| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Standard +0.3 This is a straightforward application of a paired t-test with standard bookwork explanations. Part (i) requires understanding of paired designs, part (ii) tests knowledge of t-test assumptions, and parts (iii)-(iv) involve routine calculations with given data. The question is slightly easier than average because it's a direct textbook-style application with clear structure and no conceptual surprises, though it does require multiple steps and proper statistical interpretation. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Surface | A | B | C | D | E | F | G | H | I | J |
| Old paint | 16.6 | 17.0 | 16.5 | 15.6 | 16.3 | 16.5 | 16.4 | 15.9 | 16.3 | 16.1 |
| New paint | 15.9 | 16.3 | 16.3 | 15.9 | 15.5 | 16.6 | 16.1 | 16.0 | 16.2 | 15.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| CI given by \(0.28 \pm\) | M1 | Allow c's \(\bar{x}\) |
| \(2.262\) | B1 | |
| \(\times \frac{0.3853}{\sqrt{10}}\) | M1 | Allow c's \(s_{n-1}\) |
| \(= 0.28 \pm 0.2756 = (0.0044, 0.5556)\) | A1 | c.a.o. Must be expressed as an interval. Require 3/4 dp; condone 5. If final answer centred on negative sample mean do not award final A mark. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1 B0 M1 A0. Recovery to \(t_9\) is OK. |
| [4] |
# Question 1:
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| CI given by $0.28 \pm$ | M1 | Allow c's $\bar{x}$ |
| $2.262$ | B1 | |
| $\times \frac{0.3853}{\sqrt{10}}$ | M1 | Allow c's $s_{n-1}$ |
| $= 0.28 \pm 0.2756 = (0.0044, 0.5556)$ | A1 | c.a.o. Must be expressed as an interval. Require 3/4 dp; condone 5. If final answer centred on negative sample mean do not award final A mark. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1 B0 M1 A0. Recovery to $t_9$ is OK. |
| **[4]** | | |
---1 Technologists at a company that manufactures paint are trying to develop a new type of gloss paint with a shorter drying time than the current product. In order to test whether the drying time has been reduced, the technologists paint a square metre of each of the new and old paints on each of 10 different surfaces. The lengths of time, in hours, that each square metre takes to dry are as follows.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Surface & A & B & C & D & E & F & G & H & I & J \\
\hline
Old paint & 16.6 & 17.0 & 16.5 & 15.6 & 16.3 & 16.5 & 16.4 & 15.9 & 16.3 & 16.1 \\
\hline
New paint & 15.9 & 16.3 & 16.3 & 15.9 & 15.5 & 16.6 & 16.1 & 16.0 & 16.2 & 15.6 \\
\hline
\end{tabular}
\end{center}
(i) Explain why a paired sample is used in this context.\\
(ii) The mean reduction in drying time is to be investigated. Why might a $t$ test be appropriate in this context and what assumption needs to be made?\\
(iii) Using a significance level of $5 \%$, carry out a test to see if there appears to be any reduction in mean drying time.\\
(iv) Find a 95\% confidence interval for the true mean reduction in drying time.
\hfill \mbox{\textit{OCR MEI S3 2012 Q1 [18]}}