Question 9 - A-Level Maths

CAIE FP2 2013 November — Question 9 10 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	November
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	T-tests (unknown variance)
Type	Single sample confidence interval t-distribution
Difficulty	Standard +0.3 This is a straightforward application of a one-sample t-test with standard formulas for sample mean, sample variance, and confidence intervals. While it requires multiple steps (calculate statistics, perform hypothesis test, construct confidence interval), all procedures are routine and follow directly from textbook methods with no conceptual challenges or novel problem-solving required.
Spec	5.05c Hypothesis test: normal distribution for population mean 5.05d Confidence intervals: using normal distribution

9 A random sample of 9 observations of a normally distributed random variable $X$ gave the following summarised data. $$\Sigma x = 94.5 \quad \Sigma x ^ { 2 } = 993.6$$ Test, at the $5 \%$ significance level, whether the population mean of $X$ is 10.2 . Calculate a $90 \%$ confidence interval for the population mean of $X$.

Show mark scheme Show mark scheme source

Question 9:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\bar{x} = 94.5/9 = 10.5$	M1	Calculate sample mean
$s^2 = (993.6 - 94.5^2/9)/8 = 0.16875$ or $0.4108^2$	M1	Estimate population variance (allow biased: $0.15$ or $0.3873^2$)
$H_0: \mu = 10.2,\ H_1: \mu \neq 10.2$	B1	State hypotheses (A.E.F.)
$t = (\bar{x} - 10.2)/(s/\sqrt{9}) = 2.19$	M1 *A1	Calculate value of $t$ (to 3 s.f.)
$t_{8,\ 0.975} = 2.306$	*B1	State or use correct tabular $t$ value (or compare $\bar{x}$ with $10.2 + 0.316 = 10.52$)
Population mean is $10.2$	B1	Correct conclusion (AEF, dep A1, B1)
$10.5 \pm t\sqrt{\{1.35/(8 \times 9)\}}$	M1	Find confidence interval (allow $z$ in place of $t$)
$t_{8,\ 0.95} = 1.86[0]$	A1	Use correct tabular value
$10.5 \pm 0.255$ or $[10.2,\ 10.8]$	A1	Evaluate C.I. correct to 3 s.f.

## Question 9:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 94.5/9 = 10.5$ | M1 | Calculate sample mean |
| $s^2 = (993.6 - 94.5^2/9)/8 = 0.16875$ or $0.4108^2$ | M1 | Estimate population variance (allow biased: $0.15$ or $0.3873^2$) |
| $H_0: \mu = 10.2,\ H_1: \mu \neq 10.2$ | B1 | State hypotheses (A.E.F.) |
| $t = (\bar{x} - 10.2)/(s/\sqrt{9}) = 2.19$ | M1 *A1 | Calculate value of $t$ (to 3 s.f.) |
| $t_{8,\ 0.975} = 2.306$ | *B1 | State or use correct tabular $t$ value (or compare $\bar{x}$ with $10.2 + 0.316 = 10.52$) |
| Population mean is $10.2$ | B1 | Correct conclusion (AEF, dep *A1, *B1) |
| $10.5 \pm t\sqrt{\{1.35/(8 \times 9)\}}$ | M1 | Find confidence interval (allow $z$ in place of $t$) |
| $t_{8,\ 0.95} = 1.86[0]$ | A1 | Use correct tabular value |
| $10.5 \pm 0.255$ or $[10.2,\ 10.8]$ | A1 | Evaluate C.I. correct to 3 s.f. |

---

Show LaTeX source

9 A random sample of 9 observations of a normally distributed random variable $X$ gave the following summarised data.

$$\Sigma x = 94.5 \quad \Sigma x ^ { 2 } = 993.6$$

Test, at the $5 \%$ significance level, whether the population mean of $X$ is 10.2 .

Calculate a $90 \%$ confidence interval for the population mean of $X$.

\hfill \mbox{\textit{CAIE FP2 2013 Q9 [10]}}

This paper (11 questions)

View full paper

Q2 9 Q3 9 Q4 10 Q5 12 Q6 5 Q7 7 Q8 10 Q9 10 Q10 11 Q11 EITHER Q11 OR