Question

Hi, can you help me answer this question please, thank you!

Answer 1

The formula for a margin error is given by

`ME=z\frac{\sigma}{\sqrt[]{n}}`

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

`25=1.96*\frac{300}{\sqrt[]{n}}`

Solving for n, we get our sample size.

`\begin{gathered} 25=1.96*\frac{300}{\sqrt[]{n}} \\ \sqrt[]{n}=1.96*(300)/(25) \\ \sqrt[]{n}=1.96*12 \\ \sqrt[]{n}=23.52 \\ n=553.1904 \end{gathered}`

We would need 554 students.