Question

What is the margin of error for 95% confidence for a sample of size 100 where p = 0.15?

Answer 1

Given:

`\begin{gathered} n=100 \\ p=0.15 \\ z=1.96 \end{gathered}`

Required:

To find the margin error.

Step-by-step explanation:

The formula for margin error is

`\begin{gathered} =z*\sqrt{(p(1-p))/(n)} \\ \\ =1.96*\sqrt{(0.15(1-0.15))/(100)} \\ \\ =1.96*√(0.001275) \\ \\ =1.96*0.03570 \\ \\ =0.0699 \\ \\ \approx0.0700 \end{gathered}`

Final Answer:

The option B is correct.

`0.0700`