find the mergin interval error for the 95 confidence interva

find the mergin interval error for the 95% confidence interval used to estimate the population proportion(p);n=186,x=103

Solution

n = 186

V = 103

s = sqrt(V)

s = 10.14

margin for error @ 95% confidence,

t value = 1.65

Standard Error = s/sqrt(n) = 10.14 /sqrt(186) = 0.7435

Margin of error = t * standard error = 1.65 * 0.7435

Margin of erro = 1.2267