2.18. Can have $\times 2$. Allow use of same incorrect sd as part (i) for 2nd M1 |
| $0.05\sqrt{n}/1.58 > 1.96$ or $= 1.96$ | A1 | All but $n$ numeric |
| $n = [3836,\ 3840]$ | A1 | NOT $<$ or $>$. Integer required. |
| | **[4]** | |
---
# Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n$ large enough for sample mean to approximate to normal | B1 | Must be mean. Allow $n$ large enough for CLT to apply. NOT just '$n$ large'. |
| | **[1]** | |
---4 A new computer was bought by a local council to search council records and was tested by an employee. She searched a random sample of 500 records and the sample mean search time was found to be 2.18 milliseconds and an unbiased estimate of variance was $1.58 ^ { 2 }$ milliseconds ${ } ^ { 2 }$.\\
(i) Calculate a $98 \%$ confidence interval for the population mean search time $\mu$ milliseconds.\\
(ii) It is required to obtain a sample mean time that differs from $\mu$ by less than 0.05 milliseconds with probability 0.95 . Estimate the sample size required.\\
(iii) State why it is unnecessary for the validity of your calculations that search time has a normal distribution.
\hfill \mbox{\textit{OCR S3 2013 Q4 [8]}}