| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Standard CI with summary statistics |
| Difficulty | Standard +0.3 This is a straightforward two-part confidence interval question using standard formulas. Part (a) requires a single-sample CI with given summary statistics (routine calculation). Part (b) involves a two-sample CI for difference in means with pooled variance—slightly more involved but still a textbook application of standard S4 techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) Confidence interval is given by | B1 | |
| \(\bar{x} \pm t_{0.95} \frac{s}{\sqrt{n}}\) | M1 | Use \(\bar{x} \pm t_{9\alpha}\) |
| i.e.: \(207.1 \pm 2.539 \times \sqrt{\frac{3.2}{28}}\) | A1 | All correct |
| i.e.: \(207.1 \pm 1.0156\) | A1 (4) | |
| i.e.: \((206.08..., 208.1156)\) | A1 (4) | AWRT (206, 208) |
| (b) \(s_p^2 = \frac{19 \times 3.2 + 9 \times 10.2173}{28} = 5.4565...\) | B1 M1 A1 | \(\frac{10.2173}{...}\) AWRT 10.2 or 5.46 |
| Confidence interval is given by | ||
| \(\bar{x}_C - \bar{x}_Q \pm t_{...} \times \sqrt{s.4555(\frac{1}{20} + \frac{1}{10})}\) | B1 | |
| i.e.: \((207.1 - 204.62) \pm 1.701 \sqrt{5.4555(\frac{1}{20} + \frac{1}{10})}\) | M1 A1 | Use \(\bar{x}\)-dot; \(\sqrt{s_p^2(\frac{1}{n_1} + \frac{1}{n_2})}\) |
| i.e.: \(2.48 \pm 1.538 75\) | M1 A1 | All correct |
| i.e.: \((0.94125, 4.0187)\) | A1N (a) | AWRT 0.941; 4.02 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** Confidence interval is given by | B1 | |
| $\bar{x} \pm t_{0.95} \frac{s}{\sqrt{n}}$ | M1 | Use $\bar{x} \pm t_{9\alpha}$ |
| i.e.: $207.1 \pm 2.539 \times \sqrt{\frac{3.2}{28}}$ | A1 | All correct |
| i.e.: $207.1 \pm 1.0156$ | A1 (4) | |
| i.e.: $(206.08..., 208.1156)$ | A1 (4) | AWRT (206, 208) |
| **(b)** $s_p^2 = \frac{19 \times 3.2 + 9 \times 10.2173}{28} = 5.4565...$ | B1 M1 A1 | $\frac{10.2173}{...}$ AWRT 10.2 or 5.46 |
| Confidence interval is given by | | |
| $\bar{x}_C - \bar{x}_Q \pm t_{...} \times \sqrt{s.4555(\frac{1}{20} + \frac{1}{10})}$ | B1 | |
| i.e.: $(207.1 - 204.62) \pm 1.701 \sqrt{5.4555(\frac{1}{20} + \frac{1}{10})}$ | M1 A1 | Use $\bar{x}$-dot; $\sqrt{s_p^2(\frac{1}{n_1} + \frac{1}{n_2})}$ |
| i.e.: $2.48 \pm 1.538 75$ | M1 A1 | All correct |
| i.e.: $(0.94125, 4.0187)$ | A1N (a) | AWRT 0.941; 4.02 |
---6. Brickland and Goodbrick are two manufacturers of bricks. The lengths of the bricks produced by each manufacturer can be assumed to be normally distributed. A random sample of 20 bricks is taken from Brickland and the length, $x \mathrm {~mm}$, of each brick is recorded. The mean of this sample is 207.1 mm and the variance is $3.2 \mathrm {~mm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the $98 \%$ confidence interval for the mean length of brick from Brickland.
A random sample of 10 bricks is selected from those manufactured by Goodbrick. The length of each brick, $y \mathrm {~mm}$, is recorded. The results are summarised as follows.
$$\sum y = 2046.2 \quad \sum y ^ { 2 } = 418785.4$$
The variances of the length of brick for each manufacturer are assumed to be the same.
\item Find a $90 \%$ confidence interval for the value by which the mean length of brick made by Brickland exceeds the mean length of brick made by Goodbrick.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2005 Q6 [12]}}