| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Test using proportion |
| Difficulty | Standard +0.3 This is a straightforward two-part hypothesis testing question involving proportions. Part (i) requires standard confidence interval calculation for a proportion, and part (ii) is a routine one-tailed test comparing two proportions. Both parts use well-practiced formulas with no conceptual challenges or novel problem-solving required, making it slightly easier than average for S3 level. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks |
|---|---|
| M1, B1, A1, A1 | Or /199 \((0.095, 0.175)\) to 3DP |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Pooled estimate of \(p = 35/300\) | M1, B1, A1, A1 | Correct form of sd |
| M1 | OR: \(P(z \geq 1.399) = 0.0809 < 0.10\) | |
| A1 | SR: No pooled estimate; B1M1B0B0 | |
| A1 for 1.514, M1A1 Max 5/7 |
## 5(i)
Use $p_s \pm z\sqrt{p(q/200)}$
$z = 1.645$
$s = \sqrt{0.135 \times 0.865/200}$ giving $(0.0952, 0.1747)$
| | M1, B1, A1, A1 | Or /199 $(0.095, 0.175)$ to 3DP |
| --- | --- | --- |
| | | **[4]** |
## 5(ii)
$H_0: p_1 - p_2 = 0$, $H_1: p_1 - p_2 > 0$
$\frac{27/200 - 8/100}{\sqrt{35/300 \times 265/300 \times (200^{-1} + 100^{-1})}}$
$= 1.399$
$> 1.282$
Reject $H_0$. There is sufficient evidence at the 10% significance level that the proportion of faulty bars has reduced
| Pooled estimate of $p = 35/300$ | M1, B1, A1, A1 | Correct form of sd |
| --- | --- | --- |
| | M1 | OR: $P(z \geq 1.399) = 0.0809 < 0.10$ |
| | A1 | SR: No pooled estimate; B1M1B0B0 |
| | | A1 for 1.514, M1A1 Max 5/7 |
**Total: [11]**
---5 Each of a random sample of 200 steel bars taken from a production line was examined and 27 were found to be faulty.\\
(i) Find an approximate $90 \%$ confidence interval for the proportion of faulty bars produced.
A change in the production method was introduced which, it was claimed, would reduce the proportion of faulty bars. After the change, each of a further random sample of 100 bars was examined and 8 were found to be faulty.\\
(ii) Test the claim, at the $10 \%$ significance level.
\hfill \mbox{\textit{OCR S3 2010 Q5 [11]}}