CAIE FP2 2009 June — Question 11 OR

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2009
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeTwo-sample t-test equal variance
DifficultyStandard +0.8 This is a standard two-sample t-test question from Further Maths Statistics requiring hypothesis setup, assumptions, critical region calculation, and an inverse problem finding the maximum value of a. While methodical, it requires understanding of pooled variance, degrees of freedom, and working backwards from the test statistic—more demanding than typical A-level statistics but routine for Further Maths students who have practiced this topic.
Spec5.05d Confidence intervals: using normal distribution

A study was made of the acidity levels in farmland on opposite sides of an island. The levels were measured at six randomly chosen points on the eastern side and at five randomly chosen points on the western side. The values obtained, in suitable units, are denoted by \(x _ { E }\) and \(x _ { W }\) respectively. The sample means \(\bar { x } _ { E }\) and \(\bar { x } _ { W }\), and unbiased estimates of the two population variances, \(s _ { E } ^ { 2 }\) and \(s _ { W } ^ { 2 }\), are as follows. $$\bar { x } _ { E } = 5.035 , s _ { E } ^ { 2 } = 0.0231 , \bar { x } _ { W } = 4.782 , s _ { W } ^ { 2 } = 0.0195 .$$ The population means on the eastern and western sides are denoted by \(\mu _ { E }\) and \(\mu _ { W }\) respectively. State suitable hypotheses for a test for a difference between the mean acidity levels on the two sides of the island. Stating any required assumptions, obtain the rejection region for a test at the \(5 \%\) significance level of whether the mean acidity levels differ on the two sides of the island. Give the conclusion of the test. Find the largest value of \(a\) for which the samples above provide evidence at the \(5 \%\) significance level that \(\mu _ { E } - \mu _ { W } > a\).

Question 11 (EITHER - Mechanics, 14 marks):
Part 1 (3 marks)
AnswerMarks Guidance
AnswerMark Guidance
\(T = 4mgx/a\)B1 Find tension \(T\)
\(m\,\mathrm{d}^2x/\mathrm{d}t^2 = mg - T\)M1 Apply Newton's law of motion to \(B\)
\(\mathrm{d}^2x/\mathrm{d}t^2 = -(g/a)(4x-a)\)A1 Combine. A.G.
Part 2 (3 marks)
AnswerMarks Guidance
AnswerMark Guidance
\(\mathrm{d}^2y/\mathrm{d}t^2 = -(4g/a)\,y\)M1 A1 Substitute e.g. \(y = x - \tfrac{1}{4}a\) and rearrange
\(x_c = \tfrac{1}{4}a\)A1 State centre of motion, or derive from \(y=0\)
Part 3 (8 marks)
AnswerMarks Guidance
AnswerMark Guidance
\(T = \tfrac{1}{3}mg,\; x_s = a/12\)M1 A1 Find \(x\) when \(A\) starts to slip using \(F = \mu R\)
\(y = y_{max}\cos\omega t\) *or* \(y_{max}\sin\omega t\)M1 Valid use of SHM equation to find \(t_s\) to slipping
\(y_s = y_{max}\cos\omega t_s\)M1 EITHER: Valid use of cosine form
\(t_s = t_1 - t_2,\; y_{max} = y_{max}\sin\omega t_1\)M1 OR: Valid use of sine form
\(y_s = y_{max}\sin\omega t_2\)(M1)
\(y_s/y_{max} = (x_c - x_s)/x_c\)M1 Substitute for \(y_s\), \(y_{max}\)
\(= (a/6)/(a/4) = \tfrac{2}{3}\)A1
\(t_s = (\cos^{-1}\tfrac{2}{3})/\omega\)
*or* \((\tfrac{1}{2}\pi - \sin^{-1}\tfrac{2}{3})/\omega\)M1
\(t_s = 0.421\sqrt{a/g}\)A1 Substitute \(\omega = 2\sqrt{g/a}\) and evaluate
Question 11 (OR - Statistics, 14 marks):
Part 1 (9 marks)
AnswerMarks Guidance
AnswerMark Guidance
\(H_0: \mu_E = \mu_W,\; H_1: \mu_E \neq \mu_W\)B1 State hypotheses
Two populations have Normal distributions and common varianceB1 State assumption [A.E.F.]
\(\sigma^2 = (5\times0.0231 + 4\times0.0195)/9\)M1 Estimate common variance
\(= 0.0215\) *or* \(0.1473^2\)A1
\(t_{9,\,0.975} = 2.26[2]\)B1 Use correct tabular value of \(t\)
\(\lvert\bar{X}_E - \bar{X}_W\rvert \geq t\sigma\sqrt{1/6+1/5}\)M1 Formulate rejection region (with any \(t\); allow \(>\))
\(= 0.201\)A1
\(0.253 > 0.201\sqrt{\phantom{x}}\) Compare actual sample means with region
*or* \(2.85 > 2.26[2]\sqrt{\phantom{x}}\)M1 A1 *or* compare calculated \(t\) with tabular \(t\)
Mean acidity levels do differA1 Consistent conclusion (A.E.F.; dep values above)
Part 2 (4 marks)
AnswerMarks Guidance
AnswerMark Guidance
\(\bar{X}_E - \bar{X}_W - a \geq t\sigma\sqrt{1/6+1/5}\)M1 State condition on \(a\) (with any \(t\); allow \(>\) or \(=\))
\(t_{9,\,0.95} = 1.83[3]\)A1 Use correct tabular value of \(t\)
\(a_{max} = 0.253 - 0.163 = 0.09\) (2 dp)M1 A1 Substitute to find largest value of \(a\)
# Question 11 (EITHER - Mechanics, 14 marks):

## Part 1 (3 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $T = 4mgx/a$ | B1 | Find tension $T$ |
| $m\,\mathrm{d}^2x/\mathrm{d}t^2 = mg - T$ | M1 | Apply Newton's law of motion to $B$ |
| $\mathrm{d}^2x/\mathrm{d}t^2 = -(g/a)(4x-a)$ | A1 | Combine. **A.G.** |

## Part 2 (3 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathrm{d}^2y/\mathrm{d}t^2 = -(4g/a)\,y$ | M1 A1 | Substitute e.g. $y = x - \tfrac{1}{4}a$ and rearrange |
| $x_c = \tfrac{1}{4}a$ | A1 | State centre of motion, or derive from $y=0$ |

## Part 3 (8 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $T = \tfrac{1}{3}mg,\; x_s = a/12$ | M1 A1 | Find $x$ when $A$ starts to slip using $F = \mu R$ |
| $y = y_{max}\cos\omega t$ *or* $y_{max}\sin\omega t$ | M1 | Valid use of SHM equation to find $t_s$ to slipping |
| $y_s = y_{max}\cos\omega t_s$ | M1 | EITHER: Valid use of cosine form |
| $t_s = t_1 - t_2,\; y_{max} = y_{max}\sin\omega t_1$ | M1 | OR: Valid use of sine form |
| $y_s = y_{max}\sin\omega t_2$ | (M1) | |
| $y_s/y_{max} = (x_c - x_s)/x_c$ | M1 | Substitute for $y_s$, $y_{max}$ |
| $= (a/6)/(a/4) = \tfrac{2}{3}$ | A1 | |
| $t_s = (\cos^{-1}\tfrac{2}{3})/\omega$ | | |
| *or* $(\tfrac{1}{2}\pi - \sin^{-1}\tfrac{2}{3})/\omega$ | M1 | |
| $t_s = 0.421\sqrt{a/g}$ | A1 | Substitute $\omega = 2\sqrt{g/a}$ and evaluate |

---

# Question 11 (OR - Statistics, 14 marks):

## Part 1 (9 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_E = \mu_W,\; H_1: \mu_E \neq \mu_W$ | B1 | State hypotheses |
| Two populations have Normal distributions and common variance | B1 | State assumption [A.E.F.] |
| $\sigma^2 = (5\times0.0231 + 4\times0.0195)/9$ | M1 | Estimate common variance |
| $= 0.0215$ *or* $0.1473^2$ | A1 | |
| $t_{9,\,0.975} = 2.26[2]$ | B1 | Use correct tabular value of $t$ |
| $\lvert\bar{X}_E - \bar{X}_W\rvert \geq t\sigma\sqrt{1/6+1/5}$ | M1 | Formulate rejection region (with any $t$; allow $>$) |
| $= 0.201$ | A1 | |
| $0.253 > 0.201\sqrt{\phantom{x}}$ | | Compare actual sample means with region |
| *or* $2.85 > 2.26[2]\sqrt{\phantom{x}}$ | M1 A1 | *or* compare calculated $t$ with tabular $t$ |
| Mean acidity levels **do** differ | A1 | Consistent conclusion (A.E.F.; dep values above) |

## Part 2 (4 marks)

| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{X}_E - \bar{X}_W - a \geq t\sigma\sqrt{1/6+1/5}$ | M1 | State condition on $a$ (with any $t$; allow $>$ or $=$) |
| $t_{9,\,0.95} = 1.83[3]$ | A1 | Use correct tabular value of $t$ |
| $a_{max} = 0.253 - 0.163 = 0.09$ (2 dp) | M1 A1 | Substitute to find largest value of $a$ |
A study was made of the acidity levels in farmland on opposite sides of an island. The levels were measured at six randomly chosen points on the eastern side and at five randomly chosen points on the western side. The values obtained, in suitable units, are denoted by $x _ { E }$ and $x _ { W }$ respectively. The sample means $\bar { x } _ { E }$ and $\bar { x } _ { W }$, and unbiased estimates of the two population variances, $s _ { E } ^ { 2 }$ and $s _ { W } ^ { 2 }$, are as follows.

$$\bar { x } _ { E } = 5.035 , s _ { E } ^ { 2 } = 0.0231 , \bar { x } _ { W } = 4.782 , s _ { W } ^ { 2 } = 0.0195 .$$

The population means on the eastern and western sides are denoted by $\mu _ { E }$ and $\mu _ { W }$ respectively. State suitable hypotheses for a test for a difference between the mean acidity levels on the two sides of the island.

Stating any required assumptions, obtain the rejection region for a test at the $5 \%$ significance level of whether the mean acidity levels differ on the two sides of the island. Give the conclusion of the test.

Find the largest value of $a$ for which the samples above provide evidence at the $5 \%$ significance level that $\mu _ { E } - \mu _ { W } > a$.

\hfill \mbox{\textit{CAIE FP2 2009 Q11 OR}}