Question 8 - A-Level Maths

CAIE FP2 2015 June — Question 8 12 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2015
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	T-tests (unknown variance)
Type	Single sample confidence interval t-distribution
Difficulty	Standard +0.8 This is a two-part question requiring (1) a one-sample t-test with all steps (hypotheses, test statistic calculation, critical value, conclusion) and (2) working backwards from a confidence interval to find sample mean and variance. The second part requires understanding the structure of t-intervals and solving simultaneous relationships, which goes beyond routine application. This is moderately challenging for A-level Further Maths statistics.
Spec	5.05b Unbiased estimates: of population mean and variance 5.05c Hypothesis test: normal distribution for population mean 5.05d Confidence intervals: using normal distribution

8 A large number of long jumpers are competing in a national long jump competition. The distances, in metres, jumped by a random sample of 7 competitors are as follows. $$\begin{array} { l l l l l l l } 6.25 & 7.01 & 5.74 & 6.89 & 7.24 & 5.64 & 6.52 \end{array}$$ Assuming that distances are normally distributed, test, at the $5 \%$ significance level, whether the mean distance jumped by long jumpers in this competition is greater than 6.2 metres. The distances jumped by another random sample of 8 long jumpers in this competition are recorded. Using the data from this sample of 8 long jumpers, a $95 \%$ confidence interval for the population mean, $\mu$ metres, is calculated as $5.89 < \mu < 6.75$. Find the unbiased estimates for the population mean and population variance used in this calculation.

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Question 8:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\bar{x} = 45.29/7 = 6.47$	M1	Calculate sample mean
$s_1^2 = (295.36 - 45.29^2/7)/6 = 0.389$ or $0.6236^2$	M1	Estimate population variance (allow biased: 0.3334 or $0.5774^2$)
$H_0$: $\mu = 6.2$, $H_1$: $\mu > 6.2$	B1	State hypotheses (AEF; B0 for $\bar{x}$)
$t = (\bar{x} - 6.2)/(s_1/\sqrt{7}) = 1.15$	M1 A1	Calculate value of $t$
$t_{6,\ 0.95} = 1.94$ [3]	B1	State or use correct tabular $t$-value
[Accept $H_0$] Population mean not greater than 6.2	B1$\checkmark$	Consistent conclusion
$\frac{1}{2}(5.89 + 6.75) = 6.32$	M1 A1	Find estimate of population mean
$t\sqrt{(s_2^2/8)} = \frac{1}{2}(6.75 - 5.89)$ [= 0.43]	M1	SR Allow M1 for $t\sqrt{s_2^2/7}$, max 4/5
$t_{7,\ 0.975} = 2.36$ [5]	A1	Use correct tabular value (use of 2.36 may lose final A1)
$s_2^2 = 0.2645$ or $32/121$	A1
Part marks: 7, 5	Total: 12

## Question 8:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 45.29/7 = 6.47$ | M1 | Calculate sample mean |
| $s_1^2 = (295.36 - 45.29^2/7)/6 = 0.389$ or $0.6236^2$ | M1 | Estimate population variance (allow biased: 0.3334 or $0.5774^2$) |
| $H_0$: $\mu = 6.2$, $H_1$: $\mu > 6.2$ | B1 | State hypotheses (AEF; B0 for $\bar{x}$) |
| $t = (\bar{x} - 6.2)/(s_1/\sqrt{7}) = 1.15$ | M1 A1 | Calculate value of $t$ |
| $t_{6,\ 0.95} = 1.94$ [3] | B1 | State or use correct tabular $t$-value |
| [Accept $H_0$] Population mean not greater than 6.2 | B1$\checkmark$ | Consistent conclusion |
| $\frac{1}{2}(5.89 + 6.75) = 6.32$ | M1 A1 | Find estimate of population mean |
| $t\sqrt{(s_2^2/8)} = \frac{1}{2}(6.75 - 5.89)$ [= 0.43] | M1 | **SR** Allow M1 for $t\sqrt{s_2^2/7}$, max 4/5 |
| $t_{7,\ 0.975} = 2.36$ [5] | A1 | Use correct tabular value (use of 2.36 may lose final A1) |
| $s_2^2 = 0.2645$ or $32/121$ | A1 | |

**Part marks: 7, 5 | Total: 12**

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8 A large number of long jumpers are competing in a national long jump competition. The distances, in metres, jumped by a random sample of 7 competitors are as follows.

$$\begin{array} { l l l l l l l } 
6.25 & 7.01 & 5.74 & 6.89 & 7.24 & 5.64 & 6.52
\end{array}$$

Assuming that distances are normally distributed, test, at the $5 \%$ significance level, whether the mean distance jumped by long jumpers in this competition is greater than 6.2 metres.

The distances jumped by another random sample of 8 long jumpers in this competition are recorded. Using the data from this sample of 8 long jumpers, a $95 \%$ confidence interval for the population mean, $\mu$ metres, is calculated as $5.89 < \mu < 6.75$. Find the unbiased estimates for the population mean and population variance used in this calculation.

\hfill \mbox{\textit{CAIE FP2 2015 Q8 [12]}}

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