| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a standard two-sample z-test with known population standard deviation. Students must state hypotheses, calculate sample means, apply the z-test formula, find a p-value, and interpret. While it's a Further Maths question requiring knowledge of hypothesis testing, it's a textbook application with no conceptual challenges—just careful arithmetic and table lookup. Slightly easier than average A-level difficulty overall. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu_d = \mu_F; H_1: \mu_d \neq \mu_F\) | B1 | AO3 |
| Answer | Marks | Guidance |
|---|---|---|
| Let X = male weight, Y = female weight (\(\sum x = 39.2; \sum y = 46.6\)) | B1 | AO1 |
| \(\bar{x} = 4.9\) | B1 | AO1 |
| \(\bar{y} = 4.66\) | B1 | AO1 |
| \(\text{SE of diff of means} = \sqrt{\frac{0.5^2}{8} + \frac{0.5^2}{10}} = 0.237\ldots\) | M1 | AO2 |
| \(\text{Test statistic} = \frac{4.9 - 4.66}{0.237\ldots} = 1.01\) | m1 | AO1 |
| Prob from tables = 0.1562 | A1 | AO1 |
| A1 | AO1 | From calculator, p-value = 0.3116 |
| \(p\text{-value} = 0.3124\) | B1 | AO2 |
| Insufficient evidence to conclude that there is a difference in mean weight between males and females. | B1 | AO3 |
| [10] |
## 5(a)
$H_0: \mu_d = \mu_F; H_1: \mu_d \neq \mu_F$ | B1 | AO3 |
## 5(b)
Let X = male weight, Y = female weight ($\sum x = 39.2; \sum y = 46.6$) | B1 | AO1 |
$\bar{x} = 4.9$ | B1 | AO1 |
$\bar{y} = 4.66$ | B1 | AO1 |
$\text{SE of diff of means} = \sqrt{\frac{0.5^2}{8} + \frac{0.5^2}{10}} = 0.237\ldots$ | M1 | AO2 |
$\text{Test statistic} = \frac{4.9 - 4.66}{0.237\ldots} = 1.01$ | m1 | AO1 | Award m0 if no working seen
Prob from tables = 0.1562 | A1 | AO1 | From calculator, prob = 0.1558; FT 'their' test statistic
| A1 | AO1 | From calculator, p-value = 0.3116
$p\text{-value} = 0.3124$ | B1 | AO2 | FT 'their' p-value
Insufficient evidence to conclude that there is a difference in mean weight between males and females. | B1 | AO3 |
| [10] | |A new species of animal has been found on an uninhabited island. A zoologist wishes to investigate whether or not there is a difference in the mean weights of males and females of the species. She traps some of the animals and weighs them with the following results.
\begin{align}
\text{Males (kg)} &\quad 5.3, 4.6, 5.2, 4.5, 4.3, 5.5, 5.0, 4.8 \\
\text{Females (kg)} &\quad 4.9, 5.0, 4.1, 4.6, 4.3, 5.3, 4.2, 4.5, 4.8, 4.9
\end{align}
You may assume that these are random samples from normal populations with a common standard deviation of 0.5 kg.
\begin{enumerate}[label=(\alph*)]
\item State suitable hypotheses for this investigation. [1]
\item Determine the $p$-value of these results and state your conclusion in context. [9]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 Q5 [10]}}