| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-sample t-test, variance unknown |
| Difficulty | Standard +0.3 This is a straightforward hypothesis testing question requiring standard procedures: stating hypotheses, calculating sample statistics (mean and standard deviation from given sums), performing a one-sample z-test, and making a conclusion. Part (a) requires basic understanding of test choice but is conceptually simple. The calculations are routine A-level statistics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Valid reason. e.g. A consumer, (Hopcyn), would only be concerned with whether the company was overstating and therefore only wish to use a lower tail test. | E1 | Reasonable explanations |
| Valid reason. e.g. The company would not wish to overstate the distance the car could travel because they would be liable to have claims of false advertising brought against them, nor understate the distance the car could travel because they would like to claim the greatest mileage possible. | E1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu = 123\) \(H_1: \mu < 123\) | B1 | — |
| \(\bar{x} = \frac{11007}{90} = 122.3\) | B1 | — |
| \(s^2 = \frac{1}{89} \times \left(1361913 - \frac{11007^2}{90}\right)\) | M1 | Alternative p-value method M1 for Test statistic \(= \frac{122.3 - 123}{13.30578.../\sqrt{90}}\) If standardising. A1 p-value from tables \(= 0.30854\) Alternative CV method M1 for \(c - 123 = \frac{-t \times 13.30578...}{\sqrt{90}} = -1.6449\) A1 \(c = 120.7\) B1 since \(120.7 < 122.3\) |
| \(s = 13.3...\) | A1 | — |
| p-value \(= P(\bar{X} < 122.3)\) | M1 | — |
| p-value \(= 0.30886\) | A1 | — |
| Since \(p > 0.05\) there is insufficient evidence to reject \(H_0\). | B1 | — |
| Insufficient evidence to reject the manufacturer's claim that a one hour charge gives 123 miles of travel. | B1 | — |
## (a)
Valid reason. e.g. A consumer, (Hopcyn), would only be concerned with whether the company was overstating and therefore only wish to use a lower tail test. | E1 | Reasonable explanations
Valid reason. e.g. The company would not wish to overstate the distance the car could travel because they would be liable to have claims of false advertising brought against them, nor understate the distance the car could travel because they would like to claim the greatest mileage possible. | E1 | —
## (b)(i)(ii)
$H_0: \mu = 123$ $H_1: \mu < 123$ | B1 | —
$\bar{x} = \frac{11007}{90} = 122.3$ | B1 | —
$s^2 = \frac{1}{89} \times \left(1361913 - \frac{11007^2}{90}\right)$ | M1 | Alternative p-value method M1 for Test statistic $= \frac{122.3 - 123}{13.30578.../\sqrt{90}}$ If standardising. A1 p-value from tables $= 0.30854$ Alternative CV method M1 for $c - 123 = \frac{-t \times 13.30578...}{\sqrt{90}} = -1.6449$ A1 $c = 120.7$ B1 since $120.7 < 122.3$
$s = 13.3...$ | A1 | —
p-value $= P(\bar{X} < 122.3)$ | M1 | —
p-value $= 0.30886$ | A1 | —
Since $p > 0.05$ there is insufficient evidence to reject $H_0$. | B1 | —
Insufficient evidence to reject the manufacturer's claim that a one hour charge gives 123 miles of travel. | B1 | —
**Total: [10]**
---A manufacturer of batteries for electric cars claims that an hour of charge can power a certain model of car to travel for an average of 123 miles. An electric car company and a consumer, Hopcyn, both wish to test the validity of the manufacturer's claim.
\begin{enumerate}[label=(\alph*)]
\item Explain why Hopcyn may want to use a one-sided test and why the car company may want to use a two-sided test. [2]
\end{enumerate}
To test the validity of this claim, Hopcyn collects data from a random sample of 90 drivers of this model of car to see how far they travelled, $X$ miles, on an hour of charge. He produced the following summary statistics.
$$\sum x = 11007 \quad \sum x^2 = 1361913$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item
\begin{enumerate}[label=(\roman*)]
\item Assuming Hopcyn uses a one-sided test, state the hypotheses.
\item Test at the 5\% significance level whether the manufacturer's claim is correct. [8]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 2019 Q6 [10]}}