Solve for the confidence interval estimate of the population proportion.
Mikay conducted poll survey in which 520 of 1200 randomly selected voters indicated their preference for a certain candidate. What is the true population proportion of voters who prefer the candidate using 95% confidence interval?
Sample proportion "p = \\dfrac{520}{1200} = 0.43"
"n = 1200"
Confidence interval can be calculated as:
"CI = p \\pm z \\sqrt{\\dfrac{p(1-p)}{n}}"
"= 0.43 \\pm 1.96 \\sqrt{\\dfrac{0.43 \\times 0.57}{1200}}"
= "0.43 \\pm 0.0143"
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