| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Relate two regression lines |
| Difficulty | Challenging +1.2 This question tests standard relationships between regression lines and correlation coefficients (r² = product of gradients) plus using properties that regression lines pass through means. While it requires knowledge of multiple statistical concepts and algebraic manipulation, the techniques are well-established formulas from the Further Maths Statistics syllabus with straightforward application once the relationships are recalled. |
| Spec | 5.08c Pearson: measure of straight-line fit5.08d Hypothesis test: Pearson correlation5.09c Calculate regression line |
| \(x\) | 1 | 2 | 5 | 5 | \(p\) |
| \(y\) | 5 | 3 | 4 | 2 | \(q\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R_C \cdot 3a = kW \cdot 4a\cos\theta + W \cdot 2a\cos\theta\) | M1 A1 | Take moments about \(A\) for rod to find \(R_C\) |
| \(R_C = (2k+1)(2W/5)\) | A1 | |
| \(F_A = R_C\sin\theta = (2k+1)(8W/25)\) | M1 A1 | Resolve horizontal forces on rod to find \(F_A\) |
| \(R_A = W + kW - R_C\cos\theta = (13k+19)(W/25)\) | M1 A1 | Resolve vertical forces on rod to find \(R_A\) |
| *OR:* Take moments about \(C\) for rod: | ||
| \(F_A \cdot 3a\sin\theta - R_A \cdot 3a\cos\theta = kWa\cos\theta - Wa\cos\theta\) | (M1 A1) | |
| \(4F_A - 3R_A = (k-1)W\) | (A1) | |
| Resolve forces parallel to rod: \(F_A\cos\theta + R_A\sin\theta = kW\sin\theta + W\sin\theta\) | (M1) | |
| \(3F_A + 4R_A = 4(k+1)W\) | (A1) | |
| Combine to find \(F_A\), \(R_A\): \(F_A = (2k+1)(8W/25)\) and \(R_A = (13k+19)(W/25)\) | (M1 A1) | |
| Use \(F_A \le \mu R_A\) to find \(\mu_{\min}\): \(\mu_{\min} = 8(2k+1)/(13k+19)\) A.G. | M1 A1 | (No use of inequality loses A1) |
| \(160k + 80 \le (\text{or} =)\ 117k + 171\) | M1 | Find bound on \(k\) (allow use of equality) |
| Bound \(= 91/43\) or \(2.12\) | A1 | |
| \(k \le 91/43\) or \(2.12\) | A1 | Correct use of inequality |
| Answer | Marks | Guidance |
|---|---|---|
| \(r^2 = (-0.5)(-1.2) = 0.6 \text{ or } 0.775^2\) | M1 A1 | |
| \(r = -0.775\) | *A1 | 3 marks |
| Answer | Marks |
|---|---|
| \(H_0: \rho = 0,\ H_1: \rho \neq 0\) | B1 |
| Answer | Marks |
|---|---|
| \(r_{5,\ 2.5\%} = 0.878\) | *B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept \(H_0\) if \( | r | <\) tabular value |
| Answer | Marks | Guidance |
|---|---|---|
| Coeff. does not differ from zero | A1 | 4 marks |
| Answer | Marks |
|---|---|
| \(14 + q = (-0.5)(13 + p) + 5 \times 5\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(13 + p = (-1.2)(14 + q) + 5 \times 7.6\) | |
| \(or\ (13 + p)/5 = (-1.2)(14 + q)/5 + 7.6\) | A1 |
| Answer | Marks |
|---|---|
| \(p = 7,\ q = 1\) | A1, A1 |
| Answer | Marks |
|---|---|
| \(\bar{x} = 4;\ \bar{y} = 3\) | M1 A1; A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 7,\ q = 1\) | A1, A1 | 5 marks |
## Question 11(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R_C \cdot 3a = kW \cdot 4a\cos\theta + W \cdot 2a\cos\theta$ | M1 A1 | Take moments about $A$ for rod to find $R_C$ |
| $R_C = (2k+1)(2W/5)$ | A1 | |
| $F_A = R_C\sin\theta = (2k+1)(8W/25)$ | M1 A1 | Resolve horizontal forces on rod to find $F_A$ |
| $R_A = W + kW - R_C\cos\theta = (13k+19)(W/25)$ | M1 A1 | Resolve vertical forces on rod to find $R_A$ |
| *OR:* Take moments about $C$ for rod: | | |
| $F_A \cdot 3a\sin\theta - R_A \cdot 3a\cos\theta = kWa\cos\theta - Wa\cos\theta$ | (M1 A1) | |
| $4F_A - 3R_A = (k-1)W$ | (A1) | |
| Resolve forces parallel to rod: $F_A\cos\theta + R_A\sin\theta = kW\sin\theta + W\sin\theta$ | (M1) | |
| $3F_A + 4R_A = 4(k+1)W$ | (A1) | |
| Combine to find $F_A$, $R_A$: $F_A = (2k+1)(8W/25)$ and $R_A = (13k+19)(W/25)$ | (M1 A1) | |
| Use $F_A \le \mu R_A$ to find $\mu_{\min}$: $\mu_{\min} = 8(2k+1)/(13k+19)$ **A.G.** | M1 A1 | (No use of inequality loses A1) |
| $160k + 80 \le (\text{or} =)\ 117k + 171$ | M1 | Find bound on $k$ (allow use of equality) |
| Bound $= 91/43$ or $2.12$ | A1 | |
| $k \le 91/43$ or $2.12$ | A1 | Correct use of inequality |
**Total: 12 marks**
## Question 11(b):
**Find correlation coefficient $r$:**
$r^2 = (-0.5)(-1.2) = 0.6 \text{ or } 0.775^2$ | M1 A1 |
$r = -0.775$ | *A1 | 3 marks
---
**State both hypotheses (B0 for $r$...):**
$H_0: \rho = 0,\ H_1: \rho \neq 0$ | B1 |
---
**State or use correct tabular one-tail $r$ value:**
$r_{5,\ 2.5\%} = 0.878$ | *B1 |
---
**Valid method for reaching conclusion:**
Accept $H_0$ if $|r| <$ tabular value | M1 |
---
**Correct conclusion (AEF, dep *A1, *B1):**
Coeff. does not differ from zero | A1 | 4 marks
---
**EITHER: Find two indep. eqns for $p, q$:**
$14 + q = (-0.5)(13 + p) + 5 \times 5$ | M1 A1 |
$or\ (14 + q)/5 = (-0.5)(13 + p)/5 + 5$
$13 + p = (-1.2)(14 + q) + 5 \times 7.6$ | |
$or\ (13 + p)/5 = (-1.2)(14 + q)/5 + 7.6$ | A1 |
---
**Combine to find $p, q$:**
$p = 7,\ q = 1$ | A1, A1 |
---
**OR: Combine regression lines to find $\bar{x},\ \bar{y}$:**
$\bar{x} = 4;\ \bar{y} = 3$ | M1 A1; A1 |
**Use sample data to find $p, q$:**
$p = 7,\ q = 1$ | A1, A1 | 5 marks
---
**Total: [12]**For a random sample of 5 pairs of values of $x$ and $y$, the equations of the regression lines of $y$ on $x$ and $x$ on $y$ are respectively
$$y = - 0.5 x + 5 \quad \text { and } \quad x = - 1.2 y + 7.6$$
Find the value of the product moment correlation coefficient for this sample.
Test, at the $5 \%$ significance level, whether the population product moment correlation coefficient differs from zero.
The following table shows the sample data.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | }
\hline
$x$ & 1 & 2 & 5 & 5 & $p$ \\
\hline
$y$ & 5 & 3 & 4 & 2 & $q$ \\
\hline
\end{tabular}
\end{center}
Find the values of $p$ and $q$.
\hfill \mbox{\textit{CAIE FP2 2012 Q11 OR}}