| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| 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 large samples (n=50 each) and given summary statistics. Part (a) requires routine application of the formula for difference of means with known standard deviations and z-critical value lookup. Parts (b)(i) and (b)(ii) test understanding of random sampling and CLT, which are bookwork. The large sample sizes make the calculation straightforward with no normality assumption needed. Slightly easier than average due to being entirely procedural with no problem-solving element. |
| Spec | 2.01a Population and sample: terminology5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((42.6 - 39.7) \pm z \sqrt{\frac{6.2^2}{50} + \frac{5.3^2}{50}}\) | M1 | Correct structure for two-sample CI |
| \(z = 2.326\) for 98% | B1 | Correct z-value |
| \(SE = \sqrt{\frac{38.44}{50} + \frac{28.09}{50}} = \sqrt{1.331} = 1.1536...\) | M1 | Correct standard error calculation |
| \(2.9 \pm 2.685\) | A1 | Correct unsimplified interval |
| \((0.215,\ 5.585)\) kg | A1 | Correct final interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The two samples were selected independently | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Both sample sizes are large (\(n = 50\)), so by the Central Limit Theorem the sample means are approximately normally distributed | B1+B1 | Large samples stated; CLT referenced |
## Question 4:
**(a)** 98% confidence interval for difference in mean weights (Eastern Grey − Western Grey)
| Working | Marks | Guidance |
|---------|-------|----------|
| $(42.6 - 39.7) \pm z \sqrt{\frac{6.2^2}{50} + \frac{5.3^2}{50}}$ | M1 | Correct structure for two-sample CI |
| $z = 2.326$ for 98% | B1 | Correct z-value |
| $SE = \sqrt{\frac{38.44}{50} + \frac{28.09}{50}} = \sqrt{1.331} = 1.1536...$ | M1 | Correct standard error calculation |
| $2.9 \pm 2.685$ | A1 | Correct unsimplified interval |
| $(0.215,\ 5.585)$ kg | A1 | Correct final interval |
**(b)(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| The two samples were selected independently | B1 | |
**(b)(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Both sample sizes are large ($n = 50$), so by the Central Limit Theorem the sample means are approximately normally distributed | B1+B1 | Large samples stated; CLT referenced |4 A sample of 50 male Eastern Grey kangaroos had a mean weight of 42.6 kg and a standard deviation of 6.2 kg .
A sample of 50 male Western Grey kangaroos had a mean weight of 39.7 kg and a standard deviation of 5.3 kg .
\begin{enumerate}[label=(\alph*)]
\item Construct a 98\% confidence interval for the difference between the mean weight of male Eastern Grey kangaroos and that of male Western Grey kangaroos.\\[0pt]
[5 marks]
\item \begin{enumerate}[label=(\roman*)]
\item What assumption about the selection of each of the two samples was it necessary to make in order that the confidence interval constructed in part (a) was valid?\\[0pt]
[1 mark]
\item Why was it not necessary to assume anything about the distributions of the weights of male kangaroos in order that the confidence interval constructed in part (a) was valid?\\[0pt]
[2 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2014 Q4 [8]}}