| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Standard +0.3 This is a straightforward application of standard one-sample t-test procedures with a small sample (n=8). Students must calculate sample mean and standard deviation, then construct a confidence interval and perform a hypothesis test using t-tables. While it requires careful arithmetic and understanding of t-distributions, it follows a routine algorithmic approach with no conceptual surprises or novel problem-solving—slightly easier than average due to its mechanical nature. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum x = 44.4\), \(\sum x^2 = 247.48\) | B1 | |
| \(s^2 = \frac{1}{7}\left(247.48 - \frac{44.4^2}{8}\right) = 0.1514\left(= \frac{53}{350}\right)\) | M1 | |
| CI: \(\frac{44.4}{8} \pm 1.895\sqrt{\frac{0.1514}{8}}\) | M1 | Correct formula with a \(t\) value |
| \(1.895\) | B1 | 1.895 used |
| \(5.55 \pm 0.261 = [5.29, 5.81]\) | A1 | Either form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 5.9\) and \(H_1: \mu < 5.9\) | B1 | |
| \(t = \frac{5.55 - 5.9}{\sqrt{\frac{s^2}{8}}} = -2.544\) | M1 | |
| Critical value \(t_{0.975}(7) = 2.365\); Compare \(-2.544 < -2.365\), reject \(H_0\) | M1 | Compare calculated value with \(-2.365\) and correct FT conclusion |
| Sufficient evidence to support Raman's claim | A1 | Correct conclusion, in context. Level of uncertainty in language is used |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum x = 44.4$, $\sum x^2 = 247.48$ | B1 | |
| $s^2 = \frac{1}{7}\left(247.48 - \frac{44.4^2}{8}\right) = 0.1514\left(= \frac{53}{350}\right)$ | M1 | |
| CI: $\frac{44.4}{8} \pm 1.895\sqrt{\frac{0.1514}{8}}$ | M1 | Correct formula with a $t$ value |
| $1.895$ | B1 | 1.895 used |
| $5.55 \pm 0.261 = [5.29, 5.81]$ | A1 | Either form |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 5.9$ and $H_1: \mu < 5.9$ | B1 | |
| $t = \frac{5.55 - 5.9}{\sqrt{\frac{s^2}{8}}} = -2.544$ | M1 | |
| Critical value $t_{0.975}(7) = 2.365$; Compare $-2.544 < -2.365$, reject $H_0$ | M1 | Compare calculated value with $-2.365$ and correct FT conclusion |
| Sufficient evidence to support Raman's claim | A1 | Correct conclusion, in context. Level of uncertainty in language is used |5 Raman is researching the heights of male giraffes in a particular region. Raman assumes that the heights of male giraffes in this region are normally distributed. He takes a random sample of 8 male giraffes from the region and measures the height, in metres, of each giraffe. These heights are as follows.
$$\begin{array} { c c c c c c c c }
5.2 & 5.8 & 4.9 & 6.1 & 5.5 & 5.9 & 5.4 & 5.6
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find a $90 \%$ confidence interval for the population mean height of male giraffes in this region. [5]\\
Raman claims that the population mean height of male giraffes in the region is less than 5.9 metres.
\item Test at the $2.5 \%$ significance level whether this sample provides sufficient evidence to support Raman's claim.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q5 [9]}}