Question 10 - A-Level Maths

CAIE FP2 2014 June — Question 10 11 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2014
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hypothesis test of Pearson’s product-moment correlation coefficient
Type	Calculate PMCC from summary statistics
Difficulty	Standard +0.3 This is a straightforward application of standard formulas for PMCC, hypothesis testing, and regression. While it's Further Maths content (making it slightly above average), it requires only direct substitution into well-known formulas with no conceptual challenges or problem-solving insight—purely mechanical calculation with clear procedures.
Spec	5.08a Pearson correlation: calculate pmcc 5.08d Hypothesis test: Pearson correlation 5.09c Calculate regression line 5.09e Use regression: for estimation in context

10 Samples of rock from a number of geological sites were analysed for the quantities of two types, $X$ and $Y$, of rare minerals. The results, in milligrams, for 10 randomly chosen samples, each of 10 kg , are summarised as follows. $$\Sigma x = 866 \quad \Sigma x ^ { 2 } = 121276 \quad \Sigma y = 639 \quad \Sigma y ^ { 2 } = 55991 \quad \Sigma x y = 73527$$ Find the product moment correlation coefficient. Stating your hypotheses, test at the $5 \%$ significance level whether there is non-zero correlation between quantities of the two rare minerals. Find the equation of the regression line of $x$ on $y$ in the form $x = p y + q$, where $p$ and $q$ are constants to be determined.

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Question 10:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$r = (73527 - 866 \times 639/10) / \sqrt{\{(121276 - 866^2/10)(55991 - 639^2/10)\}}$	M1 A1	A.E.F.; A0 if only 3 s.f. clearly used
$= 18189.6 / \sqrt{(46280.4 \times 15158.9)}$	A1
$= 0.687$	*A1
4 marks
$H_0$: $\rho = 0$, $H_1$: $\rho \neq 0$	B1
$r_{10, 5\%} = 0.632$	*B1
Reject $H_0$ if \(	r	>\) tabular value
There is non-zero correlation	A1	A.E.F., dep A1, B1
4 marks
$p = 18189.6 / 15158.9 = 1.20$	B1	Gradient $p$ in $x - \bar{x} = p(y - \bar{y})$
$x = 86.6 + 1.20(y - 63.9)$
$= 1.20y + 9.92$	M1 A1
3 marks, Total: 11

Question 11A (i):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$AB = \sqrt{(4a^2 + 12a^2)} = 4a$	M1 A1	A.G.
$\angle CAB = \sin^{-1} 2a\sqrt{3}/4a$ or $\cos^{-1} 2a/4a$ or $\tan^{-1} 2a\sqrt{3}/2a$
$= 60°$ so $\angle SAB = 30°$	M1 A1	A.G.
4 marks

Question 11A (ii):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\frac{1}{2}N_A + \frac{1}{2}\sqrt{3}N_B + \frac{1}{2}\sqrt{3}F_A = W$	M1 A1	Resolve vertically and horizontally; $F_A$ may be in either direction
and $\frac{1}{2}\sqrt{3}N_A = \frac{1}{2}N_B + \frac{1}{2}F_A$
$N_A = \frac{1}{2}W$	A1	A.G.
3 marks

Question 11A (iii):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$N_A = \frac{1}{2}W$	(M1 A1)	Resolve in dirn. $PQ$ to find $N_A$; A.G.
$N_B + F_A = \frac{1}{2}\sqrt{3}W$	(A1)	Second resolution e.g. in dirn. $PS$
$\frac{1}{2}\sqrt{3}W \times 3a/2 + \frac{1}{2}W \times (2\sqrt{3}-3)a = N_B \times 2a$	M1 A1 A1	Take moments e.g. about $A$; A1 for each side of equation
$N_B = \{(7\sqrt{3}-6)/8\}W$	M1 A1
$F_A = \sqrt{3}N_A - N_B$ or $\frac{1}{2}\sqrt{3}W - N_B$
$= \{3(2-\sqrt{3})/8\}W$	M1 A1	A.E.F.
7 marks, Total: 14

Question 11B:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$s_P^2 = (236.0 - 42.8^2/8)/7$
$= 351/350$ or $1.003$ or $1.001^2$	M1	allow biased: $0.8775$ or $0.9367^2$
$42.8/8 \pm t\sqrt{(s_P^2/8)}$	M1	allow $z$ in place of $t$
$t_{7,\, 0.975} = 2.365$	A1
$5.35 \pm 0.84$ or $[4.51, 6.19]$	A1
4 marks
$(5.35 - k)/\sqrt{(s_P^2/8)} \geqslant [\text{or} >] \, t$	M1	Formulate inequality for $k$
$t_{7,\, 0.9} = 1.415$	A1
$5.35 - k \geqslant 0.50$, $k_{\max} = 4.85$	A1	A0 if $=$ or $\leqslant$ was used for $k$ above
3 marks
$H_0$: $\mu_P = \mu_Q$, $H_1$: $\mu_P > \mu_Q$	B1
Normal distributions for $[P$ and$]$ $Q$ and equal variances	B1	State assumption A.E.F.
$s^2 = (7 \times 1.003 + 11 \times 1.962)/18$
$= 1.589$ or $1.261^2$	M1 A1	Pooled common variance
$t = (5.35 - 4.60)/(s\sqrt{(1/8 + 1/12)})$
$= 1.30$	M1 A1
$t < t_{18,\, 0.9} = 1.33$ so $Q$'s mean is not less than $P$'s	B1$\checkmark$	A.E.F., $\checkmark$ on $t$
7 marks, Total: 14

# Question 10:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = (73527 - 866 \times 639/10) / \sqrt{\{(121276 - 866^2/10)(55991 - 639^2/10)\}}$ | M1 A1 | A.E.F.; A0 if only 3 s.f. clearly used |
| $= 18189.6 / \sqrt{(46280.4 \times 15158.9)}$ | A1 | |
| $= 0.687$ | *A1 | |
| | | 4 marks |
| $H_0$: $\rho = 0$, $H_1$: $\rho \neq 0$ | B1 | |
| $r_{10, 5\%} = 0.632$ | *B1 | |
| Reject $H_0$ if $|r| >$ tabular value | M1 | |
| There is non-zero correlation | A1 | A.E.F., dep *A1, *B1 |
| | | 4 marks |
| $p = 18189.6 / 15158.9 = 1.20$ | B1 | Gradient $p$ in $x - \bar{x} = p(y - \bar{y})$ |
| $x = 86.6 + 1.20(y - 63.9)$ | | |
| $= 1.20y + 9.92$ | M1 A1 | |
| | | 3 marks, Total: 11 |

---

# Question 11A (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB = \sqrt{(4a^2 + 12a^2)} = 4a$ | M1 A1 | A.G. |
| $\angle CAB = \sin^{-1} 2a\sqrt{3}/4a$ or $\cos^{-1} 2a/4a$ or $\tan^{-1} 2a\sqrt{3}/2a$ | | |
| $= 60°$ so $\angle SAB = 30°$ | M1 A1 | A.G. |
| | | 4 marks |

## Question 11A (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}N_A + \frac{1}{2}\sqrt{3}N_B + \frac{1}{2}\sqrt{3}F_A = W$ | M1 A1 | Resolve vertically and horizontally; $F_A$ may be in either direction |
| and $\frac{1}{2}\sqrt{3}N_A = \frac{1}{2}N_B + \frac{1}{2}F_A$ | | |
| $N_A = \frac{1}{2}W$ | A1 | A.G. |
| | | 3 marks |

## Question 11A (iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $N_A = \frac{1}{2}W$ | (M1 A1) | Resolve in dirn. $PQ$ to find $N_A$; A.G. |
| $N_B + F_A = \frac{1}{2}\sqrt{3}W$ | (A1) | Second resolution e.g. in dirn. $PS$ |
| $\frac{1}{2}\sqrt{3}W \times 3a/2 + \frac{1}{2}W \times (2\sqrt{3}-3)a = N_B \times 2a$ | M1 A1 A1 | Take moments e.g. about $A$; A1 for each side of equation |
| $N_B = \{(7\sqrt{3}-6)/8\}W$ | M1 A1 | |
| $F_A = \sqrt{3}N_A - N_B$ or $\frac{1}{2}\sqrt{3}W - N_B$ | | |
| $= \{3(2-\sqrt{3})/8\}W$ | M1 A1 | A.E.F. |
| | | 7 marks, Total: 14 |

---

# Question 11B:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $s_P^2 = (236.0 - 42.8^2/8)/7$ | | |
| $= 351/350$ or $1.003$ or $1.001^2$ | M1 | allow biased: $0.8775$ or $0.9367^2$ |
| $42.8/8 \pm t\sqrt{(s_P^2/8)}$ | M1 | allow $z$ in place of $t$ |
| $t_{7,\, 0.975} = 2.365$ | A1 | |
| $5.35 \pm 0.84$ or $[4.51, 6.19]$ | A1 | |
| | | 4 marks |
| $(5.35 - k)/\sqrt{(s_P^2/8)} \geqslant [\text{or} >] \, t$ | M1 | Formulate inequality for $k$ |
| $t_{7,\, 0.9} = 1.415$ | A1 | |
| $5.35 - k \geqslant 0.50$, $k_{\max} = 4.85$ | A1 | A0 if $=$ or $\leqslant$ was used for $k$ above |
| | | 3 marks |
| $H_0$: $\mu_P = \mu_Q$, $H_1$: $\mu_P > \mu_Q$ | B1 | |
| Normal distributions for $[P$ and$]$ $Q$ and equal variances | B1 | State assumption A.E.F. |
| $s^2 = (7 \times 1.003 + 11 \times 1.962)/18$ | | |
| $= 1.589$ or $1.261^2$ | M1 A1 | Pooled common variance |
| $t = (5.35 - 4.60)/(s\sqrt{(1/8 + 1/12)})$ | | |
| $= 1.30$ | M1 A1 | |
| $t < t_{18,\, 0.9} = 1.33$ so $Q$'s mean is not less than $P$'s | B1$\checkmark$ | A.E.F., $\checkmark$ on $t$ |
| | | 7 marks, Total: 14 |

Show LaTeX source

10 Samples of rock from a number of geological sites were analysed for the quantities of two types, $X$ and $Y$, of rare minerals. The results, in milligrams, for 10 randomly chosen samples, each of 10 kg , are summarised as follows.

$$\Sigma x = 866 \quad \Sigma x ^ { 2 } = 121276 \quad \Sigma y = 639 \quad \Sigma y ^ { 2 } = 55991 \quad \Sigma x y = 73527$$

Find the product moment correlation coefficient.

Stating your hypotheses, test at the $5 \%$ significance level whether there is non-zero correlation between quantities of the two rare minerals.

Find the equation of the regression line of $x$ on $y$ in the form $x = p y + q$, where $p$ and $q$ are constants to be determined.

\hfill \mbox{\textit{CAIE FP2 2014 Q10 [11]}}

This paper (12 questions)

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Q1 2 Q2 8 Q3 10 Q4 10 Q5 13 Q6 5 Q7 8 Q8 9 Q9 10 Q10 11 Q11 EITHER Q11 OR

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(s_P^2 = (236.0 - 42.8^2/8)/7\)
\(= 351/350\) or \(1.003\) or \(1.001^2\)	M1	allow biased: \(0.8775\) or \(0.9367^2\)
\(42.8/8 \pm t\sqrt{(s_P^2/8)}\)	M1	allow \(z\) in place of \(t\)
\(t_{7,\, 0.975} = 2.365\)	A1
\(5.35 \pm 0.84\) or \([4.51, 6.19]\)	A1
4 marks
\((5.35 - k)/\sqrt{(s_P^2/8)} \geqslant [\text{or} >] \, t\)	M1	Formulate inequality for \(k\)
\(t_{7,\, 0.9} = 1.415\)	A1
\(5.35 - k \geqslant 0.50\), \(k_{\max} = 4.85\)	A1	A0 if \(=\) or \(\leqslant\) was used for \(k\) above
3 marks
\(H_0\): \(\mu_P = \mu_Q\), \(H_1\): \(\mu_P > \mu_Q\)	B1
Normal distributions for \([P\) and\(]\) \(Q\) and equal variances	B1	State assumption A.E.F.
\(s^2 = (7 \times 1.003 + 11 \times 1.962)/18\)
\(= 1.589\) or \(1.261^2\)	M1 A1	Pooled common variance
\(t = (5.35 - 4.60)/(s\sqrt{(1/8 + 1/12)})\)
\(= 1.30\)	M1 A1
\(t < t_{18,\, 0.9} = 1.33\) so \(Q\)'s mean is not less than \(P\)'s	B1\(\checkmark\)	A.E.F., \(\checkmark\) on \(t\)
7 marks, Total: 14