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Mathematics

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Ivan4 years ago

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The confidence interval for confidence level of $1-\alpha$ is

$\left(\overline x-Z_{\alpha/2}\dfrac\sigma{\sqrt n},\overline x+Z_{\alpha/2}\dfrac\sigma{\sqrt n}\right)$

where $\overline x$ is the sample mean, $Z_{\alpha/2}$ is the critical value for the given confidence level, $\sigma$ is the standard deviation of the population, and $n$ is the sample size. The margin of error is the $Z_{\alpha/2}\dfrac\sigma{\sqrt n}$ term.

a) For a confidence level of $1-\alpha=0.90$ , we have $Z_{\alpha/2}=Z_{0.05}\approx1.64$ . So in order to have a margin of error of at most 25 points, we have

$1.64\dfrac{250}{\sqrt n}=25\implies n\approx268.96$

so Raina should collect a sample of at least 269 students.

b) A confidence interval with a higher confidence level would more closely approximate and reflect the population, so it stands to reason that Luke should collect a larger sample than Raina to meet his 99% confidence spec.

c) For a confidence level of $1-\alpha=0.99$ , we have $Z_{\alpha/2}=Z_{0.005}\approx2.58$ . Then the margin of error would at most satisfy