Question

Sat scores. sat scores of students at an ivy league college are distributed with a standard deviation of 250 points. two statistics students, raina and luke, want to estimate the average sat score of students at this college as part of a class project. they want their margin of error to be no more than 25 points. (a) raina wants to use a 90% confidence interval. how large a sample should she collect? (b) luke wants to use a 99% confidence interval. without calculating the actual sample size, determine whether his sample should be larger or smaller than raina's, and explain your reasoning. (c) calculate the minimum required sample size for luke.

Answer 1

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

$2.58\dfrac{250}{\sqrt n}=25\implies n\approx665.64$

so that Luke should collect a sample of at least 666 students.