Question

A survey found that 89% of a random sample of 1024 American adults approved of cloning endangered animals. Find the margin of error for this survey if we want 90% confidence in our estimate of the percentage of American adults who approve of cloning endangered animals.

Answer 1

Answer:

The margin of error for the survey is 0.016

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 1024

Sample proportion:

$\hat{p} = 89\% = 0.89$

We have to find the margin of error associated with a 90% Confidence interval.

Formula for margin of error:

$z_{stat}\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.10} = 1.64$

Putting the values, we get:

$M.E = 1.64\times (\sqrt{\dfrac{0.89(1-0.89)}{1024}})\\\\M.E=0.016$

Thus, the margin of error for the survey is 0.016