Question

Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 272 with 23.2% successes at a confidence level of 95%.

Answer 1

Solution :

Given that,

n = 272

Point estimate = sample proportion = =23.2%=0.232

1 -   = 1- 0.232 =0.768

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2   * ((( * (1 - )) / n)

= 1.96 (((0.232*0.768) /272 )

E = 0.05016