| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pooled variance estimation |
| Difficulty | Standard +0.3 This is a standard two-sample pooled variance problem requiring straightforward application of formulas: calculate sample variances from summary statistics, compute pooled variance, then construct a confidence interval. While it involves multiple steps and careful arithmetic, it requires no novel insight—just systematic application of A-level Further Statistics techniques with clearly signposted parts. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s_P^2 = (1\,560\,000 - 9600^2/60)/59 = 406.78\) | M1 | Estimate \(P\)'s popln. variance to 3 d.p.; allow biased: 400 |
| \(s_Q^2 = (1\,052\,500 - 7200^2/50)/49 = 320.41\) | M1 | Estimate \(Q\)'s popln. variance to 3 d.p.; allow biased: 314 |
| \(s^2 = (59\,s_P^2 + 49\,s_Q^2)/108 = 367.6\) or \(368\) | M1 A1 | Find pooled estimate of common variance |
| OR: \(s^2 = (1\,560\,000 - 9600^2/60 + 1\,052\,500 - 7200^2/50)/108 = 367.6\) or \(368\) or \(9925/27\) | (M3 A1) | Direct method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(9600/60 - 7200/50 = 160 - 144 = 16\) | M1 A1 | Find confidence interval for the difference |
| \(\pm z\,s\sqrt{(60^{-1}+50^{-1})}\) | ||
| \(z_{0.975} = 1.96\) or \(t_{120,\ 0.975} = 1.98\) | A1 | Use appropriate tabular value to 2 d.p. |
| \(16 \pm 7.2\ [8.8,\ 23.2]\) or \(16 \pm 7.3\ [8.7,\ 23.3]\) | M1 A1 | Evaluate confidence interval (AEF, to 1 d.p.) |
| SR: Using combined variance \(s_P^2/60 + s_Q^2/50 = 13.19\): max 3/5 |
# Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s_P^2 = (1\,560\,000 - 9600^2/60)/59 = 406.78$ | M1 | Estimate $P$'s popln. variance to 3 d.p.; allow biased: 400 |
| $s_Q^2 = (1\,052\,500 - 7200^2/50)/49 = 320.41$ | M1 | Estimate $Q$'s popln. variance to 3 d.p.; allow biased: 314 |
| $s^2 = (59\,s_P^2 + 49\,s_Q^2)/108 = 367.6$ or $368$ | M1 A1 | Find pooled estimate of common variance |
| OR: $s^2 = (1\,560\,000 - 9600^2/60 + 1\,052\,500 - 7200^2/50)/108 = 367.6$ or $368$ or $9925/27$ | (M3 A1) | Direct method |
---
# Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $9600/60 - 7200/50 = 160 - 144 = 16$ | M1 A1 | Find confidence interval for the difference |
| $\pm z\,s\sqrt{(60^{-1}+50^{-1})}$ | | |
| $z_{0.975} = 1.96$ or $t_{120,\ 0.975} = 1.98$ | A1 | Use appropriate tabular value to 2 d.p. |
| $16 \pm 7.2\ [8.8,\ 23.2]$ or $16 \pm 7.3\ [8.7,\ 23.3]$ | M1 A1 | Evaluate confidence interval (AEF, to 1 d.p.) |
| SR: Using combined variance $s_P^2/60 + s_Q^2/50 = 13.19$: max 3/5 | | |
---8 The amounts spent on the weekly food shopping by families in the big city $P$ and the small town $Q$ are to be compared. The amounts spent, in dollars, in $P$ and $Q$ are denoted by $x$ and $y$ respectively. For a random sample of 60 families in $P$ and a random sample of 50 families in $Q$, the amounts are summarised as follows.
$$\Sigma x = 9600 \quad \Sigma x ^ { 2 } = 1560000 \quad \Sigma y = 7200 \quad \Sigma y ^ { 2 } = 1052500$$
Assuming a common population variance, find\\
(i) a pooled estimate for the population variance,\\
(ii) a $95 \%$ confidence interval for the difference in the population means in $P$ and $Q$.
\hfill \mbox{\textit{CAIE FP2 2016 Q8 [9]}}