Question 11 OR - A-Level Maths

CAIE FP2 2013 June — Question 11 OR

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	June
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Chi-squared test of independence
Type	Scaled sample, find minimum N
Difficulty	Challenging +1.8 This question requires understanding of chi-squared test mechanics, working backwards from critical values to find boundary conditions, and scaling analysis for sample size effects. It demands multi-step reasoning with contingency tables, hypothesis testing interpretation, and algebraic manipulation of the chi-squared statistic formula—significantly beyond routine application but within reach of well-prepared Further Maths students.
Spec	5.06a Chi-squared: contingency tables

A researcher is investigating the relationship between the political allegiance of university students and their childhood environment. He chooses a random sample of 100 students and finds that 60 have political allegiance to the Alliance party. He also classifies their childhood environment as rural or urban, and finds that 45 had a rural childhood. The researcher carries out a test, at the \(10 \%\) significance level, on this data and finds that political allegiance is independent of childhood environment. Given that \(A\) is the number of students in the sample who both support the Alliance party and have a rural childhood, find the greatest and least possible values of \(A\). A second random sample of size \(100 N\), where \(N\) is an integer, is taken from the university student population. It is found that the proportions supporting the Alliance party from urban and rural childhoods are the same as in the first sample. Given that the value of \(A\) in the first sample was 29, find the greatest possible value of \(N\) that would lead to the same conclusion (that political allegiance is independent of childhood environment) from a test, at the \(10 \%\) significance level, on this second set of data.

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Question 11:

Part (a):

Answer	Marks
Resolve horizontally and vertically: \(F_A = R_B\)	B1
Relate limiting frictions and reactions at \(A\) and \(B\): \(F_A = 2\mu R_A\) and \(F_B = \mu R_B\)	B1

EITHER:

Answer	Marks
Resolve vertically: \(F_B + R_A = W\)	B1
Combine equations to find \(R_A\): \(R_A = W/(1 + 2\mu^2)\)	M1 A1
Take moments about \(B\): \(2R_A \cos\theta - 2F_A \sin\theta = W\cos\theta\)	M1 A1

Rearrange and use \(F_A = 2\mu R_A\): \(R_A - 2\mu R_A \tan\theta = \frac{1}{2}W\)

Answer	Marks
\(R_A = \frac{1}{2}W/(1 - 2\mu\tan\theta)\)	A1
Combine to find \(\tan\theta\): \(\tan\theta = (1 - 2\mu^2)/4\mu\) A.G.	M1 A1

OR:

Answer	Marks
Resolve vertically: \(F_B + R_A = W\)	(B1)
Combine equations to find \(R_B\): \(R_B = 2\mu W/(1 + 2\mu^2)\)	(M1 A1)
Take moments about \(A\): \(2R_B \sin\theta + 2F_B \cos\theta = W\cos\theta\)	(M1 A1)

Rearrange and use \(F_B = \mu R_B\): \(R_B \tan\theta + \mu R_B = \frac{1}{2}W\)

Answer	Marks
\(R_B = \frac{1}{2}W/(\tan\theta + \mu)\)	(A1)
Combine to find \(\tan\theta\): \(\tan\theta = (1 - 2\mu^2)/4\mu\) A.G.	(M1 A1)

OR:

Answer	Marks	Guidance
Take moments about centre of rod: \((F_A + R_B)\sin\theta = (R_A - F_B)\cos\theta\)	(M2 A1 A1)
Use first 3 equations to eliminate 3 forces: \(\tan\theta = (R_B/2\mu - \mu R_B)/(R_B + R_B)\)	(M1 A1 A1)
\(= (1 - 2\mu^2)/4\mu\) A.G.	(A1)	10 marks

Find positive value of \(\mu\) when \(\theta = 45°\): \(2\mu^2 + 4\mu - 1 = 0\)

Answer	Marks	Guidance
\(\mu = \sqrt{3/2} - 1\) or \(0.225\)	A.E.F M1 A1
Find positive value of \(\mu\) when \(\theta = 0°\): \(\mu = 1/\sqrt{2}\) or \(0.707\)	A.E.F B1
State set of possible values of \(\mu\) (A.E.F.): \([\sqrt{3/2}-1, 1/\sqrt{2}]\) or \([0.225, 0.707]\)	A1	4 marks

Question 11(b):

Tabulate observed values:

Answer	Marks	Guidance
Rural	Urban
Support	\(A\)	\(60-A\)
Not support	\(45-A\)	\(A-5\)
M1 A1

Find corresponding expected values:

Answer	Marks
\(27, 33, 18, 22\)	M1 A1

Calculate value of \(\chi^2\):

Answer	Marks
\(\chi^2 = \frac{(27-A)^2}{27} + \frac{(A-27)^2}{18} + \frac{(A-27)^2}{33} + \frac{(27-A)^2}{22}\)	M1

\(= \frac{(50/297)(A-27)^2}{}\)

Answer	Marks
or \(0.168[35](A-27)^2\)	A1
State or use correct tabular \(\chi^2\) value: \(\chi^2_{\text{tab}} = \chi^2_{1,0.9} = 2.706\) (to 3 s.f.)	B1
Use conclusion of independence to find equation for \(A\): \(\frac{(50/297)(A-27)^2 < \chi^2_{\text{tab}}}{}\)	M1

Find bounds for \(A\) (to 3 s.f.) (integer valued required for A1): \((A-27)^2 < 16.07\)

Answer	Marks	Guidance
\(A_{\min} = 23\) and \(A_{\max} = 31\)	A1, A1	10 marks

Relate new value \(\chi^2_{\text{new}}\) to original \(\chi^2\):

Answer	Marks
\(\chi^2_{\text{new}} = N \times \chi^2\)	M1
Use conclusion of independence to find equation for \(A\): \(0.168[35]\, N(A-27)^2 < \chi^2_{\text{tab}}\)	M1

Find \(N_{\max}\) with \(A = 29\) (integer value required for A1): \(N < 2.706/(4 \times 0.16835) = 4.02\)

Answer	Marks	Guidance
\(N_{\max} = 4\)	A1, A1	4 marks

# Question 11:

## Part (a):
Resolve horizontally and vertically: $F_A = R_B$ | B1 |

Relate limiting frictions and reactions at $A$ and $B$: $F_A = 2\mu R_A$ *and* $F_B = \mu R_B$ | B1 |

**EITHER:**

Resolve vertically: $F_B + R_A = W$ | B1 |

Combine equations to find $R_A$: $R_A = W/(1 + 2\mu^2)$ | M1 A1 |

Take moments about $B$: $2R_A \cos\theta - 2F_A \sin\theta = W\cos\theta$ | M1 A1 |

Rearrange and use $F_A = 2\mu R_A$: $R_A - 2\mu R_A \tan\theta = \frac{1}{2}W$

$R_A = \frac{1}{2}W/(1 - 2\mu\tan\theta)$ | A1 |

Combine to find $\tan\theta$: $\tan\theta = (1 - 2\mu^2)/4\mu$ **A.G.** | M1 A1 |

**OR:**

Resolve vertically: $F_B + R_A = W$ | (B1) |

Combine equations to find $R_B$: $R_B = 2\mu W/(1 + 2\mu^2)$ | (M1 A1) |

Take moments about $A$: $2R_B \sin\theta + 2F_B \cos\theta = W\cos\theta$ | (M1 A1) |

Rearrange and use $F_B = \mu R_B$: $R_B \tan\theta + \mu R_B = \frac{1}{2}W$

$R_B = \frac{1}{2}W/(\tan\theta + \mu)$ | (A1) |

Combine to find $\tan\theta$: $\tan\theta = (1 - 2\mu^2)/4\mu$ **A.G.** | (M1 A1) |

**OR:**

Take moments about centre of rod: $(F_A + R_B)\sin\theta = (R_A - F_B)\cos\theta$ | (M2 A1 A1) |

Use first 3 equations to eliminate 3 forces: $\tan\theta = (R_B/2\mu - \mu R_B)/(R_B + R_B)$ | (M1 A1 A1) |

$= (1 - 2\mu^2)/4\mu$ **A.G.** | (A1) | 10 marks |

Find positive value of $\mu$ when $\theta = 45°$: $2\mu^2 + 4\mu - 1 = 0$

$\mu = \sqrt{3/2} - 1$ *or* $0.225$ | A.E.F M1 A1 |

Find positive value of $\mu$ when $\theta = 0°$: $\mu = 1/\sqrt{2}$ *or* $0.707$ | A.E.F B1 |

State set of possible values of $\mu$ (A.E.F.): $[\sqrt{3/2}-1, 1/\sqrt{2}]$ *or* $[0.225, 0.707]$ | A1 | 4 marks | **Total: 14**

---

# Question 11(b):

## Tabulate observed values:
| | Rural | Urban |
|---|---|---|
| Support | $A$ | $60-A$ |
| Not support | $45-A$ | $A-5$ |
| M1 A1 |

## Find corresponding expected values:
$27, 33, 18, 22$ | M1 A1 |

## Calculate value of $\chi^2$:
$\chi^2 = \frac{(27-A)^2}{27} + \frac{(A-27)^2}{18} + \frac{(A-27)^2}{33} + \frac{(27-A)^2}{22}$ | M1 |

$= \frac{(50/297)(A-27)^2}{}$

*or* $0.168[35](A-27)^2$ | A1 |

State or use correct tabular $\chi^2$ value: $\chi^2_{\text{tab}} = \chi^2_{1,0.9} = 2.706$ (to 3 s.f.) | B1 |

Use conclusion of independence to find equation for $A$: $\frac{(50/297)(A-27)^2 < \chi^2_{\text{tab}}}{}$ | M1 |

Find bounds for $A$ (to 3 s.f.) (integer valued required for A1): $(A-27)^2 < 16.07$

$A_{\min} = 23$ *and* $A_{\max} = 31$ | A1, A1 | 10 marks |

## Relate new value $\chi^2_{\text{new}}$ to original $\chi^2$:
$\chi^2_{\text{new}} = N \times \chi^2$ | M1 |

Use conclusion of independence to find equation for $A$: $0.168[35]\, N(A-27)^2 < \chi^2_{\text{tab}}$ | M1 |

Find $N_{\max}$ with $A = 29$ (integer value required for A1): $N < 2.706/(4 \times 0.16835) = 4.02$

$N_{\max} = 4$ | A1, A1 | 4 marks | **Total: 14**

Show LaTeX source

A researcher is investigating the relationship between the political allegiance of university students and their childhood environment. He chooses a random sample of 100 students and finds that 60 have political allegiance to the Alliance party. He also classifies their childhood environment as rural or urban, and finds that 45 had a rural childhood. The researcher carries out a test, at the $10 \%$ significance level, on this data and finds that political allegiance is independent of childhood environment. Given that $A$ is the number of students in the sample who both support the Alliance party and have a rural childhood, find the greatest and least possible values of $A$.

A second random sample of size $100 N$, where $N$ is an integer, is taken from the university student population. It is found that the proportions supporting the Alliance party from urban and rural childhoods are the same as in the first sample. Given that the value of $A$ in the first sample was 29, find the greatest possible value of $N$ that would lead to the same conclusion (that political allegiance is independent of childhood environment) from a test, at the $10 \%$ significance level, on this second set of data.

\hfill \mbox{\textit{CAIE FP2 2013 Q11 OR}}

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Q1 4 Q2 8 Q3 9 Q4 9 Q5 13 Q6 6 Q7 8 Q8 9 Q9 9 Q10 11 Q11 EITHER Q11 OR