CAIE Further Paper 4 2020 June — Question 4 9 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2020
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeTwo-sample z-test large samples
DifficultyStandard +0.8 This is a two-sample t-test with pooled variance requiring calculation of sample means, variances, pooled variance estimate, and test statistic from summary statistics. While the procedure is standard for Further Statistics, the multi-step calculation with potential for arithmetic errors and the need to correctly apply pooled variance formula makes it moderately challenging, though not requiring novel insight.
Spec5.05c Hypothesis test: normal distribution for population mean

4 A company has two different machines, \(X\) and \(Y\), each of which fills empty cups with coffee. The manager is investigating the volumes of coffee, \(x\) and \(y\), measured in appropriate units, in the cups filled by machines \(X\) and \(Y\) respectively. She chooses a random sample of 50 cups filled by machine \(X\) and a random sample of 40 cups filled by machine \(Y\). The volumes are summarised as follows. $$\sum x = 15.2 \quad \sum x ^ { 2 } = 5.1 \quad \sum y = 13.4 \quad \sum y ^ { 2 } = 4.8$$ The manager claims that there is no difference between the mean volume of coffee in cups filled by machine \(X\) and the mean volume of coffee in cups filled by machine \(Y\). Test the manager's claim at the \(10 \%\) significance level.

Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu_x = \mu_y\), \(H_1: \mu_x \neq \mu_y\)B1
\(s_x^2 = \frac{1}{49}\!\left(5.1 - \frac{15.2^2}{50}\right) = 0.0097796\); \(s_y^2 = \frac{1}{39}\!\left(4.8 - \frac{13.4^2}{40}\right) = 0.007974\)M1A1
\(s^2 = \frac{0.00977959}{50} + \frac{0.007974}{40} = 0.0003949\)M1A1
\(z = \dfrac{0.304 - 0.335}{\sqrt{0.0003949}} = (-)1.56\)M1A1
Compare with 1.645M1
Accept \(H_0\): insufficient evidence to reject manager's claimA1
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_x = \mu_y$, $H_1: \mu_x \neq \mu_y$ | B1 | |
| $s_x^2 = \frac{1}{49}\!\left(5.1 - \frac{15.2^2}{50}\right) = 0.0097796$; $s_y^2 = \frac{1}{39}\!\left(4.8 - \frac{13.4^2}{40}\right) = 0.007974$ | M1A1 | |
| $s^2 = \frac{0.00977959}{50} + \frac{0.007974}{40} = 0.0003949$ | M1A1 | |
| $z = \dfrac{0.304 - 0.335}{\sqrt{0.0003949}} = (-)1.56$ | M1A1 | |
| Compare with 1.645 | M1 | |
| Accept $H_0$: insufficient evidence to reject manager's claim | A1 | |

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4 A company has two different machines, $X$ and $Y$, each of which fills empty cups with coffee. The manager is investigating the volumes of coffee, $x$ and $y$, measured in appropriate units, in the cups filled by machines $X$ and $Y$ respectively. She chooses a random sample of 50 cups filled by machine $X$ and a random sample of 40 cups filled by machine $Y$. The volumes are summarised as follows.

$$\sum x = 15.2 \quad \sum x ^ { 2 } = 5.1 \quad \sum y = 13.4 \quad \sum y ^ { 2 } = 4.8$$

The manager claims that there is no difference between the mean volume of coffee in cups filled by machine $X$ and the mean volume of coffee in cups filled by machine $Y$.

Test the manager's claim at the $10 \%$ significance level.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q4 [9]}}