| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a straightforward two-part hypothesis testing question requiring standard t-test procedures. Part (a) involves a small sample (n=5) one-tailed test with basic calculations, while part (b) uses a larger sample (n=80) where the CLT applies. Both parts follow textbook procedures with no conceptual challenges—students simply need to apply the standard t-test algorithm with given summary statistics. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{471}{5} = 94.2\) | B1 | |
| \(s = 6.058\) | B1 | Or \(s^2 = 36.7\) |
| \(v = 4\) 1-tailed test | B1 | |
| \(t_{\text{crit}} = -2.132\) | B1 | Or on diagram |
| \(H_0: \mu = 100\) | B1 | |
| \(H_1: \mu < 100\) | ||
| \(t = \frac{94.2 - 100}{6.058/\sqrt{5}} = -2.14\) | M1, A1 | their \(\bar{x} - 100\) / (their \(s\))/\(\sqrt{5}\) |
| Reject \(H_0\) at 5% level of significance | A1 | On their \(t\) and critical value |
| Evidence at the 5% level of significance to support the members' belief that the batteries last less than 100 hours. | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{8080}{80} = 101\) | ||
| \(s^2 = \frac{6399}{79} = 81\) (or \(\frac{6399}{80} = 79.9875\)) | B1 | For \(s\) (or \(s^2\)) and \(\bar{x}\) |
| \(s = 9\) (or \(s = 8.944\)) | ||
| \(H_0: \mu = 100\) | B1 | |
| \(H_1: \mu \neq 100\) | ||
| \(\bar{X} \sim N\left(100, \frac{81 \text{ (or } 79.9875)}{80}\right)\) under \(H_0\) | B1 | Or \(100, \frac{9}{\sqrt{80}}\) used |
| \(z = \frac{101 - 100}{9/\sqrt{80}} = 0.99\) | M1, A1 | Allow use of \(t\) method; AWFW 0.99 to 1.00 (allow 1) |
| 2-tailed test | ||
| \(z_{\text{crit}} = \pm 1.96\) | B1 | Or \(z = 1.96\) |
| Accept \(H_0\) at 5% level of significance. | A1 | On their \(z\) and critical value; Or \(t\) |
| Sufficient evidence at the 5% level of significance to support the manufacturer's belief. | E1 |
## 6(a)
$\bar{x} = \frac{471}{5} = 94.2$ | B1 |
$s = 6.058$ | B1 | Or $s^2 = 36.7$
$v = 4$ 1-tailed test | B1 |
$t_{\text{crit}} = -2.132$ | B1 | Or on diagram
$H_0: \mu = 100$ | B1 |
$H_1: \mu < 100$ |
$t = \frac{94.2 - 100}{6.058/\sqrt{5}} = -2.14$ | M1, A1 | their $\bar{x} - 100$ / (their $s$)/$\sqrt{5}$
Reject $H_0$ at 5% level of significance | A1 | On their $t$ and critical value
Evidence at the 5% level of significance to support the members' belief that the batteries last less than 100 hours. | E1 |
## 6(b)
$\bar{x} = \frac{8080}{80} = 101$ |
$s^2 = \frac{6399}{79} = 81$ (or $\frac{6399}{80} = 79.9875$) | B1 | For $s$ (or $s^2$) and $\bar{x}$
$s = 9$ (or $s = 8.944$) |
$H_0: \mu = 100$ | B1 |
$H_1: \mu \neq 100$ |
$\bar{X} \sim N\left(100, \frac{81 \text{ (or } 79.9875)}{80}\right)$ under $H_0$ | B1 | Or $100, \frac{9}{\sqrt{80}}$ used
$z = \frac{101 - 100}{9/\sqrt{80}} = 0.99$ | M1, A1 | Allow use of $t$ method; AWFW 0.99 to 1.00 (allow 1)
2-tailed test |
$z_{\text{crit}} = \pm 1.96$ | B1 | Or $z = 1.96$
Accept $H_0$ at 5% level of significance. | A1 | On their $z$ and critical value; Or $t$
Sufficient evidence at the 5% level of significance to support the manufacturer's belief. | E1 |
---6 The lifetime, $X$ hours, of Everwhite camera batteries is normally distributed. The manufacturer claims that the mean lifetime of these batteries is 100 hours.
\begin{enumerate}[label=(\alph*)]
\item The members of a photography club suspect that the batteries do not last as long as is claimed by the manufacturer. In order to investigate their suspicion, the members test a random sample of five of these batteries and find the lifetimes, in hours, to be as follows:
$$\begin{array} { l l l l l }
85 & 92 & 100 & 95 & 99
\end{array}$$
Test the members' suspicion at the $5 \%$ level of significance.
\item The manufacturer, believing that the mean lifetime of these batteries has not changed from 100 hours, decides to determine the lifetime, $x$ hours, of each of a random sample of 80 Everwhite camera batteries. The manufacturer obtains the following results, where $\bar { x }$ denotes the sample mean:
$$\sum x = 8080 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 6399$$
Test the manufacturer's belief at the $5 \%$ level of significance.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2006 Q6 [17]}}