The confidence interval for confidence level of [tex]1-\alpha[/tex] is
[tex]\left(\overline x-Z_{\alpha/2}\dfrac\sigma{\sqrt n},\overline x+Z_{\alpha/2}\dfrac\sigma{\sqrt n}\right)[/tex]
where [tex]\overline x[/tex] is the sample mean, [tex]Z_{\alpha/2}[/tex] is the critical value for the given confidence level, [tex]\sigma[/tex] is the standard deviation of the population, and [tex]n[/tex] is the sample size. The margin of error is the [tex]Z_{\alpha/2}\dfrac\sigma{\sqrt n}[/tex] term.
a) For a confidence level of [tex]1-\alpha=0.90[/tex], we have [tex]Z_{\alpha/2}=Z_{0.05}\approx1.64[/tex]. So in order to have a margin of error of at most 25 points, we have
[tex]1.64\dfrac{250}{\sqrt n}=25\implies n\approx268.96[/tex]
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 [tex]1-\alpha=0.99[/tex], we have [tex]Z_{\alpha/2}=Z_{0.005}\approx2.58[/tex]. Then the margin of error would at most satisfy
[tex]2.58\dfrac{250}{\sqrt n}=25\implies n\approx665.64[/tex]
so that Luke should collect a sample of at least 666 students.
