| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-sample t-test with summary statistics |
| Difficulty | Challenging +1.2 This question requires understanding of confidence intervals for difference of means with known variances, and the non-standard interpretation of relating confidence interval width to a decision threshold. While the setup is multi-step and requires careful reading, the calculations are standard (finding the critical value where the CI lower bound equals 1.25), and the assumption is routine (independence of samples). The conceptual twist of working backwards from a decision criterion to find the confidence level elevates it slightly above average difficulty. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | (SE of difference of means) | M1 |
| \(= \sqrt{\frac{0.75^2}{6} + \frac{0.6^2}{5}}\) | ||
| \(= 0.407...\) | A1 | si |
| \(2.2 - k\sqrt{\frac{0.75^2}{6} + \frac{0.6^2}{5}} = 1.25\) | M1 | Condone \(>\). FT their SE provided \(\neq \sqrt{0.75^2 + 0.6^2}\) |
| \(k = 2.333...\) | A1 | cao |
| Probability from calculator \(= 0.99018...\) Or 0.99010 from tables | M1 | FT their \(k\) for M1A1 |
| Largest value of \(p\) is 98. | A1 | Accept 98.04 |
| (b) | Valid assumption e.g. The time trials are all done on the same terrain. She suffers no mechanical problems. She doesn't get quicker because she's fitter. She isn't slower because she's tired. Weather conditions are similar. Wears the same clothing. | E1 |
| Total [7] |
(a) | (SE of difference of means) | M1 | Award M1 for $\text{Var} = \frac{0.75^2}{6} + \frac{0.6^2}{5}$ |
| $= \sqrt{\frac{0.75^2}{6} + \frac{0.6^2}{5}}$ | | |
| $= 0.407...$ | A1 | si |
| $2.2 - k\sqrt{\frac{0.75^2}{6} + \frac{0.6^2}{5}} = 1.25$ | M1 | Condone $>$. FT their SE provided $\neq \sqrt{0.75^2 + 0.6^2}$ |
| $k = 2.333...$ | A1 | cao |
| Probability from calculator $= 0.99018...$ Or 0.99010 from tables | M1 | FT their $k$ for M1A1 |
| Largest value of $p$ is 98. | A1 | Accept 98.04 |
(b) | Valid assumption e.g. The time trials are all done on the same terrain. She suffers no mechanical problems. She doesn't get quicker because she's fitter. She isn't slower because she's tired. Weather conditions are similar. Wears the same clothing. | E1 | |
| **Total [7]** | | |7. Branwen intends to buy a new bike, either a Cannotrek or a Bianchondale. If there is evidence that the difference in the mean times on the two bikes over a 10 km time trial is more than 1.25 minutes, she will buy the faster bike. Otherwise, she will base her decision on other factors.
She negotiates a test period to try both bikes. The times, in minutes, taken by Branwen to complete a 10 km time trial on the Cannotrek may be modelled by a normal distribution with mean $\mu _ { C }$ and standard deviation $0 \cdot 75$.
The times, in minutes, taken by Branwen to complete a 10 km time trial on the Bianchondale may be modelled by a normal distribution with mean $\mu _ { B }$ and standard deviation $0 \cdot 6$.
During the test period, she completes 6 time trials with a mean time of 19.5 minutes on the Cannotrek, and 5 time trials with a mean time of 17.3 minutes on the Bianchondale.
She calculates a $p \%$ confidence interval for $\mu _ { C } - \mu _ { B }$.
\begin{enumerate}[label=(\alph*)]
\item What would be the largest value of $p$ that would lead Branwen to base her purchasing decision on the time trials, without considering other factors?
\item State an assumption you have made in part (a).
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 2023 Q7 [7]}}