| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| 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. Students need to calculate sample mean and standard deviation, look up a t-critical value, and apply the standard formula. Part (b) requires only a simple comparison. While it involves multiple steps, each is routine and follows a standard textbook procedure with no problem-solving or novel insight required. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{\sum x}{n} = \frac{546}{15} = \frac{182}{5} = 36.4\) | B1 | oe |
| \(s^2 = \frac{\sum(x - \bar{x})^2}{n-1} = \frac{1407.6}{14} = 100.54\) (or \(s = 10.03\)) | B1 | \(\sigma^2 = 93.84\) or \(\sigma = 9.687\) iff \(\sqrt{\frac{}{n}}\) used below |
| \(t_{crit} = \pm 2.624\) | B1 | ignore signs for \(t_{crit}\) (allow \(t = 2.62\)) (if \(z\) used then max(B1B1B0 M0A0)) |
| 98% CI for \(\mu\): | ||
| \(36.4 \pm 2.624 \times \frac{s}{\sqrt{15}}\) | M1 | their \(\bar{x} \pm t_{14} \times \frac{\text{their } s}{\sqrt{15}}\) or their \(\bar{x} \pm t_{14} \times \frac{\text{their } \sigma}{\sqrt{14}}\) (allow any of the following for \(t_{14}\): 1.345; 1.761; 2.145; 2.624; 2.977) |
| \(36.4 \pm 6.8\) | A1ft | |
| \((29.6, 43.2)\) or \(36.4 \pm 6.8\) | A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(40.0 \in \text{C.I.} \Rightarrow\) no change | E1ft | Must refer to 40 (dep M1) |
| E1ft | 2 | Dep on previous mark |
| 8 |
### 1(a)
| $\bar{x} = \frac{\sum x}{n} = \frac{546}{15} = \frac{182}{5} = 36.4$ | B1 | oe |
| --- | --- | --- |
| $s^2 = \frac{\sum(x - \bar{x})^2}{n-1} = \frac{1407.6}{14} = 100.54$ (or $s = 10.03$) | B1 | $\sigma^2 = 93.84$ or $\sigma = 9.687$ iff $\sqrt{\frac{}{n}}$ used below |
| $t_{crit} = \pm 2.624$ | B1 | ignore signs for $t_{crit}$ (allow $t = 2.62$) (if $z$ used then max(B1B1B0 M0A0)) |
| **98% CI for $\mu$:** | | |
| $36.4 \pm 2.624 \times \frac{s}{\sqrt{15}}$ | M1 | their $\bar{x} \pm t_{14} \times \frac{\text{their } s}{\sqrt{15}}$ or their $\bar{x} \pm t_{14} \times \frac{\text{their } \sigma}{\sqrt{14}}$ (allow any of the following for $t_{14}$: 1.345; 1.761; 2.145; 2.624; 2.977) |
| $36.4 \pm 6.8$ | A1ft | |
| $(29.6, 43.2)$ or $36.4 \pm 6.8$ | A1 | 6 | cao |
| | | |
### 1(b)
| $40.0 \in \text{C.I.} \Rightarrow$ no change | E1ft | Must refer to 40 (dep M1) |
| --- | --- | --- |
| | E1ft | 2 | Dep on previous mark |
| | | **8** | |1 At the start of the 2012 season, the ages of the members of the Warwickshire Acorns Cricket Club could be modelled by a normal random variable, $X$ years, with mean $\mu$ and standard deviation $\sigma$.
The ages, $x$ years, of a random sample of 15 such members are summarised below.
$$\sum x = 546 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 1407.6$$
\begin{enumerate}[label=(\alph*)]
\item Construct a $98 \%$ confidence interval for $\mu$, giving the limits to one decimal place.\\
(6 marks)
\item At the start of the 2005 season, the mean age of the members was 40.0 years.
Use your confidence interval constructed in part (a) to indicate, with a reason, whether the mean age had changed.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q1 [8]}}