Question

A poll was conducted to determine what percentage of the peoplesupport a certain candidate for the U.S. Senate.400 people were surveyed, and44% of them said they support thecandidate. If we require a confidence level of99%, what margin of error do we report?

Answer 1

Answer: 0.064

Explanation:

The formula to find the margin of error :-

`E=z_(\alpha/2)\sqrt{(p(1-p))/(n)}`

Given : Significance level :
`\alpha=1-0.99=0.01`

Critical value :
`z_(\alpha/2)=2.576`

Sample size : n=400

The proportion of people support the candidate :p=0.44

Then ,
`E=(2.576)\sqrt{(0.44(1-0.44))/(400)}\approx0.064`

Hence, Margin of error = 0.064