The number, $x$, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country A. The number, $y$, of the same type of sea shell was counted at 50 randomly chosen sites, each one metre square, along the coastline in country B. The results are summarised as follows, where $ar{x}$ and $ar{y}$ denote the sample means of $x$ and $y$ respectively. $ar{x} = 29.2 \\ \Sigma(x - \bar{x})^2 = 4341.6 \\ \bar{y} = 24.4 \\ \Sigma(y - \bar{y})^2 = 3732.0$ Find a 95% confidence interval for the difference between the mean number of sea shells, per square metre, on the coastlines in country A and in country B.

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The number, $x$, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country A - CIE - A-Level Further Maths - Question 4 - 2020 - Paper 4

Question 4

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The number, $x$, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country A. The number, $y$, of... show full transcript

Worked Solution & Example Answer:The number, $x$, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country A - CIE - A-Level Further Maths - Question 4 - 2020 - Paper 4

Step 1

Assume that population differences are normal.

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Answer

We start by stating the null and alternative hypotheses:

Null Hypothesis ( $H_0$ ): $ar{x} - ar{y} = 0$
Alternative Hypothesis ( $H_a$ ): $ar{x} - ar{y} > 0$

Step 2

Calculate test statistic.

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Answer

To find the test statistic, we first compute the variances:

Sample variance of $x$ : $s_x^2 = \frac{\Sigma(x - \bar{x})^2}{n_x - 1} = \frac{4341.6}{60 - 1} = 74.97$
Sample variance of $y$ : $s_y^2 = \frac{\Sigma(y - \bar{y})^2}{n_y - 1} = \frac{3732.0}{50 - 1} = 76.16$

The test statistic is then given by:

$t = \frac{\bar{x} - \bar{y}}{\sqrt{\frac{s_x^2}{n_x} + \frac{s_y^2}{n_y}}} = \frac{29.2 - 24.4}{\sqrt{\frac{74.97}{60} + \frac{76.16}{50}}}$

Calculating this yields:

$t = \frac{4.8}{\sqrt{1.2495 + 1.5232}} \approx \frac{4.8}{1.607} \approx 2.99$

Step 3

Use t-critical value.

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Using the degrees of freedom calculated through the Welch-Satterthwaite equation, we find:

$df \approx 70$

For a 95% confidence level, the critical value for $t$ can be found in the t-distribution table. For a one-tailed test, $t_{0.05, df} \approx 1.67$ .

Step 4

Construct Confidence Interval.

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The confidence interval for the difference of means is given by:

$CI = (\bar{x} - \bar{y}) \pm t_{critical} \cdot SE$ where $SE$ is the standard error. We have:

$SE = \sqrt{\frac{74.97}{60} + \frac{76.16}{50}} = 1.49$ Therefore the confidence interval is:

$CI = 4.8 \pm 1.67 \cdot 1.49$ Calculating the endpoint values gives:

$CI = (4.8 - 2.48, 4.8 + 2.48) = (2.32, 7.28)$

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The number, $x$, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country A - CIE - A-Level Further Maths - Question 4 - 2020 - Paper 4

Worked Solution & Example Answer:The number, $x$, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country A - CIE - A-Level Further Maths - Question 4 - 2020 - Paper 4

Assume that population differences are normal.

Calculate test statistic.

Use t-critical value.

Construct Confidence Interval.

Join the A-Level students using SimpleStudy...

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