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:
[tex]\hat{p} = 89\% = 0.89[/tex]
We have to find the margin of error associated with a 90% Confidence interval.
Formula for margin of error:
[tex]z_{stat}\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.10} = 1.64[/tex]
Putting the values, we get:
[tex]M.E = 1.64\times (\sqrt{\dfrac{0.89(1-0.89)}{1024}})\\\\M.E=0.016[/tex]
Thus, the margin of error for the survey is 0.016
