| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample z-test large samples |
| Difficulty | Standard +0.3 This is a standard two-sample t-test with summary statistics requiring calculation of sample means and variances, then applying the test procedure. The large sample sizes (n>30) invoke CLT, making it routine. While it involves multiple computational steps, it follows a well-practiced algorithm with no conceptual surprises, making it slightly easier than average for Further Maths statistics. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_x = \mu_y\); \(H_1: \mu_x < \mu_y\) | B1 | |
| \(s_x^2 = \frac{1}{79}\left(338.1 - \frac{164^2}{80}\right) [= 0.02405]\) and \(s_y^2 = \frac{1}{59}\left(261.1 - \frac{124.8^2}{60}\right) [= 0.02569]\) | B1 | Both. Implied by \(\frac{19}{790}\), \(\frac{379}{14750}\) or 3sf. |
| \(s^2 = \frac{0.02405}{80} + \frac{0.02569}{60}\) | M1 | |
| \(s^2 = 0.0007289\) or \(s = 0.026998\) or \(0.0270\) | A1 | Implied by \(z = -1.11\) |
| \(z = \dfrac{\frac{164.0}{80} - \frac{124.8}{60}}{s}\) | M1 | FT their value for \(s\). |
| \(-1.11\) | A1 | |
| \('1.11' < 1.282\); Accept \(H_0\) / not significant | M1 | Compare with correct \(z\)-value 1.282 and consistent signs. Condone 'reject \(H_1\)'. Using probabilities, \(P(Z > 1.11) = 0.1333 > 0.1\). |
| Insufficient evidence to support the inspector's suspicion / Insufficient evidence that the (mean) lengths of rods from machine \(X\) are shorter than the (mean) lengths of rods from machine \(Y\) | A1 | Correct conclusion in context, following correct work, level of uncertainty in language (not 'prove', not 'there is no evidence'), no contradictions. e.g. proves that the inspector is incorrect scores A0. A0 if hypotheses wrong way round. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Large sample sizes OR central limit theorem applies. | B1 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_x = \mu_y$; $H_1: \mu_x < \mu_y$ | B1 | |
| $s_x^2 = \frac{1}{79}\left(338.1 - \frac{164^2}{80}\right) [= 0.02405]$ and $s_y^2 = \frac{1}{59}\left(261.1 - \frac{124.8^2}{60}\right) [= 0.02569]$ | B1 | Both. Implied by $\frac{19}{790}$, $\frac{379}{14750}$ or 3sf. |
| $s^2 = \frac{0.02405}{80} + \frac{0.02569}{60}$ | M1 | |
| $s^2 = 0.0007289$ or $s = 0.026998$ or $0.0270$ | A1 | Implied by $z = -1.11$ |
| $z = \dfrac{\frac{164.0}{80} - \frac{124.8}{60}}{s}$ | M1 | FT their value for $s$. |
| $-1.11$ | A1 | |
| $'1.11' < 1.282$; Accept $H_0$ / not significant | M1 | Compare with correct $z$-value 1.282 and consistent signs. Condone 'reject $H_1$'. Using probabilities, $P(Z > 1.11) = 0.1333 > 0.1$. |
| Insufficient evidence to support the inspector's suspicion / Insufficient evidence that the (mean) lengths of rods from machine $X$ are shorter than the (mean) lengths of rods from machine $Y$ | A1 | Correct conclusion in context, following correct work, level of uncertainty in language (not 'prove', not 'there is no evidence'), no contradictions. e.g. proves that the inspector is incorrect scores A0. A0 if hypotheses wrong way round. |
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## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Large sample sizes OR central limit theorem applies. | B1 | |
---4 An inspector is checking the lengths of metal rods produced by two machines, $X$ and $Y$. These rods should be of the same length, but the inspector suspects that those made by machine $X$ are shorter, on average, than those made by machine $Y$. The inspector chooses a random sample of 80 rods made by machine $X$ and a random sample of 60 rods made by machine $Y$. The lengths of these rods are $x \mathrm {~cm}$ and $y \mathrm {~cm}$ respectively. Her results are summarised as follows.
$$\sum x = 164.0 \quad \sum x ^ { 2 } = 338.1 \quad \sum y = 124.8 \quad \sum y ^ { 2 } = 261.1$$
\begin{enumerate}[label=(\alph*)]
\item Test at the $10 \%$ significance level whether the data supports the inspector's suspicion.
\item Give a reason why it is not necessary to make any assumption about the distributions of the lengths of the rods.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q4 [9]}}