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

Step-by-step explanation:

The formula to find the margin of error :-

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

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 , $E=(2.576)\sqrt{\dfrac{0.44(1-0.44)}{400}}\approx0.064$

Hence, Margin of error = 0.064