Question

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to thegiven statistics and confidence level. Round the margin of error to four decimal places.1)99% confidence; n

Answer 1

Answer:

$E = 0.0158$

Step-by-step explanation:

Given

$n = 5900$

$x = 1770$

$CI = 99\%$

Required

The margin of error (E)

First, calculate proportion p

$p = x/n$

$p = 1770/5900$

$p = 0.3$

Given that:

$CI = 99\%$

Calculate the alpha leve;

$\alpha = 1 - CI$

$\alpha = 1- 0.99$

$\alpha= 0.01$

Divide by 2

$\alpha/2= 0.01/2$

Subtract from 1

$1 - \alpha/2= 1 - 0.01/2$

$1 - \alpha/2= 0.995$

The corresponding z value is:

$z =2.576$

So, the margin of error is:

$E = z * \sqrt{p * (1 - p)/n}$

So, we have:

$E = 2.576 * \sqrt{0.3 * (1 - 0.3)/5600}$

Using a calculator, we have:

$E = 0.01577471394$

Approximate

$E = 0.0158$