| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Standard +0.3 This is a straightforward paired t-test application with clear data and standard procedures. Students must calculate differences, perform a one-tailed test against a specific value (not just zero), and construct a confidence interval. While requiring careful attention to the hypothesis (decrease > 0.2 rather than simply ≠ 0), the mechanics are routine S3 material with no conceptual surprises or complex interpretation needed. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Bottle | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Original strength | 8.7 | 9.4 | 9.2 | 8.9 | 9.6 | 8.2 | 9.9 | 8.8 |
| Final strength | 8.1 | 9.0 | 9.0 | 8.8 | 9.3 | 8.0 | 9.5 | 8.5 |
| Answer | Marks |
|---|---|
| B1 | For both hypotheses |
| B1, M1 | B1 Use paired differences t-test |
| B1, A1, M1, A1 | Must have /8 |
| M1, A1 | OR: \(P(t \geq 2.049) = 0.0398 < 0.05\) |
| Allow M1 from \(t_{0.4} = 1.761\) | |
| SR: 2-sample test:B1B1M0B1A0 | |
| M1 using 1.761 A1 Max 4/9 |
| Answer | Marks |
|---|---|
| M1 | Allow with \(z\) but with /8 |
| B1 | Rounding to \((0.283, 0.442)\) |
| A1 |
## 6(i)
Assumes that decreases have a normal distribution
$H_0: \mu_D = 0.2$ (or $\equiv 2$), $H_1: \mu_D > 0.2$
O-F: 0.6 0.4 0.2 0.1 0.3 0.2 0.4 0.3
$\bar{D} = 0.3125$, $s^2 = 0.024107$
$\frac{(0.3125 - 0.2)\sqrt{0.024107/8}}{= 2.049}$
$> 1.895$
Reject $H_0$ – there is sufficient evidence at the 5% significance level that the reduction is more than 0.2
| B1 | For both hypotheses |
| --- | --- |
| B1, M1 | B1 Use paired differences t-test |
| B1, A1, M1, A1 | Must have /8 |
| M1, A1 | OR: $P(t \geq 2.049) = 0.0398 < 0.05$ |
| | | Allow M1 from $t_{0.4} = 1.761$ |
| | | SR: 2-sample test:B1B1M0B1A0 |
| | | M1 using 1.761 A1 Max 4/9 |
**Total: 9**
## 6(ii)
$0.3125 \pm t \sqrt{0.024107/8}$
$t = 2.365$
$(0.1827, 0.4423)$
| M1 | Allow with $z$ but with /8 |
| --- | --- |
| B1 | Rounding to $(0.283, 0.442)$ |
| A1 | |
**Total: [12]**
---6 The deterioration of a certain drug over time was investigated as follows. The drug strength was measured in each of a random sample of 8 bottles containing the drug. These were stored for two years and the strengths were then re-measured. The original and final strengths, in suitable units, are shown in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Bottle & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
Original strength & 8.7 & 9.4 & 9.2 & 8.9 & 9.6 & 8.2 & 9.9 & 8.8 \\
\hline
Final strength & 8.1 & 9.0 & 9.0 & 8.8 & 9.3 & 8.0 & 9.5 & 8.5 \\
\hline
\end{tabular}
\end{center}
(i) Stating any required assumption, test at the $5 \%$ significance level whether the mean strength has decreased by more than 0.2 over the two years.\\
(ii) Calculate a 95\% confidence interval for the mean reduction in strength over the two years.
\hfill \mbox{\textit{OCR S3 2010 Q6 [12]}}