| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Standard +0.3 This is a straightforward application of a one-sample t-test with standard bookwork questions about assumptions and confidence intervals. The calculations are routine (computing sample mean and standard deviation from 9 values, looking up t-critical values, applying formulas), and parts (i), (ii), and (iv) test standard recall of statistical concepts. While it requires multiple steps, no novel insight or complex reasoning is needed—slightly easier than average due to its textbook nature. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| In repeated sampling, 95% of all confidence intervals constructed in this way will contain the true mean. | E1, E1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| CI is given by \(7.631 \pm\) | M1 | ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to \(t_8\) is OK. Allow c's \(\bar{x}\). |
| \(2.306\) | B1 | 2.306 seen. |
| \(\times \dfrac{0.1547}{\sqrt{9}}\) | M1 | Allow c's \(s_{n-1}\). |
| \(= 7.631 \pm 0.118(9) = (7.512, 7.750)\) | A1 | c.a.o. Must be expressed as an interval. |
| [4] |
# Question 1:
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| In repeated sampling, 95% of all confidence intervals constructed in this way will contain the true mean. | E1, E1 | |
| **[2]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| CI is given by $7.631 \pm$ | M1 | ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to $t_8$ is OK. Allow c's $\bar{x}$. |
| $2.306$ | B1 | 2.306 seen. |
| $\times \dfrac{0.1547}{\sqrt{9}}$ | M1 | Allow c's $s_{n-1}$. |
| $= 7.631 \pm 0.118(9) = (7.512, 7.750)$ | A1 | c.a.o. Must be expressed as an interval. |
| **[4]** | | |
---1 A certain industrial process requires a supply of water. It has been found that, for best results, the mean water pressure in suitable units should be 7.8. The water pressure is monitored by taking measurements at regular intervals. On a particular day, a random sample of the measurements is as follows.
$$\begin{array} { l l l l l l l l l }
7.50 & 7.64 & 7.68 & 7.51 & 7.70 & 7.85 & 7.34 & 7.72 & 7.74
\end{array}$$
These data are to be used to carry out a hypothesis test concerning the mean water pressure.\\
(i) Why is a test based on the Normal distribution not appropriate in this case?\\
(ii) What distributional assumption is needed for a test based on the $t$ distribution?\\
(iii) Carry out a $t$ test, with a $2 \%$ level of significance, to see whether it is reasonable to assume that the mean pressure is 7.8 .\\
(iv) Explain what is meant by a $95 \%$ confidence interval.\\
(v) Find a $95 \%$ confidence interval for the actual mean water pressure.
\hfill \mbox{\textit{OCR MEI S3 2013 Q1 [18]}}