| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Moderate -0.3 This is a straightforward application of t-distribution confidence intervals with given summary statistics. Part (a)(i) requires standard formula substitution (calculating sample mean, standard deviation, and using t-tables), part (a)(ii) is simple interpretation, and part (b) tests basic understanding of confidence interval meaning. All steps are routine for S2 level with no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = 3.19\) and \(s^2 = \frac{1.849}{9} = 0.2054\) | B1 | Both (\(s = 0.453\)) |
| \(t_9 = 3.250\) | B1 | |
| \(3.19 \pm 3.250 \times \frac{\sqrt{0.2054}}{\sqrt{10}}\) | M1 | \(3.19 \pm (\text{their } t_9) \times \frac{\sqrt{0.2054}}{\sqrt{10}}\) |
| \(= 3.19 \pm 0.4658\) | A1ft | |
| \(= (2.72,\ 3.66)\) | A1 | 5 marks; (2.72 to 2.73, 3.65 to 3.66) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Reasonable claim, with 3.5 within the 99% confidence interval | B1, E1 | 2 marks; dep on correct CI in (a)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.01 \times 200 = 2\) | B1 | 1 mark |
## Question 5:
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 3.19$ and $s^2 = \frac{1.849}{9} = 0.2054$ | B1 | Both ($s = 0.453$) |
| $t_9 = 3.250$ | B1 | |
| $3.19 \pm 3.250 \times \frac{\sqrt{0.2054}}{\sqrt{10}}$ | M1 | $3.19 \pm (\text{their } t_9) \times \frac{\sqrt{0.2054}}{\sqrt{10}}$ |
| $= 3.19 \pm 0.4658$ | A1ft | |
| $= (2.72,\ 3.66)$ | A1 | 5 marks; (2.72 to 2.73, 3.65 to 3.66) |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Reasonable claim, with 3.5 within the 99% confidence interval | B1, E1 | 2 marks; dep on correct CI in (a)(i) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.01 \times 200 = 2$ | B1 | 1 mark |
---5 The weight of fat in a digestive biscuit is known to be normally distributed.\\
Pat conducted an experiment in which she measured the weight of fat, $x$ grams, in each of a random sample of 10 digestive biscuits, with the following results:
$$\sum x = 31.9 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 1.849$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Construct a $99 \%$ confidence interval for the mean weight of fat in digestive biscuits.
\item Comment on a claim that the mean weight of fat in digestive biscuits is 3.5 grams.
\end{enumerate}\item If 200 such $99 \%$ confidence intervals were constructed, how many would you expect not to contain the population mean?
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2008 Q5 [8]}}