The formula for a margin error is given by
[tex]ME=z\frac{\sigma}{\sqrt[]{n}}[/tex]
Where z is the z-critical(associated with the confidence), sigma is the standard deviation, n is the sample size, and ME is the interval.
From the text, we know the desired margin error is 25(ME = 25), the standard deviation is 300, and using our confidence of 95% we can calculate the z-critical value. For a 95% confidence, we have a z-critical of 1.96.
Using those values in our equation, we have
[tex]25=1.96\times\frac{300}{\sqrt[]{n}}[/tex]
Solving for n, we get our sample size.
[tex]\begin{gathered} 25=1.96\times\frac{300}{\sqrt[]{n}} \\ \sqrt[]{n}=1.96\times\frac{300}{25} \\ \sqrt[]{n}=1.96\times12 \\ \sqrt[]{n}=23.52 \\ n=553.1904 \end{gathered}[/tex]
We would need 554 students.
