Question 6 - A-Level Maths

CAIE Further Paper 4 2022 November — Question 6 9 marks

Exam Board	CAIE
Module	Further Paper 4 (Further Paper 4)
Year	2022
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	T-tests (unknown variance)
Type	Two-sample t-test equal variance
Difficulty	Standard +0.8 This is a two-sample t-test with unequal sample sizes requiring calculation of pooled variance, manual computation of summary statistics from raw data, and proper hypothesis test execution. While methodical, it demands careful arithmetic, understanding of pooled variance formula, and correct application of t-distribution critical values—more demanding than standard single-sample tests but follows established procedures without requiring novel insight.
Spec	5.05c Hypothesis test: normal distribution for population mean

6 A company manufactures copper pipes. The pipes are produced by two different machines, $A$ and $B$. An inspector claims that the mean diameter of the pipes produced by machine $A$ is greater than the mean diameter of the pipes produced by machine $B$. He takes a random sample of 12 pipes produced by machine $A$ and measures their diameters, $x \mathrm {~cm}$. His results are summarised as follows. $$\sum x = 6.24 \quad \sum x ^ { 2 } = 3.26$$ He also takes a random sample of 10 pipes produced by machine $B$ and measures their diameters in cm. His results are as follows. $$\begin{array} { l l l l l l l l l l } 0.48 & 0.53 & 0.47 & 0.54 & 0.54 & 0.55 & 0.46 & 0.55 & 0.50 & 0.48 \end{array}$$ The diameters of the pipes produced by each machine are assumed to be normally distributed with equal population variances. Test at the $2.5 \%$ significance level whether the data supports the inspector's claim.
If you use the following page to complete the answer to any question, the question number must be clearly shown.

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Question 6:

Answer	Marks	Guidance
Answer	Marks	Guidance
$s_x^2 = \frac{1}{11}\left(3.26 - \frac{6.24^2}{12}\right) = 0.001382$	M1	$\frac{19}{13750}$, $\frac{19}{15000}$
$s_y^2 = \frac{1}{9}\left(2.6124 - \frac{5.1^2}{10}\right) = 0.001267$
Both correct	A1	Both correct
$s^2 = \frac{11 \times 0.001382 + 9 \times 0.001267}{12 + 10 - 2} = 0.001330$	M1 A1	Find pooled variance
$t = \dfrac{\dfrac{6.24}{12} - \dfrac{5.1}{10}}{s\sqrt{\dfrac{1}{12} + \dfrac{1}{10}}} = 0.640$	M1 A1	Use correct formula for $t$
$H_0: \mu_a = \mu_b \quad H_1: \mu_a > \mu_b$	B1
Critical value is $2.086$ $(t_{0.975}(20))$	M1	Compare their calculated value with 2.086 and conclusion
$0.640 < 2.086$
Accept $H_0$
There is insufficient evidence to support inspector's claim	A1	All correct except possibly B1, in context, level of uncertainty in language. 'Prove' scores A0
9	Pooled variance not used: M1A1 M0A0 M0A0 B1 M1A0 max 4/9

## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_x^2 = \frac{1}{11}\left(3.26 - \frac{6.24^2}{12}\right) = 0.001382$ | M1 | $\frac{19}{13750}$, $\frac{19}{15000}$ |
| $s_y^2 = \frac{1}{9}\left(2.6124 - \frac{5.1^2}{10}\right) = 0.001267$ | | |
| Both correct | A1 | Both correct |
| $s^2 = \frac{11 \times 0.001382 + 9 \times 0.001267}{12 + 10 - 2} = 0.001330$ | M1 A1 | Find pooled variance |
| $t = \dfrac{\dfrac{6.24}{12} - \dfrac{5.1}{10}}{s\sqrt{\dfrac{1}{12} + \dfrac{1}{10}}} = 0.640$ | M1 A1 | Use correct formula for $t$ |
| $H_0: \mu_a = \mu_b \quad H_1: \mu_a > \mu_b$ | B1 | |
| Critical value is $2.086$ $(t_{0.975}(20))$ | M1 | Compare their calculated value with 2.086 and conclusion |
| $0.640 < 2.086$ | | |
| Accept $H_0$ | | |
| There is insufficient evidence to support inspector's claim | A1 | All correct except possibly B1, in context, level of uncertainty in language. 'Prove' scores A0 |
| | **9** | Pooled variance not used: M1A1 M0A0 M0A0 B1 M1A0 max 4/9 |

Show LaTeX source

6 A company manufactures copper pipes. The pipes are produced by two different machines, $A$ and $B$. An inspector claims that the mean diameter of the pipes produced by machine $A$ is greater than the mean diameter of the pipes produced by machine $B$. He takes a random sample of 12 pipes produced by machine $A$ and measures their diameters, $x \mathrm {~cm}$. His results are summarised as follows.

$$\sum x = 6.24 \quad \sum x ^ { 2 } = 3.26$$

He also takes a random sample of 10 pipes produced by machine $B$ and measures their diameters in cm. His results are as follows.

$$\begin{array} { l l l l l l l l l l } 
0.48 & 0.53 & 0.47 & 0.54 & 0.54 & 0.55 & 0.46 & 0.55 & 0.50 & 0.48
\end{array}$$

The diameters of the pipes produced by each machine are assumed to be normally distributed with equal population variances.

Test at the $2.5 \%$ significance level whether the data supports the inspector's claim.\\

If you use the following page to complete the answer to any question, the question number must be clearly shown.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q6 [9]}}

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