| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| 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 standard confidence interval formulas for proportions. Part (i) uses the normal approximation with a simple formula, part (ii) applies the margin of error formula requiring one algebraic rearrangement to solve for n, and part (iii) asks for conceptual understanding. All techniques are routine for S3 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Use \((^s/_{80})^{11/80}/80\) | B1 | Or /79 |
| \(\rho_s \pm zx\) | M1 | \(s\) of the form \(\sqrt{(\rho_s q_s/80}\) (or 79) or no \(\sqrt{}\) |
| \(z= 1.96\) | B1 | |
| \((0.0173, 0.1327)\) | A1 | 4 Accept \((0.017,0.133)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use \(z\sqrt{(p_s q_s/n)}\) | M1 | or no \(\sqrt{}\) |
| \(\leq 0.05\) | A1 | and \(z=1.96\) ,Or \(=\) |
| \(n \geq 106.6\) , least is 107 | A1 | 3 Allow 110 |
| Answer | Marks | Guidance |
|---|---|---|
| e.g Variance is an estimate OR Distribution of \(p_s\) is only approx normal | B1 | 1(8) Must state distribution of what. |
**2(i)**
Use $(^s/_{80})^{11/80}/80$ | B1 | Or /79
$\rho_s \pm zx$ | M1 | $s$ of the form $\sqrt{(\rho_s q_s/80}$ (or 79) or no $\sqrt{}$
$z= 1.96$ | B1 |
$(0.0173, 0.1327)$ | A1 | 4 Accept $(0.017,0.133)$
**2(ii)**
Use $z\sqrt{(p_s q_s/n)}$ | M1 | or no $\sqrt{}$
$\leq 0.05$ | A1 | and $z=1.96$ ,Or $=$
$n \geq 106.6$ , least is 107 | A1 | 3 Allow 110
**2(iii)**
e.g Variance is an estimate OR Distribution of $p_s$ is only approx normal | B1 | 1(8) Must state distribution of what.
---2 The population proportion of all men with red-green colour blindness is denoted by $p$. Each of a random sample of 80 men was tested and it was found that 6 had red-green colour blindness.\\
(i) Calculate an approximate $95 \%$ confidence interval for $p$.\\
(ii) For a different random sample of men, the proportion with red-green colour blindness is denoted by $p _ { s }$. Estimate the sample size required in order that $\left| p _ { s } - p \right| \leqslant 0.05$ with probability $95 \%$.\\
(iii) Give one reason why the calculated sample size is an estimate.
\hfill \mbox{\textit{OCR S3 2011 Q2 [8]}}