Question 7 - A-Level Maths

CAIE FP2 2016 June — Question 7 8 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2016
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	T-tests (unknown variance)
Type	Single sample t-test
Difficulty	Standard +0.3 This is a straightforward one-sample t-test with all necessary summary statistics provided. Students need to calculate sample mean and standard deviation, then apply the standard t-test procedure with clear hypotheses and a one-tailed test at 5% level. While it requires multiple steps (calculate statistics, find critical value, compare), each step is routine and the question structure is standard textbook format with no conceptual challenges or novel insights required.
Spec	5.05b Unbiased estimates: of population mean and variance 5.05c Hypothesis test: normal distribution for population mean

7 A random sample of 9 observations of a normal variable $X$ is taken. The results are summarised as follows. $$\Sigma x = 24.6 \quad \Sigma x ^ { 2 } = 68.5$$ Test, at the $5 \%$ significance level, whether the population mean is greater than 2.5.

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Question 7:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\bar{x} = 24.6/9$ or $41/15$ or $2.73[3]$	B1	Calculate sample mean
$s^2 = (68.5 - 24.6^2/9)/8 = 63/400$ or $0.1575$ or $0.3969^2$	M1 B1	Estimate population variance; allow biased: $0.14$ or $0.3742^2$
H$_0$: $\mu = 2.5$; H$_1$: $\mu > 2.5$	M1 A1	State hypotheses (AEF; B0 for $\bar{x}$...)
$t = (\bar{x} - 2.5)/(s/\sqrt{9}) = 1.76$	B1	Calculate value of $t$ (to 3 s.f.)
$t_{8,\,0.95} = 1.86[0]$	M1	State or use correct tabular $t$-value (to 3 s.f.); or can compare $\bar{x}$ with $2.5 + 0.246 = 2.746$
Accept H$_0$ if $t <$ tabular value		State or imply valid method for conclusion
$1.76\ [\pm 0.02] < 1.86$ so population mean not greater than $2.5$	A1	Conclusion (AEF, requires both values correct); Total: [8]

## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 24.6/9$ or $41/15$ or $2.73[3]$ | B1 | Calculate sample mean |
| $s^2 = (68.5 - 24.6^2/9)/8 = 63/400$ or $0.1575$ or $0.3969^2$ | M1 B1 | Estimate population variance; allow biased: $0.14$ or $0.3742^2$ |
| H$_0$: $\mu = 2.5$; H$_1$: $\mu > 2.5$ | M1 A1 | State hypotheses (AEF; B0 for $\bar{x}$...) |
| $t = (\bar{x} - 2.5)/(s/\sqrt{9}) = 1.76$ | B1 | Calculate value of $t$ (to 3 s.f.) |
| $t_{8,\,0.95} = 1.86[0]$ | M1 | State or use correct tabular $t$-value (to 3 s.f.); or can compare $\bar{x}$ with $2.5 + 0.246 = 2.746$ |
| Accept H$_0$ if $t <$ tabular value | | State or imply valid method for conclusion |
| $1.76\ [\pm 0.02] < 1.86$ so population mean not greater than $2.5$ | A1 | Conclusion (AEF, requires both values correct); Total: [8] |

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7 A random sample of 9 observations of a normal variable $X$ is taken. The results are summarised as follows.

$$\Sigma x = 24.6 \quad \Sigma x ^ { 2 } = 68.5$$

Test, at the $5 \%$ significance level, whether the population mean is greater than 2.5.

\hfill \mbox{\textit{CAIE FP2 2016 Q7 [8]}}

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