| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Standard CI with summary statistics |
| Difficulty | Moderate -0.3 This is a standard two-sample confidence interval question with all summary statistics provided. Students need to apply the formula for difference of means, calculate the standard error, find the critical value (z = 2.326 for 98%), and interpret. The only slight challenge is stating the independence assumption, but this is routine S3 material requiring no problem-solving insight. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| \multirow{2}{*}{} | Starting salary (€) | ||
| Sample size | Sample mean | Sample standard deviation | |
| Science graduates | 175 | 19268 | 7321 |
| Arts graduates | 225 | 17896 | 8205 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Samples are independent or random | B1 | |
| \(98\% \Rightarrow z = 2.3263\) | B1 | AWFW 2.32 to 2.33 |
| \((\bar{x}_S - \bar{x}_A) \pm z \times \sqrt{\frac{s_S^2}{n_S} + \frac{s_A^2}{n_A}}\) | M1 | Form; allow sigmas, \(A\&B\) or \(1\&2\) and \(n-1\) |
| Correct expression | A1 | |
| \((19268 - 17896) \pm 2.3263 \times \sqrt{\frac{7321^2}{175} + \frac{8205^2}{225}}\) | A1\(\checkmark\) | \(\checkmark\) on \(z\) only; \(s_P = 7830\) to \(7850\) |
| \(1372 \pm (1805\) to \(1820)\) or \((-450\) to \(-430, 3170\) to \(3200)\) | A1 | \(1372 \pm (1830\) to \(1845)\); AWFW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Confidence interval includes zero so (at 5% level) | B1\(\checkmark\) \(\uparrow\)dep\(\uparrow\) | \(\checkmark\) on CI; OE |
| Mean starting salaries may be equal | B1\(\checkmark\) | \(\checkmark\) on CI; OE |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Samples are independent or random | B1 | |
| $98\% \Rightarrow z = 2.3263$ | B1 | AWFW 2.32 to 2.33 |
| $(\bar{x}_S - \bar{x}_A) \pm z \times \sqrt{\frac{s_S^2}{n_S} + \frac{s_A^2}{n_A}}$ | M1 | Form; allow sigmas, $A\&B$ or $1\&2$ and $n-1$ |
| Correct expression | A1 | |
| $(19268 - 17896) \pm 2.3263 \times \sqrt{\frac{7321^2}{175} + \frac{8205^2}{225}}$ | A1$\checkmark$ | $\checkmark$ on $z$ only; $s_P = 7830$ to $7850$ |
| $1372 \pm (1805$ to $1820)$ or $(-450$ to $-430, 3170$ to $3200)$ | A1 | $1372 \pm (1830$ to $1845)$; AWFW |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Confidence interval includes zero so (at 5% level) | B1$\checkmark$ $\uparrow$dep$\uparrow$ | $\checkmark$ on CI; OE |
| Mean starting salaries may be equal | B1$\checkmark$ | $\checkmark$ on CI; OE |
---1 As part of an investigation into the starting salaries of graduates in a European country, the following information was collected.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
\multirow{2}{*}{} & & \multicolumn{2}{|c|}{Starting salary (€)} \\
\hline
& Sample size & Sample mean & Sample standard deviation \\
\hline
Science graduates & 175 & 19268 & 7321 \\
\hline
Arts graduates & 225 & 17896 & 8205 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Stating a necessary assumption about the samples, construct a $98 \%$ confidence interval for the difference between the mean starting salary of science graduates and that of arts graduates.
\item What can be concluded from your confidence interval?
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2007 Q1 [8]}}