| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Proportion confidence interval |
| Difficulty | Standard +0.3 This is a straightforward application of the standard proportion confidence interval formula with normal approximation. Part (i) requires plugging values into a formula, part (ii) tests understanding of sampling/polling limitations (standard textbook discussion), and part (iii) involves rearranging the width formula—all routine S3 techniques with no novel problem-solving required. Slightly above average difficulty only because it's Further Maths content and requires careful arithmetic with the 99.9% z-value. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \sqrt{\frac{0.376 \times 0.624}{1731}}\) | B1 | Allow from % throughout. |
| \(p_s \pm zs\) | M1 | \(s\) must be correct structure. |
| \(z = 3.291\) | B1 | Allow 3.29 |
| \((0.338,\ 0.414)\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 2 from e.g.: One in a thousand CI does not contain popn. proportion. / Some of the voters lied. / The CI is approx. (because a discrete distn. has been approx. by a cs. one.) or estimate. / A continuity correction has not been applied. / The popn. var. is estimated from the sample. / The distn. of \(P_s\) is only approx. normal. | B1, B1 | The sample was unrepresentative B2. Voters not independent. / Voters may have changed their minds. / Some voters forgot to vote. / Sample biased. Sample not random B0. Small sample B0. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z\sqrt{\frac{0.376 \times 0.624}{n}}\) | M1 | Allow incorrect structure if same as (i) |
| \(= 0.025\) with \(z = 3.291\) | A1 | SC 4065 with no working B2 |
| \(n = 4066\) | A1 | Must be integer. |
| [3] |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \sqrt{\frac{0.376 \times 0.624}{1731}}$ | B1 | Allow from % throughout. |
| $p_s \pm zs$ | M1 | $s$ must be correct structure. |
| $z = 3.291$ | B1 | Allow 3.29 |
| $(0.338,\ 0.414)$ | A1 | |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 2 from e.g.: One in a thousand CI does not contain popn. proportion. / Some of the voters lied. / The CI is approx. (because a discrete distn. has been approx. by a cs. one.) or estimate. / A continuity correction has not been applied. / The popn. var. is estimated from the sample. / The distn. of $P_s$ is only approx. normal. | B1, B1 | The sample was unrepresentative B2. Voters not independent. / Voters may have changed their minds. / Some voters forgot to vote. / Sample biased. Sample not random B0. Small sample B0. |
| **[2]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z\sqrt{\frac{0.376 \times 0.624}{n}}$ | M1 | Allow incorrect structure if same as (i) |
| $= 0.025$ with $z = 3.291$ | A1 | SC 4065 with no working B2 |
| $n = 4066$ | A1 | Must be integer. |
| **[3]** | | |
---5 The day before the 1992 General Election, an opinion poll showed that $37.6 \%$ of a random sample of 1731 voters intended to vote for the Conservative party.\\
(i) Calculate an approximate $99.9 \%$ confidence interval for the proportion of voters intending to vote Conservative.
The actual proportion voting Conservative was above the upper limit of the confidence interval.\\
(ii) Give two possible reasons for this occurrence.\\
(iii) What sample size would be required to produce a $99.9 \%$ confidence interval of width 0.05 ?
\hfill \mbox{\textit{OCR S3 2014 Q5 [9]}}