| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question followed by a reverse application. Part (a) requires standard t-distribution CI calculation with given summary statistics. Part (b) asks students to recognize that the greatest k is the upper bound of the 90% CI (one-tailed test at 10% corresponds to 90% CI). While it requires understanding the relationship between confidence intervals and hypothesis tests, this is a standard Further Stats topic with no novel problem-solving required. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s^2 = \frac{78.5}{9} = 8.7222\) | B1 | \(\frac{157}{18}\) |
| CI: \(25.6 \pm t\sqrt{\frac{s^2}{10}}\) | M1 | Correct expression with a \(t\) value |
| With \(t = 1.833\) | A1 | With correct \(t\) value |
| \(25.6 \pm 1.71(2)\) or \([23.9, 27.3]\) | A1 | Accept in either form, ISW. Accept inequality form |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{25.6 - k}{\sqrt{\dfrac{s^2}{10}}} \geqslant t\) | M1 | |
| With \(t = -1.383\) | A1 | Allow \(1.383\) |
| \(26.9\) | A1 | CAO |
| 3 |
**Question 1(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $s^2 = \frac{78.5}{9} = 8.7222$ | B1 | $\frac{157}{18}$ |
| CI: $25.6 \pm t\sqrt{\frac{s^2}{10}}$ | M1 | Correct expression with a $t$ value |
| With $t = 1.833$ | A1 | With correct $t$ value |
| $25.6 \pm 1.71(2)$ or $[23.9, 27.3]$ | A1 | Accept in either form, ISW. Accept inequality form |
| | **4** | |
---
**Question 1(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{25.6 - k}{\sqrt{\dfrac{s^2}{10}}} \geqslant t$ | M1 | |
| With $t = -1.383$ | A1 | Allow $1.383$ |
| $26.9$ | A1 | CAO |
| | **3** | |1 The times taken for students at a college to run 200 m have a normal distribution with mean $\mu \mathrm { s }$. The times, $x$ s, are recorded for a random sample of 10 students from the college. The results are summarised as follows, where $\bar { x }$ is the sample mean.
$$\bar { x } = 25.6 \quad \sum ( x - \bar { x } ) ^ { 2 } = 78.5$$
\begin{enumerate}[label=(\alph*)]
\item Find a 90\% confidence interval for $\mu$.\\
A test of the null hypothesis $\mu = k$ is carried out on this sample, using a $10 \%$ significance level. The test does not support the alternative hypothesis $\mu < k$.
\item Find the greatest possible value of $k$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q1 [7]}}