| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Recover sample stats from CI |
| Difficulty | Challenging +1.2 This is a two-part question requiring (a) working backwards from a confidence interval to find sample statistics, and (b) conducting a two-sample t-test with pooled variance. While it involves multiple steps and understanding of confidence intervals and hypothesis testing, these are standard Further Statistics techniques with straightforward application of formulas. The reverse-engineering in part (a) requires some algebraic manipulation but follows directly from the confidence interval formula. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{x} = \frac{1}{2}(4.22 + 3.36) = 3.79\) | B1 | |
| \(4.22 - 3.36 = \frac{2ts}{\sqrt{10}}\) | M1 | Using \(z\) value implies M0 |
| With \(t = 1.833\) | A1 | |
| \((s = 0.7418)\) variance \(= 0.55(0)\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_A = \mu_B \quad H_1: \mu_A > \mu_B\) | B1 | Not \(\bar{x}\) |
| \(s^2 = \frac{1}{7}\left(98.02 - \frac{27.86^2}{8}\right) = 0.142(507)\) | M1 | \(= \frac{1}{7}(98.02 - 97.02)\) |
| Pooled estimate \(= \frac{9 \times 0.550316 + 7 \times 0.142507}{16}\) | M1 | |
| \(= 0.372\) | A1 | |
| \(t = \dfrac{\frac{27.86}{8} - 3.79}{\sqrt{0.372}\sqrt{\frac{1}{10}+\frac{1}{8}}} = (-)1.063\) | M1A1 | |
| \(1.063 < 1.337\) oe | M1 | Compare *their* value with \(1.337\) |
| Accept \(H_0\), Nassa's claim is not supported | A1ft | Correct conclusion in context, level of uncertainty in language used. No contradictions. |
| Total | 8 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} = \frac{1}{2}(4.22 + 3.36) = 3.79$ | B1 | |
| $4.22 - 3.36 = \frac{2ts}{\sqrt{10}}$ | M1 | Using $z$ value implies M0 |
| With $t = 1.833$ | A1 | |
| $(s = 0.7418)$ variance $= 0.55(0)$ | A1 | |
| **Total** | **4** | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_A = \mu_B \quad H_1: \mu_A > \mu_B$ | B1 | Not $\bar{x}$ |
| $s^2 = \frac{1}{7}\left(98.02 - \frac{27.86^2}{8}\right) = 0.142(507)$ | M1 | $= \frac{1}{7}(98.02 - 97.02)$ |
| Pooled estimate $= \frac{9 \times 0.550316 + 7 \times 0.142507}{16}$ | M1 | |
| $= 0.372$ | A1 | |
| $t = \dfrac{\frac{27.86}{8} - 3.79}{\sqrt{0.372}\sqrt{\frac{1}{10}+\frac{1}{8}}} = (-)1.063$ | M1A1 | |
| $1.063 < 1.337$ oe | M1 | Compare *their* value with $1.337$ |
| Accept $H_0$, Nassa's claim is not supported | A1ft | Correct conclusion in context, level of uncertainty in language used. No contradictions. |
| **Total** | **8** | |6 Nassa is researching the lengths of a particular type of snake in two countries, $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item He takes a random sample of 10 snakes of this type from country $A$ and measures the length, $x \mathrm {~m}$, of each snake. He then calculates a $90 \%$ confidence interval for the population mean length, $\mu \mathrm { m }$, for snakes of this type, assuming that snake lengths have a normal distribution. This confidence interval is $3.36 \leqslant \mu \leqslant 4.22$.
Find the sample mean and an unbiased estimate for the population variance.
\item Nassa also measures the lengths, $y \mathrm {~m}$, of a random sample of 8 snakes of this type taken from country $B$. His results are summarised as follows.
$$\sum y = 27.86 \quad \sum y ^ { 2 } = 98.02$$
Nassa claims that the mean length of snakes of this type in country $B$ is less than the mean length of snakes of this type in country $A$. Nassa assumes that his sample from country $B$ also comes from a normal distribution, with the same variance as the distribution from country $A$.
Test at the $10 \%$ significance level whether there is evidence to support Nassa's claim.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q6 [12]}}