Question 5 - A-Level Maths

CAIE Further Paper 4 2024 November — Question 5 9 marks

Exam Board	CAIE
Module	Further Paper 4 (Further Paper 4)
Year	2024
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	T-tests (unknown variance)
Type	Find critical alpha or significance level
Difficulty	Challenging +1.2 This is a standard two-sample t-test requiring calculation of sample means and variances from summary statistics, followed by computing a pooled variance t-statistic and comparing to critical values. While it involves multiple computational steps and careful handling of the inequality direction for significance levels, it follows a well-established procedure taught in Further Statistics with no novel insight required. The main challenge is computational accuracy rather than conceptual difficulty.
Spec	5.05c Hypothesis test: normal distribution for population mean

5 Dev owns a small company which produces bottles of juice. He uses two machines, $X$ and $Y$, to fill empty bottles with juice. Dev is investigating the volumes of juice in the bottles. He chooses a random sample of 35 bottles filled by machine $X$ and a random sample of 60 bottles filled by machine $Y$. The volumes of juice, $x$ and $y$ respectively, measured in suitable units, are summarised by $$\sum x = 30.8 , \quad \sum x ^ { 2 } = 29.0 , \quad \sum y = 62.4 , \quad \sum y ^ { 2 } = 76.8 .$$ Dev claims that the mean volume of juice in bottles filled by machine $Y$ is greater than the mean volume of juice in bottles filled by machine $X$. A test at the $\alpha \%$ significance level suggests that there is sufficient evidence to support Dev's claim. Find the set of possible values of $\alpha$. \includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-10_2717_33_109_2014} \includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-11_2726_35_97_20}

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Question 5:

Answer	Marks	Guidance
$s_x^2=\frac{1}{34}\left(29.0-\frac{30.8^2}{35}\right)\quad[=0.055765]$	B1	$\frac{237}{4250}$
$s_y^2=\frac{1}{59}\left(76.8-\frac{62.4^2}{60}\right)\quad[=0.201763]$		$\frac{1488}{7375}$; Allow B1 for correct expression for pooled variance.
$s^2=\frac{0.055765}{35}+\frac{0.201763}{60}\quad[=0.004956]$	M1 A1	Pooled variance M0A0.
$H_0: \mu_x=\mu_y$; $H_1: \mu_x<\mu_y$	B1
$z=\dfrac{\dfrac{30.8}{35}-\dfrac{62.4}{60}}{s}=-\dfrac{0.25}{\sqrt{0.004956}}=-2.27$	M1
A1	Pooled variance M0A0.
Corresponding probability is 0.9885	M1	FT their $z$.
One tail: $100\times(1-{'}0.9885{'})=1.15$	M1	Allow for $1-{'}0.9885{'}$.
$\alpha>1.15$	A1	Allow $\geqslant$.
9

## Question 5:

$s_x^2=\frac{1}{34}\left(29.0-\frac{30.8^2}{35}\right)\quad[=0.055765]$ | B1 | $\frac{237}{4250}$

$s_y^2=\frac{1}{59}\left(76.8-\frac{62.4^2}{60}\right)\quad[=0.201763]$ | | $\frac{1488}{7375}$; Allow B1 for correct expression for pooled variance.

$s^2=\frac{0.055765}{35}+\frac{0.201763}{60}\quad[=0.004956]$ | M1 A1 | Pooled variance M0A0.

$H_0: \mu_x=\mu_y$; $H_1: \mu_x<\mu_y$ | B1 |

$z=\dfrac{\dfrac{30.8}{35}-\dfrac{62.4}{60}}{s}=-\dfrac{0.25}{\sqrt{0.004956}}=-2.27$ | M1 | 

| A1 | Pooled variance M0A0.

Corresponding probability is 0.9885 | M1 | FT their $z$.

One tail: $100\times(1-{'}0.9885{'})=1.15$ | M1 | Allow for $1-{'}0.9885{'}$.

$\alpha>1.15$ | A1 | Allow $\geqslant$.

| 9 |

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5 Dev owns a small company which produces bottles of juice. He uses two machines, $X$ and $Y$, to fill empty bottles with juice. Dev is investigating the volumes of juice in the bottles. He chooses a random sample of 35 bottles filled by machine $X$ and a random sample of 60 bottles filled by machine $Y$. The volumes of juice, $x$ and $y$ respectively, measured in suitable units, are summarised by

$$\sum x = 30.8 , \quad \sum x ^ { 2 } = 29.0 , \quad \sum y = 62.4 , \quad \sum y ^ { 2 } = 76.8 .$$

Dev claims that the mean volume of juice in bottles filled by machine $Y$ is greater than the mean volume of juice in bottles filled by machine $X$. A test at the $\alpha \%$ significance level suggests that there is sufficient evidence to support Dev's claim.

Find the set of possible values of $\alpha$.\\

\includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-10_2717_33_109_2014}\\
\includegraphics[max width=\textwidth, alt={}, center]{b9cbf607-4f40-41bb-8374-6b2c39f945ac-11_2726_35_97_20}\\

\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q5 [9]}}

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Q1 4 Q2 8 Q3 10 Q4 10 Q5 9 Q6 9

Answer	Marks	Guidance
\(s_x^2=\frac{1}{34}\left(29.0-\frac{30.8^2}{35}\right)\quad[=0.055765]\)	B1	\(\frac{237}{4250}\)
\(s_y^2=\frac{1}{59}\left(76.8-\frac{62.4^2}{60}\right)\quad[=0.201763]\)		\(\frac{1488}{7375}\); Allow B1 for correct expression for pooled variance.
\(s^2=\frac{0.055765}{35}+\frac{0.201763}{60}\quad[=0.004956]\)	M1 A1	Pooled variance M0A0.
\(H_0: \mu_x=\mu_y\); \(H_1: \mu_x<\mu_y\)	B1
\(z=\dfrac{\dfrac{30.8}{35}-\dfrac{62.4}{60}}{s}=-\dfrac{0.25}{\sqrt{0.004956}}=-2.27\)	M1
A1	Pooled variance M0A0.
Corresponding probability is 0.9885	M1	FT their \(z\).
One tail: \(100\times(1-{'}0.9885{'})=1.15\)	M1	Allow for \(1-{'}0.9885{'}\).
\(\alpha>1.15\)	A1	Allow \(\geqslant\).
9