| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Find k or boundary value from test |
| Difficulty | Standard +0.8 This is a two-part hypothesis testing question requiring calculation of sample statistics from summations, finding a critical boundary value in part (a), and performing a two-sample t-test with pooled variance in part (b). While the techniques are standard for Further Statistics, the multi-step nature, need to work backwards from significance level to find maximum k, and careful handling of pooled variance calculations make this moderately challenging but within typical Further Maths scope. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{828}{10} = 82.8\), \(s^2 = \frac{1}{9}\left(68622 - \frac{828^2}{10}\right) = 7.0667\) | B1 | |
| \(\frac{82.8 - k}{\sqrt{\frac{s^2}{10}}} \geq t\) where \(t = 1.833\) | M1A1 | M1 if equality, A1 for inequality and correct \(t\) value |
| \(k \leq 81.259 \quad k \leq 81.3\) | A1 |
| Answer | Marks |
|---|---|
| \(H_0: \mu_A = \mu_B \quad H_1: \mu_A \neq \mu_B\) | B1 |
| Pooled variance \(= \frac{9 \times 7.0667 + 7 \times 9.966}{10 + 8 - 2} = s_p^2\) | M1 |
| \(= 8.335\) | A1 |
| \(t = \frac{82.8 - 79.8}{s_p\sqrt{\frac{1}{10} + \frac{1}{8}}} = 2.19\) | M1A1 |
| Compare with \(2.12\) (\(t(16, 0.975)\)) and reject \(H_0\) | M1 |
| Population means are not the same | A1 |
## Question 5:
**Part 5(a):**
$\bar{x} = \frac{828}{10} = 82.8$, $s^2 = \frac{1}{9}\left(68622 - \frac{828^2}{10}\right) = 7.0667$ | B1 |
$\frac{82.8 - k}{\sqrt{\frac{s^2}{10}}} \geq t$ where $t = 1.833$ | M1A1 | M1 if equality, A1 for inequality and correct $t$ value |
$k \leq 81.259 \quad k \leq 81.3$ | A1 |
**Part 5(b):**
$H_0: \mu_A = \mu_B \quad H_1: \mu_A \neq \mu_B$ | B1 |
Pooled variance $= \frac{9 \times 7.0667 + 7 \times 9.966}{10 + 8 - 2} = s_p^2$ | M1 |
$= 8.335$ | A1 |
$t = \frac{82.8 - 79.8}{s_p\sqrt{\frac{1}{10} + \frac{1}{8}}} = 2.19$ | M1A1 |
Compare with $2.12$ ($t(16, 0.975)$) and reject $H_0$ | M1 |
Population means are not the same | A1 |
---5 Students at two colleges, $A$ and $B$, are competing in a computer games challenge.
\begin{enumerate}[label=(\alph*)]
\item The time taken for a randomly chosen student from college $A$ to complete the challenge has a normal distribution with mean $\mu$ minutes. The times taken, $x$ minutes, are recorded for a random sample of 10 students chosen from college $A$. The results are summarised as follows.
$$\sum x = 828 \quad \sum x ^ { 2 } = 68622$$
A test is carried out on the data at the $5 \%$ significance level and the result supports the claim that $\mu > k$.
Find the greatest possible value of $k$.
\item A random sample of 8 students is chosen from college $B$. Their times to complete the same challenge give a sample mean of 79.8 minutes and an unbiased variance estimate of 9.966 minutes ${ } ^ { 2 }$.
Use a 2 -sample test at the $5 \%$ significance level to test whether the mean time for students at college $B$ to complete the challenge is the same as the mean time for students at college $A$ to complete the challenge. You should assume that the two distributions are normal and have the same population variance.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q5 [11]}}