In a crossword competition the times, x minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows. $\sum x = 210.9$ $\sum(x - \bar{x})^2 = 151.2$ The time to complete a crossword has a normal distribution with mean $\mu$ minutes. Calculate a 95% confidence interval for $\mu$. Assume now that the standard deviation of the population is known to be 5.6 minutes. Find the smallest sample size that would lead to a 95% confidence interval for $\mu$ of width at most 5 minutes.

A-Level CIE Further Maths 18.1 Central Limit Theorem: In a crossword competition the t

In a crossword competition the times, x minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows - CIE - A-Level Further Maths - Question 8 - 2011 - Paper 1

Question 8

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In a crossword competition the times, x minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows. $\sum x = 210.9$ $\sum(x... show full transcript

Worked Solution & Example Answer:In a crossword competition the times, x minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows - CIE - A-Level Further Maths - Question 8 - 2011 - Paper 1

Step 1

Calculate sample mean:

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Answer

To find the sample mean $\bar{x}$ , we use the formula:

$\bar{x} = \frac{\sum x}{n}$

Substituting the values, we have:

$\bar{x} = \frac{210.9}{6} = 35.15$

Step 2

Estimate population variance (allow biased):

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Answer

The unbiased estimator for population variance $\sigma^2$ is given by:

$s^2 = \frac{\sum(x - \bar{x})^2}{n-1}$

Substituting the values:

$s^2 = \frac{151.2}{6 - 1} = 30.24$

Step 3

Find confidence interval (allow in place of n):

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Answer

The 95% confidence interval is given by:

$\bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

In this case, using a t-distribution with $n-1=5$ degrees of freedom, we find:

table value: $t_{0.025} = 2.571$ (2 d.p.)

Substituting the values into the formula:

$C.I. = 35.15 \pm 2.571 \left( \frac{\sqrt{30.24}}{\sqrt{6}} \right)$

Calculating gives us the interval.

Step 4

Find smallest sample size:

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Answer

Given that the population standard deviation $\sigma = 5.6$ minutes, we use the formula for sample size:

$n = \left( \frac{z_{\alpha/2} \sigma}{E} \right)^2$

Where:

$z_{\alpha/2} = 1.96$ for a 95% confidence level
$E = 2.5$ (half the desired width of the confidence interval)

Substituting values:

$n = \left( \frac{1.96 \times 5.6}{2.5} \right)^2$

Calculating gives:

$n = 19.3$

Rounding up gives the smallest sample size as $n_{min} = 20$ .

In a crossword competition the times, x minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows - CIE - A-Level Further Maths - Question 8 - 2011 - Paper 1

Worked Solution & Example Answer:In a crossword competition the times, x minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows - CIE - A-Level Further Maths - Question 8 - 2011 - Paper 1

Calculate sample mean:

Estimate population variance (allow biased):

Find confidence interval (allow in place of n):

Find smallest sample size:

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