| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Standard +0.3 This is a straightforward application of standard hypothesis testing and confidence interval procedures from S2. Part (a)(i) is a routine one-sample z-test with known variance requiring only substitution into formulas. Part (a)(ii) tests understanding of normal distribution tails. Part (b) involves a standard t-confidence interval calculation. All steps are textbook procedures with no novel problem-solving 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 |
|---|---|---|
| \(H_0: \mu = 61.4\) \(H_1: \mu \neq 61.4\) | B1 | (both) |
| \(z_{\text{calc}} = \frac{65.0 - 61.4}{7.5/\sqrt{16}} = 1.92\) | M1 | Alternative: \(P(\bar{X} > 65.0) = P(Z > 1.92) = 1 - 0.97257 = 0.02743 \geq 0.025 \therefore\) Accept \(H_0\) Use of \(t \Rightarrow \max(B1M1A1)\) |
| \(z_{\text{crit}} = \pm 1.96\) or (shown in/implied by diagram) | B1 | |
| Accept \(H_0\) | Adep1 | dep(B1M1) but not A1B1 If incorrect or no hypothesis then B0 \(\Rightarrow \max(M1A1B1)\) i.e. final Adep1Edep1 not available |
| Insufficient / No evidence (at 5% level) to suggest /show mean (age has) changed (from 61.4 years.) Mean (age) has not changed at 1% level (of significance) | Edep1 | dep(Adep1) \(z = \frac{25 - 61.4}{7.5} = -4.85\) \(\Rightarrow P(Z < -4.85) \approx 0\) \(\Rightarrow\) none aged under 25 included |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{y} = \sum y = \frac{702}{12} = 58.5\) | B1 | (\(s = 2.83\)) (\(\sigma^2 = 7.35\) or \(\sigma = 2.71\) iff \(\sigma/\sqrt{11}\) used below) |
| \(s^2 = \frac{\sum(y - \bar{y})^2}{n-1} = \frac{88.25}{11} = 8.02\) | B1 | |
| \(t_{\text{crit}} = \pm 1.796\) | B1 | Ignore signs for \(t_{\text{crit}}\) If \(z\) used then \(\max(B1B1B0M0A0)\) |
| 90% CI for \(\mu\): \(58.5 \pm 1.796 \times \frac{s}{\sqrt{12}}\) \(58.5 \pm 1.4685\) \(= 57.03, 59.97\) | M1 | (their \(\bar{y}\)) \(\pm t_{11} \times \frac{\text{their } s}{\sqrt{12}}\) OR (their \(\bar{y}\)) \(\pm t_{11} \times \frac{\text{their } \sigma}{\sqrt{11}}\) |
| \(= (57.0, 60.0)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| upper limit \(< 61.4\) \(\Rightarrow\) recruitment drive lowered the average age of the club membership | B1ft | Must refer to 61.4 (on their CI) |
## (a)(i)
| $H_0: \mu = 61.4$ $H_1: \mu \neq 61.4$ | B1 | (both) |
| $z_{\text{calc}} = \frac{65.0 - 61.4}{7.5/\sqrt{16}} = 1.92$ | M1 | **Alternative:** $P(\bar{X} > 65.0) = P(Z > 1.92) = 1 - 0.97257 = 0.02743 \geq 0.025 \therefore$ Accept $H_0$ Use of $t \Rightarrow \max(B1M1A1)$ |
| $z_{\text{crit}} = \pm 1.96$ or (shown in/implied by diagram) | B1 | |
| Accept $H_0$ | Adep1 | dep(B1M1) but not A1B1 If incorrect or no hypothesis then B0 $\Rightarrow \max(M1A1B1)$ i.e. final Adep1Edep1 not available |
| Insufficient / No evidence (at 5% level) to suggest /show **mean (age has) changed** (from 61.4 years.) **Mean (age) has not changed at 1% level (of significance)** | Edep1 | dep(Adep1) $z = \frac{25 - 61.4}{7.5} = -4.85$ $\Rightarrow P(Z < -4.85) \approx 0$ $\Rightarrow$ none aged under 25 included |
## (a)(ii)
| $\bar{y} = \sum y = \frac{702}{12} = 58.5$ | B1 | ($s = 2.83$) ($\sigma^2 = 7.35$ or $\sigma = 2.71$ iff $\sigma/\sqrt{11}$ used below) |
| $s^2 = \frac{\sum(y - \bar{y})^2}{n-1} = \frac{88.25}{11} = 8.02$ | B1 | |
| $t_{\text{crit}} = \pm 1.796$ | B1 | Ignore signs for $t_{\text{crit}}$ If $z$ used then $\max(B1B1B0M0A0)$ |
| 90% CI for $\mu$: $58.5 \pm 1.796 \times \frac{s}{\sqrt{12}}$ $58.5 \pm 1.4685$ $= 57.03, 59.97$ | M1 | (their $\bar{y}$) $\pm t_{11} \times \frac{\text{their } s}{\sqrt{12}}$ OR (their $\bar{y}$) $\pm t_{11} \times \frac{\text{their } \sigma}{\sqrt{11}}$ |
| $= (57.0, 60.0)$ | A1 | |
## (a)(ii)
| upper limit $< 61.4$ $\Rightarrow$ recruitment drive lowered the average age of the club membership | B1ft | Must refer to 61.4 (on their CI) |
**Total: 13 marks**
---2
\begin{enumerate}[label=(\alph*)]
\item A particular bowling club has a large number of members. Their ages may be modelled by a normal random variable, $X$, with standard deviation 7.5 years.
On 30 June 2010, Ted, the club secretary, concerned about the ageing membership, selected a random sample of 16 members and calculated their mean age to be 65.0 years.
\begin{enumerate}[label=(\roman*)]
\item Carry out a hypothesis test, at the $5 \%$ level of significance, to determine whether the mean age of the club's members has changed from its value of 61.4 years on 30 June 2000.
\item Comment on the likely number of members who were under the age of 25 on 30 June 2010, giving a numerical reason for your answer.
\end{enumerate}\item During 2011, in an attempt to encourage greater participation in the sport, the club ran a recruitment drive.
After the recruitment drive, the ages of members of the bowling club may be modelled by a normal random variable, $Y$ years, with mean $\mu$ and standard deviation $\sigma$. The ages, $y$ years, of a random sample of 12 such members are summarised below.
$$\sum y = 702 \quad \text { and } \quad \sum ( y - \bar { y } ) ^ { 2 } = 88.25$$
\begin{enumerate}[label=(\roman*)]
\item Construct a $90 \%$ confidence interval for $\mu$, giving the limits to one decimal place.
\item Use your confidence interval to state, with a reason, whether the recruitment drive lowered the average age of the club's members.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q2 [13]}}