Question

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.

95% confidence; n = 2388, x = 1672

Group of answer choices

a) 0.0184

b) 0.0156

c) 0.0248

d) 0.0206

Answer 1

proportion p = x / n = 1672/2388

standard error =

p* (1 1-p) n

z critical value for 95% confidence is = 1.96

margin of error = z-score * standard error

margin of error = $z*\sqrt{}\frac{p*(1-p)}{n}$

= 1.96 * sqrt((1672/2388)*(716/2388)/2388)

= 0.0184

Ans: option A ) 0.0184 is correct